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Class 12 Chemistry Unit 16 Chemistry In Every Day Life Chemistry In Every Day Life Introduction

Chemistry plays a very important role in every aspect of human life. Products we use in our day-to-day lives are all chemical compounds. Some of the commonly used ones are—

1. Cosmetics, for example., lipstick, cream
2. Cleansing agents, for example., soaps, detergents
3. Food, for example., carbohydrates, proteins
4. Preservatives, for example., sodium benzoate, sodium chloride
5. Medicines and drugs, for example., antibiotics, sulpha drugs O Fertilisers and pesticides, for example., triple superphosphate, malathion
6. Fuels, for example., petrol, LPG
7. Paper, ink, rubber, pencil, paints, natural and synthetic fibers, dyes, and almost all other items. Chemistry’s contribution towards economic growth and the betterment of human lives is widely acknowledged by all.

In this unit, we will discuss the application of chemistry in the areas of medicines and drugs, food, and cleansing agents.

Medicines And Drugs

Relation Between Drugs And Medicines

The word ‘drug’ originated from the French word ‘drogue’ meaning ‘dry herbs Drugs are chemical compounds (natural and synthetic) of low molecular mass (~ 100-500 u) that exhibit biological activities in reaction with biomolecules.

• If these activities or responses are curative in action, then corresponding chemical compounds are called medicines.
• Medicines are used for detecting, treating, and preventing diseases. When suitable chemical compounds are used to cure different diseases, it is called chemotherapy.
• Drugs used as medicines may act as poisons and even cause death when consumed above the prescribed level.

Medicines: Natural and synthetic chemical compounds having the properties are termed as ‘medicines’—

1. Used to cure diseases,
2. Safe for the human body,
3. Do not have any appreciable toxic effects or side effects and
4. Do not cause any addiction.

Drugs: Natural and synthetic chemical compounds having the properties are termed as ‘drugs’—

1. May be used to cure diseases,
2. May not be safe for humans,
3. May have toxic effects or side effects
4. May lead to addiction.

Difference between Drugs and Medicines:

Drug Designing

Synthesis of a drug for a specific medical use is known as drug designing.

1. Drug-target and
2. Drug metabolism is considered important while synthesizing a specific drug.

1. Drug-target: Drugs on entering the body usually interact with biological macromolecules (carbohydrates, proteins, nucleic acids, and lipids) known as drug-target or target molecules. The selection of appropriate target molecules is essential to get the desired therapeutic effect of the drug.
2. Drug Metabolism: A drug should be designed in such a way that it reaches the target without participating in any intermediary metabolic reaction. The drug after its curative action should also get easily excreted from the body without causing any adverse effects.

It is important to know the physiological functions of the drug targets while synthesizing the corresponding drug.

Lead compounds: Drugs are synthesized from some chemical compounds, called lead compounds, which can react with drug targets and are mostly therapeutically active.

• Lead compounds are considered to be the starting points for drug synthesis and are usually collected from natural sources like herbs, shrubs, bushes, corals, fishes, or snake venom.
• But, nowadays most of the drugs are synthesised in laboratories.

Classification Of Drugs

Drugs are classified based on four factors—

1. Based On Pharmacological Effect: This classification is very useful for doctors. They get a clear idea about probable drugs available for the treatment of a specific type of ailment.
   1. Based On Pharmacological Effect Example: Analgesics are used as painkillers, and antibiotics are used to kill and prohibit the growth of microorganisms. A drug is administered, keeping in mind the patient’s age, condition, and presence of any other disease.
2. Based On Drug Action: This classification has been done based on the drug’s action on a particular biochemical process.
   1. Based On Drug Action Example: Antihistamines act as anti-allergens and also inhibit the action of histamine which causes inflammation in the body. Depending on the nature of the allergy, an appropriate antihistamine drug is applied.
3. Based On Chemical Structure: This classification is based on the fact that drugs having common structural characteristics exhibit similar pharmacological activities.
   1. Based On Chemical Structure Example: All sulphonamides having the following structural features are antibacterial.

Based On Molecular Targets: Drugs having similar chemical structures interact with their target molecules by similar action mechanisms. This classification is extremely useful for medicinal chemists (chemists who synthesize ‘ drugs having medicinal value).

Drug-Target Interaction

Drugs interact with macromolecular biomolecules (carbohydrates, nucleic acids, lipids, proteins) which perform various functions in the body, as explained below:

1. Enzymes: Proteins acting as biological catalysts are called enzymes.
2. Receptors: Proteins that are important in maintaining the communication system of the body are called receptors.
3. Carrier Proteins: Proteins that transport polar mole¬cules across the cell membrane are called carrier proteins.
4. Nucleic Acids: Coded genetic information is present in the nucleic acids of the cell.
5. Carbohydrates And Lipids: These biomolecules play an important role during cell membrane formation.

Enzymes As The Target Of Drugs: It is important to know the catalytic action of enzymes before discussing their interaction with drugs.

Catalytic Action Of Enzymes: Enzymes acting as catalysts perform two major functions:

• Certain areas on the enzyme molecule bind the substrate molecules in such a way that it can be attacked by the reagent effectively leading to an enzymatic reaction.
• These specific areas on the enzymes are known as active sites.
• Substrate molecules bind to the active site of the enzymes through interactions like H-bonding, ionic bonding, van der Waals interaction, and dipole-dipole interaction, and eventually form an enzyme-substrate complex.

• These interactions.should be (a) strong enough for the enzymes to hold the substrate molecules and carry out the catalytic reactions and (b) weak enough so that the product molecules formed can easily detach themselves from the active site.
• Enzymes also provide functional groups that attack substrate molecules and carry out enzymatic reactions.
• Example: If an L-serine molecule is present in the active site of an enzyme, the —OH group of the molecule acts as the nucleophile leading to an enzyme-catalyzed reaction.
• Similarly, the —SH group of L-cystine and the —COOH group of L-aspartic acid in the active sites of enzymes act as nucleophiles.

After the reaction, the enzyme-substrate complex dissociates to give the corresponding products. The free enzyme is then ready to participate in some other biochemical reactions.

Drug-Enzyme Interaction: Drugs inhibit any of the above-mentioned functions of enzymes. Drugs block the active site of the enzyme, inhibiting the formation of an enzyme-substrate complex, thus preventing the catalytic action of the enzyme.

• These drugs are called enzyme inhibitors. Drugs cause hindrance in the attachment of substrates to enzymes by blocking the active sites in two ways:
• Drugs compete with natural substrates of the enzyme to occupy the enzyme’s active site. These drugs are called competitive inhibitors.

• Some drugs instead of binding to the active site of an enzyme, attach themselves to a different site of an enzyme called the allosteric site.
• This attachment leads to the modification of the shape of the active site in such a manner that the substrates are no longer able to bind themselves to the active site.

If a strong covalent bond develops between the enzyme and drug (inhibitor), then the enzyme gets permanently blocked. In this condition, the body disintegrates the enzyme-drug complex and synthesizes new enzymes.

Receptor As Target Of Drug: Proteins that are important for carrying out communications within the body are called receptors. Most of the receptors are embedded in the cell membrane in such a manner that a small portion containing the active site remains on the outer surface of the cell membrane.

Transfer Of Message To The Cell By Receptors: Certain chemical compounds in the body communicate messages between two neurons or between neurons and muscles.

• These compounds are known as chemical messengers. The chemical messengers attach themselves to the binding site of the receptor proteins.
• The receptor binding site changes shape to accommodate a messenger and thus helps in sending the message into a cell without entering it.
• After the transfer of the message is over, the messengers detach themselves from the binding site and the receptor proteins regain their original shape.

• Different types of innumerable receptors are present in our body which interact with various kinds of chemical messengers.
• These receptors exhibit selectivity for one chemical messenger over the other because the binding sites of the receptors have different shapes, structures, and amino acid compositions.

Different Types Of Chemical Messengers

There are two types of chemical messengers—

1. Hormones and
2. Neurotransmitters.

Different Types Of Chemical Messengers Hormones: These are biomolecules, secreted from the endocrine glands (ductless) such as the thyroid gland, and pituitary gland.

• They transfer messages by moving around the body via blood and activating the receptors that recognize them. Hormones do not get easily deactivated.

Different Types Of Chemical Messengers Hormones Example: Adrenaline hormone is released from the adrenal gland during extreme stress or danger. It prepares the body to adapt itself to the stressful conditions.

Different Types Of Chemical Messengers Neurotransmitters: Nerve cells transfer messages through neurotransmitters. These are small molecules like acetylcholine, dopamine, and serotonin.

• Neurotransmitters bind to the receptor target for a short time during message transfer and after it is over they detach themselves from the receptors remaining completely unchanged.
• The receptors then transfer the message to the desired cell. The neurotransmitters get easily degraded and hence, lose their ability to send messages.
• The products obtained as a result of the degradation of the neurotransmitters move toward the nerve endings and again participate in message transfer by forming active messengers.

Both hormones and neurotransmitters activate receptors through similar mechanisms. Neurotransmitters get easily deactivated but hormones do not.

Drug-Receptor Interaction

1. In some cases, drugs bind to the receptor molecules by strong covalent bonds (~ 400 kj-mol^-1) making the process irreversible.
2. But in most cases, the addition of drugs to the receptors is short-lived and reversible. This is because in these cases interactions between drugs and receptor molecules are weak such as— hydrogen bonding (~20 kj-mol^-1) and van der Waals interaction (~4kj-mol).
3. The drugs which on binding with receptors inhibit the latter’s normal activities are known as antagonists. These are required when it is necessary to stop the transfer of any message within the body.
4. Drugs that activate the receptors by imitating their natural chemical messengers are called agonists. These are useful if there is a shortage of natural receptors.
5. Receptors having minimal differences in their shapes can interact with a specific group of chemical messengers. Example: There are two types of adrenergic receptors— α-adrenergic receptor and β-adrenergic receptor.
6. Despite timing and a minimal difference in shape, these receptors add to the chemical messenger ephedrine. Hence, designing a drug that will interact with only one type of adrenergic receptor is impossible.
7. Despite having a minimal difference in shape, these receptors add to the chemical messenger ephedrine. Hence, designing a drug that will interact with only one type of adrenergic receptor is impossible.
8. Receptors Are Not Equally Distributed All Over The Body Example: α-adrenergic receptors are more concentrated in tissues whereas β-adrenergic receptors are more concentrated in the heart. Therefore, drugs that interact with β-adrenergic receptors will act better in the heart, and drugs that interact with α-adrenergic receptors will act better in tissues.

Side Effects Of Drugs

1. One of the major problems of pharmacology is that any one drug can react with many receptors displaying different functions. Generally, this type of addition harms the body since the receptors get unnecessarily blocked.
   1. Side Effects Of Drugs Example: Some antidepressant drugs are associated with tire receptor serotonin. But, if these drugs interact with histamine or acetylcholine receptors or if the drug dissociates to give biologically active substances and interact with other receptors, then some undesirable side effects may occur.
2. Suitable medicine must be chosen for treating any disease. Incorrect choice of drug may lead to harmful side effects and even can be fatal to the patient. Hence, before recommending any medicine to a patient, all clinical tests of the patient must be done to get a clear idea about the physical condition.

Antacids

Excessive production of hydrochloric acid which causes pain and irritation in the stomach is known as acidity.

• In extreme situations, acidity can lead to ulcer formation in the stomach which can be detrimental to human health.
• Till 1970, treatment of acidity- was done using antacids like sodium bicarbonate, magnesium hydroxide, magnesium carbonate, magnesium trisilicate, aluminum hydroxide gel, and aluminum phosphate.

Effects Of Using Antacids

1. Excessive use of bicarbonate will make the stomach more alkaline, inducing the production of hydrochloric acid.
2. Metal hydroxides are more effective as they being insoluble, do not increase the pH value above 7 (neutral point). This treatment only controls the symptoms but does not address the cause. Hence, the metal salts can only provide temporary relief.
3. Sometimes, ulcers can become life-threatening due to acute acidity. In that case, there remains no other alternative than to remove the affected part of the stomach by operation.

Treatment Of Hyperacidity: Stimulation of histamine to secrete pepsin and hydrochloric acid in the stomach was considered a major discovery in the treatment of hyperacidity.

• A drug named cimetidine (Tegamet) was designed which prevented the interaction between histamine and the receptors on the stomach wall.
• As a result, secretion of the hydrochloric acid decreased. This leads to a successful treatment of the hyperacidity.
• This drug remained the largest-selling drug till ranitidine (Zantac) was discovered.

Two medicines that are being currently used for treating hyperacidity other than ranitidine are omeprazole and lansoprazole.

Antihistamines

Some people are sensitive to certain classes of drugs like penicillin and sulpha drugs while some others are sensitive to dust, pollens, cat fur, certain types of food, and fabrics.

• Hypersensitivity towards certain specific compounds is known as allergy. The secretion of histamine in our body is responsible for allergic reactions.

Harmful Effects Of Histamines: Histamine being a powerful vasodilator contracts the smooth muscles of the trachea and alimentary canal and relaxes the muscles present in the walls of fine blood vessels.

• As a result, allergic responses like skin rash, breathing problems, and skin diseases like urticaria are observed. Histamine is also responsible for nasal congestion and pollen allergy.
• BEF Drugs which compete with histamines to interact with receptors and consequendy inhibit the normal functioning of histamines are called antihistamines or anti-allergic drugs.

Harmful Effects Of Histamines Example: Two commonly used antihistamine drugs are brompheniramine (Dimetapp) and terfenadine (Seldane).

Application Of Antihistamines: Antihistamines are used in treating fever with cough, cold and throat pain, conjunctivitis, irritation in the nasal mucous membrane, nausea during pregnancy and the postoperative period, and sea sickness.

Some other commonly used antihistamine drugs are diphenhydramine (Benadryl), pheniramine maleate (Avil), cetirizine, chlorpheniramine, and promethazine.

Antihistamines do not affect the ad secretion process in the stomach since antihistamines and antacids interact with different sets of receptors.

Neurologically Active Drugs

Tranquilizers and analgesics are considered neurologically active drugs.

Tranquilizers

Tranquilizers Definition: Drugs that are used in treating mental stress, depression, and mild or acute mental disorders are called tranquilizers.

• They create a sense of mental peace and stability by relieving tension, anxiety, stress, depression, and hyperexcitement and stabilize mood swings.
• Tranquilizers are vital constituents of sleeping pills. They are also called psychotherapeutic drugs.

Tranquilizers Example: Noradrenaline is a type of neurotransmitter that plays an active role in mood changes.

• A person will suffer from mental depression if noradrenaline production in the body is low. In this case, antidepressant drugs like iproniazid and phenelzine (Nardil) are used.

• These drugs inhibit the action of enzymes that act as catalysts in noradrenaline degradation. Thus, the enzyme activity decreases leading to very slow degradation of noradrenaline.
• As a result, the receptors are reactivated by noradrenaline and they participate in message communication thereby curing depression-related ailments.
• Chlordiazepoxide and meprobamate are mild tranquilizers that are used to relieve tension.
• Equanil, another tranquilizer, is used in controlling mental depression and high blood pressure. Two other important tranquilizers are valium and serotonin.

• Derivatives of barbituric acid like veronal, amytal, nembutal, luminal, and seconal belong to an important class of tranquilizers called barbiturates.
• Barbiturates are hypnotic and hence, act as sleep-inducing agents.

Analgesics: Drugs that reduce or completely cure pain without causing unconsciousness, mental confusion, paralysis, incoordination among muscles, or any other disruption in the nervous system are called analgesics.

Analgesics Are Classified Into Two Groups:

1. Non-narcotic analgesics and
2. Narcotic analgesics.

Non-narcotic Analgesics

Non-narcotic Analgesics Definition: Drugs that relieve pain without causing any after-effects like drowsiness, sleepiness, or addiction are known as non-narcotic (non-addictive) analgesics.

Non-narcotic Analgesics Example: Two extremely important non-narcotic analgesics are aspirin (2-acetoxybenzoic acid or acetylsalicylic acid) and paracetamol (4-acetamidophenol).

Non-narcotic Analgesics Application: The chemical compound prostaglandins causes inflammation of the body tissues leading to pain.

• Aspirin prevents the synthesis of prostaglandins. These drugs are effective in treating arthritis (joint pain). Apart from this, these drugs act as antipyretics (capability of reducing fever) and also prevent blood platelet coagulation.
• Aspirin is extensively used as a preventive drug for heart attacks due to its anti-blood clotting properties.

Non-narcotic Analgesics Harmful Effects: Aspirin is considered detrimental to the body since it causes liver toxicity.

• Sometimes, aspirin is responsible for bleeding in the stomach leading to irritation and pain in the stomach.
• Due to these reasons, nowadays instead of aspirin, analgesics like naproxen, ibuprofen, and diclophenac sodium are used.
• Aspirin should never be ingested on an empty stomach. It gets hydrolyzed in the stomach and produces salicylic acid and acetic acid which gets absorbed by the cell wall.
• This causes injury and bleeding in the stomach. Hence, it is advisable to take an antacid along with aspirin.
• Despite having such disadvantages, aspirin is still used for preventing heart attacks and pain during angina (chest pain due to improper blood circulation).

Narcotic Analgesics

• Narcotic Analgesics Definition: Drugs that relieve pain when applied in small doses but cause drowsiness and addiction among patients are known as narcotic analgesics.
  • Narcotic Analgesics Example: Morphine and codeine are alkaloids obtained from the opium plant which are excellent pain-relievers. Morphine diacetate commonly known as heroin, is a powerful analgesic but at the same time is extremely addictive.

Narcotic Analgesics Application: Narcotic analgesics are primarily used in relieving post-operative pain, cardiac pain, cancer pain, and pain during childbirth.

Narcotic Analgesics Harmful Effects: Intake of narcotic analgesics above the optimum level leads to senselessness, seizures, and even death of the patients. Since these drugs originate from opium, they are also known as opiates. These alkaloids, being addictive, are not sold without a prescription to prevent their misuse.

Antimicrobials

Antimicrobials Definition: Drugs used to cure diseases caused by microorganisms like bacteria, viruses, and fungi are known as antimicrobial drugs.

These are antibacterial drugs, antifungal agents or drugs, and antiviral agents.

Discovery Of Antimicrobial Drugs: In the nineteenth century, German biologist Paul Ehrlich assumed that chemicals that can create an adverse effect on disease-producing bacteria but not on the host, (like human beings or animals) may be used as drugs to cure bacteriological ailments.

• During his search for such compounds, he analyzed various arsenic-based compounds and finally was successful in developing the drug arsphenamine (also known as salvarsan). This drug is used in the treatment of syphilis.
• Paul Ehrlich was awarded the Nobel Prize for Medicine in the year 1908 for this outstanding work. Although salvarsan is lethal to human beings, it is very effective on spirochete, the bacteria responsible for syphilis.
• Ehrlich at the same time was working on azodyes and observed a structural similarity between salvarsan and azodyes. He noticed that in azodyes there is the —N=N— linkage similar to that of the — As=As—
  linkage in salvarsan. He also observed that tissues were selectively colored by dyes.
• Based on these observations, Ehrlich began looking for compounds that had structural similarities with azodyes and could selectively bind to bacteria.

In 1932, he successfully formulated an effective antimicrobial drug similar to salvarsan, known as prontosil.

• These observations led to the belief among chemists that there is a definite relation between the structure and activity of a particular chemical compound.
• It was discovered that in the body, prontosil is converted to sulphanilamide which is the true active compound.
• This led to the discovery of the sulpha drugs and eventually, a whole range of sulphonamide analogs was synthesized. Sulphapyridine is considered the most effective sulpha drug.

Control Of Microbial Diseases

1. Drugs that can kill microbes present in the body i.e., bactericidal drugs are used to control diseases caused by microorganisms.
2. Bacteriostatic drugs are also used to prevent any kind of microbial growth in the body.
3. Microbial diseases can be prevented by increasing the resistance of the body to any infection.

Primarily, three types of antibacterial drugs are used. They are

1. Antibiotics,
2. Antiseptics and
3. Disinfectants.

Antibiotics: Antibiotics constitute an important class of drugs and are termed life-saving drugs. In modern times, no treatment is considered feasible without the application of antibiotics.

• Antibiotics are chemical compounds that originate from bacteria, fungi, and molds and at low concentrations can either kill or inhibit the growth of infection-causing microorganisms.
• Nowadays, antibiotics are prepared by a completely synthetic process or by biosynthesis.

Discovery Of Antibiotics: In 1929, British scientist Alexander Fleming was the first to discover the antibiotic penicillin. In 1942, penicillin was clinically used for the first time on humans. At present, molds like Penicillium notation and Penicillium cryogenic are cultured to prepare penicillin.

Depending upon the —R group, different types of penicillin are obtained. To date, 6 types of natural penicillin have been found.

Different Types Of Natural Penicillin:

Penicillins are called β-lactam antibiotics since a 4-membered amide ring is present in their molecule. Penicillin deactivates the enzymes that participate in the synthesis of bacterial cell walls thereby destroying those bacteria.

Types Of Antibiotics

Antibiotics are of two types—

1. Bactericidal: Kills the microorganisms and
2. Bacteriostatic: Prevents the growth of microorganisms.

Types Of Antibiotics Example:

1. Bactericidal: Penicillin, Aminoglycosides or Streptomycin, Ofloxacin.
2. Bacteriostatic: Chloramphenicol, Erythromycin, Tetracyclin.

Classification Of Antibiotics According To Their Activity Range

The range of bacteria and microorganisms affected by a particular antibiotic is expressed as its spectrum of action. Antibiotics are classified into three groups based on their range of activity.

1. Broad-spectrum Antibiotics: Antibiotics that can kill or inhibit the growth of a wide range of gram-positive and gram-negative bacteria thereby curing an array of bacterial infections fall under the category of broad-spectrum antibiotics.
   1. Broad-spectrum Antibiotics Example: Tetracyclin, Ofloxacin, Chloramphenicol, Ampicillin, Amoxycillin. The last two antibiotics are synthetic modifications of penicillin.
   2. Diseases like typhoid, dysentery, urinary tract infections, meningitis, and pneumonia are treated by chloramphenicol.
2. Narrow-spectrum Antibiotics: Antibiotics that kill or inhibit either gram-positive or gram-negative bacteria are called narrow-spectrum antibiotics.
   1. Narrow-spectrum Antibiotics Example: Penicillin is used in the treatment of a wide range of infections caused by gram-positive bacteria.
   2. These antibiotics are very effective against infections caused by cocci bacteria, pneumonia, bronchitis, and throat infections.
3. Limited-spectrum Antibiotics: Drugs that are effective against a single microorganism or a disease are known as limited-spectrum antibiotics.
   1. Limited-spectrum Antibiotics Example: Dysidazirine is toxic to cancer cells only.

Antiseptics And Disinfectants

Antiseptics

Antiseptics Definition: Chemical compounds that either kill microbes or prevent their growth in an infected part of the body without causing any adverse effect on the body tissues are called antiseptic

Antiseptics Example:

1. Furacin and soframycin are antiseptic creams used on wounds, cuts, and infected skin surfaces.
2. Dettol, a mixture of chloroxylenol and terpineol is a widely used antiseptic.
3. Bithionol (also known as bitonal) is added to soaps to make them antiseptic.
4. Iodine is a powerful antiseptic. Its 2-3% solution in an alcohol-water mixture is known as a tincture of iodine and is applied on wounds.
5. Iodoform (CHI₃) is also used as an antiseptic.
6. A dilute aqueous solution of boric acid is used as an antiseptic for the eyes.
7. Savlon, a commonly used antiseptic is a solution of chlorhexidine gluconate and cetrimide.
8. 2-5% aqueous solution of merbromin, known as mercurochrome is an effective antiseptic for skin and mucous membranes.
9. Phenyl salicylate or salol is used in treating throat ailments.
10. Some oxidizing agents are also used as antiseptics. For example, a very dilute aqueous solution of KMnO₄ (-0.1%) and a dilute aqueous solution of H₂O₂ or perhydrol are used for washing wounds.

The decomposition of organic compounds in the body and mouth leads to malodor and to prevent this, mouth fresheners and breath purifiers containing antiseptics are used. Antiseptics are also added to toiletries like toothpaste, facepowder, and creams.

Disinfectants

Disinfectants Definition: Chemical compounds that destroy microorganisms and are harmful to living tissues are used to disinfect non-living objects like household floors, drainage pipes, bathrooms, and toilets are called disinfectants.

Disinfectants Example:

1. Phenyl is a very common disinfectant.
2. Gaseous formaldehyde is used as a disinfectant in hospitals.
3. Suspension of cresol in soap solution (known as lysol) acts as an excellent floor disinfectant.

Compounds Acting As Both Antiseptic And Disinfectant: A chemical compound can act as an antiseptic or a disinfectant depending on the concentration in which it is used.

Antiseptic And Disinfectant Example:

1. 0.2% phenol solution is used as an antiseptic but 1% phenol solution is used as a disinfectant.
2. 6% H₂O₂ solution is used as an antiseptic while stronger solutions (>30%) are used as bleaching agents (disinfectants).

Difference Between Antiseptics And Disinfectants:

Antifertiuty Drugs

Population is a long-standing problem in the current global scenario and is considered the root cause behind many social problems like shortage of food, pollution, and unemployment.

• Hence, family planning is essential for birth control and it poses a tough challenge for the government to execute this concept.
• In this context, antifertility drugs have gained much importance. Chemical compounds that prevent pregnancy are called antifertility drugs.
• Birth control pills or oral contraceptives are synthesized from these drugs.

Antifertiuty Drugs Example:

1. Antifertility drugs are prepared by mixing synthetic derivatives of estrogen and progesterone. These are hormones and are more powerful than natural hormones. Estrogen controls the menstrual cycle while progesterone prevents ovulation.
2. In commonly used birth control pills, norethindrone (progesterone derivative) and nostril or ethynylestradiol (estrogen derivative) are present. The unwanted side effects of these pills are the formation and enlargement of tumors and complications in the blood vessels of the heart.

Another important antifertility drug is mifepristone. It is an artificially synthesized steroid.

Different Classes Of Drugs And Their Actions:

Chemicals In Food

Food Additives

Food Additives Definition: Chemical compounds that are added to food to improve their preservation value, nutritional value, appearance, taste, and odor are called food additives.

Some Important Food Additives:

1. Pood colors;
2. Flavors and sweetening agents;
3. Fat emulsifiers and stabilising agents:
4. Antistaling agents and bleaching agents;
5. Antioxidants;
6. Preservatives;
7. Nutritional supplements like minerals, vitamins, and amino acids.

Apart from nutritional supplements, none of the mentioned food additives have any nutritional importance. These are added just to increase the shelf life of packaged food and to enhance the visual appeal of food items t.e, the cosmetic valve.

Artificial Sweeteners: Cane sugar (sucrose) is considered to be the most commonly used natural sweetener.

• However diabetic patients and people concerned about their calorie intake nowadays prefer artificial sweeteners as an alternative to cane sugar since the latter is harmful to their health.
• These sweetening agents do not participate in any biochemical reactions in the body and are excreted out in unchanged conditions. Hence, they do not contribute to calorie generation in the body and are also known as calorie-free sweeteners.

Saccharin: Orftho-sulphobenzimide, commonly known as saccharin is a very popular artificial sweetener. Saccharin is insoluble in water. Therefore, the sodium and calcium salts of saccharin are used as artificial sweeteners as they are water-soluble.

• Saccharin is a white solid with a melting point of 224°C. Its sweetness is 550 times higher than ordinary cane sugar. It is non-biodegradable and has no calorific value.
• It does not undergo any change within the body and is excreted out through urine.
• This is an appropriate sweetener for diabetic and obese people. But prolonged intake of saccharin may lead to cancer and hence, it has been banned in many countries.

Aspartame: Perhaps the most successful and widely used artificial sweetener is aspartame, the methyl ester of the dipeptide of aspartic acid [HOOC-CH₂-CH(NH₂)COOH] and phenylalanine [C₆H₅CH₂CH(NH₂)COOH].

Aspartame is 100 times sweeter than cane sugar. Since it gets decomposed at high temperatures, aspartame is mostly used in cold food and soft drinks. People suffering from phenylketonuria should avoid aspartame.

Alitame: It is more stable than aspartame and is 2000 times sweeter than cane sugar. The difficulty in using sweeteners like alitame is to control the sweetness while adding to different food items.

Sucralose: It is the trichloro derivative of sucrose. It is stable at cooking temperature and 600 times sweeter than cane sugar. It is not responsible for any calorie generation in the body and does not cause tooth decay.

Cydamate: N-cyclohexylsulphamate or cyclamate is seven times sweeter than cane sugar. A 10: 1 mixture of cyclamate and saccharin imparts more sweetness than its components and is a commonly used artificial sweetener. But this mixture can cause tumor formation in the body and hence, its use is banned.

L-Hexose: Like D-sugars, L-sugars are also sweet but they do not generate any calories or energy in the body since enzymes required for their metabolism are absent in the human body. They are excreted from the body through urine. All eight L-hexoses are used as artificial sweeteners.

L-Hexose Example:

Comparison Of Artificial Sweeteners Concerning Cane Sugar

Food Preservatives

Food Preservatives Definition: Chemical compounds that inhibit microbial (bacteria, fungus, molds) growth on food items and also prevent them from getting oxidized by air are called food preservatives.

• Compounds like common salt, sugar, and oil when added to food create an adverse environment for microbial activity, thereby preserving the food items.
• Food preservation by adding a sufficient amount of table salt is known as salting.
• Table salt is used in preserving edibles like green mango, gooseberry (commonly known as amla), tamarind, fish, meat, butter, and cheese.
• Fruits like mango, apple, and strawberry are preserved in sugar syrup. Vinegar, oils, and citric acid are used to preserve pickles, jams, and squash.

Sodium Benzoate (C₆H₅COONa): Sodium Benzoate is commonly used as a preservative in soft drinks and acidic foods like pickles, fruit juices, salad dressings, and jams. Sodium benzoate metabolizes to give hippuric acid (C₆H₅CONHCH₂COOH) inside the body and the latter is excreted from the body through urine.

Sodium Or Calcium Propionate: Sodium propionate (CH₃CH₂COONa) or calcium propionate [(CH₃CH₂COO)₂Ca] prevents bacterial and fungal growth in flour-based food items.

Sodium Metabisulphite (Na₂S₂O₅): It is used as a preservative in jams, pickles, and squashes. Dissociation of this compound gives SO₂ which dissolves in water to form sulphurous acid. This acid prevents microbial growth in food. Instead of Na₂S₂O₅, SO₂ gas is also used in preserving fruits and vegetables.

Na₂S₂O₅ → Na₂SO₃ + SO₂; SO₂ + H₂O → H₂SO₃

Sorbic Acid And Potassium Sorbate: Sorbic acid (CH₃CH=CH-CH=CHCOOH) or potassium sorbate (CH₃CH=CHCH=CHCOOK) is used as preservatives in butter, cheese, flour-based products, fish, and meat to inhibit the growth of fungus and yeast.

Epoxides: Ethylene oxide and propylene oxide are used as preservatives in dry foods (having low moisture content) like spices, nuts, and dry fruits.

P-Hydroxybenzoate Ester:  Methyl, ethyl, propyl, and heptyl esters of p-hydroxybenzoic acid are used as preservatives in cooked food, soft drinks, pickles, fish, and meat-based food items. These preservatives are less effective on bacteria compared to molds and yeasts.

Antioxidants

Antioxidants Definition: Chemical compounds that increase the shelf lives of food by preventing oxidation of unsaturated fats and oils are called antioxidants.

• Antioxidants are also a type of food preservative. Oil is a mixture of double-bonded, long-chained triglycerides. It gets oxidized by oxygen to give hydroperoxides.
• Due to this reason, food containing unsaturated oils starts to decompose in the presence of air emitting a foul smell when kept for a long time.
• Antioxidants are prohibitions of free radicals and have more affinity towards oxygen than food items. They act as preservatives by lowering the rate of self-oxidation of the oils present in food.

Antioxidants Example: Among the commonly used antioxidants, BHT (butylated hydroxytoluene) and BHA (butylated hydroxyanisole) are the most important. The structures of these two antioxidants are given below—

Cleansing Agents

Chemical compounds that decrease the surface tension of water and remove oil and fat-based dirt by forming emulsions are called surface active agents or surfactants. These are of two types:

1. Soaps and
2. Synthetic detergents.

Soaps

Sodium and potassium salts of long-chain fatty acids like lauric acid, palmitic acid, stearic acid, oleic acid and linoleic acid are called soaps. The potassium salts are referred to as soft soaps or toilet soaps while the sodium salts are known as hard soaps.

Saponification: Fats are glycerol triesters of long-chained fatty acids and these triesters are also known as glycerides. When a glyceride is heated with an aqueous solution of NaOH, it undergoes hydrolysis to form soaps. This is known as saponification.

Saponification Example:

• Soap obtained in this hydrolytic reaction remains in the colloidal state. It is precipitated by adding NaCl. The precipitated soap is separated by filtration.
• The filtrate contains glycerol that is separated by fractional distillation. Only sodium and potassium soaps are water-soluble and act as cleansing agents.
• Potassium soaps are more sensitive to the skin than sodium soap. Potassium soaps (soft soaps) are prepared by hydrolyzing fats with an aqueous solution of potassium hydroxide.

Types Of Soaps

1. Toilet Soap: High-quality fats and oils are used to prepare toilet soaps. A variety of colors and fragrances are added to make them attractive.
2. Floating Soap: Floating soaps are prepared by whisking air bubbles into the soap before its hardening, thus making the density of soap lighter than water.
3. Transparent Soap: These are prepared by dissolving the soap in ethanol and then evaporating the excess solvent.
4. Medicated Soap: Medicated soaps are prepared by adding antiseptics like Dettol and Savlon. Sometimes, deodorants like bitonal are also added to soaps.
5. Shaving Soap: Glycerol is added to these soaps to prevent rapid drying. Rosin, a type of gum is added to shaving soaps which forms sodium rosinate and has excellent foaming properties.
6. Laundry Soap: Laundry soaps are prepared by adding fillers like sodium resinate, sodium silicate, borax, and sodium carbonate.
7. Soap Chips And Granules: Soap chips are prepared by passing a thin sheet of melted soap through a cool cylinder and scouring off the soap into small pieces. Soap granules are miniature versions of soap bubbles.
8. Soap Powders And Scouring Soaps: These soaps are prepared by adding scouring agents like powdered pumice or finely divided sand and builders like sodium carbonate and trisodium phosphate to the soap.

Builders: Chemical compounds that are added to soap to make it more effective as a cleansing agent are called builders.

Builders Example: Sodium tripolyphosphate (Na₅P₃O₁₀) is an important phosphate builder Approximately, 20-45% tripolyphosphate is added to detergent powders and liquid detergents.

• It removes Ca²⁺ and Mg²⁺ from hard water by forming stable chelates with them, thus, converting it into soft water. This process is known as sequestration.
• Hence, Ca²⁺ and Mg²⁺ do not form any precipitate with soaps in the presence of Na₅P₃O₁₀.
• Hydrolysis of the anion of Na₅P₃O₁₀(P₃O₁₀^5-) makes the solution basic thus dissolving the dirt and grease and making the detergent more effective.

Advantages And Disadvantages Of Using Soap

Using Soap Advantages: Soap is an excellent cleansing agent which is completely biodegradable. Microorganisms in the dirty water oxidize soap into CO₂ and hence, soap is not responsible for any type of water pollution.

Using Soap Disadvantages: Soaps cannot be used in hard water, since, Ca²⁺ and Mg²⁺ ions present in it precipitate as white calcium and magnesium salts of fatty acids.

• This insoluble white precipitate or scum sticks to the fibers of the clothes and makes it difficult for soap to remove stains and grease from it. For the same reason, hair when washed with hard water appears dull.
• Soaps cannot be used in an acidic medium, as in the presence of H⁺ ions, they precipitate as their corresponding fatty acids.
• As a result, the precipitated fatty acid reduces the cleansing capacity of the soap, and the soap is also wasted.

Synthetic Detergents

Synthetic detergents are soapless soaps i.e., they have all the cleansing properties of soaps without containing any soap. Since these are prepared artificially, they are known as synthetic detergents or syndets.

Synthetic Detergents Definition: Ammomum, sulphonate, or sulphate salts of long-chained hydrocarbons containing 12-18 carbon atoms are called detergents.

These are non-biodegradable and cause water pollution.

Advantages And Disadvantages Of Using Detergents

Using Detergents Advantages:

1. Synthetic detergents, unlike soaps, can be used in both hard water (because their calcium and magnesium salts are soluble in water) and soft water.
2. Synthetic detergents can be used in acidic medium but soaps cannot be used in acidic medium.
3. Synthetic detergents being more soluble in water than soaps act as better foaming agents.
4. They can reduce the surface tension of water to a greater extent and hence, are considered better cleansing agents.

Using Detergents Disadvantages: Soaps are easily decomposed by microorganisms (biodegradable) but, long hydrocarbon chains containing detergents having a large number of side chains are non-biodegradable and cause water pollution.

• Synthetic detergents containing a large number of side- chains attached to the long-chained hydrocarbon structure are resistant to microbial attack and thus, are non-biodegradable.
• But, detergents having simple hydrocarbon chains are easily degraded by microbes. To avoid pollution, several side chains of the hydrocarbon chain in detergents should be reduced.
• Polyphosphates are usually added to detergents to soften water. They form soluble complexes with Ca²⁺ and Mg²⁺ ions thereby making them ineffective.

These polyphosphates nourish bacteria that grow excessively and deplete water, thus killing fish and other small aquatic animals.

Cleansing Action Of Soaps And Detergents: The working action of soaps and detergents is the same because of their structural similarity. Both remove dirt, oil, and grease by forming micelles. For more information.

Enzyme-based detergents: Enzymes are complex biomolecules that act as catalysts in different biochemical reactions. In 1914, a German chemist first observed that detergent when mixed with the enzyme trypsin, rapidly removed oil and grease from clothes.

• This led to the discovery of many enzyme-based detergents. Nowadays, detergents containing alkalase, amylase, protease, and lipase are used.
• Amylase is used for removing dirt caused by starch-based materials. Protease and lipase remove dirt caused by proteins and lipids.
• Enzyme-based detergents are more effective than soaps and ordinary detergents. Tough stains caused by oil, egg, sauce, and blood are easily removed by enzyme-based detergents.
• But stains of coffee, tea & some fruits cannot be removed by these detergents. Oxidants like perborate and percarbonate are added to detergents to remove stains by oxidation.

Classification, Synthesis, And Uses Of Detergents

Classification Of Detergents

Synthetic detergents are of three types:

1. Anionic detergents,
2. Cationic detergents,
3. Non-ionic detergents.

Anionic Detergents: In these detergents, a large part of the molecule remains in the anionic form and the anionic part of the molecule is directly involved in the cleansing action. They are classified into two groups.

Sodium Alkyl Sulphates: Long-chained alcohols  (containing 12-18 carbon atoms) react with concentrated H₂SO₄ to give the corresponding hydrogen sulphate. The latter is then neutralized by NaOH to obtain sodium alkyl sulfates.

Sodium Alkyl Sulphates Example: Sodium lauryl sulphate (C₁₁H₂₃CH₂OSO₃^–Na⁺), sodium stearyl sulphate [CH₃(CH₂)₁₆CH₂OSO₃^–Na⁺]. These detergents are completely biodegradable.

Synthesis:

Sodium Alkylbenzene Sulphonates: These are sodium salts of long-chained alkylbenzene sulphonic acid.

• Benzene undergoes Friedel-Crafts reaction in the presence of acylating agents like alkyl halides, alkenes, or alcohols to give alkylbenzenes.
• The latter undergoes sulphonation to give alkylbenzene sulphonic acid which is neutralized with NaOH to prepare sodium alkylbenzene sulphonates.

Sodium Alkylbenzene Sulphonates Example: Sodium 4-dodecylbenzene sulphonate (SDS).

Synthesis:

Another important detergent of this class is sodium 4-(1-methyiundecyl) benzenesulphonate.

Classification Of Detergents Uses: Long-chained alkylbenzene sulphonates are mostly used in household cleaning. Some anionic detergents are also used in toothpaste.

Soft Or LAS Detergents And Hard Or ABS Detergents: Benzene-sulphonates having unbranched carbon chains or single-branched carbon chains (i.e., the phenyl group is present in any secondary position) are highly biodegradable.

• These are called soft detergents. On the other hand, benzene sulphonates having highly branched carbon chains are non-biodegradable and are called hard detergents.
• Soft detergents are also called LAS detergents (linear alkylbenzene sulphonates) and hard detergents are called ABS detergents (alkylbenzene sulphonates).
• Sodium 4-(1,3,5,7-tetramethyloctyl)-benzenesulphonate is an ABS detergent.

Cationic Detergents: In cationic detergents, a large part of the molecule remains in cationic form and the cationic part of the molecule participates in cleansing action.

• These detergents are also called inverted soaps since the cationic part of the molecule is responsible for removing dirt and grease.
• Cationic detergents are quarternary ammonium salts of amines with acetates, chlorides, or bromides as anionic parts having many long-chained alkyl groups.

Cationic Detergents Example: Some important cationic detergents are:

Cationic Detergents Uses: Cationic detergents have germ-killing properties and hence, are used in manufacturing shampoos, mouthwashes, and antibacterial soaps. Cationic detergents being expensive are not used by many.

Non-Ionic Detergents

Non-ionic detergents do not have any ions and are esters of alcohols and fatty acids having high molecular masses.

Non-Ionic Detergents Example: Polyethylene glycol stearate, lauryl alcohol ethoxylate, pentaerythritol monostearate.

Non-Ionic Detergents Synthesis: Ethylene glycol reacts with ethylene oxide to give polyethylene glycol which on esterification with stearic acid gives polyethylene glycol stearate.

These types of detergents can also be prepared by reacting long-chained alcohols with excess ethylene oxide in the presence of a base, for example., lauryl alcohol ethoxylate is prepared by reacting lauryl alcohol with ethylene oxide.

These detergents dissolve in water by forming H-bonds with O-atoms in the polyether part of the compound. Ethoxylates are converted to sulphates and can be used as their Na-salts.

Pentaerythritol monostearate can be prepared by the following reaction.

Non-Ionic Detergents Uses: Non-ionic detergents are used as dish-cleaning liquids. Their cleansing action is similar to soap and they remove oils and grease by micelle formation.

Class 12 Chemistry Unit 16 Chemistry In Every Day Life Very Short Questions And Answers

Question 1. Name two alkaloids that are used as analgesics.
Answer: Codeine and morphine.

Question 2.  Name the antibiotics used for the treatment of tuberculosis and typhoid.
Answer: Streptomycin and chloramphenicol respectively.

Question 3. Name a drug which is both an analgesic & antipyretic
Answer: Aspirin

Question 4. What type of drug is chloramphenicol?
Answer: It is an antibiotic (antimicrobial).

Question 5. Why is bitonal added to cosmetic soaps?
Answer: Bithional (antiseptic) removes the malodor arising from microbial decomposition of organic matter in the skin.

Question 6. Name a broad-spectrum antibiotic and two diseases on which it is effective.
Answer: Chloramphenicol; typhoid and dysentery.

Question 7. Name a drug used in the treatment of mental disorders.
Answer:  Iproniazid.

Question 8. Give an example of an antihistamine drug.
Answer: Brompheniramine (Dimetapp).

Question 9. Name an estrogen derivative that is a part of an oral contraceptive.
Answer: Ethynylestradiol (Novestrol)

Question 10. Name an analgesic used to prevent heart attacks.
Answer:  Aspirin.

Question 11. Give an example of a sulpha drug
Answer: Sulphapyridine

Question 12. Why BHA is added to butter?
Answer: BHA, an antioxidant is added to butter to increase its storage life.

Question 13. Due to which structural feature, detergents become non-biodegradable?
Answer: In detergents, the hydrocarbon chain is highly branched which makes them non-biodegradable.

Question 14. Why non-biodegradable detergents are not used?
Answer: Non-biodegradable detergents cause water pollution.

Question 15. What type of forces are involved in the binding of substrate to the active site of the enzyme?
Answer: Van der Waals forces, ionic bonding, hydrogen bonding dipole-dipole interaction, etc.

Question 16. The methyl ester of a dipeptide is 100 times sweeter than cane sugar. What are the constituents a -amino acids of this dipeptide?
Answer: Artificial sweetener aspartame is methyl ester of the dipeptide which is derived from aspartic acid and phenylalanine.

Question 17. Hair shampoos belong to which class of synthetic detergents?
Answer: Hair shampoos are made up of cationic detergents. For example, cetyltrimethylammonium bromide.

Question 18. Which analgesics are called opiates?
Answer: Narcotic analgesics obtained from opium poppy.

Question 19. Classify the following as artificial sweeteners, preservatives, soaps, and detergents: sodium palmitate, sucralose, salt of sorbic acid, and cetyltrimethylammonium bromide.
Answer: Sodium palmitate—-soap; Sucralose—artificial sweetener; Salt of sorbic acid—preservative; cetyltrimethylammonium bromide-detergent.

Question 20. What structural unit makes detergents non-biodegradable?
Answer: Branching in the hydrocarbon chain.

Question 21. Name the artificial sweetener which is 550 times sweeter than cane sugar.
Answer: Saccharin is an artificial sweetener that is 550 times sweeter than cane sugar.

Question 22. Mention one use of borax and oil of wintergreen.
Answer: Borax is used for preparing antiseptic soaps. The oil of wintergreen is used as an analgesic.

Question 23. What is the chemical nature of common antacids?
Answer: Common antacids are metal hydroxides and bicarbonates.

Question 24. Give an example of a soap and indicate its polar and nonpolar parts.
Answer: Sodium stearate is an example of soap.

Question 25. By what type of reaction do the common antacids destroy the excess acid in the stomach?
Answer: Common antacids destroy the excess acid of the stomach through an acid-base neutralization reaction.

Question 26. What is the purpose of adding food preservatives to packaged food?
Answer: Food preservatives are added to packaged foods to prevent spoilage due to microbial (bacterial/fungal etc.) growth.

Question 27. Explain the term, target molecules, or drug targets as used in medicinal chemistry.
Answer: Target molecules or drug targets are macromolecules such as proteins, carbohydrates, lipids, and nucleic acids with which the drug interacts in our body to cause therapeutic effects.

Question 28. Name the macromolecules that are chosen as drug targets.
Answer: Proteins, carbohydrates, lipids and nucleic acids.

Question 29. Name a substance that can be used as an antiseptic as well as a disinfectant.
Answer: A 0.2% solution of phenol is used as an antiseptic while a 1% solution of phenol acts as a disinfectant.

Question 30.  What are the main constituents of Dettol?
Answer: Chloroxylenol and a-terpineol

Question 31. What is a tincture of iodine? What is its use?
Answer: A 2-3% solution of iodine in an alcohol-water mixture is called a tincture of iodine. It is used as an antiseptic.

Question 32. Why is the use of aspartame limited to cold foods and drinks?
Answer: Use of aspartame is limited only to cold foods and soft drinks as it decomposes at baking or cooking temperatures.

Question 33. Name the sweetening agent used in the preparation I of sweets for a diabetic patient..
Answer: Artificial sweeteners such as saccharin, aspartame alitame, etc., may be used.

Question 34. What problem arises in using alitame as an artificial sweetener?
Answer: Alitame is a high-potency artificial sweetener (about 2000 times sweeter than sucrose). So it is difficult to control the sweetness of the food to which it is added.

Question 35. Define the term chemotherapy.
Answer: The treatment of diseases using chemical compounds is called chemotherapy.

Question 36. How does aspirin help in preventing heart attacks?
Answer: Heart attacks are mainly caused by to clotting of blood in arteries. Aspirin stops coagulation of blood and blood clotting. This is how it prevents heart attack.

Question 37. Why sulpha drugs are not true antibiotics?
Answer: Sulpha drugs inhibit microbial growth but do not kill them, ie., these drugs are bacteriostatic. Hence, sulpha drugs are not true antibiotics.

Question 38. How does aspirin act as an analgesic?
Answer: Aspirin inhibits the synthesis of prostaglandins which stimulate inflammation of the tissue. Thus, it cures pain.

Question 39. Name an artificial sweetener that is a derivative of sucrose.
Answer: Sucralose (a trichloro derivative of sucrose) is an artificial sweetener. It is 600 times sweeter than sucrose.

Question 40. Name two α-amino acids that form a dipeptide that is 100 times sweeter than cane sugar.
Answer: Aspartame (an artificial sweetener) is the methyl ester of the dipeptide derived from aspartic acid and phenylalanine.

Question 41. Aspartame is unstable at cooking temperature, where would you suggest aspartame to be used for sweetening?
Answer: In cold foods and soft drinks.

Question 42. Sodium salts of some acids are very useful as food preservatives. Suggest a few such acids.
Answer: Sodium salts of benzoic acid, sorbic acid, and propanoic acid are used as food preservatives.

Question 43. What is the average molecular mass of drugs?
Answer: ~100-500u

Question 44. Write the uses of medicines.
Answer: Medicines are used in the diagnosis, prevention, and treatment of diseases.

Question 45. What are antiseptics?
Answer: Antiseptics are chemicals that either kill or prevent the growth of microorganisms and can be applied safely to living tissues.

Question 46. Which type of drugs come under antimicrobial drugs?
Answer: Antiseptics, disinfectants, and antibiotics.

Question 47. Where are receptors located?
Answer: Receptors are embedded on the outer surface of the cell membrane.

Question 48. What is the harmful effect of hyperacidity?
Answer: Hyperacidity causes the development of ulcers in the stomach.

Question 49. Which site of an enzyme is called an allosteric site?
Answer: Sites different from the active site of an enzyme where a molecule (called inhibitor) can bind and affect the shape of the active site are called allosteric sites.

Question 50. What type of forces are involved in the binding of substrate to the active site of the enzyme?
Answer: Ionic bonding, H-bonding, van der Waals interactions, and dipole-dipole interactions.

