WBCHSE Class 12 Chemistry Unit 12 Aldehydes Ketones And Carboxylic Acids

Class 12 Chemistry Unit 12 Aldehydes Ketones And Carboxylic Acids Aldehydes And Ketones Introduction

Aldehydes Ketones Notes:Aldehydes and ketones are the first-stage oxidation products of primary and secondary alcohols respectively. Both these classes of compounds contain the carbonyl group, so they are collectively called carbonyl compounds.

When one valency of the bivalent carbonyl group is satisfied by a hydrogen atom, a monovalent —CH=O group, called aldehyde or aldehydic group is formed. This group is also called the ‘formyl or methyl group’. For example:

Where, R=H, Alkyl or aryl group

In aldehydes, the —CHO group always occupies one terminal position in the carbon chain. For example:

When the two valencies of the carbonyl group are satisfied by alkyl, aryl or aralkyl groups, ketones are produced, i.e., ketones contain the bivalent keto or ketonic group. For example:

[Where R, R’= alkyl, aryl or aralkyl group]

If the two alkyl, aryl or aralkyl groups bonded to the carbon atom of ketones are identical (R = R’ ), then those ketones are called simple ketones while ketones containing two dissimilar groups (R* R’) are called mixed ketones. For example:

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In ketones, the keto group lies somewhere in between the two terminal carbon atoms in the carbon chain.

For example:

Both aldehydes and ketones exhibit striking similarities in their physical and chemical properties because of the presence of the carbonyl group. However, due to the presence of one H-atom attached to the carbonyl carbon in aldehydes, they differ from ketones in some properties.

Aldehydes Ketones Notes

 

When one valency of a bivalent carbonyl group is satisfied by H, alkyl or aryl group and the other by —OH, —OR (alkoxy), —X (halogen), —OCOR (acyloxy) or —NH₂ (amino) group, the compounds formed are called carboxylic acids, esters, acid halides, acid anhydrides and acid amides, respectively.

The functional group of carboxylic acids is called the carboxyl (—COOH) group. Esters, anhydrides, acid halides and acid amides are called carboxylic acid derivatives because these compounds on hydrolysis produce carboxylic acids. The functional groups present in these compounds are:

General Formula And Nomenclature

General formula of carbonyl compounds Both aldehydes and ketones are collectively called carbonyl compounds because they contain the carbonyl group. Aliphatic aldehydes and ketones both are expressed by the general formula C_nH_2nO. For example, both propionaldehyde (CH₃COCH₃) can be represented by the molecular formula C₃H₆O.

Nomenclature Of Aldehydes

Common System Of Nomenclature

The common names of aldehydes are derived by replacing the terminal ‘ic acid’ from the common names of the carboxylic acids which they yield on oxidation, with the word ‘aldehyde’.

Example: CH₃CHO on oxidation produces CH₃COOH (acetic acid). Thus, the name of this aldehyde is acetaldehyde (acetic -ic + aldehyde = acetaldehyde).

In substituted aliphatic aldehydes, the positions of the substituents are indicated by the Greek letters α, β, γ, δ, etc., with the carbon atom next to the —CHO group being designated as a.

Aldehydes Ketones Notes

The old name is used for some aromatic aldehydes. Normally they are named as the derivative of benzaldehyde (C₆H₅CHO). The positions of the substituents on the benzene ring with respect to the —CHO group are indicated by prefixes ortho(o-) for 1, 2, meta(m-) for 1, 3 and para(p-) for 1, 4. Many aldehydes, however, are called by their special names.

IUPAC System Of Nomenclature

In the IUPAC system, aliphatic aldehydes are called alkanes. The IUPAC names of aldehydes are derived by replacing the terminal ‘e’ from the name of the corresponding alkane with the suffix ‘al (i.e., Alkane-e + al = Alkanal). For substituted aldehydes, the numbering of the longest chain should be done from the side of the —CHO group, i.e., by giving its carbon the serial number 1.

Example:

The IUPAC name of the simplest aldehyde ‘benzaldehyde’ is benzene carbaldehyde. The name benzaldehyde is also adopted by the IUPAC system. Other aromatic aldehydes are named as substituted benzaldehyde or benzene carbaldehyde. The positions of the substituents in the benzene ring are indicated by placing numerals like 2, 3, 4, etc., before the names of the substituents.

Example:

.

Aldehydes Ketones Notes

When an aldehyde group is directly attached to the ring of a cycloalkane, the suffix ‘carbaldehyde’ is added to the full name of the hydrocarbon. The IUPAC names of such compounds are written as cycloalkanecarbaldehyde.

The compounds in which the aldehyde group is attached to any side chain of the aromatic ring are called aryl-substituted aliphatic aldehydes. They are named in the same way as aliphatic aldehydes.

Example:

IUPAC: 2-phenyl ethanal 2-methyl-3-phenyl propanal

Common: Phenylacetaldehyde α-methyl-β-phenyl propionaldehyde

If there is a double bond on the parent chain of an aliphatic aldehyde, then the aldehyde is named ‘enal’ along with mentioning the position of the double bond.

Example:

When an unbranched carbon chain is linked to more than two aldehyde groups, the compound is named as a derivative of the parent alkane which does not include the carbon atoms of the aldehyde groups and the compounds are considered as tri, tetra, etc. carbaldehyde of the alkane.

Example:

Common and IUPAC names of some aliphatic and aromatic aldehydes

Aldehydes Ketones Notes

Nomenclature Of Ketones

Common System Of Nomenclature

1. The common names of simple ketones (RCOR) are derived by adding the numerical prefix ‘di’ before the name of the alkyl group.
2. The common names of mixed ketones (RCOR’) are derived by adding ‘ketone’ to the names of the alkyl groups arranged alphabetically. The positions of the substituents on either side of the parent chain are indicated by α (or α’), β (or β’), γ (or γ’ ), etc.

Example:

Some aromatic ketones, however, have old names and they are called ‘Phenones’.

Example:

IUPAC System Of Nomenclature

The IUPAC names of ketones are derived by replacing the terminal e from the name of the corresponding alkane with the suffix ‘one’ (i.e., Alkane-e + one = Alkanone) and by putting the positional number (the lowest possible number) of the carbonyl carbon before ‘one’.

If the carbonyl compound contains two or more keto groups, the terminal V of the name of the corresponding alkane is retained and the positions of the keto groups on the parent chain are indicated by numbers along with the numerical prefix di, tri, etc., being used before suffix one.

Example:

The names of unsaturated ketones are derived by replacing the terminal ‘ane’ with the name of the corresponding alkane with the suffix ‘ene’. However, ‘e’ is dropped because the secondary suffix begins with a vowel. The position of the double or triple bond is indicated by numbers.

Example:

Cyclic ketones are called ‘cycloalkanones’. The position of the substituent is indicated by giving the serial number T to the carbonyl carbon.

Example:

The common names of some ketones are adopted by the IUPAC system, like acetophenone (C₆H₅COCH₃), benzophenone (C₆H₅COC₆H₅), etc.

Aldehydes Ketones Notes

Example:

Diaryl ketones are also named methanone derivatives.

Example:

Alkyl aryl ketones are named as derivatives of alkanones.

Example:

In the case of ketoaldehydes, the aldehyde group is considered the principal functional group and the keto group is regarded as a substituent. The prefix ‘oxo’ is used for the keto group and its position on the parent chain is indicated by a suitable number obtained by assigning the number T to the carbon atom of the — CHO group.

Example:

Common And IUPAC Names Of Some Ketones

Aldehydes Ketones Notes

Class 12 Chemistry Unit 12 Aldehydes Ketones And Carboxylic Acids Isomerism In Aldehydes And Ketones

Aldehydes and ketones generally exhibit four types of isomerism:

1. Chain isomerism,
2. Position isomerism,
3. Functional group isomerism and
4. Metamerism.

Chain Isomerism

Higher aldehydes containing four or more carbon atoms in their molecules and higher ketones having five or more carbon atoms in their molecules display chain isomerism.

Example:

1. Butanal and 2-methylpropanal
2. Pentan-2-one and 3-methylbutan-2-one. The members of each pair are chain isomers of each other.

Position Isomerism

The higher ketones as well as aromatic aldehydes exhibit position isomerism.

Example:

1. Pentan-2-one and Pentan-3-one
2. O-and m-hydroxybenzaldehyde,
3. 2-bromoacetophenone and 4-bromoacetophenone. The members of each pair are position isomers of each other.

Aliphatic aldehydes do not display position isomerism because the monovalent —CHO group is always bonded to the terminal carbon atom of the chain.

Functional Group Isomerism

Aldehydes and ketones having the same molecular formula exhibit functional group isomerism.

Example:

1. Propionaldehyde (CH₃CH₂CHO) and Acetone (CH₃COCH₃)
2. Butanal (CH₃CH₂CH₂CHO) and Butanone (CH₃COCH₂CH₃). The members of each pair are functional group isomers of each other.

Metamerism

Different ketones having the same molecular formula but different numbers of carbon atoms on either side of the functional group exhibit this type of isomerism.

Example:

1. Pentan-2-one (CH₃COCH₂CH₂CH₃) and pentan-3-one (CH₃CH₂COCH₂CH₃) are metamers of each other.
2. Pentan-3-one (CH₃CH₂COCH₂CH₃) and 3-methylbutan-2-one [CH₃COCH(CH₃)₂] are metamers of each other.

Class 12 Chemistry Unit 12 Aldehydes Ketones And Carboxylic Acids Structure Of Carbonyl Group

1. Like the C=C bond in alkenes, the C=O bond in aldehydes and ketones is composed of one sigma (σ) and one pi (π) -bond. Both the carbon and the oxygen atoms are sp²-hybridised.
2. The (C—O) σ-bond is formed as a result of the overlapping of a half-filled sp²-hybrid orbital of carbon and a half-filled sp²-hybrid orbital of oxygen along their axes.
3. The carbon-oxygen n-bond is formed by lateral or sidewise overlapping between two half-filled p-orbitals, each on carbon and oxygen atoms. The electron cloud of the n-bond exists above and below the σ-bond.
4. The remaining two sp²-orbitals of carbon form two σ-bonds either by head-on overlapping with Is -orbital of two H-atoms as in formaldehyde (HCHO) or with 1s -orbital of one hydrogen atom and one sp²-orbital of an alkyl group as in aldehydes (RCHO) other than formaldehyde, or with two sp³– orbitals of two alkyl groups as in ketones (RCOR). All three cr -bonds lie in the same plane and are inclined to each other at an angle of 120°.
5. The remaining two sp²-hybrid orbitals of the oxygen atom are occupied by two unshared pairs of electrons. These two sp²-orbitals lie in the same plane as that formed by the three σ-bonds.

Aldehydes Ketones Notes

Excited state electronic configuration of C-atom

Ground state electronic configuration of O-atom

In carbonyl compounds,

1. The carbonyl carbon atom,
2. The carbonyl oxygen atom,
3. The two carbon atoms or one carbon atom and one hydrogen atom or two hydrogen atoms attached to the carbonyl carbon and
4. The two unshared electron pairs on the oxygen atom lie in the same plane.

Class 12 Chemistry Unit 12 Aldehydes Ketones And Carboxylic Acids Nature Of Carbonyl Group

The electronegativity of oxygen is greater than that of the carbonyl carbon atom (C = 2.5, O = 3.5). So, the electron pair of the cr -bond in general and that of the n -bond in particular shifts more towards the highly electronegative oxygen atom. As a result, the carbonyl carbon atom acquires a partial positive charge and the oxygen atom acquires a partial negative charge. Hence, the carbonyl group behaves as a polar group.

group having unsymmetrical n-electron density

group having symmetrical n-electron density

The polar carbonyl group is generally expressed as a resonance hybrid:

The polar nature of the carbonyl group is amply supported by the fact that the values of dipole moments of aldehydes and ketones are quite high (μ= 2.3 – 2.8D).

Aldehydes Ketones Notes

Comparative Study Of C=C And C=O

Class 12 Chemistry Unit 12 Aldehydes Ketones And Carboxylic Acids General Methods Of Preparation Of Aldehydes

By Oxidation Of (1°) Aalcohols

Aldehydes are prepared by controlled oxidation of primary alcohols with acidified potassium dichromate or alkaline potassium permanganate solution.

Aldehydes are very much susceptible to oxidation as compared to primary alcohols. Due to this, the oxidising agent used is kept at a minimum possible concentration and the aldehyde is removed quickly from the reaction mixture by distillation as soon as it is formed.

In this way, the oxidation of primary alcohol is restricted at the aldehyde stage. It is possible to materialise this oxidation process because the boiling points of aldehydes are much lower than those of the corresponding alcohols. The aldehydes formed undergo ready oxidation to yield carboxylic acids if allowed to remain in the reaction mixture, i.e., in contact with the oxidising agent.

Aldehydes Ketones Notes

Aldehydes may be prepared by passing a mixture of primary alcohol (in a vapour state) and the appropriate air or oxygen over a silver powder catalyst heated at 250°C.

Aldehydes can be prepared conveniently by oxidising primary alcohols with Collin’s reagent (chromium trioxide-pyridine complex, CrO₃-2C₅H₅N ). The reaction is carried out in a non-aqueous medium such as CH₂Cl₂. This reagent restricts the oxidation to the aldehyde stage.

Aldehydes can be prepared by oxidising the primary alcohols using pyridinium chlorochromate or PCC [CrO₃. C₅H₅N.HCl] in CH₂Cl₂ solvent. This reagent restricts the oxidation to the aldehyde stage. When pyridine is used as the solvent, it is known as the Sarret method of oxidation.

Aldehydes are obtained when primary alcohols are oxidised with Jones’ reagent (mixture of CrO₃, H₂SO₄ and aqueous acetone). This process of oxidation is restricted to the aldehyde stage.

This oxidation may also be carried out by CrO₃ dissolved in acetic acid.

Allylic and benzylic alcohols can be oxidised by active MnO₂ to α, β-unsaturated aldehydes in the presence of inert solvents like CH₂Cl₂ or CCl₄.

Example:

By dehydrogenation of primary alcohols

When vapours of primary alcohols are passed over a heated copper catalyst at 300°C, aldehydes are obtained.

Aldehydes Ketones Notes

Example:

From Calcium Salts Of Carboxylic Acids

Aldehydes are obtained by dry distillation of a mixture of calcium formate and calcium salt of any other carboxylic acid in 1:1 molar ratio.

Example:

In this reaction, the yield of aldehyde (RCHO) is low because side-products like formaldehyde (HCHO) and ketone (RCOR) are produced in sufficient quantities.