Question 51. What is the commonality between the antibiotic arsphenamine and azodye?
Answer: Arsphenamine possesses (—As=As— ) linkage which resembles (—N=N— ) linkage in azodye.

Question 52. Which class of drugs is used in sleeping pills?
Answer: Tranquilizers are used in sleeping pills.

Question 53. Aspirin is a pain-relieving antipyretic drug but can be used to prevent heart attacks. Explain.
Answer: Aspirin has anti-blood clotting action because it prevents platelet coagulation. So, it is widely used to prevent heart attack.

Question 54. Both antacids and antiallergic drugs are antihistamines but they cannot replace each other. Explain why?
Answer: Since antacids and antiallergic drugs work on different receptors, they cannot replace each other. Antacids cure acidity while antihistamines prevent allergy.

Question 55. What is a soft soap?
Answer: Potassium salts of fatty acids (for example., palmitic acid, stearic acid, etc.).

Question 56. If soap has a high alkali content it irritates the skin. How can the amount of excess alkali be determined? What can be the source of excess alkali?
Answer: Acid-base titration (employing phenolphthalein as an indicator) can be used to determine the excess amount of alkali present in the soap. The excess alkali left after the alkaline hydrolysis of oil is the source of alkalinity in soap.

Question 57. Explain why sometimes foaming is seen in river water near the place where sewage water is poured after treatment.
Answer: Detergents (which are not biodegradable) persist in water even after sewage treatment and cause foaming in river water.

Question 58. Which category of synthetic detergents is used in toothpaste?
Answer: Anionic detergent.

Question 59. Hair shampoos belong to which class of synthetic detergent?
Answer: Cationic detergent.

Question 60. Dishwashing soaps are synthetic detergents. What is their chemical nature?
Answer: Non-ionic detergent

Question 61. How does the branching of the hydrocarbon chain of synthetic detergents affect their biodegradability?
Answer: Biodegradability increases as branching in the hydrocarbon chain decreases.

Question 62. Why is it safer to use soap from an environmental point of view?
Answer: Soaps are biodegradable while detergents containing branched hydrocarbon chains are quite stable (not degraded by microorganisms) thereby causing water pollution.

Question 63. What are analgesics?
Answer: Analgesics are neurologically active drugs that can reduce or abolish pain without causing impairment of consciousness, mental confusion, incoordination paralysis or some other disturbances of the nervous system.

Question 64. Which analgesics are called opiates?
Answer: Narcotic analgesics which are obtained from opium poppy are called opiates. Examples are morphine and its derivatives such as heroin and codeine.

Question 65. What is the medicinal use of narcotic drugs?
Answer: Since narcotic drugs relieve pain and produce sleep, these are used for the relief of post-operative pain, cardiac pain, pains of terminal cancer, and childbirth.

Question 66. What are antagonistic drugs?
Answer: Drugs that bind to the receptor site and inhibit its natural function are called antagonistic drugs.

Question 67. What is the mode of action of antimicrobial drugs?
Answer: Antimicrobials are drugs that can kill microorganisms such as bacteria, viruses, fungi, or other parasites. They can, alternatively, inhibit the pathogenic action of microbes.

Question 68. What is the difference between bathing soap and washing soap?
Answer: Bathing soaps are potassium salts of long-chain fatty acids and these are usually soft. On the other hand, washing soaps are sodium salts of long-chain fatty acids, and these
are usually hard.

Question 69. How are transparent soaps manufactured?
Answer: These are prepared by dissolving the soap in ethanol and evaporating the excess solvent.

Question 70. Mention one important use of meprobamate.
Answer: It is a mild tranquilizer for relieving mental stress

Question 71. Name a drug that is both an analgesic and an antipyretic.
Answer: Aspirin

Question 72. Choose the odd one out of the following list of compounds based on medicinal use. Chloroxylenol, phenol, chloramphenicol, bithional.
Answer: Chloramphenicol; since it is an antibiotic and the rest are antiseptics.

Question 73. The first antibiotic was prepared from which compound?
Answer: Penicillium notatum

Question 74. Omeprazole is used to cure which disease of the human body?
Answer: Hyperacidity

Question 75. What is added to a soap to impart antiseptic properties to it?
Answer: Bitcoin

Question 76. Give an example of a hormone and a neurotransmitter.
Answer: Adrenaline(hormone) and acetylcholine(neurotransmitter);

Question 77. Give the common name and use of morphine diacetate.
Answer:  Heroin; this is a narcotic analgesic

Question 78. Give an example of a bacteriostatic drug.
Answer: Sulphapyridine

Question 79. Give an example of an anti-fertility drug.
Answer: Mifepristone

Question 80. Name a drug that is used for curing depression.
Answer: Valium

Question 81. Give the name of a carbohydrate antibiotic.
Answer: Streptomycin

Question 82. Give an example of a narrow-spectrum antibiotic.
Answer: Penicillin

Question 83. Can chloramphenicol be termed as a broad-spectrum antibiotic?
Answer: Can be said because it either kills or inhibits the growth of a wide range of Gram-positive and Gram-negative bacteria.

Question 84. Which heterocyclic ring is present in sulphapyridine?
Answer: Pyridinering is present

Question 85. How many types of antibacterial drugs are present? What are those?
Answer: Two types: antibiotics and sulpha drugs

Question 86. Give the chemical name of ibuprofen along with its use.
Answer: 2-(4-isobutyl phenyl) propanoic acid and is a non-narcotic analgesic.

Question 87. Give an example of an antihistamine drug.
Answer: Brompheniramine (Dimetapp)

Question 88. What are tranquilizers? Give example.
Answer: Drugs that are used in relieving mental stress and tension and in treating mild or acute mental disorders are called tranquilizers, for example., serotonin.

Question 89. Name the iodine-based antiseptic obtained from the reaction between ethanol, I₂, and NaOH.
Answer: Iodoform (CHI₃)

Question 90. How many halogen atoms are present in sucralose?
Answer: 3 atoms of chlorine

Question 91. How many times alitame is sweeter than cane sugar?
Answer: Approximately 2000 times sweeter

Question 92. Name an artificial sweetener that has a ring containing sulphur.
Answer: Saccharin/Alitame

Question 93. Give an example of an antioxidant.
Answer:  BHT (Butylated hydroxytoluene)

Question 94. Mention the use of potassium sorbate.
Answer: It is a food preservative used for preserving butter, cheese, flour-based products, fish, and meat.

Question 95. Name a gas used in the preservation of fruits and vegetables.
Answer: Sulphur dioxide (SO₂)

Question 96. Name a preservative used for dry fruits.
Answer: Propylene oxide

Question 97. Mention the use of sodium metabisulphate.
Answer: It is a food preservative used for preserving jams, squashes, and pickles.

Question 98. Give an example of a soft soap.
Answer: Potassium stearate

Question 99. Cosmetic soaps are prepared by saponification of which oils?
Answer: Coconut oil

Question 100. How many types of soaps will be obtained from glyceryl oleopalmitostearate?
Answer:  3 types

Question 101. Indicate the hydrophilic and hydrophobic parts of a soap.
Answer: The hydrocarbon chain is lyophobic and —COO^–Na⁺ is lyophilic;

Question 102. Give an example of a cationic detergent.
Answer: Cetyl trimethyl ammonium bromide [(C₁₆H₃₃)N⁺(CH₃)₃]Br^–

Question 103. What is added during the preparation of shaving soaps?
Answer:  Glycerol

Question 104. Give an example of a non-biodegradable detergent.
Answer:  Sodium 4-(1,3,5,7-tetramethyloxyl) benzene sulphonate

Class 12 Chemistry Unit 16 Chemistry In Every Day Life Short Questions And Answers

Question 1. Pickles have a long shelf life and do not get spoiled for months—why?
Answer: Plenty of salt and cover of oil act as preservatives. These substances do not allow moisture and air to enter the material and hence bacteria cannot thrive on them. Therefore, pickles do not get spoiled for months together.

Question 2. What is the advantage of using antihistamines over antacids?
Answer:

• Antacids neutralize the excess acid in the stomach but do not control or cure the cause responsible for this excess secretion.
• Histamine interacts with the receptors on the stomach wall and releases pepsin and hydrochloric acid.
• However, when antihistamines are taken, they adversely affect the interaction between histamine and receptors and reduce acid secretion in the stomach or stop it completely.

Question 3. Why is paracetamol preferred over aspirin as an antipyretic?
Answer: Aspirin when ingested, hydrolyses into salicylic acid in the stomach which may lead to bleeding and finally ulcer may form.

But, paracetamol (calcium and sodium salts) being more soluble is less harmful to the body and is preferred over aspirin as an antipyretic.

Question 4. What are sulpha drugs? Give examples.
Answer:

• Drugs that are derivatives of the compound sulphanilamide are called sulpha drugs.
• They have antibacterial properties and are effective against diseases caused by cocci infections like dysentery, tuberculosis, etc.
• Some important sulpha drugs are sulphadiazine, sulphapyridine, and sulpha guanidine.

Question 5.  Boric acid is added to talcum powder (for babies) and chlorine is added to water in swimming pools. Why?
Answer:

• Boric acid is added to talcum powder which is used for babies because it is a mild antiseptic and inhibits microbial growth.
• Chlorine is a disinfectant and it kills the micro-organisms and controls algal growth in swimming pool water.

Question 6. Mention the use of Willow bark as a medicine.
Answer: Willow bark may be called nature’s aspirin. Its principal ingredient saline works the same way as aspirin by reducing inflammation and bringing down the fever (antipyretic).

Question 7. Why are detergents preferred over soaps?
Answer: Synthetic detergents can be used in hard water as well as in acidic solution and this is because sulphonic acids and their calcium and magnesium salts are soluble in water but fatty acids and their calcium and magnesium salts are insoluble in water.

Question 8. What will form aspirin on heating to about 55°C with acetic anhydride in the presence of a little cone? H₂SO₄? Write the structure of aspirin.
Answer:

 

Question 9. Mark the hydrophobic and hydrophilic parts of the following synthetic detergent.
Answer:

Question 10. On the occasion of World Health Day, Dr. Satpal organized a ‘health camp’ for the poor farmers living in a nearby village. After the check-up, he was shocked to see that most of the farmers suffered from cancer due to regular exposure to pesticides and many were diabetic. They distributed free medicines to them. Dr. Satpal immediately reported the matter to the National Human Rights Commission (NHRC). On the suggestions of NHRC, the government decided to provide medical care and financial assistance and set up super-specialty hospitals for treatment and prevention of the deadly disease in the affected villages all over India.

1. Write the values shown by Dr. Satpal NHRC.
2. What type of analgesics are chiefly used for the relief of pains of terminal cancer?
3. Give an example of an artificial sweetener that could have been recommended to diabetic patients.

Answer:

1. Dr. Satpal was concerned and distributed free medicines to the patients.
2. The valuable suggestion was offered by NHRC and so the government adopted preventive measures like medical care, financial assistance, and the setting up of super-specialty hospitals for affected villagers.

1. Aspirin,
2. Aspartame

Question 11. Ramesh went to a departmental store to purchase groceries. On one of shelves, he noticed sugar-free tablets. He decided to buy them for his grandfather who was a diabetic. There were three types of sugar-free tablets. Ramesh decided to buy sucralose which was good for his grandfather’s health.

1. Name another sugar-free table that Ramesh did not buy.
2. Was it right to purchase such medicines without a doctor’s prescription?
3. What quality of Remesh is reflected above?

Answer:

1. Aspartame,
2. No,
3. Social concern, empathy.

Question 12. Neeraj went to the departmental store to purchase groceries. On one of the shelves, he noticed sugar-free tablets. He decided to buy them for his grandfather who was a diabetic. There were three types of sugar-free tablets. He decided to buy sucralose which was good for his grandfather’s health.

1. Name another sugar-free tablet that Neeraj did not purchase.
2. Was it right to purchase such medicines without a doctor’s prescription?
3. What quality of Neeraj is reflected above?

Answer:

1. Saccharin,
2. No,
3. Social concern, social awareness.

Question 13. Due to a hectic schedule, Mr. Angad made his life full of tension and anxiety. He started taking sleeping pills to overcome the depression without consulting the doctor. Mr. Deepak, a close friend of Mr. Angad, advised him to stop taking sleeping pills and suggested to change his lifestyle by doing yoga, meditation, and some physical exercise. Mr. Angad followed his friend’s advice and after a few days, he started feeling better. After reading the above passage, answer the following:

1. What are the values (at least two) displayed by Mr. Deepak?
2. Why is it not advisable to take sleeping pills without consulting a doctor?

Answer:

1. Aware, concerned.
2. It is advisable not to take sleeping pills without consulting a doctor as they may cause severe side effects and lead to unknown health problems.

Question 14. Due to a hectic schedule, Mr. Singh started eating junk food in the lunch break and slowly became habitual of eating food irregularly to excel in his field. One day, during a meeting he felt severe chest pain and fell. Mr. Kh’anna, a close friend of Mr. Singh took him to the doctor immediately. The doctor diagnosed that Mr. Singh was suffering from acidity and prescribed some medicines. Mr. Khanna advised him to eat homemade food and change his lifestyle by doing yoga, meditation, and some physical exercise. Mr. Singh followed his friend’s advice and after a few days, he started feeling better. After reading the above passage, answer the following:

1. What are the values (at least two) displayed by Mr. Khanna?
2. Would it be advisable to take antacids for a long period? Give reason.

Answer:

1. Supportive, Aware
2. No, it is not advisable to take antacids for a long period because it would make the stomach alkaline, triggering the production of more acids.

Question 15.

1. Why is bitonal added to soap?
2. Aspartame, aspirin, sodium benzoate and paracetamol

Answer:

1. Bithional works as an antiseptic agent and reduces the odor generated by bacterial decomposition of organic matter on the skin.
2. Sodium benzoate

Question 16. Sleeping pills are recommended by doctors to patients suffering from sleeplessness but it is not advisable to take their doses without consultation with the doctor. Why?
Answer: Most of the drugs taken in doses higher than those recommended may cause harmful effects, act as potential poison, and even cause death. Therefore, it is not advisable to take its doses without consultation with the doctor.

Question 17. Concerning which classification has the statement, “ranitidine is an antacid” been given?
Answer: This statement refers to the classification according to the pharmacological effect of the drug, because any drug which is used to counteract the effect of excess acid present in the stomach is called an antacid.

Question 18. Why do we require artificial sweetening agents?
Answer: Artificial sweetening agents are used by diabetic patients and people who need to control their intake of calories. The reason is that artificial sweeteners do not participate in any biochemical reactions of the body and hence, do not contribute to calorie generation in the body.

Question 19. Write the chemical equation for preparing sodium soap from glyceryl oleate and glyceryl palmitate. The structural formulae of these compounds are given below.

1. (C₁₅H₃₁COO)₃C₃H₅ —Glyceryl palmitate
2. (C₁₇H₃₃COO)₃C₃H₅ —Glyceryl oleate

Answer:

Question 20. The following types of non-ionic detergents are present in liquid detergents, emulsifying agents, and wetting agents. Label the hydrophilic and hydrophobic parts of the molecule. Identify the functional group(s) present in the molecule.

Answer:

Functional groups are: 

1. Ether (—O—) and
2. Primary alcoholic (—OH) group.

Question 21. Why do we need to classify drugs in different ways?
Answer:

• Drugs are classified in different ways as different modes of classification are useful to different categories of persons dealing with drugs.
• For example, the classification of drugs based on pharmacological effect is most useful for doctors because it provides them with the whole range of drugs available for the treatment of a particular type of disease.
• On the other hand, the classification of drugs based on chemical structure or based on molecular targets is useful for medicinal chemists, who are involved in designing and synthesizing drugs.

Question 22. Why should not medicines be taken without consulting doctors?
Answer:

• Side effects are caused when a drug binds to more than one receptor site. Furthermore, a dose of the drug is also crucial because some drugs in higher doses may act as a poison.
• Therefore, we should consult a doctor who can diagnose the disease properly and prescribe the correct medicine at the appropriate dose.

Question 23. Which forces are involved in holding the drugs to the active site of enzymes?
Answer: Drugs bind to the active site of enzymes through a variety of forces such as H-bonding, ionic bonding, dipole-dipole interactions, or van der Waals interactions.

Question 24. While antacids and antiallergic drugs interfere with the function of histamines, why do these not interfere with the function of each other?
Answer:

• Antacids and antiallergic drugs do not interfere with the function of histamines as they work on different receptors. for example., allergy is caused by the secretion of histamine.
• It also causes acidity due to the secretion of HCl in the stomach. Since antiallergic and antacid drugs act on different receptors, therefore, antihistamines prevent allergies and antacids cure acidity.

Question 25. A low level of noradrenaline is the cause of depression. What type of drugs are needed to cure this problem? Name two drugs.
Answer:

• Noradrenaline is a type of neurotransmitter. When it is produced in low quantity in the body, it retards the process of message transfer across the body and as a result, the person suffers from mental depression.
• In this case, antidepressant drugs are required. Two important antidepressant drugs are iproniazid and phenelzine (Nardil).

Question 26. Why are cimetidine and ranitidine better antacids, than sodium hydrogen carbonate or magnesium or aluminium hydroxide?
Answer:

• The production of excessive HCl in the stomach causes acidity. The use of excessive NaHCO₃ can make the stomach alkaline and trigger the production of even more acid.
• However, metal hydroxides (insoluble compounds that do not increase the pH in the stomach) can only control the symptoms but not the cause.
• On the other hand, ranitidine and cimetidine prevent the interaction of histamine with the receptors present in the stomach wall and result in the release of a lesser amount of HCl.

Question 27. What are biodegradable and non-biodegradable detergents? Give one example of each.
Answer:

1. Detergents containing unbranched hydrocarbon chains are easily degraded by microorganisms. These are called biodegradable detergents.
   1. Example: Sodium 4-(l-dodecyl)benzenesulphonate.
2. Detergents containing branched hydrocarbon chains are not easily degraded by microorganisms. These are called non-biodegradable detergents.
   1. Example: Sodium 4-(l,3,5,7-tetramethyloctyl)benzene- sulphonate.

Question 28. Can you use soaps and synthetic detergents to check the hardness of water?
Answer:

• Soaps can be used to check the hardness of water as a sample of soft water will form a lather with soap immediately but hard water will form a lather only after precipitation of all the Ca²⁺ and Mg²⁺ ions as calcium and magnesium soaps respectively.
• Synthetic detergents cannot be used for this purpose as they produce lather with soft as well as hard water.

Question 29. Label the hydrophilic and hydrophobic parts of—

1. CH₃(CH₂)₁₀CH₂OSO₃^–Na⁺
2. CH₃(CH₂)₁₅N⁺(CH₃)₃Br^–
3. CH₃(CH₂)₁₆COO(CH₂CH₂O)BCH₂CH₂OH

Answer:

Question 30. Write the side effects of aspirin.
Answer:

Aspirin has two side effects:

1. It is toxic to the liver and
2. It undergoes hydrolysis in the stomach to produce salicylic acid which causes bleeding from the stomach walls. Thus, it acts as a gastric irritant and may produce ulcers.

Question 31. How can heroin be prepared from morphine? Give the reaction.
Answer: Heroin can be prepared by acetylation of morphin.

Question 32. What are β-lactam antibiotics? Give two examples of such antibiotics
Answer: Antibiotics having azetidine or β-lactam ring are called β-lactam antibiotics, for example., penicillin and cephalosporin.

Question 33. Serine, aspartic acid, and phenylalanine are present at the active site of an enzyme. What are the forces responsible for the binding of these amino acids to the active site?
Answer: The — OH group of serine binds through H-bonding, the —COO^– group of aspartic acid binds through ionic bonding and C₆H₅ —a group of phenylalanine binds through the van der Waals force of attraction.

Question 34. Differentiate between antioxidants and preservatives.
Answer:

• Antioxidants are chemical compounds that prevent the oxidation of food containing unsaturated fats and oils.
• Butylated hydroxyanisole (BHA) and butylated hydroxytoluene (BHT) are two commonly used antioxidants.
• Preservatives are chemical compounds that are added to food to prevent them from spoilage by microbial growth. Some commonly used food preservatives are common salt, vinegar, and sodium benzoate.

Question 35. What are alkaloids? Give two examples of alkaloids i which are used as analgesics.
Answer:

• Alkaloids are complex nitrogenous compounds that are isolated from plants.
• They have a bitter taste and when administered into the body, exhibit some physiological actions like relieving pain.
• Two alkaloids which are used as analgesics are codeine and morphine.

Question 36. How can saccharin, an important artificial sweetener, be prepared from toluene?
Answer: Saccharin can be prepared from toluene as follows:

Question 37. Sodium lauryl sulphate [CH₃(CH₂)₁₁OSO₃Na] is a more effective detergent than sodium propyl sulphate [CH₃CH₂CH₂OSO₃Na] in hard water. Why?
Answer:

• In sodium propyl sulphate [CH₃CH₂CH₂OSO₃Na], the hydrocarbon chain is only three carbon-long chains which are not sufficient for the oil and grease to dissolve well.
• But, the hydrocarbon chain of sodium lauryl sulphate [CH₃(CH₂)OSO₃^–Na⁺] is long enough for the complete dissolution of oil and grease.
• Therefore, sodium lauryl sulphate [CH₃(CH₂)OSO₃^–Na⁺] is more effective than sodium propyl sulphate as a detergent in hard water.

Question 38. What type of drugs can be prepared by the condensation of urea with malonic acid derivatives? Give examples.
Answer:

• 5,5 Derivatives of barbituric acid, called barbiturates, can be obtained from urea and malonic acid derivatives.
• Barbiturates belong to the class of tranquilizers. They are also used as sleep-inducing agents. Some examples of barbiturates are luminal, veronal, seconal, etc.

Question 39. Write structures of the artificial sweeteners aspartame and alitame with R, and S-designations of their asymmetric carbon atoms.
Answer:

Question 40. What are called sulpha drugs? What type of drug are these? Give two examples of sulpha drugs and mention their uses.
Answer:

A group of drugs that are derivatives of sulphanilamide are called sulpha drugs.

These are antibacterial drugs that have been widely used against diseases caused by cocci infections.

Examples of two sulpha drugs are

1. Sulphadiazine which is mainly used for dysentery, and urinary and respiratory infections; and
2. Sulphapyridine which is used for the treatment of pneumonia.

Question 41. Identify the compound in each case:

1. A trichloro derivative of sucrose which is 600 times sweeter than sucrose.
2. A diacetyl derivative of morphine which is a powerful analgesic and extremely addictive.
3. A derivative of phenol which is used as an antioxidant.
4. An alkylbenzene sulphonate which acts as a biodegradable detergent.
5. A non-narcotic analgesic that can be prepared from salicylic acid.
6. A cationic detergent that is used in hair conditioners.
7. An antioxidant used in wine and beers.

Answer:

1. Sucralose, a trichloro derivative of sucrose, is 600 times sweeter than sucrose.
2. Heroin, the diacetyl derivative of morphine, is a powerful analgesic and at the same time is extremely addictive.
3. Butylated hydroxytoluene (BHT), a derivative of phenol, is used as an antioxidant.
4. zenesulphonate, an alkylbenzene sulphonate, acts as a biodegradable detergent.
5. Aspirin, a non-narcotic analgesic, can be prepared from salicylic acid.
6. A cationic detergent, cetyltrimethylammonium bromide ([CH₃(CH₂)₁₅N(CH₃)₃]Br^–) is used in hair conditioners.
7. Sodium sulphite or sodium metabisulphite.

Question 42. Give an example of a derivative of barbituric acid which can act as a tranquilizer. How can it be prepared?
Answer:

Veronal, a derivative of barbituric acid, can act as a tranquilizer. It can be prepared from malonic ester as follows:

Question 43. What is the scientific explanation for the feeling of depression?
Answer: Noradrenaline is one of the neurotransmitters that play a role in mood changes. If a person has a low level of noradrenaline, then the signal-sending activity becomes low (and hence, the message transfer process becomes slow), and the person suffers from depression.

Question 44. What is the basic difference between antiseptics and disinfectants?
Answer: Antiseptics prevent the growth of microorganisms and may even kill them. They are safely applied to living tissues. Disinfectants also kill microorganisms but are not safe to be applied to living tissues. They are applied to non-living objects such as drains, toilets, floors, etc.

Question 45. Between sodium hydrogen carbonate and magnesium hydroxide which is a better antacid and why?
Answer: Magnesium hydroxide is a better antacid because being insoluble it does not allow the pH to increase above 7(neutrality). In contrast, sodium hydrogen carbonate is soluble, excess of it can make the stomach alkaline and trigger the generation of even more acid.

Question 46. What is the advantage of using antihistamines over antacids in the treatment of acidity?
Answer:

• Usual antacids control only the symptoms and not the cause of acidity.
• They work by neutralizing the acid (HCl) produced in the stomach but do not control the causes of the production of more acid. On the other hand, antihistamines are the drugs that suppress the action of histamine responsible for the secretion of pepsin and HCl in the stomach.
• Antihistamines prevent the binding of histamine with the receptors present in the stomach wall thereby suppressing the production of HCl.
• Thus, antihistamines are better drugs for the treatment of acidity.

Question 47. What are the functions performed by histamine in the body?
Answer:

1. Histamine is a potent vasodilator,
2. It contracts muscles in the gut and bronchi,
3. It relaxes some other muscles, such as those in the walls of fine blood vessels,
4. It is responsible for congestion in the nose associated with common cold and allergies,
5. It stimulates the release of pepsin and HCl in the stomach.

Question 48. With the help of an example explain how tranquilizers control the feeling of depression.
Answer:

• Noradrenaline is one of the neurotransmitters that play a role in mood changes. If a person has low levels of noradrenaline in the body, then the signal-sending activity becomes low, and the person suffers from depression. In such situations, tranquilizers (antidepressant drugs) are used.
• These drugs inhibit the enzymes which catalyze the degradation of noradrenaline.
• As a result, this neurotransmitter (noradrenaline) is slowly metabolized and can activate its receptor for longer periods, thus counteracting the effect of depression.

Question 49. Why are certain drugs called enzyme inhibitors?
Answer:

• Enzymes have active sites that bind the substrate for carrying out chemical reactions rapidly and effectively.
• The functional groups present at the active site of the enzyme interact with functional groups of substrate involving ionic bonding, H-bonding, van der Waals interactions, etc.
• Some drugs interfere with this interaction by blocking the binding site of the enzyme and preventing the binding of the actual substrate with the enzyme.
• This inhibits the catalytic activity of the enzyme and the drugs which inhibit the catalytic activity of enzymes are called enzyme inhibitors.

Question 50. What are fillers and what role do these fillers play in soap?
Answer:

• Substances that are added to soaps to modify their properties to make them useful for a particular application are called fillers. For example,
• Sodium resinate is added to laundry soaps to increase their lather-forming ability,
• Glycerol is added to shaving soaps to prevent them from drying.

Question 51. Sugar is the main source of energy as it produces energy on metabolic decomposition. But these days calorie drinks are more popular, why?
Answer:

• In such drinks, artificial sweetening agents (which are often many hundred times sweeter than sugar) are present which do not metabolize and hence, do not produce any calories (energy).
• The use of such sweetening agents is of great value to diabetic persons and people who need to control their intake of calories.

Question 52. Pickles have a long shelf life and do not get spoiled for months, why?
Answer: Plenty of salt and coating of oil act as preservatives. These do not allow air and moisture to enter the material and hence, bacteria cannot thrive on them.

Question 53. What is the difference between saccharin and saccharic acid?
Answer: Saccharin is an artificial sweetener. But, saccharic acid is a dibasic tetrahydroxy acid (obtained by oxidation of glucose using cone. HNO₃).

Question 54. Explain the role of the allosteric site in enzyme inhibition.
Answer:

• Some drugs do not bind to the enzyme’s active site but instead, bind to a different site of the enzyme, called an allosteric site.
• The binding of a drug molecule at the allosteric site changes the shape of the active site of the enzyme in such a way that the substrate cannot recognize it.
• Thus the chemical reaction is inhibited.

Question 55. How are receptor proteins located in the cell membrane?
Answer: Receptor proteins are embedded in the cell membrane in such a way that their small part possessing the active site projects out of the surface of the membrane and opens on the outside region of the cell membrane.

Question 56. What happens when the bond formed between an enzyme and an inhibitor is a strong covalent bond?
Answer:

• If a strong covalent bond is formed between an enzyme and an inhibitor and cannot be broken easily, then the enzyme is blocked permanently.
• The body then degrades the enzyme-inhibitor complex and synthesizes the enzyme.

Question 57. In what respect do prontosil and salvarsan resemble? Is there any resemblance between azodye and prontosil?
Answer:

Both salvarsan and prontosil are antibacterial drugs (antimicrobials). There is a structural similarity between these two compounds. Salvarsan contains — As=As — linkage, while prontosil contains —N=N — linkage. Both prontosil and azo dyes contain the same —N=N — linkage.

Question 58. What is the relation between D-glucose and L-glucose? Among D-glucose & L-glucose, which one provides energy to the body?
Answer: D-glucose and L-glucose are enantiomers; D glucose metabolizes in the body and hence, contributes to the body’s energy but L-glucose does not undergo metabolization and has no role in energy production.

Class 12 Chemistry Unit 16 Chemistry In Every Day Life Multiple Choice Questions And Answers

Question 1. Which is not an antacid—

1. Aluminium hydroxide
2. Cimetidine
3. Phenelzine
4. Ranitidine

Answer: 3. Phenelzine

Question 2. Antiseptics and disinfectants kill microbes or prevent their growth. Which one of the following statements related to them is not true—

1. Disinfectants are harmful to living tissues
2. A 0.2% solution of phenol is an antiseptic
3. Chlorine and iodine are used as a strong disinfectant
4. A dilute solution of boric acid and H₂O₂ is an antiseptic

Answer: 3. Chlorine and iodine are used as a strong disinfectant

Question 3. Which of the following is an analgesic—

1. Chloromycetin
2. Novalgin
3. Penicillin
4. Streptomycin

Answer: 2. Novalgin

Chloromycetin, penicillin, streptomycin: antibiotic; novalgin: analgesic.

Question 4. A mixture of chloroxylenol and terpineol acts as—

1. Antiseptic
2. Antipyretic
3. Antibiotic
4. Analgesic

Answer: 1. Antiseptic

The mixture of chloroxylenol and terpinol is known as Dettol which is an antiseptic.

Question 5. Which one of the following is not employed as an antihistamine—

1. Dimetane
2. Chloramphenicol
3. Seldane
4. Both 1 and 2

Answer: 2. Chloramphenicol

Chloramphenicol is a bacteriostatic antibiotic drug.

Question 6. Which is a bactericidal antibiotic—

1. Penicillin
2. Erythromycin
3. Tetracycline
4. Chloramphenicol

Answer: 1. Penicillin

Erythromycin, tetracycline, and chloramphenicol are bacteriostatic antibiotic drugs.

Question 7. Arsenic drugs are mainly used in the treatment of—

1. Jaundice
2. Typhoid
3. Syphilis
4. Cholera

Answer: 3. Syphilis

Arsenic drugs such as salvarsan are used for the treatment of syphilis.

Question 8. Which is incorrect—

1. Novestrol—antifertility
2. Serotonine—tranquilizer
3. Narrow spectrum—chloramphenicol
4. Rantac—antacid

Answer: 3. Narrow spectrum—chloramphenicol

Chloramphenicol is a broad-spectrum antibiotic.

Question 9. Among the following which is an artificial sweetening agent— 

1. Sucrose
2. Lactose
3. Sucralose
4. Cellulose

Answer: 3. Sucralose

Question 10. For which of the following purposes sodium benzoate is used—

1. An antioxidant
2. An analgesic
3. A tranquilizer
4. A food preservative.

Answer: 4. A food preservative.

Question 11. Which of the following is a constituent of soap—

1. Sodium stearate
2. Sodium salicylate
3. Sodium butyrate
4. Sodium benzene sulphonate

Answer: 1. Sodium stearate

Question 12. Which of the following is an antibiotic—

1. Aspirin
2. Chloramphenicol
3. Veronal
4. Forestal

Answer: 2. Chloramphenicol

Question 13. Which of the following is not a preservative— 

1. Common salt
2. Sucrose
3. Sodium benzoate
4. Sucralose

Answer: 4. Sucralose

Question 14. A compound X is used as an antiseptic in 0.2% solution and as a disinfectant in 1 % solution. Which of the following is X?

1. Phenol
2. Soframycin
3. Benzil
4. Iodoform

Answer: 1. Phenol

Question 15. Which of the following can be used as an antacid—

1. Ranitidine
2. Histamine
3. Equanil
4. Aspirin

Answer: 1. Ranitidine

Question 16. Which of the following artificial sweeteners is methyl ester of a dipeptide—

1. Aspertame
2. Sucralose
3. Saccharine
4. Alitame

Answer: 1. Aspertame

Question 17. Which of the following compounds is not a detergent?

1. CH₃(CH₂)₁₆CH₂OSO₂^–Na⁺
2. CH₃(CH₂)₁₅N⁺(CH₃)₃Br^–
3. CH₃(CH₂)₁₄CH₂NH₂

Answer: CH₃(CH₂)₁₄CH₂NH₂ is not a detergent.

Question 18. Which of the following statements is not correct—

1. Some antiseptics can be added to soaps
2. Dilute solutions of some disinfectants can be used as antiseptic
3. Disinfectants are antimicrobial drugs
4. Antiseptic medicines can be ingested

Answer: 4. Antiseptic medicines can be ingested

Question 19. What is correct about birth control pills—

1. Contain estrogen only
2. Contain progesterone only
3. Contains a mixture of estrogen and progesterone derivatives
4. Progesterone enhances ovulation

Answer: 3. Contain a mixture of estrogen and progesterone derivatives

Question 20. Which statement about aspirin is not true—

1. Aspirin belongs to narcotic analgesics
2. It is effective in relieving pain
3. It has anti-blood clotting action
4. It is a neurologically active drug

Answer: 1. Aspirin belongs to narcotic analgesics

Question 21. The most useful classification of drugs for medicinal chemists is

1. Based on the chemical structure
2. Based on drug action
3. Based on molecular targets
4. Based on pharmacological effect

Answer:  3. Based on molecular targets

Question 22. Which of the following statements is correct—

1. Some tranquilizers function by inhibiting the enzymes which catalyse the degradation of noradrenaline
2. Tranquilizers are narcotic drugs
3. Tranquilizers are chemical compounds that do not affect the message transfer from nerve to receptor
4. Tranquilizers are chemical compounds that can relieve pain and fever

Answer: 1. Some tranquilizers function by inhibiting the enzymes which catalyse the degradation of noradrenaline

Question 23. Salvarsan is an arsenic-containing drug that was first used for the treatment of

1. Syphilis
2. Typhoid
3. Meningitis
4. Dysentery

Answer: 1. Syphilis

Question 24. A narrow-spectrum antibiotic is active against

1. Gram-positive or Gram-negative bacteria.
2. Gram-negative bacteria only.
3. Single organism or one disease.
4. Both Gram-positive and Gram-negative bacteria.

Answer: 1. Gram-positive or Gram-negative bacteria.

Question 25. The compound that causes general antidepressant action on the central nervous system belongs to the class of

1. Analgesics
2. Tranquilizers
3. Narcotic analgesics
4. Antihistamines

Answer: 2. Tranquilizers

Question 26. The compound which is added to soap to impart antiseptic properties is

1. Sodium lauryl sulphate
2. Sodium dodecylbenzenesulphonate
3. Rosin
4. Bithional

Answer: 4. Bithional

Question 27. Equanilis

1. Artificial sweetener
2. Tranquilizer
3. Antihistamine
4. Antifertility drug

Answer: 2. Tranquilizer

Question 28. Which enhances the lathering property of soap—

1. Sodium carbonate
2. Sodium rosinate
3. Sodium stearate
4. Trisodium phosphate

Answer: 2. Sodium rosinate

Question 29. Glycerol is added to soap. It functions

1. As a filler.
2. To increase lathering.
3. To prevent rapid drying.
4. To make soap granules.

Answer: 3. To prevent rapid drying.

Question 30. Which of the following is an example of liquid dishwashing detergent—

Answer: 2

Question 31. Polyethyleneglycols are used in the preparation of which type of detergents—

1. Cationic detergents
2. Anionic detergents
3. Non-ionic detergents
4. Soaps

Answer: 3. Non-ionic detergents

Question 32. Which of the following is not a target molecule for drug function in the body—

1. Carbohydrates
2. Lipids
3. Vitamins
4. Proteins

Answer: 3. Vitamins

Carbohydrates, proteins, nucleic acids, and lipids are target molecules for drug function in the body.

Question 33. This is not true about enzyme inhibitors—

1. Inhibit the catalytic activity of the enzyme
2. Prevent the binding of substrate
3. Generally, a strong covalent bond is formed between an inhibitor and an enzyme
4. Inhibitors can be competitive or non-competitive

Answer: 3. Generally a strong covalent bond is formed between an inhibitor and an enzyme

weak bonds such as H-bond, van der Waals interactions, etc. are found between an inhibitor and an enzyme.

Question 34. Which of the following chemicals can be added for sweetening of food Items at cooking temperature and does not provide calories—

1. Sucrose
2. Glucose
3. Aspartame
4. Sucralose

Answer: 4. Sucralose

Both aspartame and. sucralose do not provide calories. Aspartame decomposes at cooking temperature while sucralose does not.

Question 35. Which will not enhance the nutritional value of food—

1. Minerals
2. Artificial sweeteners
3. Vitamins
4. Amino acids

Answer: 2. Artificial sweeteners

Artificial sweeteners do not enhance the nutritional value of food.

Question 36. Which of the following statements are incorrect about receptor proteins—

1. The majority of receptor proteins are embedded in the cell membranes
2. The active site of receptor proteins opens on the inside region of the cell
3. Chemical messengers are received at the binding sites of receptor proteins
4. The shape of the receptor doesn’t change during the attachment of the messenger

Answer: 2 and 4

The active site of receptor proteins opens on the outer region of the cell. Also, the shape of the receptor changes during the attachment of the messenger.

Question 37. Which are not used as food preservatives—

1. Table salt
2. Sodium hydrogen carbonate
3. Cane sugar
4. Benzoic acid

Answer: 2 and 3

Question 38. Compounds with antiseptic properties are.

1. CHCI₃
2. CHI₃
3. Boric acid
4. 0.3 ppm solution of Cl₂

Answer: 2 and 3

CHCl₃ and 0.3 ppm aqueous solution of Cl₂ are not used as antiseptic.

Question 39. Which are correct about barbiturates—

1. Hypnotics or sleep-producing agents
2. These are tranquilizers
3. Non-narcotic analgesics
4. Pain reduction without disturbing the nervous system

Answer: 1 and 2

Question 40. Which of the following are sulpha drugs—

1. Sulphapyridine
2. Prontosil
3. Salvarsan
4. Nardil

Answer: 1 and 2

Question 41. Which of the following are antidepressants—

1. Iproniazid
2. Phenelzine
3. Equanil
4. Salvarsan

Answer: 1,2 and 3

Question 42. Which of the following are incorrect about penicillin—

1. An antibacterial fungus
2. Ampicillin is its synthetic modification
3. It has a bacteriostatic effect
4. It is a broad-spectrum antibiotic

Answer: 3 and 4

Question 43. Which of the following are administered as antacids—

1. Sodium carbonate
2. Sodium hydrogen carbonate
3. Aluminium carbonate
4. Magnesium hydroxide

Answer: 2 and 4

Question 44. Amongst given antihistamines, which are antacids—

1. Ranitidine
2. Brompheniramine
3. Terfenadine
4. Cimetidine

Answer: 1 and 4

Question 45. Veronal and luminal are derivatives of barbituric acid which are

1. Tranquilizers
2. Non-narcotic analgesic
3. Antiallergic drugs
4. Neurologically active drugs

Answer: 1 and 4

Question 46. Which of the following are anionic detergents—

1. Sodium salts of sulphonated long-chain alcohol.
2. Ester of stearic acid and polyethylene glycol.
3. Quarternary ammonium salt of amine with acetate ion.
4. Sodium salts of sulfonated long-chain hydrocarbons.

Answer: 1 and 4

Question 47. Which of the following statements is correct—

1. Cationic detergents have germicidal properties
2. Bacteria can degrade the detergents containing highly branched chains.
3. Some synthetic detergents can give foam even in ice-cold water.
4. Synthetic detergents are not soaps.

Answer: 1,3 and 4

Detergents containing highly branched chains are not degraded by bacteria.

Question 48. Which of the following is used for treating malaria—

1. Aspirin
2. Penicillin
3. Chloroquin
4. Paracetamol

Answer: 3. Chloroquin

Question 49. Which of the following is an analgesic—

1. Seldane
2. Paludrin
3. Analgin
4. Iodex

Answer: 3. Analgin

Question 50. The chemical name of aspirin is—

1. Methyl benzoate
2. Ethyl salicylate
3. Acetylsalicylic acid
4. Hydroxybenzoic acid

Answer: 3. Acetylsalicylic acid

Question 51. Phenacetin is used as an—

1. Antipyretic
2. Antiseptic
3. Analgesic
4. Antimalarial

Answer: 1. Antipyretic

Question 52. Aluminum salt used to stop bleeding is—

1. Aluminium sulphate
2. Potash alum
3. Aluminium chloride
4. Aluminium fluoride

Answer: 2. Potash alum

Question 53. Which of the following is an antioxidant—

1. Sucralose
2. Butylated hydroxyanisole
3. Sorbic acid
4. None of these

Answer: 2. Butylated hydroxyanisole

Question 54. Artificial sweeteners used in cold drinks is—

1. Lactose
2. Aspartame
3. Glycerol
4. Fructose

Answer: 2. Aspartame

Question 55. Which of the following must be present in a biodegradable detergent—

1. Simple alkyl chain
2. Branched alkyl chain
3. Phenyl side chain
4. Cyclohexyl side chain

Answer: 1. Simple alkyl chain

Question 56. A compound responsible for the antiseptic effect of Dettol—

1. Chlorobenzene
2. Chloroxylenol
3. Terpineol
4. Both 2 and 3

Answer: 4. Both 2 and 3

Question 57. Which of the following is a synthetic detergent—

1. C₁₅H₃₁COOK
2. CH₃(CH₂)₆COO^–Na⁺
3. 
4. None of these

Answer: 3. None of these

Question 58. Which compounds are used as antihistamines—

1. Aspirin
2. Dimetapp
3. Promethazine
4. Seldane

Answer: 2,3 and 4

Question 59. Which regarding barbiturates are correct—

1. These are hypnotic or sleep-inducing agents
2. These are tranquilizers.
3. These are non-narcotic analgesics.
4. These are painkillers but do not affect the nervous system.

Answer: 1 and 2

Question 60. Which of the following are sulpha drugs—

1. Sulphapyridine
2. Prontosil
3. Salvarsan
4. Nardil

Answer: 1 and 2

Question 61. Which of the following pairs are bactericidal antibiotics—

1. Penicillin, streptomycin
2. Erythromycin, chloramphenicol
3. Ofloxacin, aminoglycoside
4. Tetracyclin, chloramphenicol

Answer: 1 and 3

Question 62. Which of the following compounds are used to cure mental depression—

1. Iproniazid
2. Phenelzine
3. Equanil
4. Salvarsan

Answer: 1,2 and 3

Question 63. Give one use of each of the following—

1. Chloramphenicol,
2. Streptomycin,
3. Paracetamol,
4. Bithional.

Answer:

1. In treating typhoid;
2. In treating tuberculosis;
3. In reducing arthritic pain and fever;
4. As an antiseptic

Class 12 Chemistry Unit 16 Chemistry In Every Day Life Match The Following Questions And Answers

Question 1.

Answer: 1-C, 2-D, 3-A, 4-B;

Question 2.

Answer: 1-B, 2-A, 3-D, 4-C;

Question 3.

Answer: 1-C, 2-D, 3-B, 4-A;

Question 4.

Answer: 1-C, 2-D, 3-B, 4-A;

Question 5.

Answer: 1-B, 2-D, 3-A, 4-E, 5-C;

Question 6.

Answer: 1-E, 2-F, 3-D, 4-G, 5-B, 6-A, 7-C;

Class 12 Chemistry Unit 16 Chemistry In Every Day Life Assertion-Reason Type

In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices.

1. (A) and (R) both are correct statements but (R) does not explain (A).
2. (A) and (R) both are correct and (R) explains the (A).
3. Both (A) and (R) are wrong statements.
4. (A) Is the correct statement (R) is the wrong statement.
5. (A) Is a wrong statement (R) is a correct statement.

Question 1. Assertion (A): Penicillin (G) is an antihistamine.

Reason (R): Penicillin (G) is effective against Gram¬positive as well as Gram-negative bacteria.

Answer: 3. Both (A) and (R) are wrong statements.

Question 2. Assertion (A): Sulpha drug contains sulphonamide group.

Reason (R): Salvarsan is a sulpha drug.

Answer: 4. (A) Is the correct statement (R) is the wrong statement.

Question 3. Assertion (A): Receptors are crucial to the body’s communication process.

Reason (R): Receptors are proteins.

Answer: 1. (A) and (R) both are correct statements but (R) does not explain (A).

Question 4. Assertion (A): Enzymes have active sites that hold substrate molecules for a chemical reaction.

Reason (R): Drugs compete with natural substrates by attaching covalently to the active site of an enzyme.

Answer: 4. (A) Is the correct statement (R) is the wrong statement.

Question 5. Assertion (A): Chemical messengers are chemicals that enable communication of messages between two neurons or between neurons and muscles.

Reason (R): Chemicals enter the cell through receptors.

Answer: 4. (A) Is the correct statement (R) is the wrong statement.

Question 6. Assertion (A): Transparent soaps are made by dissolving soaps in ethanol.

Reason (R): Ethanol makes things invisible.+

Answer: 4. (A) Is the correct statement (R) is the wrong statement.

Question 7. Assertion (A): Sodium chloride is added to precipitate soap after saponification.

Reason (R): Hydrolysis of esters of long-chain fatty acids by the alkali produces soap in colloidal form.

Answer: 2. (A) and (R) both are correct and (R) explains the (A).

Question 8. Assertion (A): Competitive inhibitors compete with natural substrates for their attachment to the active sites of enzymes.

Reason (R): In competitive inhibition, the inhibitor binds to the allosteric site of the enzyme.

Answer: 2. (A) and (R) both are correct and (R) explains the (A).

Question 9. Assertion (A): Non-competitive inhibitor inhibits the catalytic activity of an enzyme by binding with its active site.

Reason (R): Non-competitive inhibitor changes the shape of the active site in such a way that the substrate can’t recognize it.

Answer: 5. (A) Is the wrong statement (R) is correct?

Question 10. Assertion (A): A chemical messenger gives a message to the cell without entering the tire cell.

Reason (R): Chemical messenger is received at the binding site of receptor proteins.

Answer: 2. (A) and (R) both are correct and (R) explains the (A).

Question 11. Assertion (A): Receptor proteins show selectivity for one chemical messenger over the other.

Reason (R): Chemical messenger binds to the receptor site and inhibits its natural function.

Answer: 4. (A) Is the correct statement (R) is the wrong statement.

Question 12. Assertion (A): All chemicals added to food items are called food preservatives.

Reason (R): All these chemicals increase the nutritive value of the food.

Answer: 3. Both (A) and (R) are wrong statements.

Question 13. Assertion (A): Preservatives are added to food items.

Reason (R): Preservatives inhibit the growth of microorganisms.

Answer: 2. (A) and (R) both are correct and (R) explains the (A).

Question 14. Assertion (A): Artificial sweeteners are added to the food to control the intake of calories.

Reason (R): Most of the artificial sweeteners are inert and do not metabolize in the body.

Answer: 2. (A) and (R) both are correct and (R) explains the (A).

Class 12 Chemistry Unit 16 Chemistry In Every Day Life Fill in the blanks

Question 1. A drug prepared by acetylation of p-aminophenol which is both an analgesic and an antipyretic is____
Answer: Paracetamol;

Question 2. A powerful narcotic analgesic prepared by acetylation of morphine is____
Answer: Heroin

Question 3. ____is a limited-_spectrum antibiotic.
Answer: Dysidazirine

Question 4. Dettol is a mixture of chloroxylenol and____
Answer:  α-terpineol

Question 5. Ortho-sulphobenzimide, an artificial sweetener is commonly known as____
Answer:  Saccharin

Question 6. Artificial sweeteners used in calorie-free drinks is____
Answer: Aspartame

Question 7. ____is a natural antioxidant.
Answer: Vitamin-E

Question 8. Sodium dodecylbenzene sulphonate is a____ detergent.
Answer: Anionic

Question 9. Non-ionic detergent dissolves in water by forming____
Answer: Hydrogen bond

Class 12 Chemistry Unit 16 Chemistry In Every Day Life Warm Up Exercise

Question 1. What is the function of a receptor protein?
Answer: Receptor protein transfers information to the cell

Question 2. Give the differences between tranquilizers and analgesics.
Answer:

Question 3. Which drug is obtained when morphine undergoes acetylation?
Answer: Morphine gives heroine (a narcotic drug) when undergoes acetylation.

Question 4. Name a drug that is both an analgesic and antipyretic.
Answer: Aspirin

Question 5. Give an example of a sulpha drug.
Answer: Sulphapyridin

Question 6. Among D- and L-glucose, which is more beneficial for obese people?
Answer: L-glucose is beneficial for obese people because it cannot be digested in the body due to the absence of such enzymes and thus it is excreted from the body through urine.

Question 7. Ortho-sulphobenzimide is commonly known as which compound? What is its calorific value?
Answer: Saccharin has no calorific value

Question 8. Which detergents cause environmental pollution and why?
Answer: Detergents that are highly branched are nonbiodegradable and they cause environmental pollution

Class 12 Chemistry Unit 14 Biomolecules Introduction

A characteristic feature of living organisms is that they undergo growth and reproduction, though they are composed of atoms and molecules.