When calcium formate is subjected to dry distillation, formaldehyde is produced.

When a mixture of calcium benzoate and calcium formate is heated, benzaldehyde is obtained.

Aldehydes Ketones Notes

From Carboxylic Acids:

Aldehydes are produced by passing vapours of formic acid and any other carboxylic acid mixed in a 1:1 molar ratio over manganous oxide heated at 300°C or thorium dioxide (thoria) heated at 350°C.

Example:

In this reaction, aldehyde yield is low because formaldehyde and ketone (in this case acetone, CH₃COCH₃) are obtained as side-products.

When vapours of formic acid are passed over heated manganous oxide or thoria, formaldehyde is produced.

From acid chlorides: Rosenmund reduction

Aldehydes can be prepared by the partial reduction of acid chlorides by hydrogen in boiling xylene in the presence of a palladium catalyst deposited on barium sulphate. The reaction is named Rosenmund reduction after the name of the discoverer.

Example:

Role of BaSO₄: In this reaction, BaSO₄ acts as a poisoned catalyst and reduces the catalytic action of palladium. As a result, the aldehyde obtained is prevented from further reduction to alcohol. Moreover, a small amount of sulphur or quinoline (poison catalyst) is added to the reaction medium to decrease the catalytic efficiency of Pd.

Aldehydes Ketones Notes

Carboxylic acids cannot be directly reduced to aldehydes. So, carboxylic acids are first converted into acid chlorides and then to aldehydes by Rosenmund reduction.

Rosenmundreduction cannot be employed for preparing formaldehyde because the acid chloride involved, i.e., formyl chloride (HCOCl) is unstable at ordinary temperature. Ketones cannot be prepared by this method.

By Grignard Reagent

Aldehydes may be prepared by reacting hydrocyanic acid (HCN) with Grignard reagents (RMgX) followed by acidic hydrolysis of the resulting addition compound.

Example:

Aldehydes are produced when the additional compounds formed by the reaction between orthoformic ester [H—C(OR)₃] with Grignard reagent are hydrolysed by dilute acid.

Example:

Aldehydes are not usually prepared by treating a Grignard reagent with formic ester because the resulting aldehyde (obtained before hydrolysis) reacts with the Grignard reagent to form secondary alcohol. To avoid this problem, an ethereal solution of Grignard reagent (1 mole) is added slowly to the ethereal solution of ethyl formate (1 mole) so that the quantity of ethyl formate in the reaction mixture is always in excess.

Aldehydes Ketones Notes

Benzaldehyde can be prepared by the reaction between phenylmagnesium bromide and ethyl formate or ethyl orthoformate.

Example:

From Alkyl Cyanides: Stephen’s Method

When alkyl cyanides are reduced with stannous chloride and concentrated HCl in an ethereal solution, the corresponding aldehydes are obtained. Benzaldehyde is obtained from phenyl cyanide. However, ketones cannot be prepared by this method.

The reaction occurs through the formation of an imine.

Aldehydes are obtained by reducing the alkyl cyanides with diisobutylaluminlum hydride [(CH₃)₂CHCH₂]₂AlH, (or in short, AlH(’BU)₂ or DIBAL-H) followed by hydrolysis of the resulting imine. The double or triple bond in an unsaturated alkyl cyanide remains unaffected in this reduction process.

By Reduction Of Esters

Aldehydes are obtained when esters are reduced with diisobutylaluminium hydride (DIBAL-H).

Example:

By Ozonolysis Of Alkenes

Alkenes in which the doubly bonded carbon atoms contain one H-atom each undergoes ozonolysis to produce aldehydes.

Aldehydes Ketones Notes

From Alkenes: Wacker Process

When ethylene is passed through an acidified solution of palladium chloride mixed with cupric chloride in the presence of air or oxygen, acetaldehyde is produced. It is a one-step process which gives only acetaldehyde. A ketone is obtained when an alkene other than ethylene is used.

From Alkenes: Oxo Process

It is an important method for preparing acetaldehyde from ethylene. Aldehydes can be prepared by heating alkenes at elevated temperatures with carbon monoxide and hydrogen in the presence of octacarbonyl dicobalt [Co(CO)₄]₂ catalysts under pressure. This is also known as a hydroformylation reaction. Ketones cannot be prepared by this method.

Example:

By Hydration Of Alkynes

Acetaldehyde can be prepared by passing acetylene through 20% H₂SO₄ in the presence of mercuric sulphate or mercuric oxide as a catalyst at 60-80°C.

Aldehydes Ketones Notes

Other aldehydes cannot be prepared by this method because any other alkyne always leads to the formation of a ketone following Markownikoff’s rule.

Terminal alkynes can be converted into aldehydes by a hydroboration-oxidation process in the presence of disiamylborane, [(CH₃)₂CH—CH(CH₃)— ]₂BH.

Examples:

By Hydrolysis Of Gem-Dihalides

Aldehydes can be prepared by the alkaline hydrolysis of gem-dihalides (1,1-dihaloalkanes) using a dilute solution of NaOH or KOH or Ba(OH)₂ (baryta).

Examples:

From 1, 2-diols

Aldehydes are obtained when 1, 2-diols having appropriate structures are oxidised by lead tetraacetate [Pb(OCOCH₃)₄] or periodic acid (HIO₄).

Example:

Aromatic Aldehydes From Toluene And Its Derivatives

Aromatic aldehydes can be prepared by oxidising toluene and its derivatives with chromium trioxide in acetic anhydride. The gem-diacetate first formed is isolated and then hydrolysed with dilute H₂SO₄ or HCl to yield the corresponding aromatic aldehyde.

Aldehydes Ketones Notes

Benzaldehyde is obtained in the first step but it is not isolated because it readily reacts with acetic anhydride to form benzylidene diacetate. Since this diacetate cannot be oxidised further, the oxidation of benzaldehyde to benzoic acid is avoided.

Etard reaction: When toluene is oxidised with a solution of chromyl chloride (CrO₂Cl₂) dissolved in CS₂ a brown chromium complex is precipitated. The complex is separated and decomposed with dilute acid to give benzaldehyde. This reaction is known as the Etard reaction.

Preparation Of Aromatic Aldehydes From Benzyl Chloride

When benzyl chloride is heated with a solution of lead nitrate or copper nitrate in an atmosphere of CO₂, benzaldehyde is obtained.

Sommelet’s reaction: Benzaldehyde is produced when benzyl chloride is refluxed with hexamethylenetetramine in an aqueous ethanolic solution and the mixture is then acidified followed by steam distillation.

Aldehydes Ketones Notes

Gattermann-Koch Aldehyde Synthesis

Benzaldehyde is obtained when a mixture of carbon monoxide and hydrogen chloride is bubbled through a solution of benzene dissolved either in nitrobenzene or ether in the presence of a catalyst comprised of aluminium chloride (AlCl₃) and a small amount of cuprous chloride (Cu₂Cl₂). The reaction is known as Gattermann-Koch aldehyde synthesis.

Gattermann Aldehyde Synthesis

Benzaldehyde may be synthesised by treating benzene with a mixture of hydrogen cyanide and hydrogen chloride in the presence of aluminium chloride followed by decomposing the complex thus produced, with water. The reaction occurs through the formation of imidoformyl chloride and is commonly known as Gattermann aldehyde synthesis. The reaction occurs in two steps.

Class 12 Chemistry Unit 12 Aldehydes Ketones And Carboxylic Acids General Methods Of Preparation Of Ketones

By Oxidation Of Secondary (2°) Alcohols

Ketones can be prepared by oxidising secondary alcohols with acidified potassium dichromate or alkaline potassium permanganate.

Aldehydes Ketones Notes

Example:

Oppenauer oxidation: Ketones may be prepared by oxidising secondary alcohols with aluminium tertiary butoxide in the presence of excess acetone. Acetone is reduced to isopropyl alcohol.

Since the reaction is reversible, the ketone thus produced is continuously removed from the reaction medium by slow distillation.

Oppenauer oxidation is useful for oxidising unsaturated secondary alcohols to unsaturated ketones because the double bond remains unaffected by this oxidation.

In this process, if p-benzoquinone is used instead of acetone, primary alcohols can be oxidised to aldehydes.

Oppenauer oxidation of alcohol is the reverse of Meerwein-Ponndorf-Verleyreduction of ketone.

Aldehydes Ketones Notes

By Dehydrogenation Of Secondary (2°) Alcohols

Ketones are obtained when vapours of secondary alcohols are passed over a copper catalyst at 300°C.

Example:

From The Calcium Salts Of Carboxylic Acids

Ketones are obtained by dry distillation of the calcium salts of carboxylic acids other than formic acid.

Example:

From Carboxylic Acids

When vapours of carboxylic acids except formic acid are passed over heated MnO at 300°C, ketones are obtained.

Example:

From Acid Chlorides

Ketones can be prepared by reacting acid chlorides with dialkyl cadmium (R2Cd).

Example:

Aldehydes Ketones Notes

By Grignard Reagent

Ketones can be prepared by the action of Grignard reagents on alkyl or aryl nitriles followed by acid hydrolysis of the intermediate addition product.

Example:

By Ozonolysis Of Alkenes

Fully substituted alkenes on ozonolysis produce ketones.

From Alkenes: Wacker Process

Ketones are obtained when propene or higher alkenes are treated with an acidified solution of PdCl₂ mixed with cupric chloride in the presence of oxygen or air.

Aldehydes Ketones Notes

Example:

By Hydration Of Alkynes

Ketones can be prepared by hydration of alkynes other than acetylene in the presence of 20% H₂SO₄ containing mercuric sulphate (HgSO₄) or mercuric oxide (HgO) as a catalyst at 60-80°C. Hydration of unsymmetrical alkynes occurs according to Markownikoff’s rule.

Example:

By Hydrolysis Of Gem-Dihalides

When gem-dihalides (the compounds in which the halogen atoms are present at any carbon atom other than terminal one) are hydrolysed with dilute NaOH or KOH or Ba(OH)₂ solution, ketones are obtained.

Aldehydes Ketones Notes

Example:

From 1, 2-diol

Ketones may be prepared by oxidising 1,2-diols of suitable structure with lead tetraacetate or periodic acid.

Example:

From Acetoacetic Ester

Acetoacetic ester or any of its alkyl derivative on hydrolysis by dilute aqueous KOH solution followed by acidification with dilute HCl produces a p-keto acid. When the β-keto acid is heated, and it undergoes decarboxylation (elimination of CO₂) to yield a ketone.

Example:

By Pinacol-Pinacolone Rearrangement

When pinacol is heated with concentrated H₂SO₄, it undergoes dehydrative rearrangement to yield pinacolone or 3,3-dimethylbutan-2-one.

Preparation Of Aromatic Ketones By Friedel-Crafts Reaction

Aromatic ketones may be prepared by reacting benzene or any aromatic hydrocarbon with acid chlorides in the presence of anhydrous AlCl₃.

Example:

By Fries Rearrangement

When phenyl esters are heated with anhydrous AlCl₃ in the presence of CS₂ as a solvent, they undergo rearrangement (transfer of the acyl group from O-atom to the O-and positions of the ring) to yield a mixture of O- and P-hydroxy ketones. In general, low temperatures (<60°C) favour the formation of the p-isomer while high temperatures (>160°C) favour the formation of the O-isomer.

Example:

Class 12 Chemistry Unit 12 Aldehydes Ketones And Carboxylic Acids Physical Properties Of Aldehydes And Ketones

Physical State And Odour

Formaldehyde, the first member of the aldehyde series is a gaseous substance. Other aldehydes and ketones (up to 11 carbon atoms) are colourless liquids. Aldehydes and ketones with carbon atoms more than 11 are solids. 0 Formaldehyde is a pungent-smelling gas and acetaldehyde is a liquid with a pungent and suffocating odour. Higher aldehydes and almost all ketones possess a pleasant smell.

Aldehydes Ketones Notes

Boiling Point

The boiling points of aldehydes and ketones are higher than those of alkanes of comparable molecular masses. The alkane molecules are held together by weak van der Waals forces of attraction while the polar molecules of aldehydes and ketones are held together by much stronger dipole-dipole attractive forces. So, high thermal energy is required to separate these molecules. Hence, aldehydes and ketones have higher boiling points.

Ether molecules are much less polar than the molecules of aldehydes and ketones. So, the dipole-dipole attractive forces among ether molecules are almost negligible. Since the ether molecules are held together primarily by the van der Waal forces of attraction, the amount of heat required to separate the molecules by overcoming these forces is very small. For this reason, aldehydes and ketones have higher boiling points than ethers of comparable molecular masses.

The alcohol molecules, due to the presence of the —OH group. remain associated through the formation of intermolecular H-bonds. So, a lot of thermal energy is required to separate the molecules by breaking these stronger hydrogen bonds. Due to the absence of the —OH group in aldehyde and ketone molecules, the formation of the H-bond is not possible. Their molecules are held together by relatively weak dipole-dipole attractive forces. Hence, much less thermal energy is required to separate the molecules by overcoming these forces. So, the boiling points of aldehydes and ketones are much lower than that of alcohols of comparable molecular masses.

Molecules of carboxylic acid form dimers through inter-molecular hydrogen bonding. H-bonds in a dimer are sufficiently strong. Therefore, a large amount of thermal energy is required to break these bonds and because of this, the boiling points of carboxylic acids are much higher than those of alkanes, ethers, aldehydes, ketones and such other classes of compounds having comparable molecular masses.

in a ketone molecule C=0, the two electron-repelling (+1) alkyl groups Eire attached to the carbonyl carbon atom. Their resulting moment adds to the moment of the C=0 group and hence, ketones are relatively more polar than aldehydes containing only one alkyl group attached to the carbonyl carbon and consequently, the magnitude of dipole-dipole interactions in ketones is slightly more than that in aldehydes. Therefore, the separation of ketone molecules by overcoming these forces requires a little bit more thermal energy as compared to aldehydes. Thus, ketones boil at slightly higher temperatures than their isomeric aldehydes.

Aldehydes Ketones Notes

Boiling points of some compounds [of comparable molecular masses] of different homologous series:

Solubility

The molecules of aldehydes and ketones having lower molecular masses (up to 4 carbon atoms) can form H-bonds with water molecules. Hence, these are soluble in water.

However, with an increase in molecular mass, the size of the alkyl group of aldehydes and ketones also increases, i.e., the size of the hydrophobic hydrocarbon part present in these molecules increases and consequently, their solubility in water decreases. In practice, it is observed that aldehydes and ketones containing 5 or more C-atoms are almost insoluble in water.