• The branch of chemistry that deals with the chemical composition of living organisms and the biochemical processes brought about by those chemical compounds is called biochemistry.
• The complex organic compounds which are essential components of living beings are regarded as biomolecules. They form the structural and functional basis of life.
• Some important biomolecules are carbohydrates, proteins, enzymes, lipids, nucleic acid, hormones and vitamins.

Carbohydrates

General Discussion On Carbohydrates

Carbohydrates form an important class of biomolecules and are the main source of energy in the living body.

General Discussion On Carbohydrates Example:

1. Most of the staple food in our diet consists mainly of starch.
2. Cotton, linen and rayon fabrics are derived from the cellulose of plant cell walls.
3. Wood is the dead cells formed by the deposition of cellulose and lignin. Thus starch, cellulose, and lignin are all carbohydrates.

Plants: Primary Source Of Carbohydrates

All green plants photosynthesize to produce carbohydrates. The raw materials used are—

1. H₂O absorbed from soil,
2. Atmospheric CO₂,
3. Solar energy and
4. Chlorophyll pigment.

Definition Of Carbohydrates: The name carbohydrate was originally given to the class of compounds (containing carbon, hydrogen and oxygen) having the general formula C_x(H₂O)_y. Since these compounds contain carbon and hydrogen in the same ratio as in water, they were considered as the hydrates of carbon. But this definition did not survive long for the following reasons—

1. There are several compounds which are known to be carbohydrates by their chemical behaviour but do not possess the general formula C_x(H₂O)_y for example., rhamnose (C₆H₁₂O₅) and deoxyribose (C₅H₁₀O₄).
2. Again there are some compounds such as formaldehyde (HCHO or CH₂O), acetic acid [CH₃COOH or C₂(H₂O)₂ ] etc., which do not behave like carbohydrates but can be represented by the general formula C_x(H₂O)_y
3. Carbon is not known to form any hydrate.

Hence the above definition has been modified as given below:

Modern Definition: Carbohydrates are now defined as optically active polyhydroxy aldehydes or polyhydroxy ketones or the compounds which give these on hydrolysis.

There is no free aldehydic (—CHO) or keto (C=O) group in carbohydrates. They exist as hemiacetal or hemiketal formed by the condensation of the aldehydic or ketonic group with an alcoholic  —OH group present in that molecule.

Classification Of Carbohydrates

Classification Based On Taste

Classification Based On Taste Sugars: Water-soluble crystalline carbohydrates having a sweet taste are regarded as sugars. All monosaccharides and oligosaccharides fall under this category.

Classification Based On Taste Sugars Example: Glucose, fructose, sucrose, lactose, etc.

Relative Sweetness Of Some Sugars:

Classification Based On Taste Non-sugars: Amorphous, tasteless carbohydrates which are slightly soluble or completely insoluble in water are called non-sugars. All polysaccharides fall under this category.

Classification Based On Taste Non-sugars Example: Starch, cellulose, etc.

Classification Based On Hydrolysis

Monosaccharides

Classification Based On Hydrolysis Monosaccharides Definition: The carbohydrates which cannot be hydrolysed to produce any simpler carbohydrates are termed monosaccharides.

Classification Based On Hydrolysis Monosaccharides Example: Arabinose (C₅H₁₀O₅), ribose (C₅H₁₀O₅), glucose (C₆H₁₂O₆), fructose (C₆H₁₂O₆), etc.

Oligosaccharides Definition: The carbohydrates which on hydrolysis, produce 2-10 monosaccharide molecules are called oligosaccharides.

Depending on the number of monosaccharide units produced, oligosaccharides can be further classified into disaccharides, trisaccharides and tetrasaccharides.

Disaccharides: Carbohydrates whose hydrolytic products are two similar or different monosaccharides are called disaccharides.

Disaccharides Example: Sucrose, lactose, maltose, etc.

Trisaccharides: Carbohydrates whose hydrolytic products are three similar or different monosaccharides are called trisaccharides.

Trisaccharides Example: Raffinose.

Tetrasaccharides: Carbohydrates whose hydrolytic products are four similar or different monosaccharides are called tetrasaccharides.

Tetrasaccharides Example: Stachyose, hychoose

Polysaccharides Definition: Carbohydrates which, on hydrolysis, produce several monosaccharide units are called polysaccharides. The general formula of most polysaccharides is (C₆H₁₀O₅)_n.

Polysaccharides Example: Starch, cellulose, glycogen, etc.,

1. Homopolysaccharides (homoglycans) i.e., composed of the same monosaccharide molecules such as starch, cellulose and glycogen.
2. Heteropolysaccharides (heteroglycans), i.e.„ composed of different monosaccharide units such as insulin, heparin and hyaluronic acid. Polysaccharides are devoid of sweetness and form colloids in boiling water.

Classification Based On Reducing Property

Reducing Sugars: Carbohydrates which can reduce Fehling’s solution or Tollen’s reagent are called reducing sugars.

• All monosaccharides (aldoses or ketoses) are reducing sugars. Again, disaccharides in which the two monosaccharide units are linked by an aldehydic or ketonic group (reducing centre) also behave as reducing sugars.
• The other aldehydic or ketonic group remains as a hemiacetal or hemiketal.

Reducing Sugars Example: Monosaccharides: Arabinose, ribose, glucose, mannose, fructose, etc. Disaccharides: maltose, lactose, etc.

Non-Reducing Sugars Definition: Carbohydrates which cannot reduce Fehling’s solution or Tollen’s reagent are called non-reducing sugars.

Disaccharides whose monosaccharide units are linked by aldehydic or ketonic groups only are considered as non¬ reducing sugars.

Non-Reducing Sugars Example: Disaccharide: Sucrose. Polysaccharide: All polysaccharides (starch, cellulose, glycogen), etc.

Classification Of Carbohydrates:

• Identification of carbohydrates (Molisch’s test): In the aqueous solution of a carbohydrate, Molisch’s reagent (1% alcoholic solution of α-naphthol) is added, followed by the addition of concentrated H₂SO₄ along the side of the sloping test tube.
• A red-violet ring is produced at the junction of the two liquids (acid and the aqueous solution of carbohydrates). All carbohydrates respond to this test.

Monosaccharides

Monosaccharides are the simplest carbohydrates. They possess 3-7 carbon atoms, generally represented by C_n(H₂O)_n where n = 3-7. There are 20 monosaccharides found in nature. The nomenclature depends upon the nature of the carbonyl group present in the monosaccharide molecule.

Classification Of Monosaccharides

Classification Based On The Nature Of Carbonyl Group Aldose: Monosaccharides in which an aldehyde (— CHO) group is present are called aldoses. Hence, all the polyhydroxy aldehydes are considered aldoses.  The monovalent aldehyde group is found at the C₁-terminal position of the aldose chain.

Classification on the basis of the nature of carbonyl group Ketose: Monosaccharides in which a ketogroup is present are termed ketoses. Hence, all polyhydroxy ketones are considered as ketoses. The bivalent keto group can be found at any position other than the terminal carbon. But, naturally occurring ketoses contain a keto group at the second carbon atom.

Classification Based On Number Of Carbon Atoms

Depending upon the number of the C-atoms, mono-saccharides can be of different types—

1. Triose (containing three C-atoms),
2. Tetrose (containing four C-atoms),
3. Pentose (containing five C-atoms),
4. Hexose (containing six C-atoms) and
5. Heptose (containing seven C-atoms).

Classification Of Monosaccharides Based On The Number Of Carbon Atoms:

Nomenclature Of Monosaccharides: Monosaccharides are named according to the number of carbon atoms and the nature of the carbonyl group.

• The numeral prefixes indicating the number of carbon atoms, such as— tri- (3), tetra-(4), pent-(5), hex-(6) are succeeded by the suffix-ose.
• The terms aldo & keto are used before the numeral prefixes of the carbon atoms to indicate the presence of aldehyde and ketone groups.

Nomenclature Of Monosaccharides Example: (aldo-)+(hex-)+(-ose)=aldohexose [polyhydroxy aldehydes containing 6 C-atoms] (keto-)+(hex-)+(-ose)=ketohexose [polyhydroxy ketones containing 6 C-atoms]

Some Important Aspects Regarding Monosaccharides

Numbering The C-Chain With Consecutive Rank: The C-atom present in the aldehyde group of aldoses and the terminal C-atom nearest to the keto group of the ketoses are numbered as the first carbon or C₁.

D- and L- Configuration:

1. The D- and L- symbols refer to the configuration of monosaccharides. These do not indicate the dextro-rotatory and laevorotatory-specific rotations.
2. In the case of a monosaccharide, if the asymmetric C-atom of the highest rank is similar to the asymmetric carbon of D-glyceraldehyde (H-atom on the left, —OH group on the right) then the monosaccharide is of D-configuration. If the H-atom is on the right and the —OH group is on the left, then the monosaccharide is of L-configuration.
3. As per convention, if in a monosaccharide unit (expressed by the Fischer projection formula), the (— OH) group falls on the right of the highest ranked asymmetric carbon, then the monosaccharide has D -configuration and if the hydroxyl group falls on the left, it has L-configuration.

Use Of + And -Sign: D- and L-symbols are followed by ‘+’ and ‘-‘ in brackets. These represent the dextro- and laevo-specific rotation. There is no relation between optical activity and the nature of rotation.

Use Of + And -Sign Example: Both glucose and fructose have D-configuration, but their optical activities are positive (dextrorotatory) and negative (laevorotatory) respectively.

Epimers: If two optical isomers containing 3 or more asymmetric C-atoms differ in the configuration of only one single C-atom, then the isomers are regarded as epimers. For example, D-( + ) -glucose and D-(+)-galactose are epimers of each other. [Follow ‘ D -class of aldehydes’]

Number Of Optical Isomers Of Monosaccharides: All monosaccharides, except dihydroxyacetone, contain at least one asymmetric C -atom. The number of optical isomers of a monosaccharide containing n -number of non-identical asymmetric C -atoms = 2ⁿ, for example., there are 4 non-identical asymmetric C -atoms in an aldohexose and therefore the number of optical isomers of aldohexose = 2⁴ = 16.

D-Aldose Series:

D-Ketose Series:

Every isomer forms two diastereoisomers due to an increase of one C-atom. Hence, two 4-C, four 5-C and eight 6-C diastereoisomers are found

Pentoses

Monosaccharides containing 5 C-atoms, expressed by the formula C₅H₁₀O₅ are called pentoses.

Aldopentoses: There are four hydroxyl groups and one aldehyde group in an aldopentose molecule, i.e., aldopentose is a tetrahydroxyaldehyde. Its structural formula is:

• The number of optical isomers of an aldopentose = 2³ = 8. These are ribose, arabinose, xylose and lyxose of D and L -configurations.
• The most important aldopentose is D-(-)-ribose which is a component of ribonucleic acid (RNA).
• All aldopentoses do not take part in fermentation but aldohexoses (like glucose) do take part.
• The removal of the O-atom from the —OH group of the second C- atom of the aldopentose ribose, forms 2 -deoxyribose (CH₂OH-CHOH-CHOH-CH₂-CHO). It is an essential component of DNA.

Ketopentoses: There are four hydroxyl groups and one keto group in ketopentose molecules, ketopentoses are tetrahydroxy ketones with the structural formula:

The number of optical isomers of ketopentoses due to the presence of two different asymmetric C-atoms = 22 = 4. These are ribulose and xylulose of D- and L-configuration.

Hexoses

Monosaccharides containing six C -atoms and expressed by the formula C₆H₁₂O₆ are called hexoses.

Aldohexoses: There are five hydroxyl groups and one aldehyde group in an aldohexose i.e., aldohexoses are pentahydroxy aldehydes.

• The number of optical isomers of an aldohexose due to the presence of four different asymmetric C – atoms = 24 = 16. These are allose, altrose, glucose, mannose, gulose, iodose, galactose and tallose of D-and L-configurations. The D- and L-configurations of a particular aldohexose are mirror images of each other.
• The most important aldohexose is glucose.

D-(+)-Glucose Or Dextrose (C₆H₁₂O₆)

1. It is an important aldohexose. The term ‘glucose’ originated from the Greek word ‘glucose’ meaning sweet. It is found in ripe grapes, honey, most fruits, blood and urine of diabetic patients. Its sweetness is comparatively less than sugar.
2. Glucose is commonly found in nature as D-glucose. The asymmetric C-atom of the highest rank (C5) has the same configuration as that of D-glyceraldehyde and hence, it is called D-glucose.
3. The aqueous solution of naturally occurring glucose rotates the plane of polarised light in the clockwise direction and hence, is termed dextrorotatory (+). Therefore, naturally occurring glucose is known as D-(+)- glucose or dextrose.

Preparation Of Glucose

Laboratory Method: The hydrolysis of an alcoholic solution of sucrose (cane sugar) by dilute HCl at 50-60°C temperature produces an equimolecular mixture of glucose and fructose (The hydrolysis of sucrose is called inversion of sucrose).

• Crystalline Glucose precipitates on cooling the above mixture while fructose remains dissolved in the solution (as the solubility of glucose in an alcoholic solvent is less than that of fructose).
• Crystallised glucose is then separated by filtration.

Industrial Method: Glucose Is Prepared From Starch In Industries

Starch is hydrolysed by heating it with dilute H₂SO₄ under pressure at 120°C to obtain glucose.

Open-Chain Structure Of Glucose: One aldehyde (—CHO) group, one primary alcoholic ( —CH₂OH) group and four secondary alcoholic (—CHOH) groups are present in the open-chain structure of glucose. CHO-(CHOH)₄-CH₂OH (Glucose)

Detection Of Open Chain Structure Of Glucose By Experimental Observations

1. Molecular Formula: It can be proved that C₆H₁₂O₆ is the molecular formula of glucose by the analysis of its components and determination of its molecular weight.
2. Unbranched Chain: A hexahydric alcohol, sorbitol is obtained on reduction of glucose by Na -amalgam.

N-hexane and 2 -iodohexane are obtained on the reduction of glucose by HI and red phosphorus at 373K.

The above observations prove that six C -atoms of glucose are linked by an unbranched chain (C—C— C—C—C—C).

Presence Of Aldehyde Group: Glucose reacts with hydroxylamine (NH₂OH) and hydrogen cyanide, (HCN) to produce monoxide and cyanohydrin respectively.

• It proves the presence of one carbonyl group in glucose.
• Weak oxidant bromine-water converts glucose into gluconic acid containing the same number of C-atoms which proves the presence of the —CHO group at the terminal position (since the —CHO group is monovalent) of glucose.

Presence Of Five Hydroxyl Groups: Glucose reacts with acetic anhydride to form a pentaacetate. Hence, a glucose molecule has five hydroxyl groups.

• An organic compound containing two —OH groups at the same carbon is very unstable and readily converts into a carbonyl compound by giving off one water molecule.
• Glucose is a stable compound as its water molecules cannot be removed even by heating. Thus, it can be concluded that no single C-atom of glucose has two or more hydroxyl groups attached to it or the five hydroxyl groups are linked to five different C-atoms.

Presence Of A 1° Alcoholic Group: Both glucose and gluconic acid are oxidised by HNO₃ to produce saccharic acid or glucaric acid. This proves the presence of a 1° alcoholic group at the terminal position of the glucose chain.

Open-Chain Structure Of Glucose: The above discussion shows the presence of one aldehyde (—CHO) and primary alcoholic (—CH₂OH) group in the glucose molecule. There are five hydroxyl groups, of which one is CH₂OH and the rest four are secondary hydroxyl groups (—CHOH—) i.e., four —CHOH groups are present between —CHO and —CH₂OH. So, the open-chain structure of glucose is:

There are four different asymmetric C -atoms (starred) in a glucose molecule. Therefore, the number of stereoisomers of glucose is 2⁴ = 16. These sixteen stereoisomers form eight pairs of enantiomers, for example., naturally occurring D-glucose and its enantiomer L-glucose.

Configuration Of Natural D-(+)- Glucose: Emil Fischer (1891) determined the exact configuration of four chiral carbons of D-(+) -glucose. It can be expressed by the Fischer projection formula.

Cyclic Structure Of D-(+)- Glucose

Limitations Of Open-Chain Structure Of Glucose

Glucose does not show all the characteristic reactions of aldehydes though it has —CHO group. For example, glucose does not form addition compounds with ammonia or sodium bisulphite and does not turn Schiff’s reagent violet in cold conditions.

Though glucose reacts with hydroxylamine to form oxime, it does not react with pentaacetate hydroxylamine.

D-(+) -glucose possess two stereoisomers- α-D-(+) glucose and β-D-(+) -glucose. D-glucose, when crystallised from water at 303K temperature, forms or-D-(-t-) – glucose.

• Its melting point is 419K and the specific rotation of its aqueous solution is [α]_D = +112°. D -glucose forms β-D-(+) -glucose on crystallisation from water at 371K. Its melting point is 423K and specific rotation of its aqueous solution [a]D = +19°.

Both α-and β-D -glucose exhibit mutarotation. If any form is dissolved in water, the specific rotation in case of or-D -glucose decreases from +112° and that of β-D – glucose increases from +19° to attain an intermediate value of +52.7°.

• This phenomenon is called mutarotation. It refers to the change in the values of optical rotation, α- and β-D -glucose attains a dynamic equilibrium in an aqueous solution.
• 36% of α-D -glucose and 64% of β-D -glucose are found in an equilibrium mixture at room temperature. Hence, specific rotation decreases in the case of α-D -glucose and increases in the case of β-D -glucose.

Cyclic Structure Of D-(+)- Glucose Chain Structure Of Glucose Mutarotation: If the specific rotation of an optically active substance changes on its dissolution in a solvent and finally becomes fixed at a particular value, the phenomenon is called mutarotation.

• D(+) -glucose forms two isomeric methyl glucosides. One molecule of aldehyde reacts with two molecules of alcohol to produce an acetal.
• But, one molecule of glucose reacts with one molecule of methanol in the presence of dry HCl gas to produce a mixture of methyl α-D -glucoside (m.p. 438K; specific rotation =+158°) and methyl β-D- glucoside (m.p. 380K; specific rotation =-33°)

• Methyl glucosides do not behave like acetals. Though these are formed in anhydrous condition, they are hydrolysed in glucose and methanol acidified with dilute HCl.
• These glucosides do not reduce Tollen’s reagent or Fehling’s solution and do not react with HCN or NH₂OH. Hence, it proves the absence of free —CHO group in these glucosides.

6-Membered Ring Structure Of D-(+)-Glucose: As the open-chain structure of glucose cannot explain the above observations, scientist W. N. Haworth proposed the six-membered ring structure of glucose.

1. There is no free aldehyde (—CHO) group in this structure. The —CHO group of the glucose molecule bonds with the —OH group of C-5 of the same molecule by intramolecular reaction to produce a hemiacetal. This hemiacetal is a heterocyclic ring formed of one oxygen atom and five carbon atoms.
2. The anomeric carbon (C_-1) turns into a chiral carbon (C) after the formation of hemiacetal. The newly formed C-atom can link with the H and —OH groups in two patterns, i.e., D -glucose is found in two stereoisomeric forms of α-D-(+) -glucose and β-D-(+) -glucose.
3. Conventionally, if the —OH group is linked to C-1, the —OH group is linked to C-5 (determines the prefix D) and the O-atom of the ring in the Fischer projection formula is on the same side, the structure is referred to as α-D-(+) glucose and vice versa in case of β-D-(+) -glucose.
4. The closed ring structure of glucose is similar to the pyranring and hence, is also known as the pyranose ring. Thus, the six-membered closed-ring structure of glucose is called glucopyranose. α-and β-D-(+) -glucose are known as α- and β-D-(+) -glucopyranose respectively.

The stereoisomers α-and β-D-(+) -glucopyranose having different C-1 configurations are referred to as anomers and C-1 of glucose is called anomeric carbon.

Anomeric Carbon: In the case of any monosaccharide, the carbon atom which takes part in the process of intramolecular hemiacetal formation is called the anomeric carbon of the monosaccharide, for example., in the case of glucose, the carbon atom of the aldehyde group participates in the intramolecular reaction to form hemiacetal.

• So, the number-1 carbon atom, i.e., C₁ -atom is called anomeric carbon. Similarly, in the case of fructose, the carbon atom of the keto group, i.e., the C₂-atom takes part in the process of hemiacetal formation.
• Therefore, the number-2 carbon atom, i.e., the C₂-atom of fructose is called anomeric carbon.

Anomer: Any pair of cyclic stereoisomers which are produced from monosaccharides through intramolecular hemiacetal formation are known as anomers, for example., α-D-(+) -glucopyranose and β-D-(+) -glucopyranose form such a pair of anomers.

Explanation Of Limitations Of Open-Chain Structure With The Help Of Closed Ring Structure

• Solid D-glucose is found as a closed ring as well as α- and β-anomers. When any anomer (either the α- or β- anomer) is dissolved in water, it gradually converts into the other anomeric form, resulting in a dynamic equilibrium between the two.
• The equilibrium mixtures contain the anomers in a specific ratio and hence, their specific rotation attains a fixed value.
• In closed-chain structures (36% a-anomer and 64% β-anomer) traces of open-chain structure are also found at room temperature. The optical rotation of the equilibrium mixture is +52.7°.

Some Important Points Regarding Mutarotation:

1. All monosaccharides and some disaccharides like maltose, and lactose exhibit mutarotation.
2. The addition of an acidic or basic catalyst to the aqueous solution of glucose or other monosaccharides increases the rate of mutarotation.
3. Acidic or basic solutions do not exhibit mutarotation, for example., glucose dissolved in cresol (acidic) or pyridine (basic) do not show mutarotation. But, when it is dissolved in a mixture of cresol and pyridine, it shows mutarotation. This proves that both acidic and basic solvents are required for mutarotation.
4. Water is an amphoteric solvent. So, glucose or other monosaccharides show mutarotation in aqueous solution.
5. A rise in temperature increases the rate of mutarotation.
6. It gives an idea about the closed ring structure of monosaccharides.
7. Methyl α-D-(+)- glucoside and derivatives of other monosaccharides do not show mutarotation.
8. Ring-chain tautomerism i.e., the transformation of the closed ring to an open chain and vice-versa causes mutarotation.

The open-chain structure can be used to explain the formation of cyanohydrin, oxime and osazone and the reduction of Fehling’s solution by the aldehyde groups.

• These are irreversible reactions. Firstly, the open chain structure present in traces undergoes a chemical reaction. Its decrease in concentration disturbs the equilibrium and therefore the closed-ring structures change into open-chain forms.
• These forms of glucose produce the aforesaid compounds. On the other hand, the reactions between D-glucose and Schiff’s reagent, sodium bisulphite, 2, 4- DNP and NH₃ are reversible.
• So, a sufficient concentration of open-chain structures remains in equilibrium with the ring forms. So, it can be said that the weak reagents do not disturb the equilibrium to produce more open-chain structures. Hence, these reagents do not react with D-(+) -glucose.
• The structures of two anomeric methyl glucosides can be explained by the closed ring forms. The two anomers on separately reacting with methanol in the presence of dry HCl can produce methyl α-D -glucoside and methyl β-D- glucoside respectively.

• Glucosides are acetal compounds that are produced during reactions between hemiacetal and methanol. These are hydrolysed by dilute acids only to form the corresponding hemiacetal which further gives glucose and methanol as products.
• But, acetals are stable in an alkaline medium similar to ether i.e., no hydrolysis takes place. So, glucosides do not reduce Fehling’s solution or Tollens’ reagent.
• Glucose pentaacetate has two anomeric forms similar to methyl glucosides.

These pentaacetates do not hydrolyse easily in aqueous medium to form hemiacetal, i.e., open-chain forms. So, these compounds cannot react with hydroxylamine (NH₂OH) to form oximes like glucose.

Structures Of Glucose And Other Monosaccharides Haworth Projection Formula

1. As per the Haworth projection formula, α-D-(+) and β-D-(+) -glucopyranose are considered to be equiplanar hexagons. The O-atom is placed on the right corner of the hexagonal part, which is farthest from the reader.
2. The part closest to the reader is made bold. A perpendicular vertical straight line shows that the H-atom and —OH groups are linked to the C-atom.
3. The H-atoms and —OH groups present on the right-hand side in the Fischer projection formula are drawn below the hexagonal plane and those on the left-hand side are drawn above the hexagonal plane in the Haworth projection formula.
4. The —CH₂OH group and H-atom linked to C-5 are drawn above and below the hexagonal plane respectively.
5. The —OH group joined to the anomeric carbon C-1 of α-anomer lies below the hexagonal plane and the —OH group of C-1 of β-anomer lies above. Similarly, the structures of other monosaccharides can be drawn using the Haworth projection formula.

Fischer projection Formula Of A- And P- Anomers Of D-And L-Sugars:

Reactions Of D-(+)- Glucose:

• Glucose reacts with excess phenylhydrazine (in the 1:3 molar ratio to give a yellow crystalline solid compound, glucosamine.
• Osazone of monosaccharides is a crystalline solid having well well-defined melting point. For example, glucosamine has a melting point of 204° C.
• Thus, a monosaccharide may be identified by preparing its osazone and subsequently determining its melting point-

D-(+)-glucose reacts with the cone. HCl to form 5-hydroxymethyl furfural:

D-(+) -glucose undergoes a rearrangement (known as Lobry de Bruyn-van Ekenstein rearrangement) on treatment with aqueous NaOH to form an equilibrium mixture of D-mannose. D-fructose along with the starting material:

Glucose undergoes fermentation by die action of zymase from yeast to give ethyl alcohol with the liberation of CO2.

The reaction is anaerobic. However in the presence of air (oxygen), the alcohol may be oxidised further to acetic acid or it may produce more CO₂.

D-(-) – Fructose (C₆H₁₂O₆)

Characteristics Of D-(-)- Fructose

1. D-(-)-fructose is the most essential naturally occurring ketohexose. It is found in fruits and honey in free form and hence, is called fruit sugar.
2. D-(-) -fructose remains bonded with glucose in the sucrose and hydrolysis of sucrose yields fructose. Inulin is a naturally occurring polysaccharide formed of fructose molecules.
3. Fructose is commercially prepared by the hydrolysis of Inulin. The aqueous solution of naturally occurring fructose is optically active and laevorotatory (specific rotation is -92°). Thus, it is known as laevulose.
4. The chiral carbon atom (C-5) of the highest rank in fructose has a three-dimensional configuration similar to that of D -glyceraldehyde. So, it is placed under the D – class of ketohexoses.
5. Fructose also forms intramolecular acetal and ketal just like glucose. It exhibits mutarotation.
6. Fructose has a reducing properties.

The open-chain structure of D-(-) -fructose is depicted here.

• The O-atom of the —OH group linked to C-6 of fructose bonds with the keto group of C-2. Hence, C-2 becomes an asymmetric carbon.
• There are two possible arrangements of the —CH, OH and —OH groups. So, D-fructose can exist in two stereoisomeric forms.
• These are α-D -fructopyranose C[α]_D = -20°) and β-D-fructopyranose ([α]_D = -133°). Fructose exhibits mutarotation.
• The specific rotation of aqueous solution of naturally occurring fructose in equilibrium is -92°. Fructose is a reducing sugar.

Haworth Projection Formula Of D-(-)- Fructopyranose And D-(-)- Froctofuranose

Fructose exists as a five-membered ring when bonded to glucose in sucrose. This ring structure, being similar to a furan ring is also called a furanose ring. Similarly, the five-membered ring of fructose is called fructofuranose.

Reactions Of D-(-)-Fructose

Comparison Between Glucose And Fructose 

Osazone formation by monosaccharides involves only C-l and C-2; Thus the configurations of the other carbon atoms of the starting monosaccharide remain unchanged the configuration of C-3, C-4 and C-5 of glucose, mannose and fructose are identical because they form the same osazone on treatment with excess of phenylhydrazine.

Disaccharides

Carbohydrates whose hydrolytic products are two similar or different monosaccharides are called disaccharides. Their general formula is C₁₂H₂₂O₁₁. Three essential disaccharides are sucrose, maltose and lactose. These are hydrolysed by acids or enzymes to form 2 similar or different monosaccharides.

Two monosaccharides are bonded by ether or oxide linkage by the removal of one water molecule to form a disaccharide. This bond is known as glycosidic linkage.

Classification Of Disaccharides: Depending upon the type of linkage, disaccharides are of two types—

1. Reducing and
2. Non-reducing.

If two monosaccharides are joined by a —CO group, it forms a nonreducing disaccharide as monosaccharide units in the form of acetal (or ketal) ring cannot open up to form —CHO (or —COCH₂OH) groups (responsible for the reduction of Fehling’s solution or Tollen’s reagent).

If there is a free —CO group in a monosaccharide unit, it will act as a reducing agent as it remains in the form of hemiacetal or hemiketal which produces —CHO or —COCH₂OH.

Classification Of Disaccharides Example:

1. Reducing: Maltose and lactose.
2. Non-reducing: Sucrose, Trehalose

Sucrose Or Cane Sugar (C₁₂H₂₂O₁₁): It is a common disaccharide present in sugarcane, beet, palms, etc. A major source of sucrose is sugarcane, hence, it is called cane sugar.

It is a colourless, water-soluble, sweet, crystalline substance’ and is used in the preparation of sweetened food. Dilute acids or enzymes hydrolyse it to produce an equimolar mixture of D-(+) -glucose & D-(-)- fructose.

Disaccharides Inversion Of Cane Sugar: The specific rotation of sucrose in an aqueous solution is +66.5°. Its hydrolysis produces an equimolar mixture of D-glucose (specific rotation =+52.5°) and D-fructose (specific rotation = -92°).

• As the specific rotation of laevorotatory fructose is higher than that of dextrorotatory glucose, the specific rotation of the hydrolysed solution is laevorotatory with the value being 1/2(+52.5°-92°)=-19.75°.
• Since, the direction of optical rotation of sucrose changes (+66.5° → -19.75°), the phenomenon is called inversion of cane sugar. The mixture thus obtained is called inverted sugar. Honey is a very common example of inverted sugar.

Disaccharides Structure Of Sucrose: Sucrose is a non-reducing sugar and hence, its two monosaccharide units, glucose and fructose are linked by a —CO group.

• Structural analysis of the ring shows that glucose exists in the form of pyranose and fructose in the form of furanose.
• As the enzymes maltase (hydrolyses α-D-glucoside) and invertase (hydrolyses β-D-fructoside) hydrolyse sucrose, glucose is found in the form of α-glycoside and fructose as β-fructoside in the sucrose molecule.
• The C-1 of α-D-glucose is linked to C-2 of β-D-fructose. Therefore, the structure of sucrose is given below, which explains all its properties.

Maltose Or Malt Sugar (C₁₂H₂₂O₁₁): The enzyme diastase present in germinating barley or malt hydrolyses starch partially to produce maltose.

• 1 mol maltose, on hydrolysis, produces 2 mol of D- glucose. Maltose is a reducing sugar as it’s one glucose unit remains as hemiacetal and produces osazone on reaction with excess hydrazine.
• It shows mutarotation and reduces both Tollens’ reagent and Fehling’s solution.

Structure Of Maltose: Several experiments have proved that maltose contains two units of glucose as pyranose of which one is reducing and the other is non-reducing. The C-1 of non-reducing glucose is linked to the C-4 of reducing glucose.

Again, as maltose is hydrolysed by the enzyme maltase (that hydrolyses α-glucoside), the non-reducing glucose unit is found in the form of a-glucoside. i.e., C-1 of non-reducing α-D-glucose is linked to C-4 of the reducing glucose unit.

Lactose Or Milk Sugar (C₁₂H₂₂O₁₁): Since lactose is found in milk, it is called milk sugar. Dilute acids hydrolyse lactose to form an equimolar mixture of D-glucose and D-galactose.

It is a non-reducing sugar as its glucose unit is a hemiacetal and reacts with excess phenylhydrazine to form osazone. It shows mutarotation and reduces both Tollens’ reagent and Fehling’s solution.

Structure Of Lactose: In lactose, glucose (reducing form) and galactose (non-reducing form) remain as pyranose and C-4 of the glucose unit is linked to C-1 of the galactose unit.

Hydrolysis of lactose by emulsion (hydrolyses fi glycoside) indicates that non-reducing galactose is present in the form of β-galactoside i. e., C-1 of non-reducing galactose is linked to C-4 of reducing glucose unit.

Polysaccharides

Several monosaccharides by the removal of H₂O molecules get linked together and form a macromolecule of polysaccharide. So, polysaccharides are naturally occurring condensation polymers in which monomeric monosaccharides are joined by glycosidic linkages. Some important polysaccharides are:

1. Starch,
2. Cellulose,
3. Glycogen and
4. Dextrin. Starch and cellulose are formed of only D-glucose units.

Polysaccharides Starch, Amylum (C₆H₁₀O₅)_n: Starch is stored as a reserve food in tuber, root and seeds. The value of n may vary from 200-1000 depending on the source.

Polysaccharides Properties of starch

1. Starch is an amorphous granular solid.
2. It is insoluble in cold water. If starch is heated as a suspension in water, it condenses to a viscous milk-like solution, which when cooled thickens into a jelly-like substance.
3. It breaks into small molecules on hydrolysis. Later, it forms maltose which ultimately converts into D-glucose.

It is a non-reducing polysaccharide. It does not reduce Tollen’s reagent or Fehling’s solution and does not form osazones. This proves that the hemiacetal —OH groups of C-1 of glucose are joined by glycosidic linkage.

Polysaccharides Composition Of Starch: Starch is a mixture of amylose (15-20%) and amylopectin (80-85%).

• It reacts with I₂ to produce a violet-blue colour as amylose forms an inclusion complex with I₂. When heated, the blue colour fades away but reappears on cooling.
• Amylopectin does not produce a blue colour with I₂. Both amylose and amylopectin are polymers of α-D glucopyranose.
• One amylose molecule contains about 200- 1000 glucose units, while one amylopectin molecule contains 2000 -3000 glucose units.

Structure Of Amylose: Amylose is a linear polymer of α-D-glucose in which C-1 of one glucose unit is linked to C-4 of another by a-glycosidic linkage.

Structure Of Amylopectin: Amylopectin is a complex polymer consisting of 10-200 unbranched linear chains.

• Each chain is made up of 25-30 α-D-glucopyranose linked serially by or- 1, 4-glycosidic bonds.
• Unbranched chains are linked by α-1, 6 -the glycosidic bond between the anomeric C-1 of the glucose residue at the reducing end of one chain and the C-6 of any glucose residue of another chain, which gives rise to an unbranched molecule of amylopectin.

Distinction Between Glucose And Starch:

Cellulose (C₆H₁₀O₅)_n: The molecular mass of cellulose is 50,000 to 5,00,000. It is formed by 300-3000 β-D-glucose units. It is water-insoluble but dissolves in an ammoniacal solution of cupric hydroxide (Schweizer’s reagent).

• The addition of alcohol, acid or salt to this solution precipitates cellulose.
• Its pure form is most abundant in plants. It is the major component of the plant cell walls and mainly brightens the cotton fabric. This process is called mercerisation and the cotton thus obtained is called mercerised cotton.

Properties Of Cellulose: Cellulose is a non-reducing sugar like starch. It does not reduce Tollen’s reagent or Fehling’s solution and does not form osazones.

• It is not fermented by yeast. It cannot be hydrolysed easily like starch. But, when cellulose is heated with dilute H₂SO₄ (under pressure), it hydrolyses to produce D-glucose.
• The enzyme emulsion or cellulase (hydrolyses β-glycosidic linkage) can hydrolyse it. In addition to concentrated NaOH, a translucent jelly-like substance is formed that brightens the cotton fabric.
• This process is called mercerisation and the cotton thus obtained is called mercerised cotton.

Structure Of Cellulose: Cellulose is a linear polymer of serially arranged α-D- glucose units linked by β-1,4-glycosidic bonds.

The CH₂OH— group of one chain links to the —OH group of C-2 of another by H -bonding to form a bundle of fibres. This linear arrangement helps to form cellulose threads.

Uses Of Cellulose: It is used in the manufacture of clothes, cameras and papers. Cellulose also forms many useful compounds by treatment with suitable chemical reagents. Some examples are—

1. Celluloid, which is used in the manufacture of toys, decorative articles and photographic films.
2. Guncotton, which is an explosive.
3. Cellulose acetate is used in the manufacture of rayon (artificial silk), plastics and nail polishes.
4. Methylcellulose is used as a drug for the treatment of constipation and for the manufacture of cosmetics.
5. Ethyl cellulose is used as a thin film coating material and also as a food additive as an emulsifier.

Can cellulose be considered as food?

• The human intestine does not synthesise the enzyme cellulase which can hydrolyse cellulose into glucose. So, human beings cannot digest cellulose.
• On the other hand, herbivores like cows, goats, and deer have bacterially produced cellulase enzymes in their intestines. So, they can feed on grasses and plants.

Importance Of Carbohydrates

In Living Systems

Biofuels: Carbohydrates are very essential for the survival of plants and animals. It forms a major portion of our diet. From an early time, ayurvedic medicine has considered honey as an instant source of energy.

• Hence, carbohydrates provide the required energy for metabolism occurring inside living beings, i.e., act as biofuels.
• C₆H₁₂O₆(glueose) + 6O₂ → 6CO₂ + 6H₂O + 2880 kj.mol^-1 Polysaccharides like starch and glycogen are hydrolysed enzymatically into glucose inside living cells.
• The produced glucose is then transported to all the cells via blood. After that oxidation of glucose into CO₂ and H₂O during enzyme catalysed reactions yields energy.

Reserve Foods: The major reserve food of plants is the polysaccharide starch.

• It is stored in seeds as food for the germinating plant until it starts to photosynthesize.
• It is also stored in the form of glycogen in animal liver and muscles. During starvation or illness, this glycogen is rapidly hydrolysed to glucose which in turn, gets oxidised to produce energy.

Constituents Of Biomolecules: Two aldopentoses, D -ribose and 2 -deoxyribose are essential constituents of RNA and DNA respectively. These nucleic acids carry hereditary traits and also participate in protein synthesis.

• The monosaccharide ribose is an essential component of adenosine triphosphate (ATP).
• Energy obtained by oxidation of biomolecules such as carbohydrates, lipids etc., is stored in the cells in the form of ATP, which, in turn, helps to perform all the cellular functions (metabolism).
• This is the reason why ATP is also called the energy currency of the cells. Carbohydrates also exist in biosystems in combination with many proteins and lipids.

As Structural Components: The major structural component of plant and bacterial cell walls is cellulose. It is used in the textile industry (as cotton) and in the furniture industry (as wood).

As Industrial Material: Carbohydrate is used as a raw material in different industries such as textiles, paper, liquor, etc.

Proteins

Chemically, proteins are condensation polymers or polyamides of high molecular mass (>10000) whose monomers are α-amino acids. Proteins contain C, H, N, O, S, and sometimes P, I, and traces of metallic elements (Fe, Cu, Zn, Mn). Partial hydrolysis of proteins yields peptides of varying molecular mass and complete hydrolysis produces a mixture of α-amino acids of L-configuration.

α-Amino Acids

Amino acids have amino (—NH₂) and carboxyl ( —COOH) groups. Depending on the relative position of these groups, they are categorised as α, β, γ, δ.

• The most important of all is a -amino acid as all biological proteins yield a -amino acids on their hydrolysis.
• Hence, or-amino acids are the building blocks of proteins. These are linked together by — CO—NH— bonds in proteins. About 23 amino acids have been isolated till now.
• Amino acids differ due to the presence of different kinds of R (alkyl) groups.

Nomenclature Of Amino Acids: Amino acids can be named as per the IUPAC convention, but they are generally known by their common names.

Nomenclature Of Amino Acids Example: The amino acid H₂NCH₂COOH is known as glycine instead of 2-amino ethanoic acid (IUPAC).

The common names are assigned depending on their sources and properties.

Nomenclature Of Amino Acids Example: Glycine is named so, as it tastes sweet (Greek word ‘glykos’ means sweet); tyrosine was first isolated from cheese (Greek word ‘tyros’ means cheese).

The abbreviations are formed from the first three letters of the common name given to an amino acid.

Nomenclature Of Amino Acids Example: ‘Gly’ stands for glycine, and ‘Ala’ stands for alanine. Sometimes, a single letter is also used as a symbol.

Classification Of α-Amino Acids

Classification Based On The Number Of —NH₂ And —COOH Groups

• Neutral Amino Acids: Amino acids containing the same number of — NH₂ and —COOH groups are known as neutral amino acids.
• Neutral Amino Acids Example: Glycine, alanine, valine, etc.
• Acidic Amino Acids: Amino acids containing one — NH₂ group and several —COOH groups are acidic amino acids.
• Acidic Amino Acids Example: Aspartic acid, glutamic acid, etc.
• Basic Amino Acids: Amino acids containing one — COOH group and several — NH2 groups are basic amino acids.
• Basic Amino Acids Example: Lysine, histidine, etc.

Classification Based On Nutritional Significance

• Essential Amino Acids: The human body cannot synthesise 10 amino acids obtained from protein. So, these should be taken through diet. A deficiency of these amino acids retards the normal growth and development of the body. These 10 amino acids are called essential amino acids.
• Essential Amino Acids Example: Arginine, valine, leucine, isoleucine, phenylalanine, methionine, tryptophan, threonine, histidine and lysine.
• Non-essential Amino Acids: The amino acids which are synthesised inside the human body are called non-essential amino acids. All amino acids except the ten essential amino acids are non-essential amino acids.
• Non-essential Amino Acids Example: Glycine, alanine, serine, cysteine, cystine, tyrosine, proline, hydroxyproline, asparagine, aspartic acid, glutamine, glutamic acid, hydroxylysine.
• Complete Proteins: Proteins providing all essential amino acids in the correct proportion for the nourishment of the body are called complete proteins, for example., proteins in fish, meat, milk and eggs.
• Incomplete Proteins: Proteins which cannot provide one or more essential amino acids to the human body are called incomplete proteins for example., lysine is absent in proteins found in rice, wheat and other cereals.

Natural Amino Acids:

Optical Activity Of Amino Acids: All amino acids (except glycine) have one chiral carbon atom. So, each amino acid has two possible optically active isomers of D- and L- configuration.

All-natural amino acids are of L-configuration. The L-amino acid has a —NH₂ group on the left and a —H group on the right according to the Haworth projection formula.

Properties Of α-Amino Acids

Physical State, Solubility And Polarity

1. Amino acids are colourless, non-volatile, crystalline solids. They melt by decomposing at high temperatures.
2. Amino acids are water-soluble but insoluble in petroleum ether, benzene and ether
3. Their aqueous solutions behave like solutions of substances having high dipole moment

Zwitterion Formation

1. The presence of both acidic (—COOH) and basic (—NH₂) groups in an amino acid molecule allows the proton (H⁺) to dissociate from the —COOH group and combine with the — NH₂ group of the same molecule. Thus, the —COOH group becomes —COO^– and the —NH₂ group converts into —N⁺H₃. Thus, an amino acid behaves as an ampholyte and is regarded as a dipolar ion or zwitterion
2. Due to zwitterion formation, amino acids have high melting points and dipole moments These are insoluble in non-polar solvents but soluble in water Hence, amino acids should be expressed in the zwitterion form as
   H₃N^–—CHR—COO^–.
3. —COO^– group acts as a proton acceptor and —N⁺H₃ acts as a proton donor, i.e., —COO^– group impart* basicity and — NH₃ group imparts acid character Amino acids may exist as zwitterions in solid or polar solvents but not in a gaseous medium.

Isoelectric Point: The proton transfer of zwitterions in alkaline and acidic mediums takes place in the following way:

• Amino acids behave as cations in an acid medium and as anions in an alkaline medium. If two electrodes are dipped in the acidic solution of an amino acid and then an electric current is passed through the solution, the amino acid will move towards the cathode.
• Similarly, if current is passed through the basic solution of an amino acid, the amino acid will move towards the anode. Hence, the movement of die amino add molecules towards a particular electrode depends on the pH of the solution.
• It can be shown that an amino acid at a particular pH does not move towards any’ electrodes. Amino acids exist as neutrally charged zwitterion at that pH, called the iso-electric point or pi. Each amino acid has its specific pl.

Isoelectric Point Definition: The particular value of pH at which an amino acid exists as a neutrally charged zwitterion and does not move towards any electrodes on passing an electric current through it, is called the isoelectric point of that amino acid.

• The value of pl of neutral amino adds is slightly less than 7 [pi of glycine = 6.0); aridic amino adds is 3-5.4 [pi of aspartic add =3.0) and basic amino adds is 7.6-10.8 [pl of lysine = 9.5).
• The solubility of amino added in water is the least at the isoelectric point So, this property’ is used to separate different amino adds obtained during protein hydrolysis.

Acidity And Basicity Constants (K_a and K_b): The acidity and basicity constants of an amino acid are very low, for example., K_a of glycine = 1.6 × 10^-10 and Kb = 2.5 × 10^-12. Most carboxylic acids have a value of 10^-5 and most aliphatic amines have a K_b value of 10^-4. The reason is the presence of the — NH₃ group as a weak acid and the —COO^– group as a weak base in a-amino acids.

Identification Of α-amino Acids: The addition of ninhydrin to the aqueous solution of a -amino acid in the presence of lithium acetate (LiOAc) produces a deep bluish-violet colour. This test, known as the Ninhydrin reaction is used to detect α-amino acids.

Peptides And Their Classification

Amino acids can bond with each other to form a long molecular chain. The (—COOH) group of one amino acid links with the (—NH₂) group of another by — CO —NH— bonds when two similar or different amino acids combine. The — CO—NH— bond formed by the elimination of 1 molecule of water is called a peptide bond or peptide linkage and the compound thus obtained is called a peptide.

Classification Of Peptide: The peptide produced in the reaction between two amino acids is called a dipeptide. Similarly, the peptides which are formed in the reaction between 3, 4, and 5 amino acid molecules are termed tripeptide, tetrapeptide, pentapeptide and so on.

Classification Of Peptide Example: The —COOH group of glycine reacts with the —NH₂ group of alanine to form glycylalanine (dipeptide). Similarly, the condensation of glycine, alanine and valine produces glycylalanylvaline (tripeptide).

Similarly, several amino acids are linked together by peptide bonds to form a macromolecular chain or polymer. These polymers are called polypeptides. Conventionally, the peptide chain is expressed keeping the N-terminal amino acid or residue at the left-hand side and the C-terminal amino acid or residue at the right-hand side.

Nomenclature Of Peptide: The peptide is named according to the order of amino acids present in the peptide chain starting from the N-terminal residue.

• The ‘ine’ suffix is eliminated from the amino acids other than the C-terminal one and is replaced by the suffix ‘yl’ and the C-terminal residue is written without any change.
• There are no spaces between the names of the two amino acids.

The abbreviation of the amino acids is generally used for the nomenclature of peptides.

Nomenclature Of Peptide Example: Glycylalanine can be represented as Gly-Ala or G-A, alanyl glycyl phenylalanine as Ala-Gly-Phe or A-G-F.

• The geometry of the peptide bond: The lone electron pairs of I N-atom present in a peptide bond ( — CO—NH— ) resonates with the C—O group, for example., delocalisation of the electron pair occurs via the C=0 group. Hence, the C=N bond gains a double bond character and geometrical isomerism is observed in the peptide bonds.
• The steric hindrance between the two R—groups on the same side of the cis-isomer makes it less stable than the trans-isomer. Hence, the — CONH group is more stable in the trans-form i.e., O and H atoms lie on the opposite sides.
• Carbon, nitrogen and the atoms (O, H and two C) linked to them lie on the same plane.

Oligopeptides, Polypeptides And Proteins: Small-chain peptides of 2-10 amino acids are known as oligopeptides and long-chain peptides consisting of more than 10 amino acids are called polypeptides.

• Polypeptides having 100 or more a-amino acids and an overall molecular mass of more than 10,000 are known as proteins.
• The difference between a polypeptide and a protein is not distinct. Polypeptides of a relatively smaller number of a-amino acids bearing conformation similar to that of proteins are also regarded as proteins, for example., insulin is a protein containing 51 amino acids.
• Proteins are amphoteric. So, these can neutralise acids or bases and possess an isoelectric point like α-amino acids. The solubility of polypeptides is the least at its isoelectric point and they are separated based on this property.

Classification Of Proteins

Classification Based On Molecular Structure: Structurally, proteins are of two types:

1. Fibrous proteins and
2. Globular proteins.

Fibrous Proteins: These proteins are slender, thread-like structures and water-insoluble and lie parallel to each other like fibres.

• H-bonds and disulphide bonds join the peptide chains together.
• The strong intermolecular force makes them water-insoluble. These proteins are components of connective tissues in living systems.

Fibrous Proteins Example: It is an essential component of skin, nails, hair, etc. Collagen, α-keratin, and myosin are different examples of this class of protein. These proteins remain stable at slight variations in temperature and pH.

Globular Proteins: These proteins intertwine among themselves to form a small sphere.

• H -bonds, disulphide bonds, van der Waals force of attraction and dipolar interaction act between the polypeptide chains.
• The hydrophobic R-groups lie on the inner side and the hydrophilic polar groups on the outer side of the bundle.
• So, these are readily soluble in water. These proteins participate in different activities of plant and animal cells.

Globular Proteins Example: Enzyme (like trypsin), hormone (like insulin), transport protein (like haemoglobin), protective protein (like antibody). These proteins are more sensitive to changes in pH and temperature than the fibrous proteins.

Classification Based On The Nature Of Hydrolytic Product

1. Simple Proteins: The proteins whose hydrolytic products are α-amino acids are called simple proteins.
   1. Simple Proteins Example: Albumin in egg white, glutinine in wheat, oxygenin in rice, keratin in hair and nails.
2. Conjugated Proteins: Proteins whose hydrolytic products consist of α-amino acids and non-proteinaceous compounds are called conjugated proteins.
   1. The nonprotein part is called the prosthetic group and the remaining part excluding the prosthetic group is called the apoprotein.
   2. The main role of the prosthetic group is to control the biochemical activities of the protein. There are different conjugated proteins based on the nature of the prosthetic group.

Derived Proteins: The compounds produced on partial hydrolysis of simple or conjugated proteins by acid, base or enzyme are called derived proteins.

Derived Proteins Example: Proteose, peptones and polypeptides.

Structures Of Proteins

Several a -amino acids combine by peptide bonds to produce organic polymers of 3-D configuration. The structure of a protein can be categorised into any one of the four structural tiers.

Primary Structure Of Proteins: The arrangement or sequence of amino acids in a protein molecule refers to its primary structure. The modification of a single amino acid will change the sequence, which will form a different protein. The characteristic chemical and biological properties of different proteins are due to the differences in their primary structure.

Secondary Structure Of Proteins: The conformation of molecular chains in a protein refers to its secondary structure. Experimentally, it has been observed that a protein molecule has two conformations:

1. Right-handed α-helix and
2. β-pleated sheet.

α-Helix Conformation: A protein chain with a large R-group exists in the form of a right-handed α-helix. R-groups are extended away from the axis of the helix. This conformation arises due to the intramolecular hydrogen bond (between C=O of an amino acid residue and N—H of a fourth amino acid residue).

β-Pleated Sheet Conformation: The protein chains acquire a coiled structure to attain the β-pleated sheet conformation due to the presence of moderately sized R-groups.

• The two adjacent protein chains are linked by H-bonds. The alternate R-groups linked to the a -carbon lie on the same side. Several chains together form a sheet—either parallel or antiparallel.
• The N-terminals lie on the same side in parallel conformation. In anti-parallel conformation, the N-terminal of one chain and the C-terminal of another chain lie on the same side.

Tertiary Structure Of Proteins: Although there are only two types of conformation in a protein molecule, large protein molecules fold to form 3-D structures.

• The 3-D structure formed due to the folding of protein molecules keeping the original conformation (α-helix or β- pleated sheet) intact, is called tertiary structure.
• It is specific for each protein and is rigid and stable. The two major tertiary structures are discussed below.

Tertiary Structure Of Fibrous Proteins: An identical secondary structure is present in the complete chains of fibrous proteins, for example., α-keratin (a major protein in hair and wool) always exists as α-helix.

• Several α-helices intertwine with each other to form a rope or rod-like tertiary structure.
• The triple helical structure of collagen is shown below. α-helix is also known as 3.6₁₃ helix as 3.6 amino acids are found in one turn and one 13-membered ring is formed by H-bonds. The two adjacent turns are 54 pm away from each other.

Tertiary Structure Of Globular Proteins: Globular proteins do not possess identical secondary structures of protein chains, like fibrous proteins. Some part of the molecule has α-helix, some parts are in the form of β- pleated sheet, and some parts do not have any secondary structure.

• The latter part with no secondary structure is known as random coils. All these different parts are convoluted together to form a globular structure.
• It is possible due to the presence of different attractive forces acting among the different R-groups in the side chains. These forces are hydrogen bonds, ionic or salt bridges, disulphide bonds and van der Waals’ force of attraction.

Hydrogen Bonds: The major radicals present in the side chains of amino acids of protein molecules are —OH, — NH₂ and — COOH which exhibit H-bonds, for example., tyrosine and aspartic acid are linked by H-bonds.

Attractive Forces Due To Salt Bridge Formation: The internal neutralisation occurring between the — COOH group and — NH₂ group of side chains of amino acids a protein molecule gives rise to —COO and — NH₃ groups. A salt bridge is formed generating an attractive force between these two groups.

Van Der Waals’ Forces Of Attraction: This force acts between the non-polar (hydrocarbon) side chains of amino acids of the protein.

Disulphide Bridge: Two — SH groups of two cysteine amino acids of a protein molecule are oxidised to form a disulphide bridge.

Differences Between Globular And Fibrous Proteins:

Quaternary Structure Of Proteins: Some complex proteins are formed of two or more molecular chains of polypeptides which are called subunits or protomers.

• The complex three-dimensional structure formed by the aggregation of different protomers by H-bonds, electrostatic force, and van der Waals force of attraction is called a quaternary structure.
• Haemoglobin bears a quaternary structure of spherical-shaped tetramer consisting of four polypeptide chains (two identical α-chains of 141 amino acids each and two identical β-chains of 146 amino acids each). Each polypeptide chain bears a terminal heme group (iron- protoporphyrin complex).

Denaturation Of Proteins: Every protein in biological systems possesses a particular 3-D configuration and a specific biological activity.

• These proteins are called native proteins. If the native proteins are subjected to physical changes like temperature variation or chemical changes like pH variation or brought in contact with UV rays, the globular or helical coils are opened up and water-soluble globular proteins are coagulated into water-insoluble fibrous protein.
• This coagulation disintegrates the original structure of protein and its biological reactivity is lost. This phenomenon is called protein denaturation and the coagulated protein is called denatured protein.
• As a result of denaturation, there is no modification in primary structure but secondary and tertiary structure undergo modification. The albumin of egg white is converted into water-insoluble fibrous protein.
• Similarly, the conversion of milk into curd is also an example of protein denaturation where lactic acid secreted by bacteria is responsible for this process.
• During the preparation of cottage cheese by heating milk with lemon juice or lactic acid, globular milk protein lactalbumin is converted into fibrous protein. Here also the protein gets denatured.

• Protein denaturation may be reversible or irreversible.
• Coagulation of egg white on heating is an irreversible process.
• Some denatured proteins return to their original secondary and tertiary structures when provided with previous pH and temperature and hence, become metabolically active.
• This phenomenon is observed among enzymes and is called renaturation (the opposite process of denaturation).

Biological Functions Of Proteins

1. Structural Materials: The primary building block of any tissue is protein. Half of the total proteins in the human body act as structural components, for example., skin, hair, nails, wool, and feathers containing keratin (fibrous protein), collagen in tendons (a group of muscle fibres), myosin in muscles, fibroin in silk, etc.
2. Transport Agent: Some proteins act as transport agents or carriers which help in the transport of oxygen, metals, fatty acids and hormones, for example., haemoglobin in human blood carries oxygen from the lungs to different parts of the body. JMJ Enzymes: Some proteins act as enzymes and catalyse biochemical reactions.
3. Metabolic Regulators: Some globular proteins control the metabolism of a body, for example., insulin secreted from the pancreas controls blood glucose level, thyroglobulin, a glycoprotein secreted from the thyroid gland synthesises the amine hormone thyroxine, fibrinogen in the blood plasma converts into insoluble protein fibrin to clot blood.
4. Antibodies: It defends the body against pathogens. Certain toxins called antigens are secreted by the invading infectious viruses, bacteria or external proteins.

Antibodies are synthesised to destroy these antigens and thus prevent the body from diseases. Gamma-globulin is an antibody present in blood plasma.

Enzymes And Hormones

Enzymes

Enzymes are biological catalysts. The activation energy of a biochemical reaction is decreased in the presence of an enzyme and so the reaction occurs rapidly and easily. All enzymes are chemically protein in nature and are synthesised in living systems.

Nomenclature Of Enzymes

1. The name of an enzyme is derived by adding the suffix ‘ase’ to the name of the substrate on which that particular enzyme acts.
   1. Nomenclature Of Enzymes Example: The hydrolytic enzyme of maltose is maltase, that of arginine is arginase, cellulose is cellulose.
2. As per the IUPAC convention, the enzyme can be named depending on the nature of the chemical reaction it catalyses.
   1. Nomenclature Of Enzymes Example: Oxidoreductase for redox reactions, hydrolase for hydrolytic reactions, transferase for reactions involving functional group transfer, and isomerase in case of isomerisation reactions to exhibit catalytic activity.

Composition Of Enzymes

Enzymes Are Globular Proteins: Some enzymes act as catalysts while some others combine with a non-protein component called a co-factor to exhibit catalytic activity. Co-factors are of 2 types:

Inorganic ions: Zn²⁺, Mg²⁺, Mn²⁺, Fe²⁺, Cu²⁺, Co²⁺ Mo³⁺, Na⁺, K⁺, etc.

Composition Of Enzymes Organic Molecules

1. Some organic molecules, called coenzymes remain loosely attached to the enzyme and can be easily separated by dialysis. For example, pyridoxal phosphate (PLP) is a coenzyme originated from pyridoxin or vitamin B6.
2. Some organic molecules called prosthetic groups remain strongly attached to the enzyme by covalent bonds and can be separated only by hydrolysis. The enzyme-cofactor complex is called holoenzyme and the remaining inactive enzyme after the cofactor is excluded is called apoenzyme.

Important Characteristics Of Enzymes

1. Enzymes accelerate the rate of a biochemical reaction. It has been observed that biochemical reactions in the presence of enzymes occur approximately 10⁶ times faster than the usual rate (in the absence of enzymes).
2. A small amount (~ 10^-5 mol) of enzyme is required for catalysis.
3. A specific enzyme can catalyse a specific reaction only.
4. Enzymes remain active at an optimum temperature of 37 °C and an optimum pH of 7.0
5. Certain organic and inorganic molecules can act as enzyme inhibitors and retard enzyme activity.
6. Several enzymes show stereospecificity. They react with or produce a particular stereoisomer only, for example., stearoyl-CoA 9 -desaturase converts a derivative of stearic acid into the cis-isomer of a derivative of oleic acid (unsaturated fatty acid).

Some Important Enzymes And Their Catalytic Behaviour:

Applications Of Enzymes

Production Of Essential Substances:

1. Invertase, zymase and maltase are used in the alcoholic beverage industry.
2. The enzyme is used in making paneer.
3. α-amylase is used in the preparation of sweet syrup from starch.
4. Invertase enzyme is used to prepare invert sugar.

Cardiological Treatment: The enzyme streptokinase can decoagulate clotted blood. This enzyme can be used to minimise heart attacks caused due to blood clotting in the coronary artery.

Diseases Related To Enzyme Deficiency: Deficiency of enzymes leads to certain diseases.

1. Lack of phenylalanine hydroxylase enzyme (that converts phenylalanine into tyrosine) causes phenylketonuria (PKU). In the absence of this enzyme, phenylalanine is converted into phenylpyruvate by another enzyme. The presence of both phenylalanine and phenylpyruvate damages the brain severely and causes mental retardation.
2. The disease albinism occurs due to a deficiency of the enzyme tyrosinase. Its absence causes less amount of melanin (pigment giving colour to skin and hair) secretion. The skin and hair colour of the affected human beings and animals, as a result, turn white.

Hormones

Hormones are secreted from endocrine glands, which are transported to different organs and tissues by blood to control metabolic reactions. These are also called chemical messengers because they send messages from one cell to another.

Classification Of Hormones:

Based on chemical structures, hormones can be divided into three classes.

1. Steroids,
2. Proteins or polypeptides and
3. Amines. Steroid hormones can be further subdivided into two sub-classes

1. Sex hormones and
2. Adrenocortical hormones. Sex hormones are of two types—male sex hormones and female sex hormones

Structures Of Few Important Hormones

Sources, Chemical Nature And Functions Of Some Important Hormones:

Lipids And Vitamins

Lipids are the compounds mainly formed of carbon, hydrogen and oxygen. Sometimes sulphur, phosphorus and nitrogen are also present in the lipids. A large part of a lipid consists of hydrocarbons and hence, lipids are soluble in non-polar organic solvents.

Lipids And Its Classification

Lipids And Its Classification Definition: Certain organic compounds which are water-insoluble (like oils, fats, steroids, and terpenes) but soluble in non-polar organic solvents like chloroform, ether, carbon tetrachloride and benzene are called lipids.

Lipids And Its Classification Lipids are of two types:

1. Hydrolysable lipid and
2. Nonhydrolysable lipid.

Hydrolysable Lipids: The lipids which can be hydrolysed into smaller molecules are called hydrolysable lipids. These lipids mostly contain an ester group. They are subdivided into three classes:

1. Waxes,
2. Fats or oils or triglycerides and
3. Phospholipids. Hydrolysable lipids are also called complex lipids.

Hydrolysable Lipids Waxes: These are the simplest hydrolysable lipids. Esters (RCOOR₇) produced from alcohol (R’H) and fatty acid (RCOOH) of high molecular mass are called waxes.

Hydrolysable Lipids Importance Of Wax: The long hydrocarbon chains make waxes hydrophobic.

• They form a protective coating over the feathers of birds and prevent them from getting wet.
• The waxy layer over the leaves of trees prevents excess water loss or absorption. Spermaceti (a wax) present on the head of whales helps to control buoyancy during deep sea diving.

Hydrolysable Lipids Example: Spermaceti [CH₃(CH₂)₁₅OCO(CH₂)₁₄CH₃] and cetyl palmitate are wax. Bee wax is myristyl palmitate, [CH₃(CH₂)₁₄COO(CH₂)₂₉CH₃].

Hydrolysable Lipids Triglycerides: The most common lipids are triglycerols or triglycerides. These are a kind of triesters whose hydrolysis yields one mole of glycerol and three moles of fatty acids.

Oils and fats are triglycerols. Triglycerols which are solids at room temperature are known as fats and those which remain liquid at room temperature are called oils.

Triglycerides Example:

Importance Of Triglycerides: The principal role of triglycerides in cells is to store energy. Complete catabolism of triglycerols into CO₂ and H₂O produces a huge amount of energy.

• Carbohydrates yield energy for a short period, while triglycerols provide energy for a longer period.
• Triglycerols store ~9 kcal-g^-1 energy while carbohydrates store only -4 kcal-g^-1 energy.
• Theoretically, triglycerols can be used as fuel in automobiles because their combustion produces a lot of heat. Nowadays, plant-based oils are mixed with diesel to use as fuels.

Phospholipids: Hydrolysable lipids containing phosphate groups are called phospholipids. The common phospholipids are phosphoryl glycerol or phosphoglycerides.

• Substitution of a fatty acid group in a triglycerol by a phosphate group produces a phospholipid.
• The simplest class of phosphoglycerides is phosphatidic acid. In some phospholipids, an extra alcoholic group may form an ester with phosphoric acid. Cephalins and lecithin belong to this class.

Triglycerides Importance Of Phospholipids: Phospholipids possess a polar head and two non-polar tails. These form a lipid bilayer in aqueous solutions as found in cell membranes.

• The polar groups lie towards the inner face (cytoplasm) and outer face (extracellular fluid) and the non-polar groups lie in between the two heads.
• These non-polar tails prevent the watery part of the cytoplasm from leaking out and thus, protect the cell.

Non-Hydrolysable Lipids

The lipids which cannot be hydrolysed into smaller molecules are called non-hydrolysable lipids. These can be subdivided into four classes:

1. Lipid-soluble vitamins,
2. Eicosanoids,
3. Terpenes and
4. Steroids. Non- hydrolysable lipids are also called simple lipids.

Fat-Soluble Vitamins: Vitamins A, D, E and K are fat-soluble vitamins. Vitamin A present in the human body is converted into light-sensitive compound 11-cis-retanol, which helps in eyesight.

• Vitamin D controls the metabolism of phosphorus and calcium. Vitamin E, being an antioxidant, prevents oxidation of unsaturated fatty acid chains.
• Vitamin K controls the synthesis of prothrombin and certain proteins needed for blood clotting.

Non-Hydrolysable Lipids Eicosanoids: These are a group of organic molecules formed of 20 carbon atoms and are derived from arachidonic acid. Eicosanoids are of four types—

1. Prostaglandins,
2. Leukotrienes,
3. Thromboxanes and
4. Prostacyclins.

Eicosanoids Importance Of Eicosanoids: These are present in low concentration in cells and are localised in activity, i.e., act near the site of synthesis.

• Each eicosanoid has its specific biological activity. Prostaglandins reduce blood pressure, prevent blood platelet coagulation, control inflammation and decrease digestive juice secretion.
• Prostacyclins dilate blood vessels and prevent blood platelet coagulation. Thromboxanes constrict blood vessels and accelerate blood coagulation and leukotrienes contract smooth muscles of lungs.

Non-Hydrolysable Lipids Terpenes: Terpenes are a group of lipids which are formed of 5-C isoprene units.

These are closed or open-chain hydrocarbons, alcohols, aldehydes or ketones. Oxygen-containing terpenes are called terpenoids. Terpene is a major component of essential oils extracted by distillation of plant extracts.

Terpenes Examples: Myrcene and menthol.

Non-Hydrolysable Lipids Importance Of Terpenes: Triterpene (six isoprene units) and tetraterpene (eight isoprene units) play important biological roles, for example., triterpene squalene is a precursor of steroid molecules.

Tetraterpenes are found in the form of carotenoids such as lycopene in tomatoes and β-carotene in carrots. Vitamin A is synthesised inside the human body from β-carotene.

Non-Hydrolysable Lipids Steroids: Biomolecules with perhydro-1, 2-cyclopentanophenan threne system are known as steroids.

Steroids Example: Cholesterol, aldosterone, testosterone.

Non-Hydrolysable Lipids Importance Of Steroids: The steroid cholesterol is found abundantly in the human body. It is an essential component of the cell membrane.

• It is also a precursor of Vitamin D and other steroids. Gall gallstone formed in the gall bladder is primarily cholesterol.
• Testosterone and estradiol are the most effective natural male and female sex hormones respectively.

Vitamins

Vitamins do not provide energy or act as building blocks of tissues but play a major role in the maintenance of good health. Their deficiency leads to diseases related to nutritional deficiency.

The human body can synthesise vitamin A, some members of vitamin B and vitamin K. Plants can synthesise all vitamins.

Vitamins Definition: A special class of biomolecules other than carbohydrates, proteins and lipids, of which most are not synthesised inside the human body and are therefore, taken through diet in small quantities for proper metabolism, growth and nourishment are called vitamins.

Vitamins Sources: The foods are the main source of vitamins— milk, butter, paneer, fruits, green vegetables, meat, fish, eggs, etc. Vitamins can be synthesised in the laboratory and are available in the form of tablets or capsules in the market.

Classification Of Vitamins: Vitamins are complex organic molecules. Different vitamins possess different structures. These are designated by English alphabets like A, B, C, D, E, and K. Different sub-groups of some vitamins are also designated as B₁ B₂, B₆ and B₁₂. Around 25 vitamins are known to date. These can be classified into two classes as follows:

1. Water-soluble Vitamins: These are vitamin B-complex (B₁, B₂, B₅, i.e., nicotinic acid, B₆, B₁₂, pantothenic acid) and vitamin C. These water-soluble vitamins should be taken through our diet daily as these are excreted out in the form of urine and are not stored inside the body (except vitamin B₁₂).
2. Fat-soluble Vitamins: These are oily substances which are soluble in lipids but not in water, for example., vitamins A, D, E, and K. Excess intake of vitamins A and D is harmful and may cause hypervitaminosis. Biotin (vitamin H) is insoluble in both water and lipids. Deficiency of one or more vitamins leads to avitaminosis. It is often noticed in human beings.

Some Essential Vitamins, Their Features, Sources And Deficiency Diseases:

Differences Between Hormones And Vitamins:

Nucleic Acids

Biomolecules present in chromosomes of the cell nucleus other than protein is called nucleic acid.

Nucleic Acid Definition: The complex organic polymers formed of nucleotide monomers located in the nucleus and cytoplasm which carry the essential hereditary traits from one generation to another and take part in protein synthesis are called nucleic acid.

Types Of Nucleic Acids And Their Components

Nucleic acids are of two types:

1. Deoxyribonucleic acid (DNA) and
2. Ribonucleic acid (RNA). Both DNA and RNA are organic polymers. These are the polymers of nucleotides.

Structural Components Of Nucleotides

Three types of compounds combine to form one nucleotide—

1. A heterocyclic nitrogenous base,
2. A pentose sugar and
3. A phosphate group.

Heterocyclic Nitrogenous Base

These are of two types—

1. Purine derivatives and
2. Pyrimidine derivatives.

1. Bases of DNA: Adenine, Guanine, Cytosine and Thymine
2. Base of RNA: Adenine, Guanine, Cytosine and Uracil

Heterocyclic Nitrogenous Base Pentose Sugar: The sugar unit of DNA is deoxyribose and that of RNA is ribose. C-atoms ofpentose sugar are distinguished from the C-atoms of nitrogenous bases by marking them with numbers like 1 2′, and 3′ (read as one-prime, two-prime and so on).

Heterocyclic Nitrogenous Base Phosphate Group: Both DNA and RNA contain a phosphate group. O -atom of a hydroxyl group of pentose sugar is linked to a phosphate group.

Nucleoside And Nucleotide

Nucleoside And Nucleotide Nucleotide: A nitrogenous heterocyclic base (A, G, C, U, T) and a pentose sugar (ribose or deoxyribose) combine to form a compound called nucleoside.

• Addition of a phosphate group to a nucleoside produces a nucleotide.
• Nucleosides containing ribose and deoxyribose sugars are called ribonucleoside and deoxyribonucleoside respectively. C-1′ of sugar is bonded to the N-atom of the base in both nucleosides.

Nucleosides In RNA:

Nucleosides In DNA:

Nucleoside And Nucleotide Nucleotide: A compound containing a heterocyclic nitrogenous base (A, G, C, U, T), a pentose sugar (ribose or deoxyribose) and a phosphate group is called nucleotide. The phosphate group is linked to the C-5′ of sugar.

Nucleotide

= 1 pentose sugar + 1 heterocyclic base + 1 phosphate group

= 1 nucleoside +1 phosphate group

Di, tri-, tetra- and polynucleotides have 2, 3, 4, or several nucleotides attached. A phosphodiester bond is present between 5′-C of a sugar and 3′-C of another sugar.

• Every nucleotide contains C-1′ of pentose sugar Jinked to a nitrogenous base and C-3’to a phosphate group. C-S’of pentose sugar is linked to a —OH group which reacts with the phosphate group of the next nucleotide to form a phosphodiester bond. This bond joins two nucleotides.
• Hence, a free phosphate group at 5′ C of one nucleotide terminus and a free —OH group at 3’C of the other nucleotide terminus is found in a chain of DNA or RNA.
• In every polynucleotide chain, a 5′ -terminal bears a free phosphate radical & a 3′ – terminal bears a free —OH group. So, a polynucleotide chain is marked as 5′- 3′.

RNA [Ribonucleotide]:

DNA [Deoxyribonucleotide]:

Hydrolytic Products Of DNA And RNA:

Differences Between Nucleotide And Nucleoside:

Structures Of Nucleic Acids

The structure of nucleic acids can be discussed under the following divisions of

1. Primary structure and
2. Secondary structure.

Structure Of Nucleic Acid Primary Structure: The nucleotide sequence in a nucleic acid molecule is the primary structure of that nucleic acid.

• Nucleic acid refers to a polynucleotide chain which is formed of several pentose sugar units joining 5′-C and 3′-C by phosphodiester bonds. Hence, the sugar and phosphate units are arranged sequentially in a nucleic acid.
• Any four nitrogenous bases (A, G, C and T or A, G, C and U) remain linked consecutively with four sugar units.
• The particular sequence of four nitrogenous bases in the sugar—phosphate framework of a nucleic acid refers to the primary structure of that nucleic acid.

Secondary Structure Of DNA: Eminent scientists James Watson and Francis Crick (1953) proposed the Watson-Crick model on the physical structure of deoxyribonucleic acid (DNA) developed by X-ray diffraction analysis. From this analysis, it is found that—

1. Every DNA molecule is formed of two polynucleotide chains.
2. The two chains rotate clockwise to form a double helix.
3. The two polynucleotide chains are joined anti-parallelly as one chain runs in the 5′-3′ direction while the other in 3 – 5.
4. A polynucleotide chain has a sugar-phosphate backbone whose nitrogenous bases are projected inwards perpendicularly in a particular sequence.
5. The pyrimidine base of one chain is linked to the purine base of another chain by hydrogen bonding. The distance between two consecutive nitrogenous bases in a DNA strand is 3.4 A.
6. The diameter of the double-stranded DNA helix is 20A. A complete spiral strand of a DNA helix consists of 10 complementary base-pairs and it is 34A long.
7. In the double-helical DNA molecule, the nitrogenous base sequence in one chain gives a specific idea about the complementary base sequence in the other chain. The base A is linked by two hydrogen bonds to T and G is linked by three hydrogen bonds to C.

As per Chargaff’s rule, the sum of the purines is equal to the sum of pyrimidines, i.e., A + G = T + C is applicable to all polynucleotide strands.

Secondary Structure Of DNA Functions Of DNA

1. It is the chemical component of genes that carries biological information. It controls the transfer of hereditary traits between generations and its expression in an individual.
2. DNA transfers the characteristics from parent to daughter cell by replication.
3. The genetic information stored is expressed by RNA synthesis and polypeptide synthesis.

Different Types Of RNA

RNA is a single-stranded linear or helical polymer of ribo¬nucleotide. The pentose sugar is ribose & nitrogenous bases are adenine, guanine, cytosine and uracil. RNA are of 3 types:

1. mRNA (messenger RNA): It is synthesised from DNA. It carries genetic information from DNA to cytoplasm and takes part in protein synthesis.
2. rRNA (ribosomal RNA): It helps in the attachment of m-RNA to ribosomes, thus forming a polysome. This polysome acts as a template for the synthesis of different proteins.
3. tRNA (transfer RNA): It is a relatively small molecule. f-RNA sends a particular amino acid to the specified location to construct the correct sequence of amino acids during protein synthesis.

Biological Role Of Nucleic Acids

Two important roles of nucleic adds are—

1. Replication and
2. Protein synthesis

Replication: The process of formation of two identical daughter DNA molecules from parent DNA molecule using enzymes and proteins is called replication.

Procedure: At first, uncoiling of supercoiled DNA occurs with the help of suitable enzymes.

• The H-bonds between the DNA strands are broken, and each strand acts as a template strand.
• By following this template DNA and with the help of base-pairing, complementary deoxyribonucleotides are added to each strand.
• In this way, two daughter strands are formed.
• Each DNA molecule consists of a parent strand (original strand) and its complementary daughter strand (newly synthesised strand).
• So, this process is called semi-conservative replication. It takes place in 5,-3/ direction only

Synthesis Of Protein: Different kinds of RNA molecules (m-RNA, r-RNA and t-RNA) take part in protein synthesis. The central dogma of molecular biology describes the protein synthesis in two steps—

1. Transcription and
2. Translation by which the information in genes flows into proteins ((DNA—>RNA—^Protein). The biological information required for the synthesis of a particular protein is obtained from a DNA molecule only.

Differences Between DNA and RNA:

DNA Fingerprint

The fingerprint of every human being is unique, i.e., no two individuals bear similar fingerprints. Thus, it is in vogue to identify a person by his or her fingerprint. However, there is a possibility of modifying the skin of the fingertip by surgery.

• Hence, nowadays DNA testing is a much more surer process for the identification of a person.
• DNA fingerprint refers to the specific nucleotide sequence present in the double-stranded DNA of an individual. Every cell of a particular individual has an identical nucleotide sequence in its DNA. One cannot modify this sequence voluntarily.

Uses of DNA-fingerprint

1. To identify criminals by forensic testing.
2. Paternity test, i.e., to identify the father of an individual.
3. To identify a dead person by comparing his or her DNA with his or her parents or children.
4. To determine a racial group.

Class 12 Chemistry Unit 14 Biomolecules Very Short Questions And Answers

Question 1. Which polysaccharide is stored in animal liver?
Answer: Glycogen,

Question 2. What kind of ring is formed by fructose in sucrose?
Answer: In the form of a 5-membered fructofuranose ring.

Question 3. State the configuration of natural a -amino acid.
Answer: L-configuration.

Question 4. Name two essential amino acids.
Answer: Valine and phenylalanine.

Question 5. State nature of glycosidic bonds in starch and cellulose.
Answer: α-glycosidic bond in starch and β-glycosidic bond in cellulose.

Question 6. Which enzyme is used in the treatment of heart disease?
Answer: Streptokinase

Question 7. Which solvent allows smooth mutarotation of glucose?

1. Cresol
2. Pyridine
3. Cresol + pyridine

Answer: Cresol + pyridine.

Question 8. Which bond helps to stabilise a -helix?
Answer: Intermolecular hydrogen bond.

Question 9. State the pH at which the solubility of amino acids is the least.
Answer: At an isoelectric point, the solubility of amino acids is the least

Question 10. Which phenomenon does the coagulation of egg white refer to?
Answer: Denaturation of protein.

Question 11. State the possible formula of a tripeptide which on hydrolysis produces glycine alanine and valine.
Answer: The tripeptide may be either of the following: Gly-Ala-Val, Gly-Val-Ala, Ala-Gly-Val, Ala-Val-Gly, Val-Ala-Gly, Val-Gly-Ala.

Question 12. State the name and structural formula of bee wax.
Answer: Myrisyl palmitate [CH₃(Ch₂)₁₄COO(CH₂)₂₉CHg₃]

Question 13. How many 5-membered rings and 6-membered rings are found in steroids?
Answer: One 5 -membered ring and three 6 -membered rings.

Question 14. Which steroid is abundant in living bodies?
Answer: Cholesterol.

Question 15. Name the components of essential oil.
Answer: Monoterpene (10-C terpene), sesquiterpene (15-C terpene).

Question 16. Name the two classes of nitrogenous bases in nucleic acid.
Answer: Purine and pyrimidine.

Question 17. Name the two pentose sugars present in DNA and RNA.
Answer: β-D-2-deoxyribose in DNA and β-D-ribose in RNA.

Question 18. Define invert sugar.
Answer: The equimolar mixture of glucose and fructose was obtained due to sucrose hydrolysis.

Question 19. Give an example of an amino acid containing 2′ amine.
Answer: Proline

Question 20. Why do enzymes accelerate biochemical reactions?
Answer: The activation energy of enzyme-catalysed biochemical reactions is comparatively smaller than that of reactions taking place without enzymes.

Question 21. How many isoprene units are present in β-carotene? What kind of terpene is it?
Answer: β-carotene has 8 isoprene units. It is a tetraterpene.

Question 22. What is the structural feature characterising reducing sugars?
Answer: Reducing sugars contain free aldehydic or ketonic groups.

Question 23. Mention the names of two carbohydrates which act as biofuels.
Answer: Glycogen in animals and starch in plants act as biofuels.

Question 24. Name the enzymes present in the saliva of human beings and mention their role.
Answer: The saliva of human beings contains amylase as the enzyme. Amylase first hydrolyses starch to maltose and then to glucose.

Question 25. Why is a -helix named as 3.6₁₃ helix?
Answer: Since each turn of the α-helix has approximately 3.6 amino acids and the hydrogen bonding leads to the formation of a 13-membered ring, the α-helix is termed as 3.6₁₃ helix.

Question 26. Mention the amino acids which exhibit aromatic properties.
Answer: Phenylalanine, tryptophan, tyrosine and histidine exhibit aromatic properties.

Question 27. Indicate the total number of basic groups in the following form of lysine:

Answer: Lysine contains two basic groups. These are —COO^– and —NH₂

Question 28. Name two α-amino acids which form a dipeptide, whose methyl ester is 100 times more sweet than cane sugar.
Answer: Aspartic acid and phenylalanine, [aspartame is an artificial sweetener 100 times sweeter than cane sugar. It is the methyl ester of the dipeptide derived from aspartic acid and phenylalanine]

Question 29. Define native state concerning protein.
Answer: The native state of a protein is its properly folded and/or assembled form, which is operative and functional. They are unaltered by denaturating agents such as heat, the action of enzymes or chemicals.

Question 30. Name the enzyme that breaks large proteins into small peptides.
Answer: The enzyme pepsin plays an important role in the digestion of proteins by causing the fission of the large amino acid chains to peptides, which are short chains of 4-9 amino acids.

Question 31. What happens when L-glucose is treated with Tollens’ reagent?
Answer: L-glucose is oxidised to the corresponding gluconate ion.

Question 32. What change in free energy occurs during the conversion of ATP into ADP? Which bonds link the phosphoric acid molecules together in ATP?
Answer:

1. There is a decrease in free energy during the conversion of ATP into ADP (AG = -ve)
2. Phosphoric anhydride bonds link the phosphoric acid molecules in ATP.

Question 33. Why cannot vitamin C be stored in our bodies?
Answer: As vitamin C is water-soluble, therefore, it is readily excreted through urine and hence, cannot be stored in our body.

Question 34. What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?
Answer: The hydrolysis products are: thymine, 2-deoxy-D-ribose and phosphoric acid.

Question 35. Amino acids show amphoteric behaviour. Why?
Answer: Amino acids are amphoteric as basic (—NH₂) and acidic (—COOH) both functional groups are present,

Question 36. Which disaccharides do not exhibit mutorotation? give an example.
Answer: Disaccharides, which contain glycosidic bonds between the anomeric carbon atoms of the constituent monosaccharide units, do not exhibit mutarotation. Example: Sucrose.

Question 37. Write the products of hydrolysis of lactose.
Answer: D-glucose and D-galactose.

Question 38. What type of bonding helps stabilise the α-helix structure of proteins?
Answer: The α-helix structure of proteins is stabilised by intramolecular H-bonding between C—O of one amino acid residue and the N—H of the fourth amino acid residue within a single chain.

Question 39. Name the sugar present in milk. How many monosaccharide units are present in it? What are such oligosaccharides called?
Answer: Lactose is the sugar present in milk. It contains two monosaccharide units. Such oligosaccharides are known as disaccharides;

Question 40. In nucleoside, a base is attached at 1′ position of the sugar moiety. Nucleotide is formed by linking of phosphoric acid unit to the sugar unit of nucleoside. At which position of the sugar unit is the phosphoric acid linked in a nucleoside to give a nucleotide?
Answer: Phosphoric acid unit is linked to C₅ of sugar unit of nucleoside to give a nucleotide;

Question 41. Name the linkage connecting monosaccharide units in polysaccharides.
Answer: Glycosidic linkage;

Question 42. Under what conditions glucose is converted to gluconic and saccharic acid?
Answer: Glucose is oxidised to gluconic acid by bromine water (Br2/H2O) and to saccharic acid by the cone. HNO₃;

Question 43. Monosaccharides contain a carbonyl group and hence are classified, as aldose or ketose. The number of carbon atoms present in the monosaccharide molecule is also considered for classification. In which class of monosaccharide will you place fructose?
Answer: Fructose is a monosaccharide containing six carbon atoms including the keto group. So it is called ketohexose;

Question 44. Vitamin B complex is a combination of several other vitamins. Name the constituent vitamins.
Answer: Vitamin B is a complex vitamin and consists of vitamins B₁, B₂, B₆, B₁₂, biotin, folic acid, pantothenic acid and nicotinic acid.

Question 45. How do enzymes effectively help a substrate to be attacked by the reagent?
Answer: The active sites of enzymes hold the substrate molecule in a suitable position so that the substrate can be attacked by the reagent effectively;

Question 46. Why are naturally occurring glucose and fructose called dextrose and laevulose respectively?
Answer: The aqueous solution of naturally occurring glucose is dextrorotatory while that of fructose is laevorotatory. Hence, glucose and fructose are known as dextrose and laevulose respectively;

Question 47. Explain the role of osazone structure in identifying monosaccharides.
Answer: The Osazone product of a particular monosaccharide has a well-defined melting point which is useful for identifying monosaccharides;

Question 48. Name the enantiomer of α-D-(+) -glucose.
Answer: α-L(-)- glucose

Question 49. Which amino acid is not optically active?
Answer: Glycine

Question 50. What idea can be obtained about the structure of monosaccharides from mutarotation?
Answer: The idea of a closed ring structure of monosaccharides is obtained from mutarotation;

Question 51. Which two compounds combine to produce inverted sugar?
Answer: Glucose and fructose

Question 52. Name the disaccharide present in milk.
Answer: Lactose;

Question 53. How many molecules of phenylhydrazine react with glucose to produce osazone?
Answer: 3

Question 54. Name the closed ring structure in glucose.
Answer: Pyranose

Question 55. Name a sulphur-containing amino acid.
Answer: Cystine

Question 56. Name the hydrolytic product of cellulose.
Answer: D-glucose;

Question 57. Why human beings cannot digest cellulose?
Answer: The human intestine cannot synthesise cellulase which can hydrolyse cellulose into glucose;

Question 58. Which vitamin deficiency causes pernicious anaemia?
Answer: B₁₂

Question 59. Which enzyme decoagulates clotted blood?
Answer: Streptokinase;

Question 60. How many a-amino acid molecules are found in one insulin molecule?
Answer: 51

Question 61. State the role of insulin in the human body.
Answer: Decreases blood glucose level

Question 62. Which enzyme converts glucose into ethanol?
Answer: Zymase;

Class 12 Chemistry Unit 14 Biomolecules Short Questions And Answers

Question 1. α-amino monocarboxylic acids have two pk_a values. Explain.
Answer: One pk_a value arises from the dissociation of R—CH(NH₃)COOH and the second value arises from the dissociation of R—CH(NH₂)—COOH.

Question 2. What is the reason for the specific action of enzymes?
Answer: Each enzyme is specific for a particular reaction.

• This is because enzymes contain active sites of definite shape and size on their surfaces so that only specific substrates can fit into these active sites (just like a key fits into a lock).
• Such specific binding leads to the formation of enzyme-substrate complex which accounts for specific action of enzymes.

Question 3. Indicate the nature of changes that occur when egg protein is boiled.
Answer: Egg contains globular proteins. On boiling such proteins undergo coagulation irreversibly to form fibrous proteins.

• In other words, proteins lose their biological activity and thus get denatured.
• During denaturation, the secondary and tertiary structures of proteins are destroyed without affecting the primary structure.

Question 4. How will you prove that— The carbonyl group is present at the C-2 atom of a fructose molecule?

Answer: Hydrolysis of fructose cyanohydrin (which is formed due to the reaction between fructose and HCN) yields an acid. The acid is reduced by HI/P to form 2-methylhexanoic acid. This proves the presence of the carbonyl group at the C-2 atom of fructose.

Question 5. What is the monomer unit of protein? give one example of such a monomer which contains sulphur. Write its zwitterionic form.
Answer: The monomer unit of proteins is α-amino acid. An α-amino acid containing sulphur is cysteine.

Question 6. Identify:

1. The vitamin responsible for blood coagulation,
2. The vitamin is not stored inside the human body.
3. The vitamin whose deficiency causes scurvy
4. The vitamin whose deficiency causes beri¬beri;
5. The vitamin included under lipid;
6. The enzyme that causes hydrolysis of cellulose
7. The enzyme that can reduce chances of heart attack;
8. The enzyme used in the preparation of inverted sugar;
9. The group B vitamin is synthesised in the human body.

Answer:

1. Vitamin K;
2. Vitamin C;
3. Vitamin C;
4. Vitamin B₁
5. Vitamin A, D, E and K;
6. Cellulase;
7. Streptokinase;
8. Invertase;
9. Vitamin B₁₂

Question 7. If the base sequence of one strand of DNA double- • stranded molecule is ‘ATCGTCCA’, state the complementary base sequence.
Answer: Adenine (A) pairs with thymine (T) and guanine (G) pairs with cytosine (C) by hydrogen bonds in a DNA molecule. Therefore, the complementary base sequence will be ‘TAGCAGGT’ corresponding to ‘ATCGTCCA’.

Question 8. Indicate the nature of linkages responsible for the formation of

1. Cross-linking of polypeptide chains
2. β-sheet structure.

Answer:

1. Cross-linking of polypeptide chains is due to hydrogen bonds or disulphide bonds.
2. β-sheet structure is due to hydrogen bonds between groups belonging to different
   polypeptide chains.

Question 9. What is essentially the difference between α-form and β-form of D-glucose?
Answer: α-and β-forms of D-(+)-glucose differ only in the configuration of Cj (Le„ the configuration of anomeric carbon). In the α-form, C₁ —OH and C₂—OH are on the same side, while in the β-form, C₁ —OH and C₂—OH are on the opposite sides.

Question 10. Fructose contains a keto group but still reduces Tollens’ reagent. Explain.
Answer: Under the basic conditions of Tollens’ reagent (ammoniacal silver nitrate solution), fructose undergoes Lobry de Bruyn van Ekenstein rearrangement to form an equilibrium mixture of fructose, glucose and mannose. Since both glucose and mannose contain the free—CHO group, they reduce Tollens’ reagent.

Question 11. Glycine exists as a zwitterion but o-and paminobenzoic acids do not— Give reason.
Answer: α-and β-aminobenzoic acids are resonance hybrids. In such resonance, the lone pair on the amino nitrogen is involved in conjugation with the aromatic ring.

• This makes the —NH₂ group poorly basic and the —COOH group poorly acidic.
• Therefore the weakly acidic —COOH group cannot transfer a proton to the weakly basic —NH₂ group, obviously o-and p-aminobenzoic acids do not exist as zwitterions.

On the other hand, in glycine, no such benzene ring is present. Thus, the NH₂ group is sufficiently basic to accept a proton from the acidic —COOH group to form a zwitterion.

Question 12. A decapeptide (M-mass =796) on complete hydrolysis gives glycine, alanine and phenylalanine. Glycine contributes 47% to the total mass of the hydrolysed products. Calculate the number of glycine units present in the decapeptide.
Answer:

The molecular mass of decapeptide = 796

The molar mass of glycine (H₂NCH₂COOH) = 75

The number of H₂O molecules that will be required for the hydrolysis of the decapeptide must be 9.

Decapeptide + 9H₂O → Glycine + Alanine + Phenylalanine

Now, the total mass of all the amino acids obtained after the addition of 9 molecules of H₂O = 796 + (9 ×18) = 958

∴ The total mass of glycine in the hydrolysed products

⇒ \(=\frac{958 \times 47}{100} \approx 450\)

But the molecular mass of glycine = 75

∴ No. of glycine units in the decapeptide \(=\frac{450}{75}=6\)

Question 13. Which of the following reagents converts glucose to gluconic acid—

1. Na-Hg/H₂O
2. Br₂/H₂O
3. HNO₃
4. NaBH₄

Answer: 2. Br₂/H₂O

Question 14. Write down the structure of D-glucose. How it differ from the structure of D-fructose?
Answer:

Question 15. In which of the following peptide bonds is present—

1. CH₃CH₂CON(CH₃)₂
2. H₂NCH₂CH₂CO₂C₂H₅
3. C₆H₅CONHOC₂H₅
4. H₂NCH₂CONHCHCO₂H CH₃

Answer: 4. H₂NCH₂CONHCHCO₂H CH₃

Question 16. Which of the following bases is not present in DNA—

1. Uracil
2. Thymine
3. Guanine
4. Cytosine

Answer: 1. Uracil

Question 17. After watching a programme on TV about the presence of carcinogens (cancer-causing agents) potassium bromate and potassium iodate in bread and other bakery products, Ritu a class XII student decided to be aware others about the adverse effects of these carcinogens in foods. She consulted the school principal and requested him to instruct the canteen contractor to stop selling sandwiches, pizzas, burgers, and other bakery products to the students. The principal took immediate action and instructed the canteen contractor to replace the bakery products with some protein and vitamin-rich food like fruits, salads, sprouts etc. The decision was welcomed by the parents and students. After reading the above passage, answer the following questions.

1. What are the values (at least two) displayed by Ritu?
2. Which polysaccharide component of carbohydrates is commonly present in bread?
3. Write the two types of secondary structure of proteins.
4. Give two examples of water-soluble vitamins.

Answer:

1. Concern for students’ health
   1. Caring in nature
   2. Socially alert.
   3. Starch.
2. α-helix and β-pleated structures.
3. Vitamin B₁, B₂, B₆ and C.

Question 18. Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six-membered ring compounds) are insoluble in water. Explain.
Answer: Molecules of glucose and sucrose contain several —OH groups and hence, are capable of forming extensive H -bonds with water molecules. On the other hand, cyclohexane and benzene are hydrocarbons and hence, their molecules can not form H-bonds with H20 molecules. Thus, glucose and sucrose are soluble in water, while cyclohexane and benzene are not.

Question 19. What are the expected products of the hydrolysis of lactose?
Answer: Lactose (a disaccharide), on hydrolysis, yields a 1:1 mixture of the monosaccharides—D-(+) -glucose and D-(+) – galactose.

Question 20. How do you explain the absence of an aldehyde group in the pentaacetate of D-glucose?
Answer: In its cyclic structure, glucose does not contain any free aldehyde group because it exists in the hemiacetal form.

• When, however, it is dissolved in water, it may undergo ring opening to generate the open chain aldehydic form, thereby indicating the presence of an aldehyde group in glucose.
• When glucose (hemiacetal) is treated with acetic anhydride, the —OH group at C-1, along with four other — OH groups at C-2, C-3, C-4 and C-6 are involved in the formation of acetyl derivative.
• The resulting pentaacetate (having a cyclic structure) does not contain a free — OH group at C-1, so it can not produce an open-chain aldehydic form. This indicates that glucose pentaacetate does not contain
  an aldehyde group.

Question 21. The melting points and solubility in water of amino acids are generally higher than those of the corresponding halo acids. Explain.
Answer: Due to the presence of both acidic and basic groups in the molecule, amino acids exist as Zwitterions having dipolar structures with salt-like character.

• Because of such a Zwitterionic structure, molecules are held together by strong electrostatic forces of attraction. Thus, amino acids have high melting points like ionic compounds.
• Furthermore, due to their dipolar nature, they interact strongly with H₂O molecules and hence, amino acids are highly soluble in water.
• On the other hand, corresponding halo acids cannot exist in the Zwitterionic form and hence they behave like simple carboxylic acids having relatively lower melting points and lower solubility in water.

Question 22. Where does the water present in the egg go after boiling the egg?
Answer: When an egg is boiled, the soluble globular protein albumin present in the egg white first undergoes denaturation and then coagulation to give fibrous protein. The water present in the egg gets attached to the fibrous protein molecules through hydrogen bonds.

Question 23. When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA?
Answer: Since there is no relationship among the quantities of different bases obtained by hydrolysis of RNA, it suggests that the base-pairing principle, i.e., adenine (A) pairs with uracil (U) and cytosine (C) pairs with guanine (G) is not followed. So RNA does not exist in the form of a double strand but it has a single strand.

Question 24. What are monosaccharides?
Answer:

• Monosaccharides are simple carbohydrates which can not be hydrolysed to still simpler carbohydrates. Usually, they have the general formula (CH₂O)_n, where n = 3-7.
• These are of two types: aldoses (which contain an aldehyde group) and ketoses (which contain a keto group). Thus, glucose (C₆H₁₂O₆) is an aldose while fructose (C₆H₁₂O₆) is a ketose.

Question 25. What are reducing sugars?
Answer: The carbohydrates which can reduce Fehling’s solution to give a red precipitate of Cu₂O or reduce Tollens’ reagent to produce metallic silver are classified as reducing sugars.

• All monosaccharides (both aldoses and ketoses) and disaccharides such as maltose and lactose are reducing sugars.
• Reducing sugars contain free aldehydic or ketonic groups.

Question 26. Write two main functions of carbohydrates in plants.
Answer:

1. Structural Material For Plant Cell Walls: Cellulose (a polysaccharide) acts as the main structural material of the plant cell walls.
2. Reserve Food Material: Starch (a polysaccharide) is the chief reserve food material in the plants. It is stored in seeds and acts as the reserve food material for sapling till it is capable of preparing its food by photosynthesis.

Question 27. Classify the following into monosaccharides and disaccharides. Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.
Answer:

1. Monosaccharides: Ribose, 2-deoxyribose, galactose and fructose.
2. Disaccharides: Maltose and lactose.

Question 28. What do you understand by the term glycosidic linkage?
Answer:

• In oligosaccharides and polysaccharides, any two adjacent monosaccharide units are joined together by an ether linkage or oxide linkage formed by the loss of a molecule of water.
• Such a linkage between two monosaccharide units through an oxygen atom is called glycosidic linkage. For example, see the structure of sucrose or maltose in the text part.

Question 29. What is glycogen? How is it different from starch?
Answer:

• Glycogen is a branched polymer of α-D-glucose. Just as glucose is stored in plants as starch, it (glucose) is stored as glycogen in the liver and muscles of human beings.
• When the body needs glucose during fasting or hard work, glycogen undergoes hydrolysis by the action of enzymes to provide glucose. However, starch is not a single compound, but is a mixture of two components:
• A water-soluble component called amylose (which is a linear polymer of α-D-glucose) and
• A water-insoluble component called amylopectin (a branched polymer of α-D-glucose). Glycogen differs from amylopectin in that the former is a more highly branched polymer of α-D-glucose.

Question 30. What are the hydrolysis products of sucrose and lactose?
Answer: Sucrose (C₁₂H₂₂O₁₁, a disaccharide) undergoes hydrolysis to give a 1: 1 molar mixture of D-(+) -glucose (C₆H₁₂O₆) and D-(-) -fructose (C₆H₁₂O₆). On the other hand, lactose a disaccharide) undergoes hydrolysis to give a 1: 1 molar mixture of D-(+) -glucose and D-(+) -galactose.

Question 31. What is the basic structural difference between starch and cellulose?
Answer: Starch is not a single compound but is a mixture of two components:

• Amylose (a linear polymer of α-D-glucose and
• Amylopectin (a branched polymer of a-D-glucose ). In both these components, D-glucose units are linked through the α-glycosidic linkage between C₁ of one glucose unit with C₄ of the next glucose unit.
• Cellulose is however a single compound. It is a linear polymer of β-D-glucose. C-1 of one glucose unit is connected to C-4 of another through β-glycosidic linkage.

Question 32. What happens when D-glucose is treated with the following reagents?

1. HI
2. Bromine water
3. HNO₃

Answer: D-glucose is reduced by HI to give a mixture of n-hexane and 2-iodohexane.

Question 33. Enumerate the reactions of D-glucose which cannot be explained by its open chain structure.
Answer:

1. D-glucose does not form NaHSO₃ addition product, aldehyde ammonia adduct, 2, 4-DNP derivative and does not respond to Schiff’s reagent test.
2. When dissolved in an aqueous solution, both α-D-glucose and β-D-glucose exhibit the phenomenon of mutarotation.
3. Glucose reacts with NH₂OH to form an oxime but the pentaacetyl derivative of glucose does not react with NH₂OH.
4. D-glucose reacts with methanol in the presence of dry HCl gas to form two isomeric methyl-D-glucosides

Question 34. How do you explain the amphoteric behaviour of amino acids?
Answer: Amino acids exist as zwitterion. In the acid medium —COO^– group of the zwitterion accepts a proton to form a cationic species while in the basic medium, the —N⁺H₃ group loses a proton to form an anionic species.