The solubility of aromatic aldehydes and ketones is much lower than their corresponding aliphatic analogues as the hydro-carbon part is quite large. However, all aldehydes and ketones are fairly soluble in organic solvents (alcohol, ether, benzene).

Class 12 Chemistry Unit 12 Aldehydes Ketones And Carboxylic Acids Mechanism Of Nucleophilic Addition Reaction Of Aldehydes And Ketones

Polarity of the group

Since the electronegativity of oxygen is greater than that of the carbonyl carbon atom, the electron pair of the <r -bond and that of it -bond, In particular, are shifted more towards the carbonyl oxygen atom.

As a result of this, the carbonyl carbon atom acquires a partial positive charge and the oxygen atom acquires a partial negative charge. Due to this, the carbonyl group behaves as a polar group. The carbonyl group is represented as a resonance hybrid of the following two structures:

Aldehydes Ketones Notes

Nucleophilic Addition Reaction

Since the polarity of the carbonyl group is sufficiently high, the positively polarised carbonyl carbon atom undergoes easy attack by a nucleophile (say Z^–) resulting in the formation of a σ-bond with simultaneous transfer of the π-electrons of the carbon-oxygen double bond to the oxygen atom.

An intermediate anion is formed and the hybridisation state of the carbon changes from sp² to sp³. In the next step, the strongly basic anion picks up a positive ion (e.g., H⁺) from the solvent or the reagent to form an addition compound.

The reaction occurs in two steps. The first step is a slow one and is called the rate-determining step because this step determines the overall rate of the reaction.

If, however, the attacking nucleophile is weak (e.g., ammonia or its derivatives) the reaction of aldehydes and
ketones is usually carried out in a weakly acid medium. Such reactions are called acid-catalysed nucleophilic addition reactions. In the presence of acids, the oxygen atom of the carbonyl group undergoes reversible protonation to form a species in

Aldehydes Ketones Notes

which the +ve charge on the carbonyl carbon increases (see resonance structures), thereby making it more susceptible to nucleophilic attack. As a result, even weak nucleophiles can readily attack the carbonyl carbon atom. This step involving nucleophilic attack is still the r.d.s. of the reaction.

Both C=O and C=C undergo addition reactions. The polar attack on C=C is normally initiated only by electrophiles (since the π-electrons shield the molecule from attack by nucleophilic reagents), while the attack C=O, due to its dipolar nature, can be initiated either by electrophilic attack on oxygen (normally by H⁺ or a Lewis acid) or by nucleophilic attack (Nu⁺ or Nu ) on carbon.

Relative Reactivities Of Aldehyde And Ketone

In nucleophilic addition reactions, ketones are relatively less reactive than aldehydes due to the following reasons:

Inductive effect: The ease with which a nucleophile will attack the carbonyl carbon atom depends on the extent of electron deficiency of the carbonyl carbon, i.e., on the quantity of positive charge accumulated on that carbon atom. In ketones, the two electron-releasing alkyl (R) groups attached to the carbonyl carbon reduce the positive charge on it while in aldehydes, only one electron-releasing alkyl group attached to the carbonyl carbon reduces the positive charge on it.

Therefore, after being partially neutralised, the amount of positive charge accumulated on the carbonyl carbon of an aldehyde is greater than that accumulated on the carbonyl carbon of a ketone. For this reason, ketones are less susceptible to nucleophilic attack than aldehydes.

Thus, from the electronic point of view, the order of reactivity of a ketone (e.g., acetone), an aldehyde (e.g., acetalde¬ hyde) and formaldehyde (HCHO) may be given as:

Aldehydes Ketones Notes

Steric effect: In the rate-determining step of nucleophilic addition reaction, the trigonal substrate is converted into a relatively more crowded tetrahedral intermediate. So, the attack by the nucleophile becomes progressively hindered in the tetrahedral intermediate. The reason behind this hindrance is steric crowding which occurs due to the increase in the number and size of the alkyl groups.

The nucleophilic attack on the carbonyl carbon in aldehydes containing only one alkyl group is, therefore, sterically less hindered than the nucleophilic attack on the carbonyl carbon in ketones. For this reason, ketones are less susceptible to nucleophilic attack than aldehydes. Thus, from the steric point of view, the order of reactivity of the following carbonyl compounds can be given as:

Aromatic aldehydes and ketones are relatively less reactive than the corresponding aliphatic analogues towards nucleophilic addition reaction. In aromatic aldehydes and ketones, the benzene ring by its stronger electron-releasing resonance effect (+R effect) increases the electron density on the carbonyl carbon considerably while, in aliphatic aldehydes and ketones, the alkyl groups by their weaker electron-releasing inductive effect (+1 effect) cannot increase the electron density on carbonyl carbon to such an extent.

As a result, carbonyl carbon atoms of aromatic carbonyl compounds become relatively less positive than those of the aliphatic carbonyl compounds. The carbonyl carbon of aromatic aldehydes and ketones is less susceptible to nucleophilic attack.

Order of reactivity of some aromatic aldehydes & ketones:

Reaction Of Aldehydes And Ketones

The chemical reactions of aldehydes and ketones can be discussed under the following seven heads:

1. Nucleophilic addition reactions,
2. Formation of addition compound followed by elimination of water,
3. Oxidation reactions,
4. Reduction reactions,
5. Halogenation reactions,
6. Reactions with alkali,
7. Miscellaneous reactions.

Aldehydes Ketones Notes

Class 12 Chemistry Unit 12 Aldehydes Ketones And Carboxylic Acids Nucleophilic Addition Reactions

Formation Of Addition Compound With Hydrogen Cyanide

Both aldehydes and ketones combine with hydrogen cyanide to form cyanohydrins. All aldehydes can form cyanohydrins. Among the ketones, only acetone, butanone, pentan-3-one and pinacolone can form cyanohydrins.

Example:

As HCN is a poisonous gas, it is not used directly during the reaction. Sodium cyanide and an insufficient dilute H₂SO₄ are mixed with aldehydes or ketones. As a result, HCN produced participates in the reaction.

⇒ \(\mathrm{NaCN}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{NaHSO}_4+\mathrm{HCN}\)

The unreacted NaCN left in the reaction medium is hydrolysed and consequently, the solution becomes alkaline. The resulting OHe acts as a catalyst to liberate the CNe ion (nucleophile) from HCN. As a result the yield of cyanohydrin increases.

Aldehydes Ketones Notes

⇒ \(\mathrm{HCN}+\mathrm{OH}^{\ominus} \rightleftharpoons \mathrm{H}_2 \mathrm{O}+\mathrm{CN}^{\ominus}\)

Role of cyanohydrins in the synthesis of organic compounds: Different classes of compounds such as a -hydroxy acids, α, β-unsaturated acids, α-amino acids, β-amino alcohols, 1,2-diols, etc., can be prepared from cyanohydrins.

Formation Of Addition Compounds With Sodium Bisulphite

Both aldehydes and ketones react with a saturated solution of sodium bisulphite to form bisulphite addition compounds. Bisulphite addition compounds are colourless crystalline solids in which a carbon-sulphur bond is present.