Thus —N⁺H₃ group acts as the acidic group, while the —COO^– group acts as the basic group thereby justifying that amino acids exhibit amphoteric behaviour.

Question 35. What is the effect of denaturation on the structure of; proteins?
Answer:

• As a result of denaturation, secondary and tertiary structures of proteins are destroyed but primary structure remains unchanged.
• During denaturation, H-bonds are broken, globules unfold and helices get uncoiled to form thread-like molecules. In other words, globular proteins (which are soluble in water) undergo coagulation to give fibrous proteins (which are insoluble in water).

Question 36. Why are vitamin A and vitamin C essential to us? Give their important sources.
Answer:

• Vitamin A is essential to our body as its deficiency causes xerophthalmia (hardening of the cornea), night blindness and xerosis. Sources: Fish liver oil, carrot, butter and milk.
• Vitamin C is essential to us as its deficiency causes scurvy (bleeding gums) and pyorrhea. Sources: Citrous fruits [for example., oranges, lemon, amla, tomatoes) and green leafy vegetables.

Question 37. The two strands in DNA are not identical but are complementary. Explain.
Answer: DNA has a double helix structure in which the two strands are held together by H-bonds between the purine base of one strand and the pyrimidine base of the other and vice-versa.

• Thymine (T) pairs with adenine (A) through two H-bonds and cytosine (C) pairs with guanine (G) through three H-bonds.
• So, opposite of each adenine (A) on one strand there must be a thymine (T) on the other strand and opposite of guanine (G), there must be cytosine (C).
• In other words, the two strands of DNA are complementary and not identical.

Question 38. Write the important structural and functional differences between DNA and RNA.
Answer:

Structural Differences:

Functional Differences:

Question 39. Write the structure of serine at pH = 1 and pH = 11
Answer: Serine exists as zwitterion at neutral pH. In pH = 1, —COO^– accepts a proton to give cation and at pH = 11, —NH₃ donating a proton to the base exists as an anion.

Question 40. Draw the Fischer projection formula of the enantiomer of α-D-(+)-glucopyranose and write its name.
Answer: It is α, as oxygen at C-1 and C-5 are on the same side. It is L as C-5 oxygen is oriented towards the right. It is (-), Le., laevorotatory (enantiomer of dextrorotatory glucopyranose).

Question 41. How many chiral carbons are found in an aldohexose? How many D and L-isomers are possible in each? State the relationship between D and L-isomers of an aldohexose. Give example.
Answer: The structural formula of aldohexose is:

Number of chiral carbons in aldohexose (n) = 4; the number of 3-D isomers =2n = 24 = 16.

Out of these 16 isomers, 8 are of D- and 8 are of L-configuration. The D-and L-isomers of the same aldohexose are enantiomers, for example., D-(+)-glucose and L-(-)-glucose.

Question 42. Name the lipids having perhydro-1, 2-cydopentano- phenanthrene system. Name an important steroid formed in the gall bladder and state its function.
Answer: Lipids having a perhydro-1, 2-cyclopentane-phenanthrene system are called steroids. The important steroid formed in the gall bladder is cholesterol. It is an essential component of the cell membrane. It acts as a precursor during vitamin D and steroid synthesis.

Question 43. What happens when glucose is treated with dilute NaOH?
Answer: When glucose is treated with dilute NaOH, it forms a mixture of D-glucose, D-fructose and D-mannose due to reversible isomerisation. This reaction is generally known as Lobry de Bruyn-van Ekenstein.

Question 44. Which aldohexose other than D-glucose can produce the same dicarboxylic acid obtained due to oxidation of D-glucose by HNO₃?
Answer: D-glucose and L-gulose both yield the same dicarboxylic acid on oxidation by HNO₃.

Question 45. Different values of specific rotation of optically active amino acids are obtained at different pH. Explain.
Answer:

The values of specific rotation of dipolar zwitterion, cation (conjugate acid) and anion (conjugate base) are different as the groups linked to the chiral carbon in each are different. A specific form exists at a particular pH. So, the specific rotation varies according to the changes in the pH of amino acids.

Question 46. A tripeptide on complete hydrolysis produces glycine, alanine and phenylalanine In 1: 1: 1 molar ratio. What are the probable structures of the tripeptide?
Answer: These three amino acids can be obtained from six tripeptides and these are Gly-Ala-Phe, Gly-Phe-Ala, Ala-GlyPhe, Ala-Phe-Gly, Phe-Ala-Gly and Phe-Gly-Ala.

Question 47. Why disaccharides obtained from monosaccharides like glucose (C₆H₁₂O₆) have the formula C₁₂H₂₂On instead of C₁₂H₂₄O₁₂?
Answer: Two glucose (C₆H₁₂O₆) units are joined along with the elimination of a water molecule (H₂O) to form a disaccharide. Hence, the resulting disaccharide has the formula C₁₂H₂₂O₁₁ rather than C₁₂H₂₄O₁₂.

Question 48. Write down two differences between amylopectin and cellulose based on their structure.
Answer: Two differences between amylopectin and cellulose are—

Question 49. A DNA molecule has a higher melting point having more GC pairs than another DNA molecule which has a greater number of AT pairs. What conclusion can be drawn from this fact?
Answer: The DNA molecule with a higher number of GC pairs will have a higher melting point than the DNA molecule with a higher number of AT pairs (i.e., a lower number of GC pairs).

• This is due to the presence of three H-bonds in the GC base pair as compared to two H-bonds in the AT base pair.
• Hence, stronger hydrogen bonding results in a higher melting point of the DNA molecule which has more GC pairs.

Question 50. Why do glucose and fructose produce the same osazone?
Answer: During the reaction of glucose and fructose with excess phenylhydrazine to form osazone, only the C-1 and C-2 atoms of glucose and fructose participate in the reaction. The rest of the molecule remains intact. Hence, glucose and fructose produce the same osazone.

Question 51. 2 DNA samples, X and Y have m.p 340 and 350K respectively. Which sample has a higher CG base pair & why?
Answer: CG base pairing occurs through 3 H-bonds while AT base pairing occurs through 2 H-bonds. Thus, CG base pairing is stronger than AT base pairing. Hence, a sample containing more CG base pairs is more compact and has a higher melting point. So, the sample Y has higher CG content.

Question 52. The structure of aspartame (a peptide and an artificial sweetener) is given below:

1. Name the functional groups present in aspartame.
2. Give the zwitterionic structure.
3. Name the amino acids obtained from hydrolysis of aspartame

Answer: The functional groups present in aspartame are amine (—NH₂), carboxylic acid (—COOH), amide

The amino acids obtained on hydrolysis of aspartame are aspartic acid and phenylalanine.

Question 53. Name the following:

1. Acidic group in DNA.
2. A vitamin which is neither fat-soluble nor water-soluble.
3. Disease caused due to deficiency of tyrosinase.
4. An or- amino acid with no chiral carbon atom.
5. A vitamin which prevents hair loss.
6. The nucleic acid base with two possible binding sites.

Answer:

1. The phosphate group.
2. Vitamin H.
3. Albinism
4. Glycine.
5. Vitamin H (Biotin).
6. Thymine.

Question 54. Explain why the ka and kb values of α-amino acids are very low.
Answer: The ka values of α-amino acids are very low because, in α-amino acids, the acidic group is —NH₃ instead of —COOH group in carboxylic acids. Similarly, the kb values of α-amino acids are very low because the basic group is e —COO instead of —NH₂ group in aliphatic amines.

Question 55. Distinguish between anomer and epimer with suitable examples.?
Answer: Carbohydrate molecules which differ in configuration at the glycosidic or anomeric carbon are called anomers while those that differ in configuration at any carbon other than anomeric carbon are called epimers.

• For example, α-D-glucose and β-D-glucose are anomers of each other as they differ in configuration at C-1 (glycosidic or anomeric carbon).
• Glucose and galactose are epimers of each other since they differ in configuration at C-4 (other than the glycosidic carbon C-1).

Question 56. An optically active amino acid (A) having molecular formula C₅H₁₁NO₂ can exist in three forms depending upon the pH of the medium. Write the structure of (A) in an aqueous medium. What are such ions called? In which medium do the cationic and the anionic forms of the amino acid (A) exist and towards which electrode do they migrate in an electric field?
Answer: The optically active amino acid (A) is valine Me₂CH—CH(NH₂)—COOH. Depending upon the pH of the medium, valine can exist in the following three forms:

In aqueous medium, valine exists as a zwitterion. In an acidic medium, it will exist; in an electric field, it will migrate towards the cathode. In a basic medium, it will exist; in an electric field, it will migrate towards the anode.

Question 57. A tetrapeptide on partial hydrolysis produces three dipeptides such as Ser-Thr, Thr-Hyp and Pro-Ser. Identify the tetrapeptide. Write its structure.
Answer: Tetrapeptlde can be obtained by adding 3 dipeptides:

Therefore, the tetrapeptide is (Pro—Ser—Thr—Hyp) prolylserylthreonylhydroxyproline. Its structure is:

Question 58. The letters ‘D’ or ‘L’ before the name of a stereoisomer of a compound indicate the correlation of the configuration of that particular stereoisomer. This refers to their relation with one of the isomers of glyceraldehyde. Predict whether the following compound has a ‘D’ or ‘L’ configuration.

Answer: The highest numbered asymmetric centre [i.e., C₅ in this case) of the given compound contains a hydroxyl group on the left side, therefore, the compound has Lconfiguration;

Question 59. Aldopentoses named ribose and 2-deoxyribose are found in nucleic acids. What is their relative configuration?
Answer: In the Fischer projection formulae of ribose and 2-deoxyribose, the OH group is on the right side of the highest numbered asymmetric centre [i.e., C-4) and hence both the compounds have D-configuration;

Question 60. Which sugar is called inverted sugar? Why is it called so?
Answer:

• An equimolar mixture of D-(+) -glucose and D-(-) -fructose (which are obtained by hydrolysis of sucrose) is called inverted sugar. Sucrose is dextrorotatory with a specific rotation of +66.5°.
• On hydrolysis it produces a 1: 1 mixture of glucose (+52.5°) and fructose (-92°) and this mixture has a specific rotation of -19.75° i.e., the mixture is laevorotatory.
• Because of such change in the sign of optical rotation, the hydrolysis of sucrose is called inversion of sucrose and the mixture of glucose and fructose obtained by hydrolysis is called inverted sugar;

Question 61. Amino acids can be classified as α-, β-, γ-, δ- and so on depending upon the relative position of the amino group concerning the carboxyl group. Which type of amino acids form polypeptide chains in proteins?
Answer: α-Amino acids form polypeptide chains in proteins;

Question 62. α-Helix is a secondary structure of proteins formed by twisting the polypeptide chain into right-handed screw-like structures. Which type of interactions are responsible for making the α-helix structure stable?
Answer: In α-helix structure, a polypeptide chain is stabilised by the formation of H-bonds between the group of one amino acid residue and the group of the fourth amino acid residue in the chain;

Question 63. Some enzymes are named after the reaction, where they are used. What name is given to the class of enzymes which catalyse the oxidation of one substrate with the simultaneous reduction of another substrate?
Answer: Enzymes which catalyse the oxidation of one substrate with simultaneous reduction of another substrate are called oxidoreductase;

Question 64. During the curdling of milk, what happens to the sugar present in it?
Answer: During the curdling of milk, the milk sugar (i.e lactose) is converted to lactic acid by the bacteria present in milk;

Question 65. How do you explain the presence of five—OH groups in glucose molecules?

Answer: Glucose reacts with an excess of acetic anhydride to form a pentadactyl derivative thereby showing the presence of five hydroxyl groups in the molecule;

Question 66. Why does compound (A) given below not form an oxime?
Answer: Glucose pentaacetate (structure A) does not have a free hydroxyl ( —OH) group at the anomeric carbon C-1 and hence, it cannot be converted to the open-chain form having a free aldehyde ( — CHO) group. Thus, it does not form an oxime;

Question 67. Why must vitamin C be supplied regularly in the diet?
Answer: Vitamin C is a water-soluble substance so excess of it is readily excreted in urine and cannot be stored in our body. Consequently, it should be supplied regularly in the diet;

Question 68. Sucrose is dextrorotatory but the mixture obtained after hydrolysis is laevorotatory. Explain.
Answer: On hydrolysis, sucrose (dextrorotatory) gives a 1:1 mixture of glucose (dextrorotatory, +52.5° ) and fructose (laevorotatory, -92°). Since the laevorotation of fructose is greater than the dextrorotation of glucose, the mixture is laevorotatory;

Question 69. Amino acids behave like salts rather than simple amines or carboxylic acids. Explain.
Answer: In an aqueous solution, the carboxyl ( — COOH) group loses a proton and the amino ( — NH₂) group accepts a proton to form a zwitterion. Thus amino acids behave like salts;

Question 70. Structures of glycine and alanine are given below. Show the peptide linkage in glycyl alanine.

Answer: In the formation of glycyialaninc, the carboxyl group of glycine interacts with an amino group of alanine to form a peptide bond.

Question 71. A protein found in a biological system with a unique three-dimensional structure and biological activity is called a native protein. When a protein in its native form, is subjected to a physical change like a temperature change or a chemical change like, a change in pH, denaturation of protein takes place. Explain the cause.
Answer:

• Due to a physical or chemical change, hydrogen bonds in proteins are disturbed, globules unfold and helices get uncoiled to form a thread-like structure.
• Consequently, secondary and tertiary structures are destroyed and the protein loses its biological activity. This is called the denaturation of proteins;

Question 72. The activation energy for the acid-catalysed hydrolysis of sucrose is 6.22kJ.mol^-1, while the activation energy is only 2.15kJ-mol^-1 when hydrolysis is catalysed by the enzyme sucrase. Explain.
Answer:

• Enzymes are biocatalysts. They reduce the magnitude of activation energy by providing an alternative path. In the hydrolysis of sucrose the enzyme sucrase reduces the activation energy from 6.22 kj- mol-1 to 2.15 kj mol^-1.
• Thus enzyme catalysed reactions proceed at a much faster rate than ordinary chemical reactions using conventional catalysts;

Question 73. How do you explain the presence of an aldehydic group in a glucose molecule?
Answer:

• Glucose reacts with hydroxylamine (NH₂OH) to form a monoxime and adds one molecule of hydrogen cyanide (HCN) to give a cyanohydrin. Thus, glucose contains a carbonyl group, which may be aldehydic or ketonic.
• However, glucose (C₆H₁₂O₆) undergoes oxidation by bromine-water (a mild oxidising agent) to give gluconic acid (C₆H₁₂O₇) with the same number of carbon atoms.
• This indicates that the carbonyl group present in glucose is an aldehydic group;

Question 74. Which moieties of nucleosides are involved in the formation of phosphodiester linkages present in dinucleotides? What does the word diester in the name of linkage indicate? Which acid is involved in the formation of this linkage?
Answer:

• In the formation of dinucleotides, the 3′ — OH group of the pentose sugar of one nucleotide unit and the 5′ — OH group of the pentose sugar of the other nucleotide unit are involved in generating the phosphodiester linkage.
• Phosphoric acid is involved in the formation of phosphodiester linkage.
• The word ‘diester’ in this linkage indicates that two —OH groups of phosphoric acid are involved in the formation of two ester linkages,

Question 75. What are glycosidic linkages? In which type of biomolecules are they present?
Answer:

• In oligosaccharides and polysaccharides, any two adjacent monosaccharide units are joined together by an ether linkage or oxide linkage formed by the loss of a molecule of water.
• Such a linkage between two monosaccharide units through an oxygen atom is called glycosidic linkage.

Question 76. Which monosaccharide units are present in starch, cellulose and glucose and which linkages link these units?
Answer:

• All these polysaccharides (i.e., starch, cellulose and glycogen) are composed of glucose units. In starch and glycogen, α-glycosidic linkages are present.
• On the other hand in cellulose, β-glycosidic linkages are present between glucose units;

Question 77. Describe the term D- and L- configuration used for amino acids with examples.
Answer: Consider the following configurations of the a -amino acid,

• R—CH(NH₂)—COOH, in which the main carbon chain is shown vertically keeping the —COOH group at the top. If the — NH₂ group is on the right side of the α-carbon (as in str. 1), it is referred to as D-amino acid.
• On the other hand, if the — NH₂ group is on the left side of the α-carbon (as in str. 2), it is referred to as L-amino acid;

Question 78. Coagulation of egg white on boiling is an example of denaturation of protein. Explain it in terms of structural changes.
Answer:

• When egg white is boiled, the soluble globular protein albumin present in it is denatured resulting in the formation of insoluble fibrous protein.
• During this process, secondary and tertiary structures of the protein (albumin) are destroyed but the primary structure (i.e„ the sequence of α-amino acids) remains unchanged;

Question 79. Carbohydrates are essential for life in both plants and animals. Name the carbohydrates used as storage molecules in plants and animals, and also name the carbohydrate present in wood or fibre of cotton cloth.
Answer:

Starch’ is the carbohydrate which is used as a storage molecule in plants, while ‘glycogen’ is the carbohydrate which is used as storage material in animals. Cellulose is present in wood or the fibre of cotton cloth;

Class 12 Chemistry Unit 14 Biomolecules Multiple Choice Questions And Answers

Question 1. In an aqueous solution, glucose remains as—

1. Only in the open-chain form
2. Only in the pyranose form
3. Only in furanose form
4. In all three forms of equilibrium

Answer: 4. In all three forms of equilibrium

The six-membered cyclic form is generally referred to as the ‘pyranose’ form, and the five-membered cyclic form is called the ‘furanose’ form.

Closure of the ring creates a chiral centre at C- 1, resulting in two diastereomers (sometimes called “anomers”)-the alpha (α) and beta (β) forms.

Question 2. Which one is not a constituent of nucleic acid—

1. Uracil
2. Guanidine
3. Phosphoric acid
4. Ribose sugar

Answer: 2. Guanidine

Nucleic acid consists of bases, sugars and phosphate groups. Thus, guanidine is not a constituent of nucleic acid.

Question 3. The correct structure of the dipeptide gly-ala is—

Answer: 3

Question 4. Ribose and 2-deoxyribose can be differentiated by—

1. Fehling’s reagent
2. Tollens’ reagent
3. Barfoed’s reagent
4. Osazone formation

Answer: 4. Osazone formation

2-deoxyribose has no ketonic or oxidisable hydroxy group and hence, it does not respond in osazone reaction.

Question 5. The number of amino acids and number of peptide bonds in a linear tetrapeptide (made of different amino acids) are respectively—

1. 4 and 4
2. 5 and 5
3. 5 and 4
4. 4 and 3

Answer: 4. 4 and 3

For tetrapeptide, there are four amino acids linked together by (4- 1) = 3 amide or peptide linkages.

Question 6. Among the following statements about the molecules X and Y, the one(s) which is (are) correct is (are)—

1. X and Y are diastereoisomers
2. X and Y are enantiomers
3. X and Y are aldohexoses
4. X is a D-sugar and Y is a L-sugar

Answer: 2,3 and 4

• X and Y are mirror images of each other. Thus, these are enantiomeric pairs having four stereogenic centres.
• In both, X and Y, the aldehyde functional group (—CHO) is present having a total no. of 6 carbon in each of them.
• Thus, both of them are aldohexose. X is D-glucose thus its mirror image, Y is L-glucose.

Question 7. Within the list shown below, the correct pair of structures of alanine in the pH range 2-4 and 9-11 is—

1. H₃N-CH(CH₃)COOH
2. H₂NCH(CH₃)COO
3. H₃N—CH(CH₃)COO
4. H₂NCH(CH₃)COOH

1. 1, 2
2. 1, 3
3. 2, 3
4. 3, 4

Answer: 1. H₃N-CH(CH₃)COOH

Question 8. ADP and ATP differ in the number of—

1. Phosphate units
2. Ribose units
3. Adenine base
4. Nitrogen atom

Answer: 1. Phosphate units

ADP is adenosine diphosphate, thus it contains 2 phosphates. On the other hand, ATP is adenosine triphosphate, thus, ADP and ATP differ in the no. of phosphate units.

Question 9. In photosynthesis, the synthesis of each glucose molecule is related to—

1. 8 molecules of ATP
2. 6 molecules of ATP
3. 18 molecules of ATP
4. 10 molecules of ATP

Answer: 3. 18 molecules of ATP

Question 10. Thiol group is present in—

1. Cytosine
2. Cystine
3. Cysteine
4. Methionine

Answer: 3. Cysteine

Question 11. Which of the following compounds will behave as reducing sugar in an aqueous KOH solution—

Answer: 3. The —OCOCH₃ group, attached to the anomeric carbon in the compound, on hydrolysis in basic medium (aq. KOH), gets converted into the —OH group. The —OH group, present in the compound shows its reducing property.

Question 12. The predominant form of histamine present in human blood is (pK_a, Histidine = 6.0 )—

Answer: 4

Question 13. Glucose on prolonged heating with HI gives—

1. N-hexane
2. 1 -hexene
3. Hexanoic acid
4. 6-iodohexanal

Answer: 1. N-hexane

Question 14. The difference between amylose and amylopectin is—

1. Amylopectin has 1→ 4 α-linkage and 1 → 6 α-linkage
2. Amylose has 1→ 4 α-linkage and 1→ 6 β-linkage
3. Amylopectin has 1 → 4 α-linkage and 1 → 6 β-linkage
4. Amylose is made up of glucose and galactose

Answer: 1. Amylopectin has 1→ 4 α-linkage and 1 → 6 α-linkage

Amylose and amylopectin both are polymers of α-D-(+) glucose. In amylose, the glycosidic linkage is formed between C-l of a glucose unit and C-4 of another glucose unit. Again, amylopectin is a branched polymer where glycosidic linkages are present between C-1-C-4 and C-1-C-6.

Question 15. Find the hydrolysis product of maltose—

1. α-D-glucose + α-D-glucose
2. α-D-glucose + α-D-fructose
3. α-D-glucose + α-D-galactose
4. α-D-glucose + α-D-glucose

Answer: 1. α-D-glucose + α-D-glucose

Question 16. Lysine is least soluble in water in the pH range—

1. 3 to 4
2. 5 to 6
3. 6 to 7
4. 8 to 9

Answer: 4. 8 to 9

Any amino acid has its lowest solubility at its isoelectric point.

Question 17. Thymine is—

1. 5-methyl uracil
2. 4-methyIuracil
3. 3-methyluradl
4. 1-methyl uracil

Answer: 1. 5-methyl uracil

Thymine is also known as 5-methyluracil.

Question 18.

1. Heptanoic acid
2. 2-isohexane
3. Heptane.
4. Heptanol

Answer: 1. Heptanoic acid

Question 19. Arrange X, Y and Z in order of increasing acidic strengths—

1. X > Z > Y
2. Z < X > Y
3. X > Y > Z
4. Z > X > Y

Answer: 1. X > Z > Y

• Carboxylic acid is a stronger acid than NH3, therefore, X is the strongest acid.
• Since —COOH has a -I effect which decreases with distance, therefore, -I effect is more pronounced on Z than on Y.
• As a result, Z is more acidic than Y, Thus, the overall order of decreasing acidic strength is X > Z > Y.

Question 20. Glycogen is a branched chain polymer of α-D-glucose units in which the chain is formed by C1—C4 glycosidic linkage whereas branching occurs by the formation of Cl—C6 glycosidic linkage. The structure of glycogen is similar to.

1. Amylose
2. Amylopectin
3. Cellulose
4. Glucose

Answer: 2. Amylopectin

Question 21. Which of the following polymers is stored in the Uver of animals—

1. Amylose
2. Cellulose
3. Amylopectin
4. Glycogen

Answer: 4. Glycogen

Question 22. Which of the following pairs represents anomers-

Answer: 3

Question 23. Proteins are found to have two different types of secondary structures viz. α-helix and β-pleated sheet structure, α-helix structure of protein is stabilised by—

1. Peptide bonds
2. Van der Waals forces
3. Hydrogen bonds
4. Dipole-dipole interactions

Answer: 3. Hydrogen bonds

Question 24. In disaccharides, if the reducing groups of monosaccharides, i.e., aldehydic or ketonic groups are bonded, these are non-reducing sugars. Which of the following disaccharide is a non-reducing sugar—
Answer:

Answer: 2

Question 25. Which of the following acids is a vitamin—

1. Aspartic acid
2. Ascorbic acid
3. Adipic acid
4. Saccharic acid

Answer: 2. Ascorbic acid

Question 26. Dinucleotide is obtained by joining two nucleotides together by phosphodiester linkage. Between which carbon atoms of pentose sugars of nucleotides are these linkages present—

1. 5′ and 3′
2. 1′ and 5′
3. 5′ and 5′
4. 3′ and 3′

Answer: 1. 5′ and 3′

Question 27. Nucleic acids are the polymers of

1. Nucleosides
2. Nucleotides
3. Bases
4. Sugars

Answer: 2. Nucleotides

Question 28. Which of the following statements is not true about glucose—

1. It is an aldohexose
2. On heating with HI, it forms n-hexane
3. It is present in furanose form
4. It does not give a 2,4-DNP test

Answer: 3. It is present in furanose form

Question 29. Each polypeptide in a protein has amino acids linked with each other in a specific sequence. This sequence of amino acids is said to be

1. Primary structure of proteins
2. Secondary structure of proteins
3. Tertiary structure of proteins
4. Quaternary structure of proteins

Answer: 1. Primary structure of proteins

Question 30. DNA and RNA contain four bases each. Which of the following bases is not present in RNA—

1. Adenine
2. Uracil
3. Thymine
4. Cytosine

Answer: 3. Thymine

Question 31. Which of the following B group vitamins can be stored in our body—

1. Vitamin B₁
2. Vitamin B₂
3. Vitamin B₆
4. Vitamin B₁₂

Answer: 4. Vitamin B₁₂

Question 32. Which of the following bases is not present in DNA—

1. Adenine
2. Thymine
3. Cytosine
4. Uracil

Answer: 4. Uracil

Question 33. Three cyclic structures of monosaccharides are given below. Which of these are anomers—

Answer: 1

Question 34. Which of the following reactions of glucose can be explained only by its cyclic structure—

1. Glucose forms pentaacetate
2. Glucose reacts with hydroxylamine to form an oxime
3. Pentaacetate of glucose does not react with hydroxylamine
4. Glucose is oxidised by nitric acid to gluconic acid

Answer: 3. Pentaacetate of glucose does not react with hydroxylamine

Question 35. Optical rotations of some compounds along with their structures are given below. Which of them have D- configuration—

1. 1,2,3
2. 2,3
3. 1,2
4. 3

Answer: 1. 1,2,3

Question 36. The structure of a disaccharide formed by glucose and fructose is given below. Identify anomeric carbon atoms in monosaccharide units—

1. ‘a’ carbon of glucose and ‘a’ carbon of fructose
2. ‘a’ carbon of glucose and V carbon of fructose
3. ‘a’ carbon of glucose and ‘b’ carbon of fructose
4. ‘f’ carbon of glucose and ‘/ carbon of fructose

Answer: 3. ‘a’ carbon of glucose and ‘b’ carbon of fructose

Question 37. Three structures are given below in which two glucose units are linked. Which of these linkages between glucose units are between C-1 and C-4 and which linkages are between C-1 and C-6—

1. (A) is between C-1 and C-4, (B) and (C) is between C-1 and C-6
2. (A) and (B) are between C-1 and C-4, (C) is between C-1 and C-6
3. (A) and (C) are between C-1 and C-4, (B) is between C-1 and C-6
4. (A) and (C) are between C-1 and C-6, (B) is between C-1 and C-4

Answer: 3. (A) and (C) are between C-1 and C-4, (B) is between C-1 and C-6

Question 38. Carbohydrates are classified based on their behaviour on hydrolysis and also as reducing or nonreducing sugar. Sucrose is a.

1. Monosaccharide
2. Disaccharide
3. Reducing sugar
4. Non-reducing sugar

Answer: 2 and 4

Question 39. Proteins can be classified into two types based on their molecular shape, i.e., fibrous proteins and globular proteins. Examples of globular proteins are—

1. Insulin
2. Keratin
3. Albumin
4. Myosin

Answer: 1 and 3

Question 40. Which of the following carbohydrates are branched polymers of glucose—

1. Amylose
2. Amylopectin
3. Cellulose
4. Glycogen

Answer: 2 and 4

Question 41. Amino acids are classified as acidic, basic or neutral depending upon the relative number of amino and carboxyl groups in their molecule. Which of the following are acidic—

Answer: 2 and 4

Question 42. Lysine, H₂N—(CH₂)₂—CH—COOH is.

1. α-amino acid
2. Basic amino acid
3. Amino acid synthesised in the body
4. β-amino acid

Answer: 1 and 2

Question 43. Which of the following monosaccharides are present as a five-membered cyclic structure (furanose structure)—

1. Ribose
2. Glucose
3. Fructose
4. Galactose

Answer: 1 and 3

Question 44. In fibrous proteins, polypeptide chains are held together by

1. Van der Waals forces
2. Disulphide linkage
3. Electrostatic forces of attraction
4. Hydrogen bonds

Answer: 2 and 4

Question 45. Which of the following are purine bases—

1. Guanin
2. Adenine
3. Thymine
4. Uracil

Answer: 1 and 2

Question 46. Which of the following terms are correct about enzyme—

1. Proteins
2. Dinucleotides
3. Nucleic acids
4. Biocatalysts

Answer: 1 and 4

Question 47. Alteration of which is responsible for DNA mutation—

1. Ribose unit
2. Nitrogenous base
3. Phosphate
4. None of the above

Answer: 4. None of the above

Question 48. Which one of the following is a globular protein—

1. Collagen
2. Myoglobin
3. Myosin
4. Fibroin

Answer: 2. Myoglobin

Question 49. The end product of protein metabolism is—

1. Peptide
2. Peptone
3. Proton
4. α-amino acid

Answer: 4. a -amino acid

Question 50. Complete hydrolysis of cellulose yields—

1. D-fructose
2. D-ribose
3. D-glucose
4. L-glucose

Answer: 3. D-glucose

Question 51. Which one of the following is non-reducing sugar—

1. Glucose
2. Fructose
3. Lactose
4. Sucrose

Answer: 4. Sucrose

Question 52. Which compound does not exhibit mutarotation—

1. Sucrose
2. D-glucose
3. L-galactose
4. None

Answer: 1. Sucrose

Question 53. The participating group in the disulphide bond of proteins is—

1. Thioether
2. Thioester
3. Thiol
4. Thiolactose

Answer: 3. Thiol

Question 54. The sequence in a nucleotide of nucleic acid is—

1. Phosphate-base
2. Sugar-base-phosphate
3. Base-sugar-phosphate
4. Base-phosphate-sugar

Answer: 2. Sugar-base-phosphate

Question 55. Which chemical compounds act as an emulsifier—

1. Phosphoric acid
2. Fatty acid
3. Bile acids
4. Mineral acids

Answer: 3. Bile acids

Question 56. The sequence in which amino acids are arranged in a protein molecule refers to its—

1. Primary structure
2. Secondary structure
3. Tertiary structure
4. Quaternary structure

Answer: 1. Primary structure

Question 57. Stability is imparted to the protein helix by—

1. Dipeptide bond
2. Hydrogen bond
3. Ether bond
4. Peptide bond

Answer: 2. Hydrogen bond

Question 58. The monomeric unit of starch is—

1. Glucose
2. Fructose
3. Glucose and fructose
4. Mannose

Answer: 1. Glucose

Question 59. Which of the following responds to Molisch’s reagent—

1. All carbohydrates
2. Sucrose
3. Fructose
4. Glucose

Answer: 1. All carbohydrates

Question 60. Carboxylic acid and amino groups of an amino acid are ionised at pK_a1 = 2.34 and pK_a2 = 9.60. The pH at which the amino acid will attain its isoelectric point is—

1. 5.97
2. 2.34
3. 9.60
4. 6.97

Answer: 1. 5.97

Question 61. The percentage of α-D -glucose and β-D-glucose in D-glucose is—

1. 50%
2. 4% and 36%
3. 36% and 64%
4. 33% each along with open open-chain structure

Answer: 3. 36% and 64%

Question 62. Glucaric acid [HOOC(CHOH)₄COOH] on the reaction with HIO₄ produces—

1. 4HCOOH+2CHO
2. 3HCOOH + HCHO + OHC — COOH
3. 2HC00H + 2OHC —COOH
4. 6HCOOH

Answer: 2. 3HCOOH + HCHO + OHC — COOH

Question 63. Which amino acid does not contain any chiral carbon—

1. Histidine
2. Glycine
3. α-alanine
4. Threonine

Answer: 2. Glycine

Question 64. Which responds to Benedict’s reagent but not to ninhydrin—

1. Protein
2. Monosaccharide
3. Lipid
4. Amino acid

Answer: 2. Monosaccharide

Question 65. The hydrolytic reaction of sucrose is called—

1. Inhibition
2. Inversion
3. Saponification
4. Hydration

Answer: 2. Inversion

Question 66. Which structural characteristic distinguishes proline from other amino acids—

1. Optical inactivity
2. Presence of aromatic group
3. The presence of two hydroxylic group
4. It is a secondary amine group

Answer: 4. It is a secondary amine group

Question 67. A compound (C₆H₁₂O₆) in reaction with phenylhydrazine gives a yellow ppt. and with Na produces a mixture of sorbitol and mannitol. The compound is—

1. Fructose
2. Glucose
3. Mannose
4. Sucrose

Answer: 1. Fructose

Question 68. Compounds X and Y obtained in the following reaction are— 

1. C₆H₅NH₂ and NH₂OH
2. C₆H₅NH₂ and NH₃
3. NH₂OH and H₂O
4. C₆H₅NHNHOH and NH₃

Answer: 2. C₆H₅NH₂ and NH₃

Question 69. Which of the following hexoses form the same osazones on reaction with phenylhydrazine—

1. D-glucose, D-galactose and D-talose
2. D-fructose, D-mannose and D-galactose
3. D-glucose, D-mannose and D-galactose
4. D-glucose, D-fructose and D-mannose

Answer: 4. D-glucose, D-fructose and D-mannose

Question 70. The least solubility of amino acids in water is at—

1. pH = 7
2. pH > 7
3. pH <7
4. pi

Answer: 4. pi

Question 71. P and Q are—

1. Invertase and zymase
2. Amylase and maltase
3. Diastase and lipase
4. Pepsin and trypsin

Answer: 2. Amylase and maltase

Question 72. The positions at which base and phosphate groups are linked in DNA and RNA are—

1. C’ and C’
2. C’ and C’
3. C’ and C.

Answer: 3. C’ and C.

Question 73. Which one of the given is the C-2 epimer of D-glucose—

1. D -galactose
2. L -glucose
3. D-mannose
4. D-fructose

Answer: 3. D-mannose

Question 74. Which one of the following is a zwitterion—

1. Urea
2. Glycine hydrochloride
3. Ammonium acetate
4. L-alanine

Answer: 4. L-alanine

Question 75. The hydrolytic product of sucrose is—

1. Galactose
2. Glucose
3. Fructose
4. Ribose

Answer: 2 and 3

Question 76. Which compound shows mutarotation—

1. Glucose
2. Fructose
3. Sucrose
4. Starch

Answer: 1 and 2

Question 77. Which amino acid is chiral—

1. Alanine
2. Glycine
3. Phenylalanine
4. Glutamine

Answer: 1 and 3

Question 78. Which compound possesses a transitional element—

1. Vitamin B₁₂
2. Chlorophyll
3. Haemoglobin
4. DNA

Answer: 1,3 and 4

Question 79. Globular protein is absent in—

1. Blood
2. Keratin
3. Egg
4. Muscles

Answer: 2 and 4

Question 80. Which of the given statements is true—

1. An amino acid contains an amino group and a carboxylic group
2. Amino acids are the structural components of peptide and protein
3. Amino acids exist as zwitterions
4. Amino acids are negatively charged in an alkaline medium

Answer: 1,2,3 and 4

Class 12 Chemistry Unit 14 Biomolecules Match The Following Questions And Answers

Question 1.

Answer: 1-C, F,  2-G, 3-A, 4-H, 5-D, 6-E, 6-B

Question 2.

Answer: 1-D, 2-C, 3-E, 4-A, 5-B

Class 12 Chemistry Unit 14 Biomolecules Assertion-Reason Type

In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices.

1. (A) and (R) both are correct statements and (R) is the correct explanation for (A).
2. (A) and (R) both are correct statements but (R) is not a correct explanation for (A).
3. (A) is a correct statement but (R) is a wrong statement.
4. (A) and (R) both are incorrect statements.
5. (A) is a wrong statement but (R) is a correct statement.

Question 1. Assertion (A): D (+) – glucose is dextrorotatory.

Reason (R): ‘D’ represents its dextrorotatory nature.

Answer: 3. (A) is the correct statement but (R) is the wrong statement.

Question 2. Assertion (A): Vitamin D can be stored in our body.

Reason (R): Vitamin D is a fat-soluble vitamin.

Answer: 1. (A) and (R) both are correct statements and (R) is a correct explanation for (A).

Question 3. Assertion (A): ft -glycosidic linkage is present in maltose

Reason (R): Maltose is composed of two glucose units in which C-l of one glucose unit is linked to C-4 of another glucose unit.

Answer: 4. (A) and (R) both are incorrect statements.

Question 4. Assertion (A): AU naturally occurring or-amino acids except glycine are optically active.

Reason (R): Most naturally occurring amino acids have L-configuration.

Answer: 5. (A) is a wrong statement but (R) is the correct statement.

Question 5. Assertion (A): Deoxyribose, C₅H₁₀O₄ is not a carbohydrate.

Reason (R): Carbohydrates are hydrates of carbon so compounds which follow the C_x(H₂O)_y formula are carbohydrates.

Answer: 2. (A) and (R) both are correct statements but (R) is not the correct explanation for (A).

Question 6. Assertion (A): Glycine must be taken through diet.

Reason (R): It is an essential amino acid.

Answer: 2. (A) and (R) both are correct statements but (R) is not a correct explanation for (A).

Question 7. Assertion (A): In the presence of an enzyme, the reagent can effectively attack the substrate molecules.

Reason (R): Active sites of enzymes hold the substrate molecule in a suitable position.

Answer: 1. (A) and (R) both are correct statements and (R) is the correct explanation for (A).

Class 12 Chemistry Unit 14 Biomolecules Fill in the blanks

Question 1. Polysaccharides mix in boiling water to form___
Answer: Colloids;

Question 2. An increase in temperature____ the mutarotation rate.
Answer: Increases;

Question 3. The simplest amino acid is____
Answer: Glycine;

Question 4. Sucrose hydrolysis is called____
Answer: Inversion of sucrose

Question 5. The heterocyclic bases present in nucleic acids are called____
Answer: Purine and pyrimidine;

Question 6. Polyhydroxy ketones are known as____
Answer: Ketoses

Question 7. Amylopectin is a branched polymer of____
Answer: Glucose

Question 8. Phospholipids are mixed glycerides of and____
Answer: Fatty acid, phosphoric acid

Question 9. Thyroxine is a____ hormone.
Answer: Amine

Question 10. The scurvy disease occurs due to a deficiency of____
Answer: Vitamin C

Question 11. The chemical name of Vitamin B₁₂ is____
Answer: Cyanocobalamine

Question 12. Lecithin is a____
Answer: Phospholipid

Question 13. The acidic property of glycine pertains to____
Answer: —COOH

Question 14. The power-house of animal cells is called____
Answer: Mitochondria

Class 12 Chemistry Unit 13 Organic Compounds Containing Nitrogen

The class of organic compounds having functional groups containing nitrogen is termed as ‘nitrogenous organic compounds.’ This chapter is devoted to the discussion of this special class of compounds, and their classifications with the corresponding functional groups in an elucidated way. Classification of important nitrogenous organic compounds and the related functional groups are as follows:

Alkyl Cyanide (R-CN)

Derivatives of hydrogen cyanide are called alkyl cyanide. These are obtained by replacing a hydrogen atom of a hydrogen cyanide molecule with an alkyl group.

In the molecules of alkyl cyanide, the alkyl group is directly attached to a carbon atom of the cyano group. Both the carbon and nitrogen atoms of the cyano group are sp- hybridised.

Nomenclature Of Alkyl Cyanides

Alkyl Cyanides General method (Common system): These are named by using the suffix ‘cyanide’ after the name of the alkyl group.

The compounds belonging to the class of cyanides are named by adding the suffix ‘nitrile’ in place of ‘ic-acid’ to the name of the corresponding acid produced by hydrolysis of the alkyl cyanide.

Alkyl Cyanides General method Example: Hydrolysis of methyl cyanide (CH₃CN) gives acetic acid. So the other name of methyl cyanide is acetonitrile. Similarly, propionic acid is obtained by hydrolysis of ethyl cyanide (CH₃CH₂CN). Thus the other name of ethyl cyanide is propiononitrile.

IUPAC Method: In this system, cyanides are named alkane nitriles. The carbon atom of the — CN group is also counted as belonging to the parent chain.

• The positions of various substituents are indicated by numbering the carbon atoms in the longest parent chain starting from the carbon atom of the —CN group.
• These compounds are thus named by adding the suffix nitrile, to the names of the parent alkanes.

If the molecule of a compound contains two cyano groups ( —CN), ‘dinitrile’ is written after the name of the parent alkane.

IUPAC Method Example: \(\mathrm{N} \stackrel{1}{\mathrm{C}}-\stackrel{2}{\mathrm{C}} \mathrm{H}_2 \stackrel{3}{\mathrm{C}} \mathrm{H}_2 \stackrel{4}{\mathrm{C}} \mathrm{H}_2 \stackrel{5}{\mathrm{C}} \mathrm{H}_2-\stackrel{6}{\mathrm{C}} \mathrm{N} \text { (Hexanedinitrile) }\)

• When a molecule of a compound contains three or more cyano groups, the —CN group is treated as the substituent.
• The suffixes ‘tricarbonitrile’, ‘tetracarbonitrile’, etc. are used after the name of the alkane along with the positions of cyano groups in the chain of carbon atoms of that alkane.
• In this case, carbon atoms of cyano groups are not included in the chain of carbon atoms of the main alkane.

Example:

In the case of alicyclic, aromatic cyano compounds, the suffix ‘carbonitrile’ is added after the name of the parent hydrocarbon.

Example:

Preparations Of Alkyl Cyanides

Alkyl Cynides From Alkyl Halides: Primary alkyl halides (preferably bromides and iodides) when heated with an alcoholic solution of NaCN or KCN, yield alkyl cyanides. In this reaction, alkyl isocyanide (RNC) is also obtained as a by-product.

Alkyl Cynides From Alkyl Halides Example:

Preparation Of Aryl Cyanide: In the reaction of diazonium salt with CuCN (extended Sandmeyer reaction), aryl cyanide is obtained.

Aryl Cyanide Reaction Mechanism: CN^– ion, produced from KCN reacts both with 1° alkyl halides in aqueous solution via S_N2 mechanism. Relatively larger and less electronegative carbon atom (electron cloud is less polarisable) of the ambident nucleophile attacks or -carbon atom of the alkyl halide, giving alkyl cyanide.

Secondary alkyl halides (preferably bromides or iodides) also react with alcoholic solution of NaCN or KCN to give alkyl isocyanide as the major product.

The reaction proceeds through the S_N2 mechanism.

From Acid Amides: When acid amide is heated in the presence of phosphorus pentoxide (P₂O₅) or thionyl chloride (SOCl₂), alkyl cyanide is obtained. In this process, alkyl isocyanide is not obtained as a by-product.

Example:

From Ammonium Salts Of Carboxylic Acid: Ammonium salt of carboxylic acid when distilled in the presence of phosphorus pentoxide gives alkyl cyanide.

Carboxylic Acid Example:

Dehydration Of Aldoximes: Alkyl cyanide is obtained by heating a mixture of aldoxime and P205 or acetic anhydride.

Dehydration Of Aldoximes Example:

From Carboxylic Acids: Alkyl cyanide is produced on a large scale by passing a mixture of carboxylic acid vapours and ammonia over alumina (Al₂O₃) heated to 500°C.

Preparation of alkyl cyanide with the same number of carbon atoms from an aldehyde: Aldoxime, derived in the reaction between aldehyde and hydroxylamine, yields alkyl cyanide on dehydration.

Preparation of alkyl cyanide with the same number of carbon atoms from a carboxylic acid: Carboxylic acid is successively converted into acid chloride and acetamide and the latter is dehydrated by heating with P₂O₅.

From Grignard Reagent: Cyanogen chloride (ClCN) reacts with Grignard reagent to give alkyl cyanide. This method is highly suitable for the preparation of tertiary alkyl cyanide.

Grignard Reagent Example:

Properties And Uses Of Alkyl Cyanides

Physical Properties

1. Lower members of alkyl cyanides are colourless liquids but higher members are crystalline solids.
2. Generally, these alkyl cyanides are quite stable, have a sweet smell and are not poisonous like hydrogen cyanide (HCN).
3. Molecules of alkyl cyanides have high dipole moments due to the presence of a polar cyano group. The boiling point of alkyl cyanides is higher than those of alkyl halides of comparable molecular mass. It is due to a stronger intermolecular force of attraction involving dipole-dipole attractive force.

CH₃CN (Boilingpoint= 82°C);  CH₃CI (Boilingpoint= – 24°C)

Alkyl cyanides of lower molecular masses are soluble in water due to the formation of hydrogen bonds between the molecules of alkyl cyanides and water.

• With the increase in molecular mass, the size of the alkyl group present in an alkyl cyanide increases, i.e., the effect of the non-polar hydrocarbon part predominates. Consequently, solubility in water decreases.
• Alkyl cyanides are easily soluble in organic solvents such as alcohol, ether, benzene etc.

Chemical Properties

Chemical Properties Hydrolysis: Alkyl cyanides, when refluxed in the presence of dilute acid solution, undergo hydrolysis and produce carboxylic acids.

If an alkyl cyanide is refluxed with dilute alkali (for example., NaOH) it gets hydrolysed forming sodium salt of carboxylic acid (RCOONa). Acidification of the reaction mixture liberates free carboxylic acid.

Alkyl cyanides (RCN) undergo hydrolysis to produce carboxylic acids. The carbon atom of the carboxylic acid group is directly linked to the alkyl group R, so, the carbon atom of the —CN group in RCN is directly attached to the R group.

Partial hydrolysis: When alkyl cyanide is dissolved in concentrated H₂SO₄ and added to cold water or it is shaken with cold and cone. HCl, it suffers incomplete hydrolysis producing acid amide.

Formation Of Acid Amide: Alkyl cyanide reacts with an alkaline H₂O₄ solution to give the acid amide.

Acid Amide Reduction: Alkyl cyanides are reduced by hydrogen in the presence of Pt or Ni catalyst or (sodium+alcohol) or lithium aluminium hydride (LiAlH₄), to give primary amines with the same number of carbon atoms.

Stephen Process: In this process, alkyl cyanides are reduced by stannous chloride and concentrated HCl in ether medium to form a complex which on hydrolysis gives aldehydes.

Formation Of Esters: Esters are formed when an alcoholic solution of alkyl cyanides is heated with concentrated H₂SO₄ or HCl.

Reaction With Grignard Reagent: Alkyl cyanide reacts with Grignard reagent to form a complex, which on hydrolysis with dilute acid gives a ketone.

Uses Of Alkyl Cyanides

1. Some of the alkyl cyanides, particularly methyl cyanide (acetonitrile), is used as a solvent.
2. Alkyl cyanides are used in the preparation of nitrile rubber and in the cotton industry.
3. The process of conversion of a lower homologue into its higher homologue is accomplished through the formation of alkyl cyanide.
4. In the synthesis of organic compounds such as amine, aldehyde, ketone, acid, ester, amide etc., alkyl cyanide is used as an intermediate.

Alkyl Isocyanide (R-NC)

Alkyl isocyanide is an isomer of alkyl cyanide. These compounds are also known as alkyl isonitrile or arylamine. The alkyl group present in a molecule of alkyl isocyanide is attached to a carbon atom through a nitrogen atom.

Nomenclature Of Alkyl Isocyanides

General method (Common system): In this system, compounds belonging to this class are named as isonitriles or isocyanides or carbylamines. The system of nomenclature for iso-nitrite is similar to that used for isomeric alkyl cyanide. In the case of names such as isocyanide or arylamine, the suffix ‘isocyanide’ or ‘arylamine’ is added to the alkyl group present in the molecules under consideration.

IUPAC Method: In this system, compounds belonging to this class are named iso carbonitride alkane or alkane isocarbonitrile.

Preparation Of Alkyl Isocyanides

Secondary alkyl halides (preferably bromides or iodides) react with an alcoholic solution of silver cyanide (AgCN) to give alkyl isocyanide as the major product. [If tertiary alkyl halide is used as the substrate, then an alkene is formed as the major product via elimination reaction.] In this reaction, alkyl cyanide is formed as a minor product.

Example:

Reacted Mechanism: In the presence of AgCN, silver halide is precipitated from alkyl halide producing a carbocation. Consequently, the reaction follows an S_N1 path. Here cyanide ion is an ambident nucleophile in which Natom is relatively smaller and more electronegative. Hence N-atom attacks the carbocation forming alkyl isocyanide as the main product.

Primary alkyl halides (preferably bromides and iodides) also react with an alcoholic solution of silver cyanide to give alkyl cyanide as the major product.

The reaction occurs by the S_N2 mechanism.

From Primary Amines: When primary amine is heated in the presence of chloroform and alcoholic potassium hydroxide (KOH) solution, alkyl isocyanide is obtained. This reaction is called arylamine reaction.

From N-alkyl Formamide: When IV-alkyl formamide is heated with POCl₃ in the presence of pyridine, alkyl isocyanide is formed.