Example:

Although most aldehydes are capable of forming such addition compounds, all ketones do not form these types of compounds due to steric hindrance. Only normal ketones, i.e., unbranched methyl ketones (e.g., CH₃COCH₃, CH₃COCH₂CH₃) produce bisulphite addition compounds. Diethyl ketone, methyl ketone, methyl tertbutyl ketone, acetophenone, benzophenone, etc., do not yield bisulphite addition compounds.

Aldehydes Ketones Notes

Utility of the formation of bisulphite addition compounds: The bisulphite addition compounds are crystalline solids. This addition reaction is reversible and because of this, these addition compounds decompose to regenerate the original aldehyde or ketone when heated with dilute mineral acids or aqueous alkalies.

Therefore, aldehydes and ketones can be purified and separated from non-carbonyl impurities with the help of two opposite reactions, viz., formation of bisulphite addition compound and decomposition of that addition compound with acid or base.

Addition reaction with Grignard reagents (RMgX)

Both aldehydes and ketones produce additional compounds with Grignard reagents. When these additional compounds are hydrolysed with water or dilute mineral acids, alcohols are obtained.

Reaction With Alcohols

Aldehydes react with alcohols in the presence of dry HCl gas to produce acetals (gem-alkoxy compounds). The reaction occurs in two steps. In the first step, aldehydes combine with one molecule of alcohol to yield unstable alkoxy alcohol intermediates called hemiacetals. In the second step, these hemiacetals react with one molecule of alcohol to give stable acetals.

If one molecule of dihydric alcohol is used instead of two molecules of monohydric alcohol, a cyclic acetal is obtained. Generally, p-toluene sulphonic acid (PTS) is used as a catalyst In this case.

Example:

Aldehydes Ketones Notes

Ketones do not normally react with monohydric alcohols, but, they react with dihydric alcohols to form cyclic ketals.

Example:

Benzaldehyde reacts with alcohol in the presence of HCl gas to form first hemiacetal which then reacts with one more molecule of alcohol to form acetal.

Example:

Hydrolysis of acetals and ketals: Reactions leading to the formation of acetals or ketals from carbonyl compounds are reversible. So, acetals and ketals undergo hydrolysis to regenerate the parent carbonyl compounds I when heated with dil. HCl.

Utility of the formation of acetals and ketals: Although acetals and ketals are decomposed in dilute acid, they are quite stable in alkaline solutions. Therefore, the carbonyl groups of aldehydes or ketones can be protected in an alkaline medium through the formation of these compounds, e.g., 2,3- di-hydroxypropyl is prepared from acetaldehyde by protecting the -CHO group to avoid oxidation by permanganate.

Aldehydes Ketones Notes

Class 12 Chemistry Unit 12 Aldehydes Ketones And Carboxylic Acids Formation Of Addition Compounds And Subsequent Elimination Of Water Molecule

Reaction With Different Ammonia Derivatives

Aldehydes and ketones react with some ammonia derivatives, such as hydroxylamine (NH₂OH), hydrazine (NH₂NH₂), phenylhydrazine (C₆H₅NHNH₂), 2,4-dinitrophenylhydrazine, semicarbazide, etc., in weakly acidic medium. At first, an unstable addition compound is formed and then a molecule of water is eliminated to yield a stable compound containing a carbon-nitrogen double bond.

These derivatives of aldehydes and ketones are crystalline solids with specific melting points. Therefore, the starting aldehydes or ketones can be identified by determining their melting points.

Products obtained by reactions of ammonia derivatives with aldehydes and ketones:

Aldehydes Ketones Notes

Reaction with hydroxylamine (NH₂OH)

Both aldehydes and ketones react with hydroxylamine to form aldoximes and ketoxime, respectively. The optimum pH for this reaction Is 3.5. If the concentration of acid Is much higher, NH₂OH combines with a proton to form N⁺H₃ —OH and as a result It cannot act as a nucleophile.

Example:

The sodium acetate-acetic acid buffer sodium offers an optimum pH for this condensation reaction.

Aldehydes Ketones Notes

It is to be noted that when aldoximes and ketoxime are refluxed with dilute HCl, they give back the original aldehydes and ketones, respectively.

Reaction With Hydrazine (NH₂NH₂)

Both aldehydes and ketones react with hydrazine to produce hydrazones.

Example:

Like oximes, hydrazones on being refluxed with dilute HCI, regenerate the initial aldehydes and ketones.

Aldehydes Ketones Notes

 

Like oximes, hydrazones on being refluxed with dilute HCl, regenerate the initial aldehydes and ketones.

Reaction With Phenylhydrazine (C₆H₅NHNH₂)

Both aldehydes and ketones react with phenylhydrazine to produce phenylhydrazones.

Example:

Aldehydes Ketones Notes

Reaction with 2,4-dinitrophenylhydrazine

Both aldehydes and ketones react with 2,4-dinitrophenyl¬ hydrazine (DNP), or Brady’s reagent to form a yellow, orange or red precipitate of 2,4-dinitrophenyl hydrazone.

Example:

Aldehydes Ketones Notes

Since 2,4-dinitrophenylhydrazones are crystalline solids having well-defined melting points, the parent aldehydes and ketones can be identified by determining the melting points of the corresponding DNP derivatives.

Brady’s reagent: This reagent can be prepared by dissolving 2,4-dinitrophenylhydrazine in a mixture of cones. H₂SO₄ and methanol. It is used for the identification of aldehydes and ketones.

Reaction With Semicarbazide (NH₂NHCONH₂)

Both aldehydes and ketones react with semicarbazide to produce semicarbazones.

Example:

Aldehydes Ketones Notes

Ammonia derivatives like hydroxylamine, hydrazine, phenylhydrazine, semicarbazide, etc., are very susceptible to aerial oxidation. To protect these basic compounds from aerial oxidation, they are converted into their solid hydrochloride salts. These salts do not get easily oxidised in the presence of air.

These salts do not normally react with carbonyl compounds but, in the presence of sodium acetate, they do. The role of sodium acetate (CH₃COONa) is that it reacts with these salts and releases the free ammonia derivatives (NH₂Z).

Example: The reaction between hydroxylamine hydrochloride and sodium acetate can be expressed by the following equations:

Overall reaction:

The liberated acetic acid acts as a catalyst by offering an optimum pH in the presence of sodium acetate. Other hydrochloride salts react similarly.

Although two N-atoms contain an unshared pair of electrons in phenylhydrazine, semicarbazide and 2,4- dinitrophenylhydrazine molecules, the N-atom which is not adjacent to the aromatic ring or the group becomes involved in the formation of a bond with the carbonyl carbon atom. This is because the unshared pair of electrons on the N-atom adjacent to the aromatic ring or group becomes involved in resonance interaction with these groups and hence, are less available for bond formation.

Reaction With Ammonia (NH₃)

Aldehydes except formaldehyde and ketones except acetone react with ammonia to form imines.

Example:

Aldehydes Ketones Notes

Benzaldehyde reacts with ammonia to form a complex compound known as hydrobenzamide instead of a simple condensation product. The reaction, however, proceeds through the formation of benzaldimine.

It has been established that the aldehyde-ammonia addition compounds exist as cyclic trimers. For example:

Formaldehyde reacts with ammonia to give a white crystalline solid hexamethylenetetramine or protopine.

Aldehydes Ketones Notes

Acetone reacts with ammonia at ordinary temperature to form diacetone amine.

Reaction With Primary (1°) Amine

Aldehydes react with primary amines to form imines. These compounds are called Schiff bases. Imines obtained from aliphatic aldehydes are not sufficiently stable but, imines obtained from aromatic aldehydes are stable compounds.

Example:

Ketones react with 1° amines to yield unstable imines.

Class 12 Chemistry Unit 12 Aldehydes Ketones And Carboxylic Acids Oxidation Reactions

Oxidation Of Aldehydes

Aliphatic and aromatic aldehydes undergo easy oxidation to yield carboxylic acids containing the same number of carbon atoms as in the parent aldehyde.

⇒ \(\mathrm{CH}_3-\mathrm{CHO} \stackrel{[\mathrm{O}]}{\longrightarrow} \mathrm{CH}_3-\mathrm{COOH}\)

⇒ \(\mathrm{C}_6 \mathrm{H}_5-\mathrm{CHO} \stackrel{[\mathrm{O}]}{\longrightarrow} \mathrm{C}_6 \mathrm{H}_5-\mathrm{COOH}\)

In the presence of oxidising agents, the H-atom on the carbonyl carbon of aldehyde molecules easily gets converted to an —OH group. So, aldehydes are oxidised not only by strong oxidising agents like acidified K₂Cr₂O₇ or alkaline KMnO₄ but also by weak oxidising agents like Br₂ -water, Ag⁺ ions, Cu²⁺ ions, etc., i.e., aldehydes behave as strong reducing agents.

In contact with air, benzaldehyde undergoes slow oxidation to form benzoic acid. In this case, it combines with one molecule of oxygen to form perbenzoic acid. This perbenzoic acid oxidises another molecule of benzaldehyde to benzoic acid and itself is reduced to benzoic acid.

Aldehydes Ketones Notes

The white crystalline substance found to be present at the neck of the bottle containing benzaldehyde is, in fact, a mixture of benzoic acid and perbenzoic acid. This auto¬ oxidation of benzaldehyde may be prevented by keeping it in an air-tight bottle or by adding a small amount of hydroquinone as an antioxidant.

Reducing Properties

Reduction of Tollens’ reagent: Tollens’ reagent is an ammoniacal silver nitrate solution. When the mixture of aldehyde and Tollens’ reagent is warmed, the aldehyde reduces silver nitrate (Tollens’ reagent) to metallic silver which forms a silver mirror on the walls of the test tube.

The presence of an aldehyde (—CHO) group in any organic compound can be detected by this experiment. Ketones do not respond to this test.

Preparation of Tollens’ reagent: When 2-3 drops of 10% NaOH solution are added to 2-3 mL of AgNO₃ solution taken in a test tube, a grey precipitate of Ag₂O is obtained. ’ Dilute NH4OH solution is then added to the test tube dropwise till the grey precipitate dissolves. The resulting; solution obtained is known as Tollens’ reagent and it contains the complex compound [ Ag(NH₃)₂]⁺OH^–.

Reduction of Fehling’s solution: When aliphatic aldehydes are heated with Fehling’s solution, the latter is reduced to give a red precipitate of cuprous oxide (Cu2O). This experiment proves the presence of the aliphatic —CHO group in any organic compound. Aromatic aldehydes do not reduce Fehling’s solution.

In brief: \(\mathrm{RCHO}+2 \mathrm{CuO} \longrightarrow \mathrm{RCOOH}+\mathrm{Cu}_2 \mathrm{O} \downarrow\)

Fehling’s Solution: A mixture of ‘Fehling A’ and ‘Fehling B, in equal volumes is called Fehling’s solution. ‘Fehllng A’ is an aqueous solution of copper sulphate (blue) while ‘Fehling B’ is a colourless alkaline (NaOH) solution of sodium potassium tartrate or Rochelle salt. When equal volumes of ‘Fehling A’ and ‘B’ solution are mixed, deep blue Fehling’s solution is obtained.

Reduction Of Benedict’s Solution: When aliphatic aldehydes are heated with Benedict’s solution, the latter is reduced to give a red precipitate of cuprous oxide.

1. This experiment proves the presence of the aliphatic —CHO group in any organic compound. However, ketones and aromatic aldehydes (e.g., C₆H₅CHO ) cannot reduce Benedict’s solution.
2. Since ketones cannot reduce Tollens’ reagent, Fehling’s solution and Benedict’s solution, aliphatic aldehydes and ketones can be distinguished by using these reagents.
3. For the same reason, aliphatic aldehydes can be distinguished from aromatic aldehydes with the help of Fehling’s and Benedict’s solution.

Benedict’s Solution: It is an aqueous solution of copper sulphate, sodium citrate and sodium carbonate. Fehling’s solution contains Cu²⁺ complexed with tartrate ions while Benedict’s solution contains Cu²⁺ complexed with citrate ions.

Aldehydes Ketones Notes

Oxidation Of Ketones

Ketones cannot be oxidised by mild oxidising agents because no H-atom is directly attached to the carbonyl carbon of ketones. Only strong oxidising agents (e.g., K₂Cr₂O₇ + conc.H₂SO₄, KMnO₄ + NaOH, cone. HNO₃, etc.) can oxidise ketones through cleavage of C—C bond and as a result, carboxylic acids having fewer number of carbon atoms than the starting ketones are produced.

In this oxidation process, any bond between the carbonyl carbon and a -carbon may be cleaved. In the case of unsymmetrical ketones, the cleavage of the C —C bond occurs in such a way that the smaller alkyl group remains preferentially attached to the carbonyl carbon (Popoff’s rule). It becomes clear by the following example.

Oxidation of aldehydes and ketones by SeO₂

Aldehydes and ketones in which one methyl (— CH₃) or methylene (—CH₂— ) group is attached to the carbonyl carbon undergo oxidation by selenium dioxide (SeO₂) to form a -dicarbonyl compound.

Example:

Oxidation of aldehydes and ketones by sodium hypohalite (NaOX or X₂ + NaOH )

Among aldehydes, only acetaldehyde (CH₃CHO) and among ketones, only methyl ketones (R—COCH₃) participate in this reaction. It is known as a haloform reaction. To carry out the reaction, aldehydes or ketones having suitable structures are heated in the presence of excess halogen (Cl₂, Br₂ or I₂ ) in an alkaline medium. Chloroform, bromoform or iodoform are obtained as products in this reaction.

Example:

Aldehydes Ketones Notes

Baeyer-Villiger Oxidation Of Ketones

Ketones on being oxidised by organic peracids (peracetic acid, perbenzoic acid, trifluoroacetic acid, etc.) produce ethers directly. This reaction is known as Baeyer-Villiger oxidation. Aldehydes do not participate in this reaction.

Example:

In a crude sense, it can be stated that in the Baeyer-Villiger oxidation of ketone, the net result is the conversion of one of the alkyl groups of the parent ketone to an alkoxy group. , It has been found that the alkyl group having greater migratory aptitude is converted to an alkoxy group. The 1 order of migratory aptitude of some typical alkyl/aryl groups is as follows:

⇒ \(\mathrm{Me}_3 \mathrm{C} \rightarrow \mathrm{Me}_2 \mathrm{CH}-\simeq \mathrm{Ph} \rightarrow \mathrm{MeCH}_2 \rightarrow \mathrm{Me}\)

Class 12 Chemistry Unit 12 Aldehydes Ketones And Carboxylic Acids Reduction Reactions

Conversions Into Alcohols By Reduction

Aldehydes, as well as ketones, are reduced by hydrogen in the presence of catalysts like Ni, Pt, Pd or lithium aluminium hydride (LiAlH₄) or sodium borohydride (NaBH₄) to yield primary and secondary alcohols, respectively.