Properties And Uses Of Alkyl Isocyanides

1. Alkyl isocyanides are poisonous liquids having obnoxious smell.
2. The boiling point of an alkyl isocyanide is lower than that of the isomeric alkyl cyanide due to its lower polarity. CH₃CN (boiling. = 82°C); CH₃NC (boiling. = 49°C)
3. Alkyl isocyanides are almost insoluble in water because of their inability to form hydrogen bonds with water molecules. But they are fairly soluble in organic solvents like ether, benzene etc.

Alkyl Isocyanides  Chemical Properties

Alkyl Isocyanides Chemical Properties Hydrolysis: Alkyl isocyanides when shaken with dilute acids, are hydrolysed even in cold conditions producing primary amine and formic acid (different from alkyl cyanide).

In the hydrolysis of alkyl isocyanide, no acid other than formic acid is produced. Alkyl isocyanides are not hydrolysed by alkaline solutions.

In an isocyanate group ( — N⁺ =C^–) both the N and C-atoms have filled octets of electrons. So, the e-nucleophile (O^–H) can attack neither the C-atom nor the Natom of the isocyanide molecule.

Alkyl Isocyanides Chemical Properties Reduction: Catalytic hydrogenation of alkyl isocyanide or its reduction with sodium and alcohol, leads to the formation of a 2° amine (note the difference from alkyl cyanide). The 2° amine, produced contains a methyl group attached to the N-atom.

Alkyl Isocyanides Chemical Properties Example:

Isomerisation: On prolonged heating, alkyl isocyanide isomerises to form more stable isomeric alkyl cyanide.

Isomerisation Reducing Action: The carbon atom of the isocyanide group in alkyl isocyanide contains a lone pair of electrons. Due to the presence of a lone pair of electrons, alkyl isocyanides act as reducing agents.

Isomerisation Example: Methyl isocyanide reduces HgO to Hg, itself being oxidised to methyl Isocyanate.

Use Of Alkyl Isocyanide: It is used in the preparation of secondary amines with N-methyl group (R—NH—CH₃).

Distinction Between Alkyl Cyanide And Alkyl Isocyanide:

Nitroparaffin Or Nitroalkane (R-NO₂)

The organic compound obtained by the replacement of one hydrogen atom from a molecule of saturated hydrocarbon or alkane is called nitroalkane or nitroparaffin.

Nitroalkanes are classified as primary, secondary or tertiary based on the nature of the carbon atom (i.e., 1°, 2° or 3°) to which the nitro group (—NO₂) is attached.

Nomenclature Of Nitroalkane

In both the trivial and IUPAC systems of nomenclature, nitroalkanes are named as derivatives of alkanes. The position of the nitro group in the longest chain of carbon atoms is marked by the lowest possible number.

Example:

Preparation Of Nitroalkanes

From Alkyl Halides: Primary alkyl halides (preferably bromides or iodides i.e., RCH₂Br and RCH₂I) dissolved in dimethyl formamide (DMF) as solvent give nitroalkane on reaction with NaNO₂ or KNO₂. To increase the solubility of nitrite salts, some urea is added to the reaction mixture. If dimethyl sulphoxide (DMSO) is used as a solvent, the addition of urea is unnecessary.

Alkyl nitrite is formed in small amounts as a by-product.

Reaction Mechanism: In the solvent DMF or DMSO, the reaction proceeds through the S_N2 path. In the given ambident nucleophile, the N-atom is relatively larger and less electronegative, having a more polarizable electron cloud O^– — N=O↔O=N—O^–. Hence N-atom of the nitrite ion attacks the carbon atom of the alkyl halide forming nitroalkane as the major product.

The reaction of ethanolic silver nitrate solution with a primary alkyl halide (preferably bromide or iodide) gives a satisfactory yield of primary nitroalkane.

Here also the reaction proceeds through the S_N2 mechanism and the O^–—N:=O (nitrite) ion acts as the nucleophile.

From Tertiary (3°) Alkyl Amines

Oxidation of a primary amine-containing tertiary alkyl group with potassium permanganate gives a 3° nitroalkane.

Example:

By Vapour Phase Nitration Of Alkanes: A mixture of gaseous alkane and HNO₃ vapour when heated to 400-475°C, yields nitroalkanes.

Example:

Nitration of alkanes containing two or more carbon atoms at high temperatures brings about the cleavage of the C—C bond, consequently producing a mixture of nitroalkanes.

Example:

From α-Halogeno Acids: After heating sodium or potassium salt of α-halogen acid with a solution of sodium nitrite, the reaction mixture is acidified to produce α-nitro acid, which eliminates CO₂ gas forming nitroalkanes.

Example:

Properties And Uses Of Nitroalkanes

Physical Properties

1. In the pure state, nitroalkanes are colourless liquids with a pleasant smell.
2. Nitroalkanes are sparingly soluble in water, lower nitroalkanes being relatively more soluble. In organic solvents, they dissolve easily.
3. The dipole moments of nitroalkanes, (μ – 3-4D) are very high because the molecules remain strongly bound by dipole-dipole attractive forces. Therefore, the boiling point of nitroalkanes is much higher than that of alkanes with comparable molecular mass.

The polarity of nitroalkanes is much higher relative to isomeric alkyl nitrites. So the nitroalkanes have much higher boiling points than the isomeric alkyl nitrites. For example, the boiling points of nitroethane and ethyl nitrite are 115°C and -12°C respectively.

Chemical Properties

Reduction: Nitroalkanes are reduced involving the following stages—

The nature of the product obtained by reduction depends on the nature of the reducing agent and the pH of the reaction medium.

Reduction in acid medium: Nitroalkanes are reduced by tin and HCl or iron and HCl or zinc and HCl to give primary amines. This reaction proves that the nitrogen atom of the nitro group is directly bonded to the alkyl group.

Reduction in neutral medium: Nitroalkane, when heated with zinc dust and ammonium chloride solution in the presence of a few drops of alcohol yields alkyl-substituted hydroxylamine.

Identification Of Nitroalkanes By Muliiken-Barker Test: The nitro compound is boiled with zinc dust and aqueous NH₄Cl solution in the presence of a few drops of ethyl alcohol.

The resulting solution is filtered into a freshly prepared Tollens reagent. The appearance of a grey precipitate indicates the presence of the —NO₂ group.

Reduction By Hydrogen: Nitroalkanes are reduced by H₂ in the presence of Ni, Pt or, Pd catalyst to give 1° amines.

Reduction by LiAlH₄: Nitroalkanes are converted into primary amines by reduction with LiAlH₄.

Hydrolysis: If primary nitroalkane is treated with boiling concentrated HCl or 85% concentrated H₂SO₄, it undergoes hydrolysis yielding carboxylic acid and hydroxylamine. This reaction is the basis of industrial production of hydroxylamine.

Secondary nitroalkanes are reduced by boiling with concentrated HCl to produce ketone and nitrous oxide.

Tertiary nitroalkanes are not hydrolysed by concentrated HCl or concentrated H₂SO₄.

Nitro-Aci-Nitro Tautomerism

Nitroalkanes containing ar-H atom, (i.e., primary and secondary nitroalkanes) display tautomerism.

Example: One H-atom from a -carbon atom of nitroethane detaches and gets itself linked to an O-atom of —NO₂, giving rise to an isomeric compound with a completely different structure. The form of the compound is called nitro-form and the form is called an acid-nitro-form.

Out of the two tautomers of nitroethane (1 and 2), nitroform Is more stable than aci-nitro-form because of the resonance stabilisation of nitro-form

Hence, in the tautomeric mixture, the concentration of nitro-form is quite greater than that of aci-nitro-form 2.

Tertiary alkanes do not exhibit tautomerism (no α-H ).

Acidic Nature

1. Nitroalkanes have no action on litmus paper as they are neutral compounds. But primary and secondary nitroalkanes which have ar-H atom, dissolve in alkali forming salts. From this, it is evident that primary and secondary nitroalkanes act as mild acids.
2. Primary and secondary nitroalkanes behave as acids only in the presence of alkali, so these are called pseudo acids. These compounds are converted into aciform before they are reacted with alkalies.
3. The nitro-form of nitroalkanes is called ‘pseudo acid’ and the aci-nitro-form is called ‘nitronic acid’. Aci-microforms are crystalline solids which dissolve in sodium hydroxide to give red solutions.

Owing to the absence of any α-H atom in tertiary nitroalkanes, they cannot exist in aci-nitro-form. As a result, tertiary nitroalkanes do not react with alkalies and hence, do not show any acidic properties.

Reaction With Nitrous Acid: Depending on their behaviour towards HNO₂, primary, secondary and tertiary nitroalkanes can be differentiated by their reactions with nitrous acid as follows:

1° nitroalkanes react with nitrous acid, forming a crystalline compound, nitrolic acid, which dissolves in NaOH to give a red solution.

2° nitroalkanes react with HN02 forming a crystalline compound, pseudonitrole, which forms a blue solution with NaOH, indicating the presence of nitroso group.

3° nitroalkanes do not contain any ar-H atom. So they do not react with nitrous acid.

Halogenation: Primary and secondary nitroalkanes react with alkalies in the presence of halogen-producing halogen nitroalkanes. In this reaction, α-H atoms of nitroalkanes are successively replaced by halogen atoms.

Example:

When nitromethane is allowed to react with an excess chlorine, trichloro nitromethane or chloropicrin is formed. This Is also known as tear gas.

Uses Of Nitroalkanes

1. Lower nitroalkanes are used as solvents for oil, fats, resin and paints.
2. Nitroalkanes are used as intermediate compounds for the preparation of detergent, propellant, etc.
3. Chloropicrin, a derivative of nitromethane, finds use as an insecticide and tear gas.
4. In the manufacture of hydroxylamine, primary nitroalkane is used.

Alkyl Nitrite (R-ONA)

Alkyl nitrite is an ester of the inorganic acid, HNO₂ (nitrous acid). Their general formula: is R—O—N=O. They are well-known as isomers of nitroalkanes. The alkyl group present in the molecule of alkyl nitrite is bonded to the nitrogen atom through an oxygen atom. Among the compounds belonging to this class, ethyl nitrite and isoamyl nitrite are important.

Preparations Of Alkyl Nitrites

From Alkyl Halides: In the reaction of 2° and 3° alkyl halides (preferably bromides and iodides) with silver nitrate solution, alkyl nitrites are obtained as the major product.

Reaction Mechanism: In the reaction between silver nitrite and alkyl halide, alkyl carbocation is produced.

• As a result, the reaction follows the S_N1 path. Here nitrite ion (NO^–₂) is an ambident nucleophile in which O-atom is relatively smaller and more electronegative having higher electron density.
• Hence O-atom attacks the carbon atom of the carbocation, resulting in the formation of alkyl nitrite as the major product.

If a secondary or tertiary alkyl halide (preferably bromide or iodide) is treated with an aqueous NaNO₂ solution, then also alkyl nitrile is formed as the major product.

From Alcohols: On adding concentrated HCl or concentrated H₂SO₄ to a solution of C₂H₅OH and aqueous NaNO₂, ethyl nitrite is obtained.

Isoamyl nitrite is produced when concentrated HCl is added to a solution of isopentyl alcohol and NaNO₂.

Properties And Uses Of Alkyl Nitrites

Physical Properties: Ethyl nitrites and isoamyl nitrites are liquids with pleasant smell. Their boiling points are 17°C and 99°C, respectively.

Chemical Properties

Hydrolysis: Alkyl nitrite, when heated with aqueous solution of alkali undergoes hydrolysis to yield alcohol and nitrite salt.

Reduction: When alkyl nitrite is reduced by (Sn + HCl) or (Zn + HCl), the major products obtained are alcohol and hydroxylamine. This reaction proves that in alkyl nitrite, the alkyl group is attached to the N-atom through an O-atom.

Uses Of Alkyl Nitrites

1. In the preparation of nitrous acid in an anhydrous medium, alkyl nitrites are used.
2. A 4% alcoholic solution of ethyl nitrite is known as the sweet spirit of nitre. It is used as a heart stimulant and diuretic.

Distinction Between Nitroalkane And Alkyl Nitrite:

Aromatic Nitro Compounds

Aromatic nitro compounds are formed by the replacement of one or more hydrogen atoms in the benzene ring. Nitrobenzene is an ideal representative of aromatic nitro compounds.

Nomenclature Of Aromatic Nitro Compounds

Aromatic nitro compounds are named as nitroarenes.

Nitro compounds Example:

Preparation Of Aromatic Nitro Compounds

With the help of nitrating reagents Aromatic nitro compounds are prepared by the reaction of a suitable aromatic compound with any one of the following nitrating reagents—

1. Mixed acid (concentrated HNO₃ + concentrated H₂SO₄) or (fuming nitric acid + concentrated H₂SO₄) or (fuming nitric acid + fuming H₂SO₄),
2. Concentrated HNO₃ dissolved in glacial acetic acid or nitromethane,
3. Acetyl nitrate (concentrated HNO₃ dissolved in acetic anhydride),
4. Nitronium salt dissolved in organic solvents [for example., nitronium perchlorate, (NO⁺₂ClO^–₄), nitronium tetrafluoroborate (NO⁺₂BF^–₄).

Preparation Of Aromatic Nitro Compound Using Mixed Acid

Nitrobenzene is prepared by heating benzene with a mixture of concentrated (HNO₃ + H₂SO₄) at 50-60°C.

During the reaction, temperature is strictly controlled as at higher temperatures (>60°C), meta-dinitrobenzene is produced.

Nitrobenzene Reaction Mechanism: In the reaction between concentrated HNO₃ and concentrated H₂SO₄ (nitrating reagent), the nitronium cation (NO₂) acts as an electrophile.

It is known from experiments that the nitrations of C₆H₆ and C₆D₆ occur at the same rate. So the second step (in which the C—H or C—D bond cleaves), is not the rate-determining step. Hence, the first step which involves the formation of σ-complex between the electrophile and the substrate is the rate-determining step.

Nitration of toluene with mixed acids at ordinary temperature gives a mixture of o-and p-nitrotoluene.

Nitration of chlorobenzene with mixed acid at 100°C, gives a mixture o-and of p-chloronitro benzenes.

In the reaction of mixed acid with acetanilide, p-isomer is produced as a major product.

(—CH₃, —Cl, —NHCOCH₃ groups are o-/p-directing.)

When nitrobenzene is heated with a mixture of fuming nitric acid and concentrated H₂SO₄ in a boiling water bath, m-dinitrobenzene is obtained.

2,4-dinitrotoluene, obtained in the nitration of toluene, is subjected to further nitration at high temperature in the presence of fuming nitric acid and fuming H₂SO₄, when 2, 4, 6-trinitrotoluene (TNT) is obtained.

Oxidation of 2,4,6-trinitrotoluene (TNT), obtained in the nitration of toluene, gives rise to 2,4,6-trinitrobenzoic acid, which on decarboxylation yields 1,3,5-trinitrobenzene.

If fuming nitric acid is used during nitration, NO^–₂ ion is produced.

In the case of nitration with N₂O₅ dissolved in a polar solvent, the NO⁺₂ ion is formed according to the following equilibrium.

Preparation of aromatic nitro compound using dil. HNO₃

The presence of hydroxyl group ( —OH) in the benzene ring activates the ring to such an extent that treatment of phenol even with dilute HNO₃ at ordinary temperature gives a mixture of o- and p- nitrophenols. A small amount of phenol is, however, oxidised by HNO₃ where nitric acid is itself reduced to nitrous acid.

Preparation of aromatic nitro compound Reaction Mechanism: In the case of nitration with dilute HNO₃, the reaction is initiated through the formation of nitrosonium ions. The nitroso compound so produced undergoes oxidation to form the corresponding nitro compound.

From Diazonium Salts: Fluoroboric acid reacts with arene diazonium chloride to form arene diazonium fluoroborate. To replace the —N^–₂BF⁺₄ group with the nitro group, the salt so produced is decomposed in the presence of an aqueous solution of NaNO₂ and Cu powder.

Diazonium Salts Example:

From Aromatic Amines: Aromatic amines are oxidised by trifluoroacetic acid to their corresponding nitro compounds.

Aromatic Amines Example:

Properties And Uses Of Aromatic Nitro Compounds

The physical and chemical properties of nitrobenzene as an ideal representative of aromatic compounds, are discussed below:

Nitro Compounds Physical Properties:

1. Nitrobenzene is a light yellow oily liquid. Its commercial name is ‘oil of mirabane’. It has a characteristic smell of bitter almonds. It is immiscible in water.
2. The boiling point of nitrobenzene is 211°C. It is solidified by cooling on ice. Solid nitrobenzene melts at 5.8°C.
3. Nitrobenzene is heavier than water (sp. gravity: 1.204). It is steam volatile and has poisonous vapours.
4. Nitro group present in nitrobenzene withdraws electrons from the ring through -I and -R effects. Consequently, the ring acquires a partial positive charge while the —NO₂ group acquires a partial negative charge, which makes the molecule sufficiently polar (dipole moment: 3.95D).

Such a high dipole moment is responsible for the dipole-dipole interactions among the molecules, making nitrobenzene a high boiling liquid (b.p. 211°C ).

Chemical Properties: Nitrobenzene is a stable compound, generally not attacked by acids, alkalis or oxidising agents. So, it is used as an effective solvent In various oxidation reactions. The reactions of nitrobenzene may be classified under two heads—

1. Reactions of nitro group (—NO₂) (where the benzene ring remains unaffected) and
2. Substitution reactions in the benzene ring (where the nitro group remains unaffected).

Reactions Of Nitro Group: The most important reaction of the nitro group in nitrobenzene is its reduction. As the nitro group is easily reduced, it is frequently used as an oxidising agent. The reduction of nitrobenzene takes place through the following steps:

Depending on the nature of the reducing agent and the concentration of hydrogen ions in the reduction medium, the products of reduction are found to be different.

Reduction in acid medium: Nitrobenzene, when reduced by tin, zinc or iron and concentrated HCl or zinc and acetic acid, gives aniline. All aromatic nitro compounds can be similarly reduced to give primary amines.

Reduction by metal in a strong acidic medium forms an intermediate compound, phenylhydroxylamine (C₆H₅NHOH) which undergoes rearrangement producing p-aminophenol.

Reduction In Neutral Medium: When nitrobenzene dissolved in 50% alcohol is warmed with zinc dust and an aqueous solution of ammonium chloride, it is reduced to phenylhydroxylamine.

Electrolytic reduction of nitrobenzene in acetic acid and aqueous solution of sodium acetate produces phenylhydroxylamine.

Reduction In Alkaline Medium: Depending on the nature of reducing agents, different products such as azoxybenzene, azobenzene and hydrazobenzene are formed in an alkaline medium.

when nitrobenzene is reduced by zinc dust and a methanolic solution of NaOH, at first azoxybenzene and then azobenzene is formed.

Nitrobenzene is reduced by Zn-dust an aqueous solution of NaOH to form hydrazobenzene.

Reduction by LiAIH₄: In the reduction of nitrobenzene by lithium aluminium hydride, azobenzene is produced.

It is interesting to note that the reduction of aliphatic nitro compounds with lithium aluminium hydride gives the corresponding amines.

Electrolytic Reduction In Acidic Medium: Electrolytic reduction of nitrobenzene in a mild acidic medium gives aniline. On the other hand, electrolytic reduction of nitrobenzene in the strong acidic medium at first produces phenylhydroxylamine, which on rearrangement gives p-aminophenol.

Selective Reduction: m-dinitrobenzene on partial reduction by NH₄HS gives m-nitroaniline.

m-nitroaniline is also obtained by controlled reduction of nitrobenzene using Na₂S or (NH₄)₂S.

Hydrogenation: Nitrobenzene on catalytic hydrogenation (in the presence of Raney nickel, Pd or, Pt-C ) under 30 atm pressure gives aniline.

Substitution Reaction In Benzene Ring

Electrophilic substitution reaction: As the nitro group (—NO₂) is meta-orienting in the electrophilic substitution reactions, the incoming substituent mainly enters the meta-position.

In the presence of electron attracting nitro group, the electron density of the benzene ring decreases and consequently, the rate of electrophilic substitution in nitrobenzene is much slower than that in benzene. Different types of electrophilic substitution reactions of nitrobenzene are given below:

Nitration: Nitrobenzene when heated with fuming nitric acid and fuming sulphuric acid in a boiling water bath, gives a deep yellow liquid, meta-dinitrobenzene.

If nitrobenzene is refluxed with fuming nitric acid and fuming sulphuric acid, 1,3,5-trinitrobenzene (TNB) is produced. The reaction takes five days, as it is extremely difficult to introduce the third nitro group. TNB is a highly explosive substance.

Chlorination: When chlorine gas is passed into hot nitrobenzene in the presence of iron powder or aluminium chloride, meta-chloronltrobenzene is formed.

Similarly, bromine reacts with nitrobenzene in the presence of iron powder to give meta-bromonitrobenzene.

Sulphonation: On heating with fuming sulphuric acid, nitrobenzene gives meta-nitrobenzene sulphonic acid.

Friedel-Crafts reaction: The electrophilic species involved in Friedel-Crafts reaction are very weak. Nitrobenzene fails to undergo substitution reactions with such electrophiles because the ring system of this molecule is highly electron deficient. This can be attributed to the -I and -R effects of the —NO₂ group. Hence, nitrobenzene does not participate in the Friedel-Crafts reaction.

Nucleophilic Substitution Reaction: Nitro group (—NO₂) present in nitrobenzene decreases the electron density of ortho- and para-positions to a greater extent, relative to meta-position. Hence, the meta-position becomes comparatively electron-rich, while the ortho- and parapositions are reduced to electron-deficient sites. Therefore, ortho- and para-positions of the molecule of nitrobenzene are easily attacked by nucleophiles.

Nucleophilic Substitution Example: When nitrobenzene is fused with caustic potash in the presence of air, it mainly gives orthonitrophenol as its potassium salt, which on subsequent acidification produces o-nitrophenol.

Uses Of Nitrobenzene

1. Nitrobenzene is used
2. As a high-boiling solvent,
3. As a mild oxidising agent in organic synthesis
4. In the preparation of aniline, benzidine and some azo-dyes,
5. In boot polish,
6. In the polishing of the floor using wax,
7. In lowgrade scented soaps and
8. In the preparation of explosives such as TNT, TNB, etc.

Tests For Nitro Group: Identification Of Nitrobenzene

Identification Nitrobenzene Reduction Test: Nitrobenzene when heated with Sn and concentrated HCl is reduced to aniline.

• The resulting solution is cooled (0-5°C) and treated with dilute HCI and dilute NaNO₂ solution, producing benzenediazonium chloride.
• The addition of a few drops of this solution to a cold alkaline p-naphthol solution gives a bright, scarlet red azo dye.

• It is a test for the identification of the aromatic primary amino (—NH₂) group. As the —NO₂ group is reduced to the —NH₂ group, it may be regarded as an indirect test for the detection of the —NO₂ group in the benzene ring.
• This test can be applied to identify the —NO₂ group in the absence of the —NH₂ group in the benzene ring.

Mulliken-Barker Test: Nitrobenzene is boiled with zinc dust and an aqueous solution of NH₄Cl in 50% C₂H₅OH, producing phenylhydroxylamine.

• The resulting solution is filtered into a freshly prepared Tollens’ reagent (ammoniacal silver nitrate solution).
• The appearance of a black or grey precipitate of silver indicates the presence of the —NO₂ group.
• This test can be applied to identify the —NO₂ group even in the presence of the —NH₂ group in the benzene ring.
• Hence, this reaction Is a confirmatory test for the detection of the —NO₂ group, when the —NH₂ group is also present in the benzene ring.

Nitro (—NO₂) group present in any organic compound (aliphatic or aromatic) is identified with the help of this test.

Limitations of Mulliken-Barker test: This test for the detection of the —NO₂ group is not applicable when an organic compound already contains any other reducible functional group.

• For example, if an aldehyde (—CHO) group or α-hydroxyketo [—CH(OH)CO—] group is present in any organic compound, then, the Mulliken-Barker test for the identification of the —NO₂ group cannot be used.
• Because aldehydes and α-hydroxyketones reduce Tollens’ reagent to give a precipitate of metallic silver (Ag).

Amines Introduction

Amines are considered as an important class of organic compounds. Amines are derived by the replacement of one or more H -atoms of ammonia molecules by alkyl or aryl groups.

• These are commonly found in nature as proteins, vitamins, hormones, alkaloids, etc.
• A large number of artificially prepared amino compounds are used as polymers, dyes and drugs. Adrenaline (a hormone) and ephedrine (a drug) in which a secondary amino group is present are used to increase blood pressure.
• Novocaine, an artificially prepared amino compound is used in dental treatment as an anaesthetic agent. Antihistamine drug viz. benadryl contains a tertiary amino group. Quaternary ammonium salts are widely used as surface active agents.
• Diazonium salts find extensive application as intermediates in the preparation of aromatic compounds and dyes.

Classification And Structure Of Amines

Classification Of Amines

Primary, Secondary And Tertiary Amines: Aliphatic amines are regarded as derivatives of ammonia. Amines are divided into three classes—primary (1°), secondary (2°) and tertiary (3°). Replacement of one, two or three H -atoms of ammonia molecule by alkyl or aryl groups produces primary (1°), secondary (2°) and tertiary (3°) amines, respectively.

R = alkyl or aryl group ( —CH₃, —C₂H₅, —C₆H₅, etc.)

Functional groups present in 1°, 2° and 3° amines are:-

Aliphatic And Aromatic Amines

Aliphatic Amines: Aliphatic amines are derived by the replacement of one or more H -atoms of ammonia molecules by alkyl groups.

Aliphatic Amines Example:

Aromatic amines: In aromatic amines, at least one aryl group is attached to the amino nitrogen atom.

Aromatic amines Example:

The amines in which the N-atom is linked to the side chain of the aromatic ring are called aryl-substituted aliphatic amines.

Aromatic amines Example:

Simple And Mixed Amines

Simple Amines: If the alkyl or aryl groups attached to the N-atom are identical, then such amines are known as simple amines.

Simple Amines Example:

Mixed Amines: If the alkyl or aryl groups bonded to the N-atom are different, then such amines are known as mixed amines.

Mixed Amines Example:

Quaternary Ammonium Salts: Besides these three types of amines, there is another class of nitrogenous compounds containing quaternary N-atom.

• These compounds are known as tetraalkyl ammonium salts or quaternary ammonium salts.
• These compounds are produced by the substitution of 4 H -atoms of the ammonium salts by the alkyl group.

Quaternary Ammonium Salts Example: [(CH₃)₄N]⁺Cl^– (tetramethylammonium chloride)

[(C₂H₅)₂N(CH₃)₂]⁺OH^– (diethyl dimethylammonium

Structure Of Amines: Like ammonia molecules, the structure of amines is pyramidal. The central N-atom is sp³-hybridised.

• The sp³-hybrid orbitals of N-atom form three σ-bonds with H-atom or an alkyl group and the fourth sp³-hybrid orbital contains a lone pair of electrons.
• Since Ip-bp repulsion is more than bp-bp repulsion, the angle between any two H-atoms or alkyl groups is less than the expected value (109°28′)- The bond angle is generally 107-108°.

• Despite having structural chirality, 3° amines with formula R¹R²R³N: do not display optical activity.
• This is because, with the exchange of a small amount of energy (~25kJ-mol^-1), rapid interconversion between a pair of enantiomers occurs and hence, they exist as ± or dl -mixture.

• Rapid interconversion between a pair of enantiomers
• However, ammonium salts with the formula R¹R²R³R⁴N⁺X^– exhibit optical activity due to their chiral structure.

Nomenclature Of Amino Compounds

Nomenclature Of Aliphatic Amines

General Method (Common System): According to this system, the amines are named using the suffix ‘amine’ after the name of the alkyl group(s) present in the amine.

General Method Example:

In the case of simple secondary and tertiary amines, the prefixes di- and tri-respectively are added before the name of the alkyl group.

General Method Example:

In the case of mixed secondary and tertiary amines, the names of the alkyl groups attached to the N-atom are arranged in alphabetical order.

General Method Example:

IUPAC Method: According to this system, the amines are named by replacing ‘e’ from the name of the parent alkane with the suffix ‘amine’ i.e., primary amines are regarded as alkanamine. They are named by replacing ‘and’ from the name of the alkane derived based on the number of carbon atoms in the longest carbon chain containing the amino group ( —NH₂) with ‘anamine’.

IUPAC Method Example:

Secondary or tertiary amines are considered N-substituted derivatives of primary amines. The longest carbon chain attached to the nitrogen atom is taken as the alkyl group of primary amine. The other alkyl
groups are written before the name of the parent primary amine (1°) with the prefix ‘N’.

IUPAC Method Example:

Nomenclature Of Aromatic Amines: According to the conventional system of nomenclature, aromatic amines are called arylamines. The simplest aromatic amine is called aniline.

• Generally, substituted aromatic amines are considered derivatives of aniline. In some cases, special names are also used.

Aromatic Amines Example: o-/m-/p- methyl anilines are called o-/m-/ptoluidines while o-/m-/p- methoxy anilines are known as anisidines.

• According to the IUPAC system, the suffix ‘e’ of the arene is replaced by ‘amine’.

Aromatic Amines Example: Aminobenzene is named as benzenamine. The name Aniline is, however, accepted by IUPAC.

Isomerism In Amino Compounds

Amino Compounds Chain Isomerism: This type of isomerism arises due to the difference in the carbon chain attached to the amino group.

Amino Compounds Chain Isomerism Example:

Amino Compounds Position Isomerism: This type of isomerism occurs due to the difference in the position of the —NH₂ group i.e., a functional group in the carbon chain.

Amino Compounds Position Isomerism Example:

Functional Group Isomerism: In compounds having the same molecular formula, the presence of different classes of amino groups (1°, 2° or 3°) gives rise to this type of isomerism.

Example: The functional group isomers of C₃H₉N are:

Metamerism: In the compounds having the same molecular formula and belonging to the same class of amino compounds, the presence of different alkyl groups bonded to N-atom gives rise to this type of isomerism. Hence, secondary and tertiary amines exhibit this type of isomerism.

Example:

Methods Of Preparation Of Mixture Of Amines

By Ammonolysis: Hofmann’s Method

An alcoholic solution of NH₃, when heated with an alkyl halide in a closed glass tube at 100°C, produces a mixture of primary, secondary and tertiary amines along with quaternary ammonium salts. This reaction is known as ammonolysis.

Hofmann’s Method Example:

⇒ \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{I}+\mathrm{NH}_3 \longrightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2+\mathrm{HI}\)

⇒ \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2+\mathrm{C}_2 \mathrm{H}_5 \mathrm{I} \longrightarrow\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}+\mathrm{HI}\)

⇒ \(\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{NH}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{I} \longrightarrow\left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{~N}+\mathrm{HI}\)

⇒ \(\left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{~N}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{I} \longrightarrow\left[\left(\mathrm{C}_2 \mathrm{H}_5\right)_4 \mathrm{~N}\right]^{+} \mathrm{I}^{-}\)

1. In this reaction, amine (base) and HI (acid) combine C₂H₅NH₂ + HI → C₂H₅NH⁺₃I^– ). If the acid present in the reaction mixture is neutralised with the addition of excess base and then the resulting solution is distilled, a mixture of primary, secondary and tertiary amines is obtained in the receiving flask.
2. Primary, secondary and tertiary amines are separated from their mixture by fractional distillation or Hinsberg’s method.
3. In ammonolysis, the composition of the final mixture is determined by the initial mole ratio of the reactants—alkyl halide and ammonia. If excess ammonia is used, primary amine is obtained as the major product. If an excess of alkyl halide is used, tertiary amine is obtained as the major product.
4. In ammonolysis reaction, the order of reactivity of alkyl halides: R—I > R—Br > R—Cl
5. Ammonolysis is not effective in preparing arylamine due to the low reactivity of aryl halides towards nucleophilic substituents.

From Alcohols: Aliphatic amines of low molecular mass may be prepared industrially by passing a mixture of alcohol and ammonia in the vapour phase at high pressure over heated alumina or copper chromite as catalyst at 300° – 400°C to give a mixture of primary, secondary and tertiary amines.

From Alcohols Example: CH₃OH + NH₃ → CH₃NH₂ + H₂O

CH₃NH₂ + CH₃OH → (CH₃)₂NH + H₂O

(CH₃)₂NH + CH₃OH → (CH3)3N + H₂O

In this process, quaternary ammonium salt is not produced. If ammonia is used in excess, primary amine is obtained as the major product.

General Methods Of Preparation Of Amines

Preparation Of Primary Amines

Preparation Of Primary Amines By Reduction Of Nitro Compounds: Reduction of nitro compounds by Sn/HCl, Zn/HCl, Fe/HCl, H2/Ni or LiAlH₄ gives primary amines.

Preparation Of Primary Amines Nitro Compounds Example:

Preparation Of Primary Amines By Reduction Of Alkyl Cyanides: Alkyl cyanides on being reduced by H₂/Ni, LiAlH₄ or (Na + C₂H₅OH) yield primary amines.

Preparation Of Primary Amines Alkyl Cyanides Example:

Preparation Of Primary Amines By Reduction Of Acid Amides: Acid amides are reduced by sodium and ethanol or LiAlH₄ to give primary amines.

Preparation Of Primary Amines Example Acid Amides:

Preparation Of Primary Amines By Reduction Of Aldoximes Or Ketoximes: When aldoximes or Ketoximes are reduced by (Na + C₂H₅OH) or LiAlH₄, primary amines are produced.

Preparation Of Primary Amines By Reductive Amination Of Aldehydes And Ketones: Aldehydes or ketones react with a mixture of excess NH₃ and H₂ at 140°-150°C under high pressure, in the presence of Raney Ni, to form primary amines.

• The reaction occurs in two, steps. In the first step, amine is produced which is reduced by H₂ to yield primary amine.
• This process of converting a carbonyl compound into imine by treatment with ammonia and its subsequent reduction is called reductive amination.

Preparation Of Primary Amines By Hofmann Degradation Or Hofmann Bromamide Reaction: Acid amides (RCONH₂) react with Br₂ in the presence of alkali (NaOH, ArCONH₂ or KOH) at about 70°C to give primary amines. The amine formed has one C-atom less than the parent acid amide. The reaction also occurs in the presence of sodium or potassium hypobromite (NaOBr or KOBr).

Hofmann Bromamide Reaction Example:

This reaction is used to prepare lower members of different homologous series.

From Alkyl Halides: Gabriel Phthalimide Synthesis:

• This is a method of converting an alkyl halide to a 1° amine free from 2° and 3° amines. In this process, phthalimide is first converted into potassium phthalimide by reacting with ethanolic KOH.
• Potassium phthalimide on treatment with alkyl halide yields IV-alkyl phthalimide, which is hydrolysed by boiling with NaOH or KOH to give pure primary amines.

N-alkyl phthalimide can also be hydrolysed by HCl under heat and pressure to produce primary amine. Primary amines are also prepared by hydrazinolysis (cleavage by hydrazine) of N-alkylphthalimide. This method is more effective and efficient than acidic or alkaline hydrolysis.

It Is important to note that in Gabriel phthalimide synthesis, aromatic amines cannot be prepared using aryl halide instead of alkyl halide because aryl halide does not participate in nucleophilic substitution reaction.

Gabriel Phthalimide Synthesis Example: This method can be used to prepare α-amino acids.

From Grignard Reagent: Grignard reagent reacts with chloramine to form primary amine.

R—MgBr + Cl—NH₂ R —NH₂ + MgBrCI

It is an effective method for preparing primary amine in which the —NH₂ group is attached to a tertiary carbon atom.

From Grignard Reagent Example:

By Schmidt Reaction: Carboxylic acid reacts with hydrazoic acid (HN₃) in the presence of concentrated H₂SO₄ to give a primary amine, which contains one carbon atom less than the carboxylic acid.

By Schmidt Reaction Example:

By Hydrolysis Of Alkyl Isocyanides: At ordinary temperature, hydrolysis of alky! isocyanide by dilute HCl gives primary amine.

By Curtius Rearrangement: Acyl azide, on heating in an inert solvent (benzene, chloroform, etc.) gives alkyl isocyanate. The latter on hydrolysis yields lcamine. This reaction is called the Curtius reaction.

By Curtius Rearrangement Reaction Mechanism

By Lossen rearrangement: When heated with concentrated HCl or XaOH, hydroxamic acids undergo Lossen rearrangement forming a primary amine which involves the formation of an intermediate, alkyl isocyanate.

By Lossen rearrangement Example:

Hydroxamic acids exhibit tautomerism; keto form 1 is called hydroxamic form and enol form 2, a hydroximic form.

Preparation Of Secondary Amines

Secondary Amines From Alkyl Isocyanides: Alkyl isocyanides are reduced by H₂/Ni or (Na + C₂H₅OH) to form secondary amines.

From Primary Amines: Secondary amines are prepared by heating primary amines with the requisite amount of alkyl halides (preferably alkyl iodides).

From Primary Amines Example:

From Alkyl Halides: Alkyl halides when heated with aniline form dialkyl aniline. Dialkyl aniline on treatment with nitrous acid gives p nitroso-N, N-dialkyl aniline which on alkaline hydrolysis yields secondary amine.

From Alkyl Halides Example:

Using this process, secondary amines, free from primary and tertiary amines are produced.

Preparation Of Diethylamine From Ethyl Iodide:

Preparation Of Tertiary Amines

From Alkyl Halides: Tertiary amines are prepared by heating an excess amount of alkyl halide with an alcoholic solution of ammonia. In this case, the quantity of alkyl halide to be used should be more than the stoichiometric amount.

From Quaternary Ammonium Hydroxide: Quaternary ammonium hydroxide on heating gives tertiary amines.

Quaternary Ammonium Hydroxide Example:

Separation Of Primary, Secondary And Tertiary Amines By Hinsberg’s Method

This method is also used to distinguish between primary, secondary and tertiary amines. The mixture of amines when reacted with benzene sulphonyl chloride (Hinsberg’s reagent), primary and secondary amines form N-alkylbenzene sulphonamide and N, N-dialkyl benzene sulphonamide, respectively but tertiary amines do not react.

• The resulting mixture is made alkaline by adding a KOH solution. Consequently, N-alkylbenzene sulphonamide forms potassium salt which remains dissolved in the reaction mixture.
• N, N-dialkylbenzene sulphonamide does not react with KOH but remains in the mixture as an insoluble compound.

C₆H₅—SO₂—NHR + KOH → C₆H₅ —SO₂ —NKR + H₂O (soluble potassium salt)

C₆H₅—SO₂—NR₂ + KOH→ No reaction

• The alkaline mixture on distillation gives tertiary amine which separates as the distillate.
• The residual mixture left in the distillation flask is filtered and lV,iV-dialkylbenzene sulphonamide is obtained as residue. The filtrate on subsequent acidification gives Nalkylbenzenesulphonamide.

C₆H₅—SO₂—NKR + HCl → C₆H₅—SO₂—NHR + KCl

• N-alkylbenzene sulphonamide and N, A-dialkylbenzene sulphonamide are separately hydrolysed by 20% HCl or 70% H₂SO₄, to give primary and secondary amines, respectively.

At present, in the separation of an amine mixture, para toluenesulphonyl chloride is used, instead of benzenesulphonyl chloride.

Physical Properties Of Amine

Odour And Nature

Among aliphatic amines, lower members (for example., methylamine, ethylamine, dimethylamine) are gases having an ammoniacal smell but the higher members are volatile liquids with a fishy odour.

Boiling Point: Due to the presence of polar N—H bonds, all amines, except tertiary amines, are capable of forming H -bonds.

• The electron density of the N-atom in the secondary amine is more than that of the N-atom in the primary amine.
• This is due to the presence of two electron-repelling (+1 effect) alkyl groups attached to the N -atom in 2° amine.
• As a result greater polarity of N—H bond in primary amines is observed. Consequently, primary amines form stronger intermolecular H-bonds and have higher boiling points than secondary amines.
• The isomeric tertiary amines have the lowest boiling points as they cannot participate in intermolecular H-bond formation.

Intermolecular Primary amines (consequently effective molecular mass increases) Due to the presence of polar N—H bonds in their molecules, 1° and 2° amines (except 3° amines) can form H -bonds.

• The N—H bond is less polar than the O—H bond. So, intermolecular H -bonds in amines are weaker than those in alcohols and carboxylic acids.
• Therefore, the boiling points of amines are comparatively lower than alcohols and carboxylic acids of comparable molecular mass. But their boiling points are higher than those of the alkanes and ethers of comparable molecular mass.

Solubility: Amines of lower molecular masses are water-soluble because their molecules can form H-bonds with water molecules.

• With the Increase In molecular mass, the size of the hydrocarbon part of the amines becomes larger. Consequently, their solubility in water decreases.
• When several carbon atoms in an amine exceed 6, then the amine becomes insoluble In water. AmlncN of higher molecular masses is, however, soluble In organic solvents (e.g., alcohol, ether, benzene, etc.).

Basic Character Of Amines

Amines are regarded as organic bases. N-atoms present in their molecules contain lone pairs of electrons and hence, can accept protons. In reactions with water, they produce OH^– ions. In fact, they are as stronger bases than water.

⇒ \(\mathrm{R} \ddot{\mathrm{N}} \mathrm{H}_2+\mathrm{H}^{\oplus} \rightleftharpoons \mathrm{R}-\stackrel{\oplus}{\mathrm{N}} \mathrm{H}_3\)

⇒ \(\mathrm{RNH}_2+\mathrm{H}-\mathrm{OH} \rightleftharpoons \mathrm{R} \stackrel{\oplus}{\mathrm{N}} \mathrm{H}_3+\mathrm{OH}^{\ominus}\)

Strength Of Bases In Terms Of K_b and pK_b: In an aqueous solution, any base (B:) can establish the following equilibrium.

⇒ \(\mathrm{B}:+\mathrm{H}-\mathrm{OH} \rightleftharpoons \mathrm{BH}^{\oplus}+\mathrm{OH}^{\ominus}\).

∴ Equilibrium constant,

⇒ \(K=\frac{\left[\mathrm{BH}^{\oplus}\right]\left[\mathrm{OH}^{\ominus}\right]}{[\mathrm{B} ;][\mathrm{HOH}]}\)

⇒ \(K \times\left[\mathrm{H}_2 \mathrm{O}\right]=\frac{\left[\mathrm{BH}^{\oplus}\right]\left[\mathrm{OH}^{\ominus}\right]}{[\mathrm{B}:]} \text { or, } K_b=\frac{\left[\mathrm{BH}^{\oplus}\right]\left[\mathrm{OH}^{\ominus}\right]}{[\mathrm{B}:]}\)…1

[H₂O is present in large quantities in solution and it does not suffer any change quantitatively. So [H₂O] can be treated as a constant. In that case AT[H₂O] = constant (Kb) .]

The constant (K_b) in equation (1) is called the basicity constant. Taking negative logarithms on both sides of the equation,

⇒ \(p K_b=\log \frac{[\mathrm{B} ;]}{\left[\mathrm{BH}^{\oplus}\right]\left[\mathrm{OH}^{\Theta}\right]} \quad \cdots(2)\)

⇒ \(\left[p K_b=-\log K_b\right]\)

From equation (1), we see that the higher the value of K_b, the greater the concentration of OH^– ions and the stronger the base.

Similarly, the lower the value of K_b, the weaker the base. From equation (2) It is clear that the higher the value of pK_b , the weaker the base and vice-versa.

If the acidity constant of the conjugate acid (BH⁺) of any base (B:) is K_a, then it can be shown that with a decrease in the value of pK_a, the strength of the corresponding base decreases and vice-versa.

pk_b And pk_a Values Of Amines In Aqueous Solutions:

Basic Character Of Aliphatic Amines

Basic Of Amines In Aprotic Solvents: In nitrogenous bases, the higher the electron density on the N-atom, the more easily the nitrogen atom donates its lone pair of electrons to the proton. So, with an increase in electron density, the basicity of amine increases.

As the number of electron-repelling alkyl (methyl) groups attached to the N -atom in methylamine, dimethylamine and trimethylamine increases, basicity also increases, i.e., the increasing order of basicity is—

The basicity of the amines dissolved in aprotic solvents (for example., chlorobenzene) also increases with an increase in the number of alkyl groups attached to the N-atom. However, in an aqueous solution or any other hydroxylic or protein solvent, this trend is not observed.

Basicity Of Amines In Aqueous Solution: In this case, basicity depends on two factors. Firstly, the higher the electron density of the N-atom in the amine molecule, the greater the basicity of the amine.

• In other words, an increase in the number of electron-repelling alkyl groups linked to the N-atom results in an increase in the basicity of the amines.
• Hence, based on relative electron density, the basicity of ammonia and primary, secondary and tertiary amines is in the following order:

Secondly, the basicity of amines depends on the relative stability of the conjugate acids (cations) formed by the combination of the amines with protons. The greater the stability of the conjugate acid, the greater the basicity of the amine.

The conjugate acid produced from the primary amine attains maximum stability through intermolecular H-bond formation with water molecules while the cation formed by tertiary amine has the least stability.

• So, the order of stability of the conjugate acid (cation) in an aqueous solution is— RN⁺H₃ > R₂N⁺H₂ > R₃N⁺H. Thus, based on the stability of the cation, the basicity of the amines follows the order RNH₂> R₂NH>R₃N.
• The difference in basicity among primary, secondary and tertian’ amines can be explained based on the two opposing factors mentioned above, viz. electron density on N-atom and stability of the conjugate acid (cation).
• Considering these two factors, it has been observed that in aqueous solution the secondary amines are always stronger bases than both primary and tertiary amines.
• The difference between the basicities of primary and tertiary amines is relatively small. In some cases, the basicity of primary amines is more than that of tertiary amines while in some cases, the reverse order is observed.

For example, in an aqueous solution, the respective order of basicity of the methylamine and ethylamine series including ammonia, is as follows:

(CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃(C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃

Basic Character Of Aromatic Amines: Aromatic amines (for example., aniline, pK_b = 9.38 ) are much weaker bases than ammonia (pK_b = 4.75) and aliphatic amines (for example., ethylamine, pK_b = 3.33).

The decreased basicity of aromatic amines may be explained in the following way—

The Hybridisation Of The C-atom Attached To The Amino Group: The N-atom of an aromatic amine is bonded to an sp²-hybridised carbon atom of the aromatic ring.

• But in the case of aliphatic amines, the N-atom is linked to an sp³-hybridised carbon atom.
• It should be noted that the order of electron-attracting property and electronegativity of carbon atoms based on hybridisation is— C(sp)>C(sp²)>C(sp³).

Effect Of Resonance: The lone pair of electrons N-atom of aniline (the simplest member of aromatic amines) takes part in resonance or delocalisation with the electrons of a benzene ring. Consequently, the electron pair on ammo nitrogen becomes available to a lesser extent to combine with a proton. This reduces the basic character to a large extent.

Decrease In The Stability Of The Conjugate Acid Relative To The Aromatic Amine: As the electron pair on N-atom in conjugate acid formed by protonation of aniline is not available, it cannot participate in delocalisation with the π-electrons of the benzene ring. So, aniline exhibits the least tendency to combine with a proton.

Due to the absence of the above effects in aliphatic amines, their basicity is found to be much higher than aromatic amines. It is interesting to note that cyclohexyl amine —NH₂ having no aromatic ring displays strong basic properties (pK_b = 3.32) like aliphatic amines.

Comparison Of Basicity Of Different Amino Compounds

Diphenyl And Triphenyl Amines: With the increase in the number of aromatic rings attached to the N-atom of an amino group, electron density on the N-atom decreases with a consequent gradual decrease in the basicity of the amines. This is because the lone pair on N-atom participates in delocalisation with π-electrons associated with a large number of aromatic rings.

Benzylamine And Methylamine: In benzylamine, the amino group is connected to the benzene ring through the —CH₂ group. For this reason, it behaves as a strong base like aliphatic amines. Due to the -I effect of the phenyl group, it is a weaker base than methylamine.

N-methyl Aniline And N, N-dimethylaniline: As the number of electron-releasing methyl groups attached to the N-atom of aniline increases, electron density also increases, leading to an increase in basicity.

The Basic Strength Of o-, m- and p-substituted Aromatic Amines: The presence of electron-donating groups [for example., —CH₃, —NH₂, —OCH₃, —OH) in the aromatic ring increases the basic strength of the corresponding aromatic amines. This is because, these groups, with a few exceptions, increase the electron density on the N-atom or any nearest atom.

On the other hand, electron-attracting groups [for example., —NO₂, —CN, —X(halo) ] present in the aromatic ring decreases the basicity of the aromatic amines. This is because the electron-attracting groups diminish the electron density of the N-atom in the amino group.

Ortho-, meta- and para-toluidine (or methyl aniline): Due to ortho-effect, the basicity of ortho-toluidine is less than aniline. On the other hand, due to the +1 effect of the —CH₃ group, its meta-isomer i.e.„ meta-toluidine is slightly more basic than aniline.

Owing to the +1 effect coupled with the hyperconjugation effect of a methyl group, para-toluidine becomes more basic than the meta-isomer i.e., m-toluidine

ortho-, meta- and para-anisidine (or methoxy aniline): Due to the ortho-effect orthomethoxyaiuline is less basic than aniline.

• Again +R effect of the — OCH₃ group cannot influence the basic strength of meta-methoxy aniline because it causes no increase in electron density on the ring carbon attached to the —NH₂ group.
• However, only the -I effect of the —OCH₃ group is active in the case of meta-isomer. The net outcome is the least basicity of meta-isomer.
• In para-isomer, due to greater distance, the -I effect of the —OCH₃ group is not perceptible to an appreciable extent. However, due to the +R effect, it is found to be the most basic.

Ortho-, meta- and para-nitroaniline: Given the -I and -R effect of the —NO₂ group, nitroanilines are always found to be less basic than aniline.

• In the case of the o-isomer, due to the shorter distance, the -I effect of the NO₂ group is most effective.
• Besides, because of the electron-attracting -R effect and H bond formation, as depicted below, the o-isomer displays the least basicity.

• Due to the combined -I (more effective at a shorter distance) and -R effects of the —NO₂ group, the o-isomer is less basic than the p-isomer.
• In the case of the m-isomer, the -R effect of the —NO₂ group does not cause any reduction of basic character but due to the -I effect, basicity decreases significantly.
• Despite this, it exhibits more basic character than the p-isomer.

Chemical Properties Of Amines

Chemical Properties Of Primary Amines

Chemical Properties Reaction With Mineral Acids: Mineral acids react with primary amines to form salts.

Chemical Properties Reaction With Mineral Acids Example:

Chemical Properties Reaction With Alkyl Halide: Primary amines when treated with excess alkyl halides give successively secondary and tertiary amines. In the presence of a large excess of alkyl halide, quaternary ammonium salts are produced. This reaction is known as the alkylation of amine.

Chemical Properties Reaction With Alkyl Halide Example:

Reaction With Acetyl Chloride Gr Acetic Anhydride: Primary amines on treatment with acetyl chloride (CH₃COCI) or acetic anhydride [(CH₃CO)₂O] give acetyl derivatives. In this reaction, one H-atom of the amino group is replaced by the acetyl group. So, it is called acetylation reaction.

Acetic Anhydride Example:

Reaction With Benzoyl Chloride (Benzoylation): In an alkaline medium, primary amines react with benzoyl chloride where one H-atom of the —NH₂ group is replaced by a benzoyl (—COC₆H₅) group.

Reaction With Benzoyl Chloride Example:

Reaction With Benzenesulphonyl Chloride And P-Toluenesulphonyl Chloride: In the reaction of benzenesulphonyl chloride or paratoluenesulphonyl chloride with primary amines, N-alkyl sulphonamide is formed. These sulphonamides dissolve in KOH or NaOH forming soluble sodium or potassium salts.

Reaction With Benzenesulphonyl Chloride Example:

Reaction With Nitrous Acid: Nitrous acid (HNO₂) is an unstable acid, produced in situ in the reaction medium by the action of sodium nitrite and dilute HCl. Aliphatic primary amine in reaction with nitrous acid gives alcohol and N₂ gas.

Nitrous Acid Example:

Aromatic primary amines react with nitrous acid at low temperatures to form diazonium salts.

Nitrous Acid Example:

Reaction With Carbon Disulphide (CS₂): In the reaction of carbon disulphide with primary amines, dithiocarbamic acid is produced.

• This decomposition with mercuric chloride (HgCl₂) yields alkyl isothiocyanate (RNCS).
• Alkyl isothiocyanate has a pungent smell like mustard oil and hence, this reaction is known as the Hofmann mustard oil reaction.
• This reaction is used as an identification test of primary amines.

Carbylamine Reaction: Primary amine on heating with chloroform and alcoholic KOH solution yields alkyl isocyanide (RNC) or arylamine. This reaction is called arylamine reaction. Alkyl isocyanides have an extremely unpleasant smell. So, a primary amine can be easily detected by this reaction.

Carbylamine Reaction Example:

Reaction With Grignard Reagent: The two H-atoms attached to the N-atom of primary amines are highly reactive concerning the Grignard reagent. So, each molecule of primary amine reacts with two molecules of Grignard reagent to form two molecules of alkane.

RNH₂ + 2CH₃MgI → 2CH₄ (Methane) + RN(MgI)₂

Reaction With Aldehydes: The reaction of primary amines with aldehydes produces imines. The imine thus formed is called Schiff’s base.

Catalytic hydrogenation (reduction in the presence of Ni catalyst) of Schiff’s base gives secondary amines.

Methylation Of Primary Amines

Eschweiler-Clarke Reaction: When a 1° amine is heated with a mixture of formaldehyde and formic acid at 100°C, one H -atom of the amino group is replaced by a methyl group. Here, a 2° amine is obtained where a methyl group is attached to the N -atom.

This reaction is known as Eschweiler-Clarke methylation.

Oxidation: Primary amines are oxidised by KMnO₄ to form aldimine ketimine or nitroalkane, depending upon their structures.

Aldimines or ketimines on hydrolysis with dilute acid regenerate aldehydes and ketones.

Reaction With Transition Metal Ion: Primary and secondary amines react with transition metal ions to produce soluble coordination compounds.

Transition Metal Ion Example: AgCl dissolves in methylamine to form a complex

The electrophilic substitution reactions of aromatic 1° amine have been discussed later.

Chemical Properties Of Secondary Amines

Reaction With Mineral Acids: Like primary amines, secondary amines also react with mineral acids to form salts.

Reaction With Mineral Acids Example:

Reaction With Alkyl Halides: The reaction of secondary amines, with alkyl halides produces tertiary amines In the presence of an excess of alkyl halides, quaternary ammonium salts are formed.

Reaction With Alkyl Halides Example:

Reaction With Acetyl Chloride And Acetic Anhydride: Like primary amines, 2° amines also react with acetyl chloride or acetic anhydride to produce acetyl derivatives.

Acetyl Chloride And Acetic Anhydride Example:

Reaction With Benzenesulphonyl Chloride And Para-Toluenesulphonyl Chloride: Secondary amines react with benzene sulphonyl chloride or para-toluene sulphonyl chloride to form, N, N-dialkyl sulphonamide. These sulphonamides are insoluble in alkali because there is no H-atom attached to their N-atom.

Reaction With Benzenesulphonyl Chloride Example:

Reaction With Nitrous Acid (HNO₂): Secondary amines react with nitrous acid to form a yellow oily compound, N-nitrosamine. In this reaction, nitrogen gas does not evolve.

R₂N H + HO —N=O R₂N —N=O(N-nitrosoamine)+ H₂O

Reaction With Nitrous Acid Example:

N-nitrosamine on heating with dilute HCl decomposes to reproduce the secondary amine.

Reaction With Carbon Disulphide: Secondary amines react with carbon disulphide (CS₂) to form dithiocarbamic acid but unlike primary amines, it is not decomposed by mercuric chloride(HgCl₂).

Carbylamine Reaction: Secondary amines do not participate in this reaction.

Reaction with Grignard reagent: The H-atom attached to the N-atom of secondary amines is highly reactive with respect to the Grignard reagent. So, one molecule of Grignard reagent reacts with one molecule of secondary amine to liberate one molecule of alkane.

Reaction with aldehyde and ketone: Aldehyde and ketone having α-H react with secondary amines to produce enamine.

Methylation Of Secondary Amine

Eschweiler-Clarke Reaction: 2° amines are methylated on heating with a mixture of formaldehyde and formic acid at 100°C. One H-atom of the amino group is replaced by a methyl group. As a result, a tertiary amine is obtained where a methyl group is attached to the N-atom.

Oxidation: Secondary amines on oxidation by potassium permanganate give tetra alkylhydrazine.

2° amines when oxidised by Caro’s acid (H₂SO₅) give N, N-dialkylhydroxylamine.

Chemical Properties Of Tertiary Amines:

Reaction With Mineral Acids: Like primary and secondary amines, tertiary amines also react with mineral acids to form salts.

⇒ \(\mathrm{R}_3 \mathrm{~N}+\mathrm{HCl} \longrightarrow \mathrm{R}_3 \stackrel{\oplus}{\mathrm{N}} \mathrm{H} \stackrel{\ominus}{\mathrm{C}}\)

Reaction With Mineral Acids Example:

Reaction With Alkyl Halides: In the reaction of tertiary amines with alkyl halides, quaternary ammonium salts are formed.

Reaction With Alkyl Halides Example:

Reaction With Nitrous Acid (HNO₂): Tertiary amines dissolve in cold nitrous acid producing nitrite salts. Nitrogen gas is not evolved in this reaction.

Oxidation: Tertiary amines are not oxidised by KMnO₄ but are oxidised by Caro’s acid (H₂SO₅) to form amine oxide.

Due to the absence of any H-atom attached to N-atom, tertiary amines do not react with the following reagents:

1. Acetyl chloride and acetic anhydride,
2. Benzenesulphonyl chloride and para-toluene sulphonyl chloride,
3. Carbon disulphide,
4. Chloroform in the presence of alcoholic KOH (Carbylamine reaction)
5. Grignard reagent
6. A mixture of formaldehyde and formic acid (Eschweiler-Clarke reaction)
7. Aldehyde.

Exhaustive Methylation: When primary, secondary and tertiary amines are reacted with excess methyl iodide, quaternary ammonium salt is obtained as the end product. This process is known as exhaustive methylation.

Quaternary ammonium iodide reacts with moist silver oxide (AgOH) to give quaternary ammonium hydroxide which on heating decomposes to give an alkene and a tertiary amine. By identifying products, the initial amine can be determined.

Identification Of Different Types Of Amines

Different Types Of Amines Hinsberg’s Test: In this test, Hinsberg’s reagent, i.e., benzene sulphonyl chloride (C₆H₅SO₂CI) is added to the sample of amine. If a precipitate appears, then the reaction mixture is made alkaline with a KOH solution. Consequently, primary, secondary and tertiary amines display different chemical reactions.

Primary Amines react with the Hinsberg reagent to give a precipitate (N-alkyl benzene sulphonamide) which dissolves in the KOH solution.

With C₆H₅SO₂Cl, secondary amines give a precipitate (N, N-dialkyl sulphonamide), which remains insoluble in the KOH solution.

Tertiary amines do not react with C₆H₅SO₂Cl. Hence, no precipitate is formed.

Identification Of Primary Amines

Carbylamine Test: This test is employed to identify both aliphatic and aromatic primary amines. In this test, a sample of primary amine is warmed with chloroform and alcoholic KOH solution when alkyl isocyanide or arylamine (RNC) having an extremely unpleasant smell is produced.

Hofmann Mustard Oil Reaction: When a 1° amine is warmed with alcoholic CS₂ solution, followed by heating with HgCl₂, an oily liquid (alkyl isothiocyanate) having the pungent smell of mustard oil is formed.

Identification Of Secondary Amines

Liebermann’s Nitroso Test

1. In cold conditions, dilute HCl and NaNO₂ are added to a sample of secondary amine when a yellow oily liquid, Nnitrosoamine, is formed.
2. The oily substance is separated and heated with a small amount of phenol and a few drops of concentrated H₂SO₄. The mixture turns green.
3. The solution on dilution with water becomes red. The solution when made alkaline, with an aqueous solution of sodium hydroxide, turns deep blue.

R₂N — H + HO — N = O →  R₂N — N=O + H₂O

Comparison Among Primary, Secondary And Tertiary Amines:

Aromatic Amine: Aniline (C₆H₅NH₂)

When —NH₂, —NHR or —NR₂ group (where R = alkyl or aryl group) is directly attached to the aromatic ring, then the compounds formed are known as primary, secondary and tertiary aromatic amines, respectively.

• Aniline (C₆H₅NH₂) is considered the simplest member among all aromatic primary amines. In 1826, Unverdorben first prepared aniline by destructive distillation of a mixture of indigo and lime.
• The Portuguese name of indigo is anil and hence, the compound was named aniline in 1841.

Preparation Of Aniline

Laboratory Preparation Of Aniline

Principle: In the laboratory, aniline is prepared by reducing nitrobenzene with tin and concentrated HCl.

2C₆H5NO₂ + 3Sn + 12HCl→ 2C₆H₅NH₂ + 3SnCl₄ + 4H₂O

In the presence of excess acid, aniline exists as its hydrochloride salt [C₆H₅NH₂.HCl]. To get free aniline, the reaction mixture is made alkaline with an excess of sodium hydroxide.

Industrial Preparation Of Aniline

From Chlorobenzene: Aniline is commercially produced by heating a mixture of chlorobenzene and excess aqueous solution of ammonia to 250°-350°C in the presence of cuprous oxide as a catalyst, under high pressure (about 60 atm pressure). This is known as the Dow process. The ammonolysis reaction involves nucleophilic substitution via the formation of a ‘benzyne’ intermediate.

From Nitrobenzene: Aniline is also produced industrially by the

1. Reduction of nitrobenzene with iron, 30% HCl solution and
2. Catalytic (Raney nickel) hydrogenation of nitrobenzene.

Example:

Other Methods Of Preparation Of Aniline

From Phenol: Ammonolysis of phenol in a closed vessel in the presence of anhy. zinc chloride catalyst, at high temperature yields aniline.

From Benzamide: Benzamide on heating with bromine and NaOH (or KOH ) solution gives aniline. This is known as ‘Hofmann rearrangement’ or ‘Hofmann degradation.

Properties And Uses Of Aniune

Physical Properties Of Aniline

1. Freshly distilled aniline is a colourless oily liquid with an unpleasant smell. It has a boiling point of 184°C and is poisonous.
2. Aniline is almost insoluble in water but dissolves in organic solvents like alcohol, ether, benzene, etc.
3. Aniline cannot turn moist red litmus paper blue. So, the litmus experiment suggests that aniline is a neutral liquid.

Freshly distilled aniline is oxidised in the presence of light and air to form different coloured compounds and slowly assumes a brown colour.

• The boiling points of these compounds are much higher than aniline. When impure aniline is subjected to distillation, colourless pure aniline is obtained, leaving behind the coloured compounds in the distillation flask.

Comparison Of Solubility Of Aniline And Aliphatic Primary Amine In Water: A lone pair of electrons on the N-atom of the amino group conjugates with π-electrons of benzene ring through resonance, i.e., delocalisation of electron pair occurs.

• Consequently, the —NH₂ group becomes partially positively charged and the ring acquires partial negative charge. So aniline is a polar molecule.
• Its dipole moment is 1.70D. Owing to this delocalisation, the availability of lone electron pair on N-atom decreases. So aniline is incapable of forming an effective H-bond with water. Hence, aniline is almost insoluble in water.

• On the other hand, in any aliphatic amine for example., methylamine (CH₃NH₂), the lone electron pair on the nitrogen atom does not participate in delocalisation.
• Consequently, methylamine can form strong H -bonds with water through this lone pair of electrons.
• This causes high solubility of methylamine in water. (But electronegativity of nitrogen atom being high, relatively weaker moment due to -I effect acts in the direction of —NH₂ group.
• This moment neutralises to some extent the strong moment, caused by the +R effect which acts in the direction of the group towards the ring.)

Chemical Properties Of Aniline: Due to resonance, aniline displays the following chemical properties—O As a result of the +R effect in the amino ( —NH₂) group, the electron density at ortho and para positions of the ring, relative to meta-position increases.

• Hence, electrophilic substitution takes place mainly at ortho- and para-positions relative to the —NH₂ group and this reaction occurs more easily than that in benzene.
• The benzene ring is electron-rich due to resonance and hence, aniline is easily oxidised, producing varieties of coloured organic compounds. As a result of resonance, aniline is converted into a weaker base than aliphatic 1° amine (for example., methylamine, ethylamine, etc.).
• Any electron-releasing group such as —CH₃, —NH₂ present in the ring increases the basicity of aromatic amines while the presence of any electron-attracting group, e.g., —NO₂, —CHO decreases the basicity of aromatic amines.
• The electron-releasing groups push the electrons towards nitrogen and hence, the lone pair of electrons on the N-atom, necessary for bond formation with proton (H+) is more easily available than benzene.
• So an aromatic amine having an electron-releasing group is found to be more basic than aniline.
• Conversely, the electron-attracting group tends to shift the lone pair of electrons from N-atom.
• Consequently, the lone pair of electrons on the N-atom required for bond formation with proton is not so easily available as in the case of aniline.
• So, the basicity of an aromatic amine-containing electron-attracting group is less than that of aniline.

The chemical reactions of aniline may be divided into two classes, viz.,

1. Reactions of the amino group and
2. Reactions of the benzene ring. But as the benzene ring influences the reactions of the —NH₂ group, the reactions of the benzene ring are also influenced by the —NH2 group.

For example, the presence of a benzene ring enables aniline to form diazonium salt. —NH₂ group being present in the ring, increases the stability and in substitution reactions, H-atoms of ortho-/para-positions are substituted leading to ortfto-and para- compounds.

Reactions Of Amino (-NH₂) Group

Reaction Of Amino Salt Formation: Aniline being a weak base reacts with strong mineral acids to produce salts.

Salt Formation Example: In the reaction of aniline with cone. HCl and cone. H₂SO₄, crystalline solid aniline hydrochloride and aniline sulphate salts are respectively formed.

• Aniline does not dissolve in water, but these salts obtained from aniline are soluble in water. So, aniline dissolves in a dilute aqueous solution of HCl or H₂SO₄ with the formation of aniline salt.
• As these salts are produced in the reaction between weak base and strong acid, they undergo hydrolysis in aq. the solution, consequently producing an acidic solution.
• Aniline is liberated when an excess sodium hydroxide solution is added to these salts in cold conditions.

N-alkylation: When a mixture of aniline and methyl chloride is heated in a closed reaction vessel, hydrogen atoms of the —NH₂ group are replaced by methyl groups successively.

• This results in the consecutive formation of N-methylaniline (secondary amine), N, N-dimethylaniline (tertiary amine) and finally N, N, N-trimethylanilinium chloride (quaternary ammonium salt).
• This process is called N-methylation. Other alkyl halides also react similarly with aniline.

This quaternary salt on heating at 300°C undergoes rearrangement (Hofmann-Martius reaction) to form 2,4,6- trimethylanilinium chloride.

N-arylation: Aniline on being heated to 250°C with aniline hydrochloride in a closed vessel yields diphenylamine.

Acetylation: At ordinary temperature, aniline reacts with acetyl chloride when one H-atom of the —NH₂ group is replaced by an acetyl group (—COCH₃) forming acetanilide, a white crystalline solid with a melting point of 114°C. Acetanilide may also be produced when aniline is heated with acetic anhydride or glacial acetic acid. This reaction is called the acetylation reaction.

Benzoylation: When aniline is shaken with benzoyl chloride and excess sodium hydroxide solution, one H-atom of the —NH₂ group is replaced by a benzoyl (—COC₆H₅) group to form white crystalline benzanilide. This reaction is known as benzoylation. It is also known as the Schotten-Baumann reaction.

Formation Of Schiff Base: Aniline participates in the condensation reaction with aromatic aldehydes to form a class of compounds, called anils or Schiff bases.

Schiff Base Example: Heating a mixture of aniline and benzaldehyde to 125°C gives benzylidene aniline (a Schiff base).

Schiff bases are easily hydrolysed. So, in certain reactions, (for example., nitration), the —NH₂ group may be protected by forming Schiffbase.

Reaction With Phosgene: Aniline reacts with phosgene to give phenyl isocyanate.

Reaction With Carbon Disulphide: When aniline is heated with caustic potash powder and an ethanolic solution of carbon disulphide, N, N’-diphenyl thiourea (thiocarbamide) is produced.

Carbylamine Reaction: When aniline is heated with a mixture of chloroform and alcoholic KOH solution, phenyl isocyanide or phenyl carbylamine having nauseating and unpleasant smell is formed. This reaction is called arylamine reaction. Aniline may be detected with the help of this reaction.

Diazotisation: When a cold aqueous solution of sodium nitrite is added to a cold acidified solution of any primary aromatic amine, benzene diazonium salt is produced.

Diazotisation Example: When aniline dissolved in dilute hydrochloric acid is cooled to 0-5°C and sodium nitrite solution is added to it slowly with constant stirring, a pale yellow-coloured solution of benzenediazonium chloride is formed. This reaction is called diazotisation. Nitrous acid produced in the reaction between NaNO₂ and HCl reacts with aniline.

In the reaction of nitrous acid with aliphatic primary amine (RNH₂), alcohol is formed with the liberation of nitrogen gas.

Oxidation: As the benzene ring of aniline is electron-rich, it is easily oxidised by various oxidising agents. The nature of the products obtained on oxidation depends on the oxidising agents.

Oxidation Example: Aniline when oxidised by chromic acid or MnO₂/ H₂SO₄ yields p-benzoquinone. Trifluoroperacetic acid oxidises aniline to nitrobenzene while personal sulphuric acid (Caro’s acid) oxidises it to nitrosobenzene.

Electrophilic Substitution Reaction Of Benzene Ring

Halogenation Amino (—NH₂) group sufficiently increases electron density at ortho- and para-positions of the ring through the +R effect. So, the ring becomes so active towards electrophilic substitution reaction that all the H-atoms of ortho- and para-positions can be substituted by halogens without any halogen carrier.

Chlorination: At ordinary temperature, chlorine water reacts with aniline to give a white precipitate of 2,4,6- trichloroaniline.

Bromination: At ordinary temperature, bromine water reacts with aniline to give a white precipitate of 2,4,6-trihydroaniline.

lodination: Iodine reacts with aniline in the presence of aq. sodium bicarbonate solution to form p-iodoaniline.

Substitution in aniline ring in a particular position (Selective substitution): For preferential substitution at a particular position of aniline ring, the high reactivity of the ring must be reduced. For this purpose, the —NH₂ group is converted into —NHCOCH₃ by acetylation.

• The lone pair of electrons on the nitrogen atom of the —NHCOCH₃ group is attracted by the adjacent carbonyl (C=O) group through resonance.
• This consequently interferes with its participation in resonance with π-electrons in the ring, thereby reducing its reactivity.

Substitution in aniline ring Example: Preparation of p-bromoaniline:

Preparation Of o-bromoaniline:

Preparation Of m-bromoaniline:

Nitration: The benzene ring in aniline is highly activated due to the presence of the — NH₂ group. Hence, direct nitration of aniline with a mixture of cone. HNO₃ and cone. H₂SO₄ gives rise to a complex mixture of mono-, di- and trinitro compounds.

• Moreover, the electron-rich aniline ring being highly susceptible to oxidation, gets oxidised by a strong oxidising agent like HNO₃, to yield different products. So, direct nitration of aniline is not carried out.
• To prevent oxidation by deactivating the benzene ring, aniline is first converted to acetanilide by acetylation.
• Nitration of acetanilide with mixed acid (concentrated HNO₃ and concentrated H₂SO₄) results in the formation of ortho- and para-nitroacetanilide.
• Due to the large size of —NHCOCH₃ (steric hindrance), p-nitro acetanilide (90%) Is obtained as the major product.
• The two isomers thus obtained are separated by crystallisation from ethanol. The para-isomer on being hydrolysed by acid or alkali produces p-nitroaniline.

Thus amino groups should be protected by acetylation before carrying out nitration of aniline.

Preparation Of O-nitroaniline: When the p-position of acetanilide is blocked by sulphonation and nitration, the product formed is hydrolysed and desulphonated by heating with dilute H₂SO₄, to obtain o-nitroaniline.

Preparation Of M-nitroaniline: m-nitroaniline is produced by selective reduction of one nitro group of dinitrobenzene using an ethanolic solution of ammonium hydrogen sulphide, m-nitroaniline may also be prepared by nitration of aniline in presence of 98% H₂SO₄.

In the presence of conc.H₂SO₄, the —NH₂ group acts as an electron-attracting (-I effect) & mete-directing —N⁺H₃ group, consequently producing m-nitroaniline as the chief product.

Sulphonation: When aniline is heated with fuming sulphuric acid (sulphuric acid containing 10% SO₃) at 180°C or with concentrated H₂SO₄ at 200°C, p-amino benzene sulphonic acid or sulphanilic acid is produced as the major product.

In this reaction, anilinium hydrogen sulphate salt is first formed, which eliminates a molecule of water to form phenylsulphamic acid, which rearranges to give sulphanilic acid.

Uses Of Aniline

1. Aniline is widely used in the preparation of various azo dyes and as an antioxidant and vulcanisation accelerator in the rubber industry.
2. It is also extensively used in the synthesis of a wide range of important organic compounds like hydroquinone, acetanilide, sulphanilic acid, nitroaniline, etc.
3. It is also used in the pharmaceutical and plastic industries.

Tests For Aromatic Amino Group: Identification Of Aniline

Azo-Dye Test: Aniline dissolved in dilute HCl solution is cooled to 0-5°C, to which, aqueous sodium nitrite solution is added. A few drops of this solution are then slowly added to a cold solution of β-naphthol dissolved in sodium hydroxide. As a result, a scarlet red azo-dye is formed,

Azo-Dye Test Reaction Mechanism: It is an electrophilic substitution reaction, where benzene diazonium ion, C₆H₅N⁺₂ (produced in the reaction of aniline with NaNO₂/HCl; i.e., HNO₂) acts as an electrophile.

Distinguishing Test For Aliphatic And Aromatic 1° Amine:

Carbyiamine Or Isocyanide Test: A few drops of aniline, mixed with a few drops of chloroform and alcoholic caustic potash solution is heated in a test tube. Phenyl isocyanide or phenyl carbamide having an obnoxious smell, is produced. (This test is also applicable for the aliphatic —NH₂ group.)

Bromine Water Test: A few drops of aniline are taken in a test tube to which excess bromine water is added. The mixture on shaking gives a white precipitate of 2,4,6-tribromoaniline.

Diazonium Salts (ArN₂X)

A primary aromatic amine reacts with nitrous acid (NaNO₂ and HCl) in cold conditions (0-5°C) to form an unstable compound known as diazonium salt.

• The general formula of this class of compounds is ArN⁺₂X^–, where Ar = aryl group for example., phenyl (C₆H₅—), p -nitrophenyl (p-NO₂C₆H₄—) etc., and X^– = acid anion, for example., Cl^–, HSO^–₄, BF^–₄, Br^–, etc.
• The functional group of this class of compounds is diazonium ion (—N⁺=N). Since the molecules of these compounds contain two nitrogen atoms and their nature is very much similar to that of ammonium salts, these compounds are known as ‘diazonium salts’.

Diazonium Salts Example:

Preparations Of Diazonium Salts

Laboratory preparation of C₆H₅N₂CI

Principle: Aniline, dissolved in dilute hydrochloric acid is cooled to 0-5°C to which, a cold aqueous solution of sodium nitrite is added. A reaction occurs to form benzene diazonium chloride.

• The reaction involves two steps. In the first step, nitrous acid is produced in the reaction between sodium nitrite and HCl.
• In the second step, aniline reacts with nitrous acid and HCl to yield benzene diazonium chloride.

Laboratory preparation Of Principle Reaction Mechanism

Since diazonium salts slowly decompose even at low temperatures (0-5°C), they are used immediately after preparation.

Diazotisation: The process which involves the reaction of primary aromatic amines with dilute mineral acids and sodium nitrite in ice-cold conditions (0-5°C) to form diazonium salts is called diazotisation. In 1858, Peter Griess discovered this reaction in Hofmann’s laboratory in London.

Causes Of Stability Of Aromatic Diazonium Salts: Diazonium salts, both aliphatic and aromatic, are unstable because of the extra stability of nitrogen gas (N₂) produced when they are decomposed.

• But all the aromatic diazonium salts including the benzenediazonium chloride, are much more stable than the aliphatic diazonium salts.
• So, in cold conditions, aromatic diazonium salts can be prepared but not aliphatic diazonium salts.
• The dispersal of the positive charge of diazonium cation into the benzene ring brings stability to the diazonium salts, i.e., these salts are stabilised by resonance.

The resonating structures of benzene diazonium cation may be represented as follows:

Aliphatic Diazonium Salts Cannot Be Stabilised Through Resonance. So, they are extremely unstable and easily decompose with the evolution of N₂ gas-forming carbocation. In the reaction of this carbocation with water, alcohol is produced.

Aliphatic Diazonium Salts Example: Ethylamine reacts with NaNO₂/HCl to give ethanol.

If aliphatic amines contain electron-withdrawing groups like —CN, —COR, and — COOR they can be converted into aliphatic diazo compounds. Thus, ethyldiazoacetate (CHN₂COOC₂H₅) may be readily obtained by treating a cold solution of the hydrochloride of ethylglycine ester with a cold NaNO₂ solution.

C^–lH₃N⁺CH₂COOC₂H₅ + NaNO₂ → CHN₂COOC₂H₅ + NaCl + 2H₂O

Properties Of Diazonium Salts

Physical Properties Of Diazonium Salts

1. Benzene diazonium salts are generally colourless crystalline solids.
2. Most of the diazonium salts, especially nitrates, are explosives in the solid state. They decompose with explosions when heated or at the slightest impact. So, these salts, except aryldiazonium fluoroborate (ArN⁺₂ B^–F₄), cannot be separated in the dry state. Hence, all important reactions of diazonium salts are carried out in aqueous medium.
3. They are highly soluble in water and dissociate into Ions in aqueous solution.

Chemical Properties Of Diazonium Salts: Diazonium salts are chemically very reactive. Like Grignard reagent, it is a very important synthetic reagent, being the starting material for the preparation of various aromatic compounds, dyes and drugs. The reactions of these salts are of three types. They are discussed in the following table—

Substitution Reaction

Substitution By Hydroxyl (—OH) Group: When diazonium salt solution is slowly added to boiling water acidified with dilute sulphuric acid or when highly acidified diazonium salt solution is heated, the salt is hydrolysed leading to replacement of the diazo group (—N⁺₂) by a hydroxyl group (—OH). For example, in this reaction, benzene diazonium chloride is converted into phenol.

Substitution By Hydroxyl Group Example:

Substitution By Hydroxyl Group Reaction Mechanism: It is an aromatic S_N1 reaction. At higher temperatures, benzene diazonium cation decomposes to liberate N₂ gas and thereby produces phenyl cation which in turn combines with water to form phenol.

Substitution By Hydrogen Atom: When heated in the presence of hypophosphorus acid (H3P02), die diazonium salt is reduced and the diazo group (—N₂) is replaced by hydrogen. For example, benzene diazonium chloride is converted into benzene in this reaction.

Hydrogen Atom Example:

The Diazo group may also be replaced by hydrogen by heating diazonium salts with ethanol.

Hydrogen Atom Example:

Deamination: The removal of the —NH₂ group from the benzene ring by diazotisation of primary aromatic amine, followed by reduction (substitution of diazonium group by H-atom) is called deamination. This process is extensively used in the synthesis of different organic compounds.

Deamination Example: Synthesis of 1,3,5-tribromobenzene from aniline.

Substitution By Halogen: Sandmeyer Reaction

Substitution By Chlorine Or Bromine: On treating aromatic diazonium salt solution with a solution of cuprous chloride dissolved in hydrochloric acid or a solution of cuprous bromide dissolved in hydrobromic acid, diazo group (—N⁺₂) is replaced by chlorine or bromine and N₂ gas is liberated.

• In this case, cuprous salt acts as the catalyst.
• This Reaction Is Called The Sandmeyer Reaction.

Chlorine Or Bromine Example: When benzene diazonium chloride solution is added to cuprous chloride dissolved in HCl, chlorobenzene is obtained as the product. Similarly, benzene diazonium chloride when added to cuprous bromide dissolved in HBr, bromobenzene is obtained as the product.

Gattermann reaction Diazonium salt when heated with MCI or HBr in the presence of copper powder gives chlorobenzene or bromobenzene. This reaction is called the Gattermann reaction which is a simplified form of the Sandmeyer reaction.

• Substitution by fluorine: Arenediazonium fluoroborate on gentle heating yields aryl fluoride. This is the best method for preparing aryl fluoride. This reaction is known as the Schiemann reaction.
• When fluoroboric acid is added to a diazonium salt solution, insoluble fluoroborate is precipitated.
• It is separated, dried and then heated slowly when aryl fluoride is obtained.

Chlorine Or Bromine Example: Benzenediazonium fluoroborate obtained from benzenediazonium chloride is heated gently to form fluorobenzene.

Substitution By Iodine: Benzene diazonium salt solution when heated with KI solution, produces aryl iodide. This is the best method for introducing iodine atoms Into the benzene ring.

Substitution By Iodine Example: Benzene diazonium chloride solution when heated with KI solution, produces iodobenzene.

Substitution By Cyano (—CN) Group: In the reaction of diazonium salt solution with cuprous cyanide dissolved in an aqueous solution of KCN or with an aqueous solution of KCN in the presence of copper powder, the diazo group is replaced by cyano group and aryl cyanide is produced.

• This substitution reaction by the cyano group is, in fact, a special form of Sandmeyer and Gattermann reactions.
• Cyanobenzene (benzonitrile) may be prepared from benzene diazonium chloride or sulphate with the help of this reaction.

Substitution By Cyano Group Example:

Preparation Of Benzoic Acid From Aniline:

Substitution By Nitro (—NO₂) Group: Diazotisation of primary aromatic amine with fluoroboric acid and sodium nitrite gives rise to arene diazonium fluoroborate salt. This salt decomposes In the presence of an aqueous solution of sodium nitrite and copper powder, thus replacing the —N₂BF₄ group with the nitro group.

Substitution By Nitro Group Example: This reaction is used to prepare o and p-dinitro benzene which cannot be produced easily by general methods.

Synthesis Of P-dinltrobenzene From Aniline:

Substitution By Aryl (— Ar) Group: This type of substitution may be carried out by the addition of an aromatic compound to the alkaline solution of diazonium salt or ethanol and copper to diazonium hydrogen sulphate solution. The first method is known as the Gomberg reaction.

Substitution By Aryl Group Example:

Schematic Diagram Of Different Substitution Reactions Of Diazonium Salts

Reduction Reactions

Reduction By SnCl₂ In Acidic Medium: Diazonium salt when reduced by SnCl₂/HCl or sodium sulphite (Na₂SO₃) gives aryl hydrazine hydrochloride.

Acidic Medium Example: Benzene diazonium chloride in this reaction produces phenylhydrazine hydrochloride.

Reduction By Zn/Concentrated HCl: Reduction of the N—with N present agent diazoniums like zinc/conc.salt with HCl the formation of aromatic amine and NH₃.

Reduction By Zn/Concentrated HCl Example: Benzene diazonium chloride on reduction gives aniline and NH₃

Coupling Reactions: Diazonium salts readily combine with phenols, naphthols and aromatic amines to form brightly coloured (orange, red or yellow) azo compounds.

• In azo compounds, two aromatic rings are linked together by the diazo group (—N=N—). These reactions are called coupling reactions.
• Phenols couple readily in weakly alkaline solution (the rate of coupling increases with a change in pH from 5 to 8).
• Coupling reactions of aromatic amines are carried out in weakly acidic solutions, but not in strongly acidic solutions (the rate of coupling decreases as pH changes from 6 to 2).

Coupling Reaction With Phenol: When a cold solution of benzene diazonium chloride is added to a cold alkaline solution of phenol, a coupling reaction occurs and an orange azo compound called p-hydroxy azobenzene is produced.

Due to steric hindrance, coupling preferentially occurs in the p-position of the hydroxyl group. But if this position is blocked, then o-coupling occurs. For example, o-cresol gives an o-azo compound. If both the positions are blocked, then coupling reactions cannot occur.

Coupling Reaction With β-Naphthol: When a cold solution of benzenediazonium chloride is added to a cold alkaline solution of β-naphthol a brilliant red azo-dye is produced.

Coupling occurs at a or 1-position of α-naphthol. In an alkaline solution, this azo dye exists as sodium salt. Primary aromatic amines can be easily identified with the help of this reaction.

Reaction Mechanism: The coupling reaction is an electrophilic substitution reaction where the diazonium cation acts as an electrophile.

Coupling Reactions With Amines: Coupling with tertiary amine: In neutral or weakly acidic solution, N, N -dimethylaniline undergoes a coupling reaction with benzene diazonium chloride.

• The coupling occurs at the carbon atom (C-coupling) of the ring to form p-( N, N -dimethylamino) azobenzene.
• As the nitrogen atom of N, N-dimethylaniline does not contain any hydrogen atom, N-coupling does not occur.

Coupling With Secondary Amine: Coupling with secondary amine generally occurs at the nitrogen atom of the amine i.e., N-coupling occurs. Some C-coupling also occurs.

Coupling With Secondary Amine Example: N-methyldiazoaminobenzene is produced in the reaction between N-methylene and benzene diazonium chloride. At the same time, due to the C-coupling in the ring, some amount of aminoazobenzene is also produced.

It should be noted that o-phenylenediamine (1,2-diaminobenzene) reacts with NaNO₂/HCl at low temperature when one —NH₂ group is diazotised, which undergoes N-coupling with the other —NH₂ group leading to the formation of cyclic diazoamino compounds.

Coupling With Primary Amine: In the case of 1° amines, coupling occurs mainly at N-atom (AT-coupling).

Coupling With Primary Amine Example: In the reaction of benzene diazonium chloride with aniline, diazoaminobenzene is produced.

• During the diazotization reaction, if the acidity of the solution is low, then auto-coupling occurs and yellow diazoaminobenzene Is precipitated.
• Diazoaminobenzene when hydrochloride or with HCl, undergoes rearrangement producing p-aminoazobenzene (C -coupling).

Coupling With Primary Amine Reaction Mechanism: In the case of tertiary amine, C-coupling is an electrophilic substitution reaction.

For primary and secondary amines, N-coupling occurs in the following way:

Transformations

Distinctive Chemical Tests

Methyl Cyanide And Methyl Isocyanide:

Methyl Cyanide And Acetamide:

Ethylamine And Dimethylamine:

Methylamine And Aniline:

Dimethylamine And Trimethylamine:

Nitromethane And Methylamine:

Aniline Hydrochloride And P-chloroaniline:

Benzamide And Acetanilide:

Acetamide And Benzamide:

Cyanobenzene And Aniline:

Aniline And Benzylamine:

Nitrobenzene And Aniline:

Class 12 Chemistry Unit 13 Organic Compounds Containing Nitrogen Very Short Questions And Answers

Question 1. Write the structural formula of vinyl cyanide and its IUPAC name.
Answer:

Structural formula: CH₂=CH—CN,

IUPAC name: prop-2-enenitrile

Question 2. Can 3° amine participate in the acetylation reaction?
Answer:

The tertiary or 3°amine (R₃N) molecule does not contain any replaceable hydrogen atom attached to the N-atom. Hence, 3° amine does not take part in acetylation reaction.

Question 3. Why does the preparation of nitrobenzene necessitate strict control of temperature?
Answer:

In the preparation of nitrobenzene, it is essential to control the temperature because if the temperature exceeds 60°C, m -dinitrobenzene, instead of nitrobenzene is obtained.

Question 4. How will you convert p-H₂NC₆H₄NH₂ into p-O₂NC₆H₄NO₂ in a single step?
Answer:

p-H₂NC₆H₄NH₂ can be converted into p-O₂NC₆H₄NO₂ In a single step by oxidation with peroxy trifluoroacetic acid

Question 5. Name the experiment through which aliphatic and aromatic amines can be distinguished.
Answer:

By the azo-dye test, aliphatic and aromatic amines can be distinguished from each other.

Question 6. Write the IUPAC name of an alkyl cyanide containing 5 carbon atoms and only primary hydrogens.
Answer: 2,2-dimethylpropanenitrile.

Question 7. Why tetramethylammonium iodide is not basic?
Answer: There is no lone pair of electrons on N -atom

Question 8. What is enamine?
Answer: Compound formed by the reaction between secondary amine and aldehyde having α-H atom

Question 9. Which class of amine is formed by Gabriel phthalimide synthesis?
Answer: Primary amine

Question 10. Which class of amine is obtained by reducing alkyl isocyanide?
Answer: Secondary amine

Question 11. Ethyl nitrite belongs to which class of compounds?
Answer: Ester (involving mineral acid)

Question 12. Give examples of 1° amine containing 2° and 3° alkyl groups
Answer: (CH₃)₂CH —NH₂,(CH₃)₃C—NH₂

Question 13. Why does tertiary nitroalkane not exhibit acidic properties?
Answer: Due to the absence of α-H, they do not exist as acids

Question 14. Give an example of a compound where the Mulliken-Barker test can not be applied for the detection of the nitro group.
Answer: p-nitrobenzaldehyde (p-O₂NC₆H₄CHO)

Question 15. Diazonium salts are not generally separated in a solid state—why?
Answer: Solid diazonium salts are explosive

Question 16. How aliphatic and aromatic 1° amines are distinguished?
Answer: Azo-dye test

Question 17. How will you convert isocyanide into its isomeric cyanide?
Answer: Prolonged heating (rearrangement)

Question 18. What happens when quaternary ammonium hydroxide is heated?
Answer: Tertiary amine

Question 19. What is Hinsberg reagent?
Answer: Benzenesulphonyl chloride (C₆H₅SO₂Cl);

Question 20. Why is benzene diazonium chloride not stored and used immediately after its preparation?
Answer: Benzene-diazonium chloride is unstable. So it is not stored but used immediately after its preparation.

Question 21. Why does acetylation of the — NH₂ group of aniline reduce its activating effect?
Answer:

On acetylation, aniline forms acetanilide (C₆H₅NHCOCH₃) in which the lone pair on N is involved in delocalization not only with the benzene ring but also with the adjacent carbonyl group. Thus, acetylation of the —NH₂ group of aniline reduces its activating effect.

Question 22. Explain why MeNH₂ is a stronger base than MeOH.
Answer:

Nitrogen is less electronegative than oxygen. So, a lone pair of electrons on N is more easily available for donation to a proton. Hence, MeNH₂ is more basic than MeOH.

Question 23. What is the role of pyridine in the acylation reaction of amines?
Answer: Pyridine acts as a basic catalyst and removes HCl.

Question 24. What is the structure and IUPAC name of the compound, allyl amine?
Answer:

Question 25. Write down the IUPAC name

Answer: N,N -dimethylbenzenamine.

Question 26. A compound Z with molecular formula C₃H₉N reacts with C₆H₅SO₂Cl to give a solid, insoluble alkali. Identify Z.
Answer:

• The compound Z (C₃H₉N) reacts with C₆H₅SO₂Cl to form a solid (i.e., sulphonamide) which is insoluble in alkali.
• This shows that there is no H-atom on the N-atom of the sulphonamide. Thus ‘Z’ is a 2° amine with the structure CH₃CH₂NHCH₃ (N-methylethanamine).

Question 27. A primary amine, RNH₂ can be reacted with CH₂X to get a secondary amine, RNHCH₃ but the only disadvantage is that 3° amine and quaternary ammonium salts are also obtained as side products. Can you suggest a method where RNH₂ forms only 2° amine?
Answer:

Question 28. Complete the following reaction,
Answer:

Question 29. Why is aniline soluble in aqueous HCl?
Answer: Aniline (base) reacts with HCl to form the salt, anilinium chloride, which is water soluble

Question 30. Suggest a route by which the following conversion can be accomplished.
Answer:

Question 31. Identify A and B in the reaction.
Answer:

Class 12 Chemistry Unit 13 Organic Compounds Containing Nitrogen Short Questions And Answers

Question 1. An organic compound with formula C₃H₉N reacts with Hinsberg’s reagent to give a product, which is insoluble in alkali but dissolves in ether. Identify the compound.
Answer:

A secondary amine reacts with Hinsberg’s reagent to form N, N-dialkyl sulphonamide which is insoluble in alkali but soluble in ether. Hence, the compound with the formula C₃H₉N is CH₃CH₂NHCH₃ (Ethylmethylamine).

Question 2. How will you prepare ethylamine, free-form diethyl and triethyl amine from ethyl iodide?
Answer:

By Gabriel-phthalimide synthesis, ethylamine, not contaminated by diethyl and triethyl amine, is prepared.

Question 3. Why is methylamine a stronger base than ammonia?
Answer:

• Molecules of both ammonia and methylamine contain a lone pair of electrons on the N-atom. So both act as Lewis bases.
• But due to the presence of an electron-repelling methyl group, attached to the N-atom in methylamine, the electron density on the N-atom increases, i.e., the availability of electrons on the N-atom increases.
• Hence, CH₃NH₂ can form a bond with the proton by donating its lone pair to N-atom with greater ease than NH₃. So CH₃NH₂ behaves as a stronger base than NH₃

Question 4. Silver chloride is insoluble in water but soluble in ethylamine—explain.
Answer:

Silver chloride is insoluble in water but when it is heated with ethylamine, it forms a complex salt which is soluble in water.

Question 5. Write the names and structural formulae of the isomeric amines having molecular formula C₄H₁₁N.
Answers:

Question 6. Starting from n -butyric acid how will you prepare n-propylamine?
Answer:

Question 7. How will you transform methylamine into dimethyl amine, free from trimethyl amine?
Answer:

Question 8. Between methyl cyanide and methyl isocyanate which one is more soluble in water and why?
Answer:

N-atom in the molecule of methyl cyanide is partially negatively charged. So, it can easily participate in the hydrogen bond formation with the water molecules. Hence, methyl cyanide shows sufficient solubility in water.

On the contrary, the N-atom in the molecule of methyl isocyanide becomes partially positively charged and hence, it fails to form an H-bond with water molecules. So it is almost insoluble in water.

Question 9. Can acetonitrile exhibit acidic properties? If so, explain with an example.
Answer:

Acetonitrile reacts with a strong base like sodamide, thus removing a proton from α-carbon. In the presence of a strong base, acetonitrile (methyl cyanide) displays its mild acidic character. The mild acidic character of acetonitrile is due to the stability of its conjugate base through resonance.

Question 10. Why does an aqueous solution of methylamine give a precipitate of ferric hydroxide with ferric chloride?
Answer:

In aqueous solution, methylamine establishes the following equilibrium:

OH^– ion present in the solution reacts with Fe³⁺ ion, produced by the dissociation of FeCl₃, to give a brown precipitate of ferric hydroxide.

Question 11. Why is it difficult to prepare pure primary amine by ammonolysis of alkyl halide?
Answer:

Ammonolysis of alkyl halides produces a mixture of primary, secondary, tertiary and quaternary ammonium salts. Isolation of primary amine from the mixture is very difficult. Hence, the preparation of pure primary amine by the ammonolysis of alkyl halide is quite difficult.

Question 12. Give a suitable method for the preparation of tertiary butylamine from tertiary butyl bromide.
Answer:

Tertiary butyl bromide is first converted into Grignard reagent which on subsequent treatment with chloramine produces tertiary butylamine.

Question 13. Out of methyl cyanide and methylamine, which one is more basic and why?
Answer:

• Nitrogen atoms of methyl cyanide and methylamine are respectively sp- and sp³-hybridised. The electronegativity of sp-hybridized N -atom is greater than that of sp³-hybridised N -atom.
• So, the electron density on the N-atom of methylamine (CH₃NH₂) is more than that on the N-atom of methyl cyanide (CH₃—C=N:).
• Moreover, due to the attachment of an electron-repelling methyl group directly to the N-atom in methylamine, the electron density of the N-atom is further enhanced.
• This concerted effect eventually imparts a more basic character to methylamine than methyl cyanide.

Question 14. What happens when the product obtained in the reaction of tetramethyl ammonium iodide with moist silver oxide is heated?
Answer:

When tetramethyl ammonium iodide reacts with moist silver oxide, tetramethyl ammonium hydroxide is obtained. The latter when heated in dry conditions, produces trimethylamine and methyl alcohol.

Question 15. Although aniline is a colorless liquid, it is generally found as a brown-colored liquid—why? How is pure colorless aniline obtained from this impure variety of aniline?
Answer:

• Any organic compound having an electron-rich center is prone to oxidation by aerial oxygen. The nitrogen atom of aniline contains a lone pair of electrons and the —NH₂ group enhances the electron density of the ring through resonance.
• So both —the NH₂ group and the ring are susceptible to oxidation. Therefore, aniline gets oxidized easily, producing different colored oxidized products.
• The colored aniline when distilled using an air condenser gives pure, colorless aniline as a distillate.

Question 16. What happens when a cupric chloride or copper sulfate solution is mixed with methylamine?
Answer:

When cupric chloride or copper sulfate solution is mixed with methylamine, a deep blue complex salt is formed and the solution turns blue.

1. CuCl₂ + 4CH₃NH₂ → [CU(CH₃NH₂)4]Cl₂ (deep blue)
2. CuSO₄ + 4CH₃NH₂ → [Cu(CH₃NH₂)₄]SO₄ (deep blue)

Question 17. The observed boiling point of nitrobenzene is much higher than that expected based on its molecular mass—why?
Answer:

• Nitrobenzene is sufficiently polar (dipole moment μ= 3.95D ). So, its molecules experience strong dipole-dipole attraction.
• Consequently, separation of the molecules requires higher thermal energy, and hence, the actual boiling point of nitrobenzene is much higher than that expected based on its molecular mass.

Question 18. Which is more basic out of benzylamine (C₆H₅CH₂NH₂) and isomeric p -toluidine (p-CH₃C₆H₄NH₃)? Explain.
Answer:

• As the — NH₂ group in p -toluidine is directly attached to the benzene ring, the lone pair of electrons of the N-atom undergoes delocalization with n -electrons of the aromatic ring. This results in a decrease in electron density on N-atom.
• On the other hand, the — NH₂ group in benzylamine is not directly linked to the benzene ring and so the lone pair of electrons on N-atom do not participate in resonance.
• Hence, the electron density on N-atom is not reduced. Therefore, benzylamine is more basic than p -toluidine.

Question 19. Why is nitrobenzene not easily oxidized? Mention j 1 one practical application based on this property.
Answer:

• Due to the strong electron-attracting property of the — NO₂ group, the electron density in the benzene ring is drastically reduced i.e., the ring becomes highly deactivated.
• So it is not oxidized even by a strong oxidant. Its boiling point is sufficiently high. Hence, nitrobenzene is used as an ideal solvent for the oxidation reactions occurring at high temperatures.

Question 20. How will you prepare isopropyl amine directly from acetone? Which class does it belong to?
Answer:

When acetone vapor mixed with excess ammonia and H₂ gas is passed over Raney nickel (catalyst) under high pressure at 140°-150°C, isopropylamine is produced. Isopropylamine belongs to primary amines containing a secondary alkyl group.

Question 21. Nitration of toluene occurs very easily and at a much faster rate relative to benzene. But nitration of nitrobenzene is very difficult and takes place at a much slower rate concerning benzene. Explain.
Answer:

• An increase in electron density in the benzene ring increases the reactivity of the compound while a decrease in electron density reduces its reactivity. The —CH₃ group present in the toluene molecule is electron-repelling.
• By its +1 effect and hyperconjugation, the electron density in the ring is significantly enhanced. On the other hand, in the nitrobenzene molecule, the —NO₂ group present is a strong electron-attracting group.
• It considerably decreases the electron density in the ring with the help of its -I effect as well as the -R effect.
• Given the reasons mentioned above, nitration of toluene (an electrophilic substitution reaction) can be accomplished very easily and with greater ease than benzene. But nitration of nitrobenzene becomes very difficult and proceeds at a much slower rate than benzene.

Question 22. Arrange in the order of increasing basicity (with reasons): aniline, p-nitroaniline, p-toluidine, N, N,4- trimethylamine
Answer:

• Each of the given compounds is an aromatic amine. An increase in the electron density of the amino-nitrogen increases the basicity of the corresponding compound.
• So if the electron-donating alkyl group is attached to the amino N -atom, the basicity of the amine increases.
• Moreover, if the electron-attracting nitro group (having -I and -R effect) is present in the ring, the basicity of the amine decreases.
• Based on the above information, the given compounds can be arranged in the following order of their increasing basicity:

Question 23. Why does the nitration of chlorobenzene occur at a slower rate compared to benzene?
Answer:

• Cl -atom present in chlorobenzene increases the electron density of the ring through the +R effect but it draws the electron from the ring through its -I effect.
• In this case, as the -I effect is stronger, it outweighs the +R effect. So, the electron density of the ring eventually decreases. Hence, nitration of chlorobenzene occurs slowly concerning benzene.

Question 24. What happens when: one. HCl cone. H₂SO₄ are added to aniline? In each case, what happens if the addition of acid is followed by the addition of water?
Answer:

1. When cone. HCl is added to aniline, and crystalline solid aniline hydrochloride is precipitated.
2. The addition of cone H₂SO₄ to aniline gives crystalline solid aniline sulfate. The salts are ionic. So when water is added, these salts dissolve giving a colourless solution.

Question 25. How will you prove that aniline is basic?
Answer:

Aniline is insoluble in water but soluble in HCl as, aniline reacts with the acid (HCl) to form a salt, aniline hydrochloride which is soluble in water. This suggests that aniline is basic.

Question 26. Why is Gabriel phthalimide synthesis unsuitable for the preparation of aniline?
Answer:

• In the preparation of aniline by Gabriel-phthalimide synthesis, the first step requires a nucleophilic substitution reaction of a halobenzene using potassiophthalimide as the nucleophile.
• However, aryl halides (for example., C₆HgCl ) generally do not participate in nucleophilic substitution reactions. So, it is not possible to prepare aniline by Gabriel’s phthalimide synthesis.

Question 27. How will you prepare paracetamol, a widely used antipyretic drug from nitrobenzene?
Answer:

At first, nitrobenzene is reduced to phenylhydroxylamine, which in the presence of dil. HCl rearranges to p-aminophenol. Then it reacts with acetic anhydride to form paracetamol.

Question 28. How is the —NO₂ group in an aromatic compound identified in the presence of the —NH₂ group? Give example.
Answer:

• Nitro group in an aromatic compound in the presence of the —NH₂ group can be detected by the Mulliken-Barker test.
• In this test, the aromatic compound is heated with Zn-dust, NH₄Cl, and 50% aqueous alcohol and the resulting mixture is filteredin Tollens’ reagent.
• The appearance of a grey or black precipitate indicates the presence of a nitro group in the compound.

Question 29. Why is acetamide more acidic than ethylamine?
Answer:

• If one proton is lost from the N-atom of the acetamide molecule, the anion produced i.e., conjugate base acquires stability through resonance.
• But in ethylamine, if one proton is lost from the N-atom conjugate base cannot be stabilized through resonance.
• So acetamide has a greater tendency to lose proton than ethylamine and thus, behaves as a stronger acid than ethylamine.

Question 30. Why is aniline less basic than methylamine?
Answer:

• The basicity of a nitrogenous base depends on the ease with which the N-atom present in the molecule of the base can donate its lone pair of electrons to a proton, i.e, the effective electron density on the N-atom present in the molecule of the base decides its basicity.
• In an aniline molecule, the lone pair of electrons present on the N-atom of the —NH, the group participates in resonance with the n electrons of the benzene ring, causing delocalization of the lone pair of electrons.
• Consequently, electron density on the N-atom diminishes. So, the N-atom cannot donate its electron pair to the proton and thus, aniline behaves as a weak base.

• On the other hand, in methylamine, the lone pair of electrons on the N-atom can not be involved in delocalization.
• Hence, the lone pair resides entirely on the N atom itself Besides this, due to the effect of — CH, group the electron j density on the N-atom increases.
• So, the N-atom can donate Us lone pair of electrons to a proton. Therefore, methylamine is a stronger base than aniline.

Question 31. How can CH₃CN be used for the synthesis of CH₃NH₂ and CH₃CH₂NH₂?
Answer:

Question 32. Identify A, B, C, and D In the following reactions:

Answer:

Question 33. Mention the reagents for the following conversion in a single step:

Answer:

The reagent is trifluoroacetic acid (CF₃CO₃H).

Question 34. Consider the compounds and answer the questions:

Which on treatment with aqueous NaNO₂ dill. HCI followed by heating with SnCI₂ /cone. HCI produces a hydrazine derivative? Write the structure of the hydrazine derivative.
Answer:

Question 35. Write the structures of A to C In the following reaction:

Answer:

Question 36. An organic compound (A) is soluble in water. Its aqueous solution liberates carbon dioxide from NaHCO₃, forms a white precipitate with aqueous BaCl₂ solution, and responds to azo-dye test. Which is (A) among the following—

Answer: 3.

Question 37. Which is produced when benzene-diazonium chloride is coupled with phenol in an alkaline medium—

Answer: 1.

Question 38. An organic compound (A) of molecular formula C₇H₇NO, on treatment with P₂O₅ provides (B). The reaction of both (A) and (B) with LiAlH₄ gives (C). Acid hydrolysis of both (A) and (B) affords benzoic acid. Identify (A), (B), and (C) with reason.

OR, Identify A, B, C, D, E, and F in the following reactions:

Answer: 1.

Question 39. Which of the following compounds will be formed when aniline reacts with H₂S₂O₅—

Answer: 3.

Question 40. Arrange the following compounds in decreasing order of their basicity:

Write the arrow-head equation for the following reaction: Aniline is refluxed with glacial acetic acid or, Write the organic products in the following reactions:

Answer:

Question 41. Which of the following is to not an aminium salt

1. CH₃CH₂N⁺H₃Cl^–
2. (CH₃CH₂)₂N+H₂Br^–
3. (CH₃)₃N⁺HI^–
4. (CH₃)₄N⁺Br^–

Answer: 4. (CH₃)₄N⁺Br^–

Question 42. Write down the reagents for the following reactions:

Answer:

Question 43. Which of the following compounds is the most basic

Answer: 4

Question 44. An organic compound (A) of molecular formula C₇H₅N in reaction with alkaline H₂O₂ furnishes (B). Acid hydrolysis of both (A) and (B) produces the same aromatic acid (C) of molecular formula C₇H₆O₂. The reaction of both (A) and (B) with LiAlH₄ gives a primary amine (B) of the molecular formula C₇H₉N. Identify (A), (B), (C), and (D). Write arrowhead equations for the formation of (B) from (A) and (D) from (A).

OR, Write the reagents for the following conversions:

Distinguish between the following two compounds by a chemical reaction:Answer:

Question 45. Write chemical equations for the following conversions:

1. Nitrobenzene to benzoic acid
2. Benzyl chloride to 2-phenylethanolamine
3. Aniline to benzyl alcohol

Answer:

Question 46. Arrange in the decreasing order of basic strength in aqueous solutions: CH₃NH₂. (CH₃)₂NH, (CH₃)₃N, NH₃.
Answer: NH₃<(CH₃)₃N<CH₃NH₂<(CH₃)₂NH

Question 47. The conversion of primary aromatic amines into diazonium salts is known as
Answer: Diazotisation.

Question 48. Account for the following: Aniline does not undergo Friedel-Crafts reaction.
Answer:

Aniline is a Lewis base and reacts with a Lewis acid, AlCl₃ to form a salt. As a result, the N-atom of aniline + acquires a positive charge. The —N⁺H₃ group is a deactivating group for electrophilic substitution reaction. C₆H₅NH₂ + AlCl₃ → C₆H₅-N⁺H₂– A^–lCl₃ Hence, aniline does not undergo Friedel-Craft’s reaction.

Question 49. Give the structures of A, B, and C:

Answer:

Question 50. Complete the reaction: C₆H₅N₂Cl + H₃PO₂ + H₂O→
Answer:

Question 51. Arrange the following in increasing order of basic strength: Aniline, p-nitroaniline, and p-toluidine
Answer: p-nitroaniline < Aniline < p-toluidine

Question 52. Write the IUPAC name of the given compound:
Answer: 2,4,6-tribromoaniline / 2,4,6-tribromobenzenamine

Question 53. Write the structures of A, B, and C In the following:

Answer:

Question 54. Write the IUPAC name of the given compound:
Answer: N-phenylethylamine

Question 11. Complete the following reactions:

Answer:

Question 55. Write the IUPAC name of the following compound: (CH₃CH₂)NCH₃
Answer: N-ethyl-N-methylethanamine.

Question 56. Give reasons:

1. Acetylation of aniline reduces its activation effect.
2. CH₃NH₂ is more basic than C₆H₅NH₂.
3. Although — NH₂ iS o/p directing group, aniline on nitration gives a significant amount of m-nitroaniline.

Answer: Due to resonance, the lone pair of electrons on the nitrogen of acetanilide gets delocalized towards the carbonyl group.

• Hence the electrons are less available for donation to the benzene ring by resonance. Therefore, the activation effect of aniline is reduced.
• Nitration is carried out in an acidic medium. In an acidic medium, aniline is protonated to form the anilinium ion (which is mem-directing).
• For this reason, aniline on nitration gives a substantial amount of nitroaniline.

Question 57. Write the structure of 2,4-dinitrochlorobenzene.
Answer:

Question 58. Write the IUPAC name of the following compound: CH₃NHCH(CH₃)₂
Answer: N-methyldopa-2-amine

Question 59. Write the IUPAC name of the following compound: (CH₃)₂N—CH₂CH₃
Answer: N, N-dimethylethanamine.

Question 60. Write the structures of compounds A, B, and C In the following reactions:

Answer:

1. A = CH₃CONH₂, B = CH₃NH₂, C = CH₃NC
2. A = C₆H₅NO₂, B = C₆H₅NH₂, C = C₆H₅NHCOCH₃

Question 61. Classify the following amines as primary, secondary, or tertiary:

Answer:

1. Primary,
2. Tertiary,
3. Primary,
4. Secondary

Question 62. Write structures of different isomeric amines corresponding to the molecular formula, C₄HnN.

1. Write the IUPAC names of all the isomers.
2. What type of isomerism is exhibited by different pairs of amines?

Answer:

Primary amines:

IUPAC Names:

1. Butan-1-amine;
2. Butan-2-amine;
3. 2-methylpropan-l-amine;
4. 2-methylpropan-2-amine;
5. N-methylpropan-l-amine;
6. N-methylpropan-2-amine;
7. N-ethylethanamine;
8. N, N-dimethylethanolamine;

1. Position Isomers: 1 and 2; 5 and 6
2. Chain Isomers: 1 and 3; 1 and 4; 2 and 3; 2 and 4
3. Metamers: 5 and 7; 6 and 7

Functional isomers: All primary amines are functional isomers of secondary and tertiary amines and vice versa. On the other hand, secondary and tertiary amines are functional isomers of each other.

Question 63. Write chemical equations for the following reactions:

1. Reaction of ethanolic NH₃ with C₂H₅Cl.
2. Ammonolysis of benzyl chloride and the reaction of amine so formed with two moles of CH₃Cl.

Answer:

Question 64.  Write chemical equations for the following conversions:

1. CH₃—CH₂—Cl into CH₃ —CH₂—CH₂—NH₂
2. C₆H₅—CH₂—Cl into C₆H₅—CH₂—CH₂—NH₂

Answer:

Question 65. Write structures and IUPAC names of

1. The amide which gives propanamide by Hofmann bromamide reaction,
2. The amine produced by the Hofmann degradation of benzamide

Answer:

1. In the Hofmann bromamide reaction, R—CONH₂ is converted to R—NH₂.
2. Thus the structure of the amide that produces propanamine (CH₃CH₂CH₂NH₂) is CH₃CH₂CH₂CONH₂ and its IUPAC name is butanamide.
3. Benzamide is C₆H₅CONH₂. So the amine produced by Hofmann degradation ofbenzamide is C₆H₅NH₂ and its IUPAC name is aniline or benzenamine.

Question 66. How will you convert— O Benzene into aniline,

1. Benzene into N, N-dimethylaniline,
2. Cl—(CH₂)₄—Cl into hexan-1, 6-diamine?

Answer:

Question 67. Arrange the following in decreasing order of their basic strength: C₆H₅NH₂, C₂H₅NH₂. (C₂H₅)₂NH, NH₃
Answer:

• Out of the given amines, C₆H₅NH₂ is the weakest base because the lone pair on the amino nitrogen is involved in delocalization with the aromatic ring.
• The presence of two electron-donating C₂H₅ groups on the amino nitrogen of diethylamine makes it a stronger base than ethylamine (which contains only one C₂H₅ group).
• Basicity of NH3 is less than that of ethylamine and diethylamine. Thus, the order of basicity is (C₂H₅)₂NH > C₂H₅NH₂ > NH₃ > C₆H₅NH₂.

Question 68. Arrange the following in increasing order of their basic strength:

1. C₂H₅NH2, C₆H₅NH₂, NH₃> C₆H₅CH₂NH₂ and (C₂H₅)₂NH
2. C₂H₅NH₂> (C₂H₅)₂NH, (C₂H₅)₃N, C₆H₅NH₂
3. CH₃NH₂, (CH₃)₂NH, (CH₃)₃N, C₆H₅NH₂, C₆H₅CH₂NH₂

Answer:

1. C₆H₅ —NH₂ is the weakest base because the lone pair on amino nitrogen is involved in resonance with the aromatic ring. (C₂H5)₂NH is stronger than C₂H₅NH₂ due to the presence of an additional C₂H5 group (+1 effect).
   • NH₃ is weaker than either of these amines due to the absence of an alkyl group.
   • C₆H₅CH₂NH₂ is weaker than NH₃ because of the electron-withdrawing -I effect of the C₆H₅ group. However, it is stronger than aniline because the C₆H₅ group is not directly attached to the amino group.
   • Thus the sequence of basicity is, C₆H₅NH₂ < C₆H₅CH₂NH₂ < NH₃ < C₂H₅NH₂ < (C₂H₅)₂NH.
2. The basic, strength of 1°, 2° and 3° alkylamines in an aqueous medium is determined by two factors: (a) electron donation by increasing the number of alkyl groups tends to increase the basicity, (b) stability through solvation of the conjugate acid formed by uptake of a proton (which decreases in the sequence: C₂H₅NH₃ > (C₂H₅)₂NH₂ > (C₂H₅)₃NH ).
   • Combining these two factors, the sequence of basicity is found to be, (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂. Thus the order of basicity is, C₆H₅NH₂ < C₂H₅NH₂ < (C₂H₅)₃N<(C₂H₅)₂NH.
3. Alkyl amines (1°, 2°, and 3°) are stronger bases than C₆H₅CH₂NH₂ because the latter contains electron withdrawing —C₆H₅ group. Based on the electron-donating +1 effect of alkyl groups on the amino nitrogen and relative stabilization of the conjugate acids of the alkyl amines, it is found that the basicity follows the sequence: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N.

Thus the order ofbasicity is, C₆H₅NH₂ < C₆H₅CH₂NH₂ < (CH₃)₃N < CH₃NH₂ < (CH₃)₂NH .

Question 69. Complete the following acid-base reactions and name the products:

1. CH₃CH₂CH₂NH₂+ HCl→
2. (C₂H₅)₃N + HCl→

Answer:

Question 70. Write reactions of the final alkylation product of aniline with excess methyl iodide in the presence of sodium carbonate solution.
Answer:

Question 71. Write the chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.
Answer:

Question 72. Write structures of different isomers corresponding to the molecular formula, C₃HgN. Write IUPAC names of the isomers that will liberate nitrogen gas on treatment with nitrous acid.
Answer:

C₃H₉N has four isomers:

Only 1° amines are found to react with HNO₂ to liberate N₂ gas, according to the equation:

⇒ \(\mathrm{R}-\mathrm{NH}_2+\mathrm{HNO}_2 \rightarrow \mathrm{ROH}+\mathrm{N}_2 \uparrow+\mathrm{H}_2 \mathrm{O}\)

Thus the isomers that liberate nitrogen gas are— CH₃CH₂CH₂NH₂ (Propan-1-amine) and

Question 73. How will you convert 4-nitrotoluene to 2-bromobenzoic acid?
Answer:

Question 74. Convert:

1. 3-methylaniline into 3-nitrotoluene.
2. Aniline Into 1,3,5-tribromobenzene.

Answer:

Question 75. Write IUPAC names of the following compounds and classify them into primary, secondary, and tertiary amines.

1. (CH₃)₂CHNH₂
2. CH₃(CH₂)₂NH₂
3. CH₃NHCH(CH₃)₂
4. (CH₃)₃CNH₂
5. C₆H₅NHCH₃
6. (CH₃CH₂)₂NCH₃
7. m-BrC₆H₄NH₂

Answer:

Question 76. Give one chemical test to distinguish between the following pairs of compounds.

1. Methylamine and dimethylamine
2. Secondary and tertiary amines
3. Ethylamine and aniline
4. Aniline and benzylamine
5. Aniline and Nmethylaniline.

Answer:

When heated with an alcoholic solution of KOH and chloroform (CHCl₃), methylamine (1° amine) gives an offensive smell of methyl isocyanide (CH₃NC). Dimethylamine (2° amine) does not respond to this test.

2° amines react with HNO₂ (i.e, NaNO₂ + HCl ) to give yellow coloured oily N-nitrosoamines.

R₂NH + HNO₂ → R₂N—N=O + H₂O

• Nitrosoamine, when warmed with phenol and cone. H₂SO₄ gives a green solution that turns deep blue when made alkaline with aq. NaOH. It becomes red on dilution with water. Tertiary amines do not respond to this test.
• It is similar to the distinction between methylamine and aniline (see ‘Distinguish between’ section).
• See the ‘Distinguish between’ section.
• When heated with an alcoholic solution of KOH and CHCl₃, aniline (1° amine) gives an offensive smell of phenyl isocyanide. N-methyl aniline (2° amine) does not respond to this test.

Question 77. Account for the following:

1. pK_b of aniline is more than that of methylamine.
2. Ethylamine is soluble in water whereas aniline is not.
3. Methylamine In water reacts with ferric chloride to precipitate hydrated ferric oxide.
4. Although the amino group Is o- and p -directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
5. Aniline does not undergo Friedel-Crafts reaction.
6. Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
7. Gabrlelphthalimide synthesis is preferred for synthesizing primary amines.

Answer:

• In aniline, the lone pair on amino nitrogen is involved in delocalization with an aromatic ring so it is less available for combination with a proton. In contrast, in CH₃NH₂, the +I effect of the CH₃ group increases the electron density on the N-atom.
• So the lone pair of N-atom is more available for protonation. Thus, aniline is a weaker base and its pK_b value is greater than that of methylamine.
• Ethylamine is soluble in water because it can form intermolecular H-bonds with H₂O molecules:

• But in the case of aniline, due to the large lyophobic part (i.e., hydrocarbon part), the extent of intermolecular H-bonding with water decreases considerably, and hence, aniline is almost insoluble in water.
• Methylamine reacts with water forming OH-ions according to the following acid-base equilibrium:

These OH- ions combine with Fe³⁺ ions to form a brown precipitate of hydrated ferric oxide:

• Nitration is usually carried out using mixed acid consisting of a mixture of cones. HNO₃ + cone. H₂SO₄. In the presence of these acids, a large portion of the aniline molecule gets protonated and forms an anilinium ion, and an equilibrium is established.
• Aniline in which the —NH₂ group is ortho/para-orienting undergoes nitration and forms a mixture of ortho- and para-nitro aniline, while anilinium ion in which the —N⁺H₃ group is meta-orienting undergoes nitration to form the meta isomer.

• In the Friedel-Crafts reaction, AlCl₃ is used as a catalyst. This is a Lewis acid that combines with the basic compound aniline (substrate) and forms a complex, in which the amino nitrogen becomes positively charged.
• Consequently, the aromatic ring gets highly deactivated and the Friedel-Crafts reaction (a type of electrophilic substitution) fails.

Diazonium salts derived from aromatic amines are more stable due to the dispersal of the +ve charge through delocalization involving the aromatic ring system.

Gabriel-phthalimide synthesis involves nucleophilic attack of the N-atom of potassiophthalimide on the alkyl group of an alkyl halide to form N-alkyl phthalimide (in fact, the alkyl group gets attached to the attacking nitrogen atom).

This N-atom (which is not attached to any H-atom) can no longer be involved in further nucleophilic attack on a second molecule of alkyl halide, and hence, pure primary amines (devoid of any secondary or tertiary amines) are formed in this reaction.

Question 78. Arrange the following:

1. In decreasing order of pKb values: C₂H₅NH₂, C₆H₅NHCH₃, (C₂H₅)₂NH and C₆H₅NH₂
2. In increasing order of basic strength: C₆H5NH₂, C₆H₅N(CH₃)₂, (C₂H₅)₂NH and CH₃NH₂ In increasing order of basic strength: (a) Aniline, p -nitroaniline and p-toluidine (b) C₆H₅NH₂, C₆H₅NHCH₃, C₆H₅CH₂NH₂
3. In decreasing order of basic strength in gas phase: C₂H5NH₂< (C₂H5)₂NH, (C₂H5)₃N and NH₃ In increasing order of boiling point: C₂H5OH, (CH₃)₂NH, C₂HgNH₂ In increasing order of solubility in C₆H₅NH₂, (C₂H5)₂NH, C₂H5NH₂.

Answer:

• Aromatic amines are weaker bases than aliphatic amines because the lone pair on the N-atom of the former is involved in delocalization with the aromatic ring system and hence, less available for protonation.
• C₆H₅NHCH₃ is a stronger base than C₆H₅NH₂ because the +1 effect of the CH₃ group causes an increase in electron density on the N-atom.
• Similarly (C₂H₅)₂NH is stronger than C₂H₅NH₂ because of greater electron donation by an additional C₂H₅ group in the former. Thus the order of increasing basic strength or decreasing pK_b values is, C₆H₅NH₂, C₆H₅NHCH₃, C₂H₅NH₂, (C₂H₅)₂NH.
• It has already been explained in Ans. Q that the basic strength of (C₂H₅)₂NH is greater than that of C₆H₅NH₂. C₆H₅N(CH₃)₂ is a stronger base than CgHgNH2 because the +1 effects of two CH₃ groups cause a greater increase in electron density on the N-atom of the former.
• But (C₂H5)₂NH is more basic than CH₃NH₂ primarily due to the greater +1 effect of the two C₂H₅ groups (in the former) over one CH₃ group (in the latter). Thus increasing order of basic strength of the given compounds is, C₆H₅NH₂ <C₆H₅N(CH₃)₂ <CH₃NH₂ <(C₂H₅)₂NH

In aromatic amines, the presence of an electron-withdrawing group in the ring system causes a decrease in basic strength, while the presence of an electron-donating group causes an increase in basic strength.

• Thus, p-p-nitroaniline (containing electron-withdrawing —NO₂) is a weaker base than aniline, while p-toluidine (containing electron-donating CH₃ group) is a stronger base than aniline.
• Thus, the order of basicity is, p-nitroaniline < aniline < p-toluidine (i.e., p methylaniline).
• In C₆H₅NH₂ and C₆H₅NHCH₃, the amino nitrogen is directly attached to the benzene ring. So the lone pair of electrons on the N-atom is delocalized over the ring system.
• Thus both C₆H₅NH₂ and C₆H₅NHCH₃ are weaker bases than C₆H₅CH₂NH₂ (in which the N-atom is attached to the ring via a CH₂ group).
• Due to the +1 effect of the CH₃ group, C₆H₅NHCH₃ is a stronger base than C₆H₅NH₂. So the basic strength increases in the order: C₆H₅NH₂ < C₆H₅NHCH₃ < C₆H₅CH₂NH₂.
• In the gas phase, we need not consider solvent effects since, there is no question of stabilization of the conjugate acids through solvation involving, H-bonding.

So basic strengths of the given aliphatic amines will be determined by the number of electron-donating alkyl groups on the amino nitrogen.

• The basic strength increases as the number of alkyl groups on the amino nitrogen increases. In other words, basic strength in the gas phase decreases in the sequence: (C₂H₅)3N > (C₂H₅)₂NH > C₂H₅NH₂.
• Since oxygen is more electronegative than nitrogen, alcohols form stronger H-bonds than amines. So the boiling points of alcohols are higher than those of amines having comparable molecular masses. Therefore, C₂H₅OH (MW = 46) has a higher boiling point than (CH₃)₂NH and C₂H₅NH₂ (each having MW = 45).
• The extent of H-bonding depends on the number of H-atoms on the N-atom. Therefore, C₂H₅NH₂ (having two H-atoms attached to N) has a higher boiling point than (C₂H₅)₂NH (having one H-atom attached to N). Thus boiling points of the given compounds follow the sequence: (CH₃)₂NH < C₂H₅NH₂ < C₂H₅OH.
• Solubility of amines decreases with an increase in molecular mass due to an increase in the size of the hydrophobic hydrocarbon part and also with a decrease in the number of H-atoms (attached to N-atom) that take part in H-bonding.
• Thus, C₆H₅NH₂ (MW = 93) has the lowest solubility. Out of (C₂H₅)₂NH and C₂H₅NH₂, the latter has higher solubility because the extent of H-bonding is greater, and also the size of the hydrophobic hydrocarbon part is smaller. In other words, the solubility increases in the order: of C₆H₅NH₂ < (C₂H₅)₂NH < C₂H₅NH₂.

Question 79. How will you convert:

1. Ethanoic acid into methenamine
2. Hexanenitrile into 1-amino pentane
3. Methanol to ethanoic acid
4. Ethanamine into methenamine
5. Ethanoic acid into propanoic acid
6. Methenamine into ethanolamine
7. Nitromethane Into dimethylamine
8. Propanoic acid into ethanoic acid?

Answer:

Question 80. Accomplish the following conversions:

1. Aniline to 2,4,6-tribromofluorobenzene
2. Benzyl chloride to 2-phenylethanolamine
3. Chlorobenzene to p-chloroaniline
4. Aniline to p-bromoaniline
5. Benzamide to toluene
6. Aniline to benzyl alcohol.

Answer:

Question 81. Give the structures of A, B, and C in the following reactions:

Answer:

Question 82. An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br₂ and KOH forms a compound ‘C’ of molecular formula C₆H₇N. Write the structures and IUPAC names of compounds A, B, and C.
Answer:

Compound ‘C’ (C₆H₇N) is obtained by heating compound ‘B’ with Br₂ and KOH. This suggests that the reaction is Hofmann degradation and ‘C’ is a primary amine. Its molecular formula shows that it is aniline (C₆H₅NH₂). So ‘B’ is an amide with the formula C₆H₅CONH, (benzamide).

Since the treatment of compound ‘A’ with aqueous ammonia, followed by heating gives the amide B (C₆H₅CONH₂), it indicates that ‘A’ must be benzoic acid (C₆H₅COOH).

Question 83. Complete the following reactions:

1. C₆H₅NH₂ + CHClg + alc.KOH→
2. C₆H₅N₂CI + H₃PO₂ + H₂O→
3. C₆H₅NH₂ + H₂SO₄(conc.)→
4. C₆H₅N₂Cl + C₂H₅OH→
5. C₆H₅NH₂ + Br₂(aq)→
6. C₆H₅NH₂ + (CH₃CO)₂O→
7. 

Answer:

Question 84. Why cannot aromatic primary amines be prepared by Gabriel-phthalimide synthesis?
Answer:

The most important step in Gabriel phthalimide synthesis is the S_N2 reaction in which the nucleophile, phthalimide anion displaces the halide ion from alkyl halide to form Nalkylphthalimide. This upon subsequent hydrolysis gives the corresponding aliphatic primary amine.

Since aryl halides are reluctant to undergo nucleophilic substitution reactions, aromatic primary amines (i.e., aryl amines) cannot be prepared by Gabriel-phthalimide synthesis.

Question 85. Write the reactions of

1. Aromatic and
2. Aliphatic primary amines with nitrous acid.

Answer:

Aromatic 1° amines react with HNO₂ in the presence of HCl or H₂SO₄ at 0-5°Cto form diazonium salts.

Aliphatic primary amines also react with HNO₂ at 0-5°C to form diazonium salts, which are unstable and thus decompose readily to form alcohol and N₂ gas.

Question 86. Give a plausible explanation for each of the following:

1. Why are amines less acidic than alcohols of comparable molecular masses?
2. Why do primary amines have higher boiling points than tertiary amines?
3. Why are aliphatic amines stronger bases than aromatic amines?

Answer:

• Since oxygen is more electronegative than nitrogen, the conjugate base (RO^–)of an alcohol (ROH) is more stable than the conjugate base (RNH^–) of an amine (RNH₂).
• So, the loss of a proton from alcohol is more favorable than that from an amine. Hence, alcohols are more acidic than amines. Conversely, amines are less acidic than alcohols.

• Due to the presence of two H-atoms on the N-atom of primary amines, they undergo association through the intermolecular H-bonding.
• However tertiary amines do not undergo such association because they cannot form intermolecular H-bonds due to the absence of a H-atom on the N-atom. Thus, primary amines have higher boiling points than tertiary amines of comparable molecular masses.

• In aromatic amines, the lone pair on the amino nitrogen is involved in delocalization with the aromatic ring and hence, is less available for protonation.
• In aliphatic amines, the lone pair on the amino nitrogen is not delocalized and hence, fully available for protonation. Because of this reason, aliphatic amines are stronger bases than aromatic amines.

Question 87. What is the role of HNO₃ in the nitrating mixture used for the nitration of benzene?
Answer:

It acts as a base in the nitrating mixture and generates the effective electrophile, NO₂ (nitronium ion).

Question 88. Why is the NH₂ group of aniline acetylated before carrying out nitration?
Answer:

• Nitration is carried out using mixed acid (cone. HNO₂ cone. H₂SO₄ ). If aniline is subjected to nitration directly, it undergoes protonation to form an anilinium cation, in e which the -NH₃ group is m-directing.
• This causes the formation of m-nitro derivatives mainly, instead of o-and p-nitro derivatives. So, before carrying out the nitration of aniline, its basic character is eliminated by subjecting it to an acetylation reaction.

—NHCOCH₃ group is o -Ip -directing and moderately activating. C₆H₅NHCOCH₃ undergoes nitration by mixed acid to give a mixture of o and pnitroacetanilides. These on hydrolysis produce o- and p-nitroanilines.

Question 89. What is the product when C₆H₅CH₂NH₂ reacts with HNO₂?
Answer: C₆H₅CH₂NH₂ (an aryl-substituted aliphatic 1° amine) reacts with HNO₂ to give C₆H₅CH₂OH.

Question 90. What is the best reagent to convert nitrile to primary amine?
Answer: Na+C₂H₅OH, or LiAlH₄

Question 91. Give the structure of ‘A’ in the following reaction.

Answer:

Question 92. Under what reaction conditions (acidic/basic), the coupling reaction of aryldiazonium chloride with aniline is carried out?
Answer:

• The reaction is not carried out under basic conditions because even under weakly basic conditions a few diazonium cations form diazohydroxide, thereby lowering the concentration of the diazonium cations.
• The reaction cannot be carried out under strongly acidic conditions, because most of the aniline undergoes protonation to form anilinium cation, and hence, the coupling reaction fails.

Therefore the coupling reaction is carried out under mild acidic conditions (pH≅4-5 ) so that the concentration of Ar-N⁺₂ is maximum and also the concentration of C₆H₅NH₂ (very weak base) is sufficiently high. Under such a condition coupling occurs smoothly.

Question 92. Predict the product of the reaction of aniline with bromine in a non-polar solvent such as CS₂.
Answer:

Explanation: In a non-polar solvent (CS₂), the contribution of the resonance structures of aniline involving separation of charge is small, i.e., in a non-polar solvent, the activating effect of the —NH₂ group of aniline is reduced. Hence, only mono-bromoanilines are formed.

Question 93. Arrange the following in increasing order of dipole moment. CH₃CH₂CH₃, CH₃CH₂NH₂, CH₃CH₂OH
Answer:

• Hydrocarbon molecules are almost non-polar and hence, propane (CH₃CH₂CH₃) has the least dipole moment.
• Out of CH₃CH₂NH₂ and CH₃CH₂OH, the latter has a greater dipole moment because O is more electronegative than N.
• So dipole moment increases in the sequence: CH₃CH₂CH₃ < CH₃CH₂NH₂ < CH₃CH₂OH.

Question 94. How will you carry out the following conversions?

1. Toluene → p-toluidine
2. P-toluidine diazonium chloride → p-toluic acid

Answer:

Question 95. Write the conversions:

1. Nitrobenzene → acetanilide
2. Acetanilide → p-nitroaniline.

Answer:

Question 96. A solution contains 1 gmol. each of p-toluene- diazonium chloride and p-nitrophenyl diazonium chloride. To this 1gmol. of an alkaline solution of phenol is added. Predict the major product. Explain.
Answer:

• In an alkaline medium, phenol is converted to phenoxide ion, which undergoes a coupling reaction (a type of electrophilic substitution) with diazonium cation.
• The reaction will be favored by increasing the strength of the electrophile (diazonium cation).
• Due to the electron-withdrawing -I and -R effects of the NO₂ group, the electrophilic character of p-nitro-benzene diazonium chloride favors the formation of p’-nitrobenzene as the major product.

Question 97. How will you bring out the following conversion?

Answer:

Question 98. How will you carry out the following conversion?

Answer:

Question 99. How will you carry out the following conversion?

Answer:

Question 100. How will you carry out the following conversions?

Answer:

Class 12 Chemistry Unit 13 Organic Compounds Containing Nitrogen Long Answers Questions And Answers

Question 1. Alkyl cyanides undergo hydrolysis in both uikuiinc and acidic media but hydrolysis of alkyl isocyanides occurs only In acidic medium. Explain
Answer:

In alkyl cyanide, due to electron deficiency in the carbon atom of the protonated alkyl cyanide (acid medium), the Catom becomes susceptible to nucleophilic attack (H₂O). So, hydrolysis is feasible. In an alkaline medium, the partially positively charged carbon atom in the alkyl cyanide is attacked by the strong nucleophile OH^– which makes the hydrolysis feasible.

However, in alkyl isocyanide due to electron deficiency in the carbon atom of the protonated alkyl isocyanide (acid medium) carbon is easily attacked by a nucleophile (H₂O). So, hydrolysis is feasible. But in an alkaline medium, the negatively charged carbon atom or N-atom with its octet filled with electrons is not attacked by the nucleophile and hence, hydrolysis is not possible.

Question 2. Why are diazo reactions carried out at low temperatures (0-5°C)?
Answer:

Diazonium salts are not sufficiently stable and decompose very fast at ordinary temperatures liberating N₂ gas. So, the reactions are carried out at low temperatures (0-5°C).

Question 3. How will you identify whether a compound is aniline hydrochloride or not?
Answer:

A compound can be identified as aniline hydrochloride if it fulfills the following criteria:

1. It should be soluble in water.
2. It should form a curdy white precipitate of silver chloride (AgCl) when silver nitrate solution is added to the aqueous solution of the compound.
3. It must respond to the azo-dye test. In this test a brilliant scarlet red azo-dye is produced when a cold solution of sodium nitrite (NaNO₂) is added to an aqueous solution of the compound under consideration (0-5°C), followed by the addition of this cold solution to an alkaline solution of β-naphthol (coupling reaction).

The reasons for the above observations are as follows:

• Aniline hydrochloride is an ionic compound that readily dissolves in water.
• The chlorine atom in aniline hydrochloride exists as a Clion. So in an aqueous solution, the free Clion reacts with silver nitrate to give a white precipitate of silver chloride:

Aniline hydrochloride is a salt of weak base & strong acid. It gets hydrolyzed in aqueous solution to produce HCl.

• So, an aqueous solution of aniline hydrochloride undergoes a diazotization reaction in the presence of NaNO₂ only (without adding any acid from an external source).
• The diazonium chloride produced reacts with an alkaline solution of β-naphthol to give a red azo dye.

Question 4. Starting with a suitable reactant how will you prepare methylamine by Curtius rearrangement and Lossen rearrangement reaction?
Answer:

Preparation of methylamine by Curtius rearrangement reaction: When acetyl azide (CH₃CON₃) is heated using benzene or chloroform as a solvent, methyl isocyanate is produced, which on hydrolysis gives methylamine.

Preparation of methylamine by Lossen rearrangement reaction: When ethane hydroxamic acid is heated with a cone. HCl, methyl isocyanate is obtained, which on hydrolysis gives methylamine.

Question 5. Arrange in the correct order of basicity with proper reasons. Aniline, p -nitroaniline and m -nitroaniline.
Answer:

The correct order of basicity of the given compounds:- Aniline > m -nitroaniline > p -nitroaniline

• This is because the —NO₂ group is an electron-attracting group. If it occupies para-position in the ring concerning —NH₂, then with the help of -R and -I effects it can draw the lone pair of electrons on nitrogen and bond pair of electrons in the C—N bond, respectively.
• As a result, the electron density of N-atom in the —NH₂ group is so drastically reduced that the basicity of p -nitroaniline becomes much less than that of aniline.
• But if the —NO₂ group resides in the m-position concerning the —NH₂ group, it causes a decrease in electron density on the N-atom with the help of the -I effect only. Due to the lack of proper conjugation between the two groups, the -R effect does not come into play.
• Consequently, the electron density of N-atom in the — NH₂ group does not decrease appreciably. In other words, the electron density diminishes only to some extent.
• Thus, the basicity of m -nitroaniline is somewhat less than that of aniline but much more than that of p -nitroaniline.

Question 6.  In the reaction between aniline and nitrous acid in cold conditions, diazonium salt is produced while ethanol is obtained in the reaction between nitrous acid and ethylamine. Explain.
Answer:

Benzene diazonium cation attains its stability through resonance. So, it does not decompose easily evolving N₂ gas. So, benzene diazonium salt (C₆H₅NCl^–) is prepared by carrying out the reaction with HNO₂ in cold conditions.

On the other hand, ethyl diazonium cation (C₂H₅N⁺₂) finds no scope for stabilization through resonance. So it is very unstable. It decomposes easily evolving N₂ gas and forming ethyl cation. This cation reacts readily with water to produce ethyl alcohol.

Question 7. Two isomeric compounds A and B have molecular formula C₃H₅N. Determine the structural formula of A and B based on the information given below:

1. A and B do not react with HNO₂ and CH₃COCl.
2. On refluxing A and B with diLHCl, two monocarboxylic acids C and D are respectively obtained.
3. The molecular mass of D is 74.
4. C reduces Tollens’ reagent but D does not. Write the structural formula of C and D. What compounds will be produced when A and B are reduced?

Answer:

Two isomers A and B (C₃H₅N) do not react with HNO₂ or CH₃COCl. So the compounds are not amino compounds. A and B when refluxed with dilute HCl give two monobasic carboxylic acids C and D, respectively. So the compounds may be alkyl cyanide or isocyanide.

The molecular mass of monobasic carboxylic acid obtained due to hydrolysis of B is 74. Suppose, the molecular formula of D is CnH_{2n + 1}COOH.

∴ 12 × n +(2n + 1) + 12 + 2 × 16 +1 = 74

or, 14n + 46 = 74 or n = 2

• So the monobasic carboxylic acid D is C₂H₅COOH or CH₃CH₂COOH (propionic acid).
• Since hydrolysis of B (C₃H₅N), produces D (CH₃CH₂COOH), B is an alkyl cyanide having structural formula CH₃CH₂CN (ethyl cyanide).
• Since the monobasic carboxylic acid, C reduces Tollens’ reagent, C is formic acid (HCOOH).
• Hydrolysis of A produces formic acid (C) and A is an isomer of B (CH₃CH₂CN). Hence, ‘A’ is an alkyl isocyanide having the structural formula CH₃CH₂NC (ethyl isocyanide).

Reactions Involving Hydrolysis Of A and B :

Reduction Reactions Of A and B:

Question 8. A compound ‘A’ with molecular formula C₃H₉N, is dissolved in HCl and is reacted with NaNO₂ solution. A colorless, odorless, and noninflammable gas is evolved. After completion of the reaction, the mixture on distillation produces a liquid organic compound (B). The compound (B) when heated with I₂ and NaOH solution gives a yellow crystalline precipitate. Identify the compound A and write the equations of the reactions which occurred.
Answer:

The compound ‘A’ dissolves in HCl. Hence, it is a basic compound. This basic compound containing carbon, hydrogen, and nitrogen reacts with (NaNO₂ + HCl) liberating a colorless, odorless, and non-inflammable gas (possibly N₂). Therefore, ‘A’ is a primary amine. Two primary amines can be expressed with the help of the formula C₃H₉N:

• The liquid compound ‘B’, produced besides N₂, in the reaction with HNO₂, responds to the iodoform reaction.
• Hence, the compound having the structural formula does not represent the compound ‘ A ’ because the reaction of with HNO₂, N₂, and CH₃CH₂CH₂OH (which does not respond to iodoform reaction) is obtained.
• On the other hand, ‘A’ can be expressed with the structural formula because, in the reaction of HNO₂, N₂, and CH₃—CH(OH)CH₃ (which gives a positive iodoform test) is obtained.

Question 9. Why is it not possible to prepare tertiary butylamine by reaction of ammonia with tertiary butyl bromide?
Answer:

Preparation of tertiary butylamine from tertiary butyl bromide, (a tertiary butyl halide) is a nucleophilic substitution reaction.

To eliminate the steric effect, tertiary butyl halides undergo an elimination reaction rather than a substitution reaction under the influence of a basic reactant like NH₃, to produce an alkene as the major product and form tertiary butylamine as a minor product in a displacement reaction.

Question 10. A nitro compound with molecular formula C₃H₇O₂N (A) reacts with nitrous acid producing a nitroso compound. The product dissolves in alkali turning the solution red. Identify the starting compound.
Answer:

• Primary nitroalkane (RCH₂NO₂) reacts with nitrous acid to give a nitroso compound which develops a red color on dissolution in alkali.
• From this, it is evident that the compound A, having the molecular formula C₃H₇O₂N is a primary nitroalkane that stands for only one primary nitroalkane which is— CH₃CH₂CH₂NO₂ (1-nitropropane)

Question 11. A nitro compound ‘A’ with molecular formula C4H902N reacts with nitrous acid forming a nitroso compound that dissolves in alkali turning the solution blue. Identify the starting compound.
Answer:

Secondary nitroalkanes react with nitrous acid (HNO₂) to give nitroso compounds which dissolve in alkali resulting in a blue solution. Hence, the compound ‘A ‘ with molecular formula C₄H₉O₂N is a secondary nitroalkane. Only one secondary nitroalkane (2°) can be expressed with the formula C₄H₉O₂N which may be represented as CH₃CH(NO₂)CH₂CH₃ (2-nitrobutane).

Question 12. A, B, C, D, E, F, G, H—These amine compounds form hydrochloride salts, the chlorine content of each being 32.42%. In the reaction with nitrous acid A, B, C, and D liberate N₂ but E, F, G, and H do not. Write the structure of each of the amino compounds from A to H, giving a reasonable explanation.
Answer:

The general formula of primary, secondary, and tertiary amines is C_nH_2n+3 N. So the general formula of the hydrochloride salts of amines is C_nH_2n+4 NCl. The percentage of chlorine in these hydrochloride salts is—

⇒ \(=\frac{35.5 \times 100}{12 \times n+(2 n+4)+14+35.5}=\frac{35.5 \times 100}{14 n+53.5}\)

According to the question, \(\frac{35.5 \times 100}{14 n+53.5}=32.42\) or,n=4

∴ The molecular formula of the amines is C₄H₁₁N.

A, B, C, and D are primary amines as they liberate N₂ gas in reaction with

Thus, the structural formulae of these four primary amines viz. A, B, C, and D are—

1. CH₃CH₂CH₂CH₂NH₂(Butan-1-amine (A))
2. CH₃CH₂CH(CH₃)NH₂(Butan-2-amine (B))

The amines E, F, G, and H do not evolve N₂ gas reacting with HNO₂. Hence, they are either secondary amines or tertiary amines. So the structural formula of these are—

Question 13. An organic compound ‘A’ (C₃HgN) when boiled with NaOH solution liberates NH₃ along with the formation of the sodium salt ‘B’ of a carboxylic acid, (C₃H₆O₂). On reduction of ‘A’, ‘C’ (C₃H₉N) is produced. ‘C’ on reaction with HNO₂, gives ‘D’. Identify A, B, C, D.
Answer:

• An organic compound containing carbon, hydrogen, and nitrogen ‘A’ (C₃H₅N) on boiling with NaOH solution liberates ammonia gas (NH₃) and forms sodium salt of the carboxylic acid B.
• Hence, ‘ A’ is an alkyl cyanide (RCN) and B is the sodium salt of a carboxylic acid (RCOONa). The only alkyl cyanide having the formula C₃H₅N is CH₃CH₂CN i.e., the compound A is ethyl cyanide (CH₃CH₂CN). The compound B produced by hydrolysis of compound A is sodium propanoate (CH₃CH₂COONa).
• The compound C obtained by reduction of alkyl cyanide must be a 1° amine. Thus compound’ C’ is CH₃CH₂CH₂NH₂ (n-propylamine).
• In the reaction between nitrous acid and 1° amine, a 1° alcohol with the same number of C-atoms and N₂ gas is produced. So the compound D is CH₃CH₂CH₂OH (n-propyl alcohol). The chemical reactions involved are—

Question 14. An organic compound with molecular formula C₆H₅O₂N when reduced by Sn and HCl gives another compound B whose molecular formula is C₆H₇N. A cold solution of NaNO₂ is added to the compound B dissolved in HCl. The product obtained in the solution is first made to react with KCN in the presence of CuCN and then subjected to hydrolysis, when the compound C is obtained. The compound, if heated with soda lime yields benzene. Identify the compounds A, B, and C.
Answer:

• The ultimate product obtained from the compound ‘A ‘ through a series of chemical reactions is benzene. Thus compound A, is a benzene derivative containing nitrogen.
• From the given formula C₆H₅O₂N, it can easily be understood that the compound is nitrobenzene (C₆H₅NO₂). Nitrobenzene is reduced by Sn and HCl to form aniline (C₆H₅NH₂).
• So the compound B with molecular formula C₆H₇N is aniline. When a cold solution of NaNO₂ is added to aniline dissolved in a cold dilute solution of HCl, benzenediazonium chloride is formed, which reacts with KCN in the presence of CuCN to yield cyanobenzene or phenyl cyanide (C₆H₅CN).
• Hydrolysis of phenyl cyanide produces benzoic acid. So compound C is benzoic acid. Thus, A : Nitrobenzene (C₆H₅NO₂), B: Aniline (C₆H₅NH₂), C: Benzoic acid (C₆H₅COOH)

Reactions Involved:

Question 15. A compound of molecular formula C₆H₈NCl is water soluble. Its aqueous solution turns blue litmus red and gives a curdy white precipitate with AgNO₃. If the alkaline solution of β-naphthol is added to an aqueous solution of the compound, premixed with NaNO₂ solution, a red precipitate is obtained. Identify the compound.
Answer:

• As the compound in an aqueous solution gives a curdy white precipitate with stiver nitrate (AgNO₃) solution, the compound is a water-soluble chloride salt. Its aqueous solution is acidic as it turns blue litmus paper red.
• This shows that the compound is a salt of a strong acid HCl and a weak base. Since one of the constituents of the weak organic base is nitrogen, so it is a hydrogen chloride salt of a weak organic base, which on hydrolysis produces an acidic solution.
• The reaction between the aqueous solution of the compound and NaNO₂ and β-naphthol indicates that the compound in question is a primary aromatic amine.
• Since, it contains six carbon atoms and one nitrogen atom, the given compound is aniline (C₆H₅NH₂) i.e., the compound with molecular formula C₆H₈NCl is aniline hydrochloride.
• This structural formula conforms with the reactions given by the compound. For instance, when NaNO₂ is added to an aqueous solution of aniline hydrochloride, nitrous acid is produced.
• It converts free aniline into benzene diazonium chloride. When a solution of benzene diazonium chloride is added to an alkaline β-naphthol solution, it undergoes coupling and gives red azo-dye.

Question 16. Organic compound ‘A’ of molecular formula C₂H₃N is reduced to give compound ‘B’ which reacts with NaNO₂ solution producing ethanol. When ‘B’ is heated in the presence of chloroform and ethanolic KOH, compound ‘C with a foul smell is evolved. Identify A, B, and C.
Answer:

• ‘B’ reacts with HNO₂ to produce ethyl alcohol (CH₃CH₂OH). Therefore ‘B’ is a 1° amine whose formula is CH₃CH₂NH₂ (ethanamine).
• ‘B’ on heating with chloroform and alcoholic KOH (arylamine test) generates compound ‘C’ having an unpleasant smell.
• Thus, the compound C is an alkyl isocyanide having the structural formula CH₃CH₂NC (ethyl isocyanide). Since, compound A (C₂H₃N) is composed of C, H, and N, which on reduction gives ethylamine, ‘A ‘ is an alkyl cyanide, represented by formula CH₃C=N (methyl cyanide).

Question 17. An optically active organic compound (A) with molecular formula C₄HnN dissolves in dilute HCl. It liberates N₂ if it is allowed to react with nitrous acid. Determine the structural formula of the compound A.
Answer:

• The given organic compound dissolves in HCl and hence, it is basic. Since the compound consisting of C, H, and N reacts with HNO₂ to evolve N₂, it is a 1° amine.
• Moreover, an optically active compound should contain an asymmetric C-atom. Thus, only 1° amine of molecular formula C₄H₁₁N and having an asymmetric C-atom is butan-2-amine.

Question 18. A compound with molecular formula C4HnN reacts with HNO₂ liberating N₂ and produces the compound B having molecular formul