Aldehydes Ketones Notes

Reduction by LiAlH₄ and NaBH₄ is carried out in ether medium and methanolic medium, respectively.

Example:

⇒ \(\mathrm{HCHO} \text { (Methanal) }+\mathrm{H}_2 \stackrel{\mathrm{Ni}}{\longrightarrow} \mathrm{CH}_3 \mathrm{OH} \text { (Methanol) }\)

⇒ \(\mathrm{CH}_3 \mathrm{CHO} \text { (Ethanal) } \stackrel{\mathrm{LAlH}_4}{\longrightarrow} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} \text { (Ethanol) }\)

Some important information regarding the reduction of aldehydes and ketones.

1. H₂ gas acts as a strong reducing agent in the presence of a metallic catalyst. It reduces both the C=C and C=0 groups of unsaturated aldehydes and ketones to yield saturated alcohols.

2. LiAlH₄ is a stronger reducing agent than NaBH₄ and it reduces both the C=C and C=O groups of unsaturated aldehydes and ketones. However, NaBH₄ reduces only the C=O group.

Aldehydes Ketones Notes

3. LiAlH₄ reduces both the carbonyl and ester groups while NaBH₄ reduces only the carbonyl group.

4. Ketones are reduced by aluminium isopropoxide in isopropyl alcohol to the corresponding secondary alcohols. If any double or triple bond is present in the ketone molecule, it remains intact.

Aldehydes Ketones Notes

This reaction is known as Meerwein-Ponndorf-Verley reduction or MPV reduction.

Conversion Into Alkanes By Reduction

The carbonyl group of aldehydes and ketones can be reduced to methylene group ( —CH₂— ) by different methods and as a result, alkanes are produced.

Clemmensen Reduction

In this reaction, aldehydes and ketones are reduced by Zn-Hg and cone. HCl to yield saturated hydrocarbons (alkanes).

Example:

Wolff-Kishner Reduction

In this reaction, aldehydes or ketones are first converted into the corresponding hydrazones by reacting them with hydrazine. When hydrazones thus formed are heated with KOH or C₂H₅ONa at 180°C in ethylene glycol solvent, alkanes are obtained. The reaction is carried out In an atmosphere of nitrogen gas.

As Wolff-Kishner reduction requires a higher temperature (≅180°-200°C), so thermal stability of the carbonyl compound under consideration must be sufficiently high.

Example:

Aldehydes Ketones Notes

Example:

Conversion Into Pinacol By Reduction

When ketones are reduced with Mg-Hg in the presence of benzene (used as solvent) and the resulting solutions are acidified with dilute acid or water, symmetrical 1,2-diols or pinacols are produced. Aldehydes do not form pinacols.

Example:

Reduction With Red Phosphorus And Concentrated HI

When aldehydes and ketones are heated with red P and cone. HI at 150°C, alkanes are produced.

Aldehydes Ketones Notes

It is to be noted that if the carbonyl compound contains another functional group which is susceptible to acidic reagents, the conversion is to be carried out by using Wolff-Kishner reduction. On the other hand, if the other functional group is susceptible to basic reagents, the  conversion is to be carried out with the help of Clemmensen reduction.

Example:

Class 12 Chemistry Unit 12 Aldehydes Ketones And Carboxylic Acids Halogenation Reactions

Aldehydes and ketones containing O-H atoms react with halogens under different conditions. In these reactions, one or more α-H atoms are replaced by halogen atoms.

Example:

When acetone is mixed with bromine in equimolar proportion in glacial acetic acid Solution, monobromo acetone is obtained.

In alkaline solution, acetone reacts with bromine to yield tribromo acetone.

However, in the presence of excess alkali, bromoform (CHBr₃) is obtained:

⇒ \(\mathrm{CH}_3 \mathrm{COCBr}_3+\mathrm{NaOH} \rightarrow \mathrm{CH}_3 \mathrm{COONa}+\mathrm{CHBr}_3\)

Class 12 Chemistry Unit 12 Aldehydes Ketones And Carboxylic Acids Reactions With Alkali

Aldol Condensation

In the presence of dilute alkali (e.g., aqueous NaOH, Na₂CO₃, Ba(OH)₂, etc.) two molecules of aldehyde or ketone containing α-H atom combine to form β-hydroxyaldehyde or β-hydroxyketone. The reaction is called aldol condensation. In this reaction, a -carbon of one molecule becomes attached to the carbonyl carbon of another molecule to form a C —C σ-bond.

Example: Two molecules of acetaldehyde react with each other in the presence of dilute alkali to form aldol.

Aldehydes Ketones Notes

Two molecules of acetone react with each other in the presence of baryta or barium hydroxide [Ba(OH)₂ ] to yield diacetone alcohol.

Aldol condensation reaction was first accomplished using acetaldehyde as the reactant. Since the product of this reaction is both an aldehyde and an alcohol, it has been given the common name aldol (aid + ol = aldol).

Reaction Mechanism

The reaction is reversible and occurs in three steps.

First step: The hydroxide ion abstracts a proton from the a -carbon of one aldehyde molecule to give a resonance-stabilised carbanion. It thus follows that the α-hydrogen of aldehyde is acidic.

Second step: The resulting carbanion acts as a nucleophile and attacks the carbonyl carbon atom of a second molecule of aldehyde to form the alkoxide ion of aldol. AC —C σ-bond is formed in this step.

Third step: The alkoxide ion abstracts a proton from water to form aldol and OH~ ion is released.

Dehydration Of Aldol

Dehydration of aldol involves the elimination of one H-atom from a -carbon and one —OH from the β-carbon.

Dehydration is carried out by the following procedures:

Example:

When aldol is heated, one water molecule gets eliminated to form crotonaldehyde.

Aldehydes Ketones Notes

When diacetone alcohol is heated in the presence of an iodine catalyst, it undergoes dehydration to form an α, β-unsaturated ketone known as mesityl oxide along with a small amount of 4-methylpent-4-en-2.

When the product (β-hydroxy aldehyde or β-hydroxy ketone) of aldol condensation is heated in the presence of an acid catalyst, it undergoes dehydration to yield an α, β-unsaturated aldehyde or ketone.

Aldehydes Ketones Notes

When the product of aldol condensation is heated in the presence of alkali, it undergoes dehydration.

Crossed Aldol Condensation

Two different aldehydes or two different ketones or an aldehyde and a ketone having α-H atom can participate in an aldol condensation reaction in the presence of a basic catalyst. Aldol condensation of this type is called crossed aldol condensation. In this case, more than one product is obtained.

Example: In the reaction between ethanal (CH₃CHO) and propanal (CH₃CH₂CHO), a mixture of four aldols is produced.

Aldehydes Ketones Notes

In this case,

1. Two molecules of ethanal react to give the aldol-1,
2. Two molecules of propanal react to give aldol-2,
3. The carbonyl carbon of ethanal becomes attached with the or -carbon of propanal to yield the aldol-3 and
4. The carbonyl carbon of propanal becomes attached with the a -carbon of ethanal to produce the aldol-4.

Utility Of Crossed Aldol Condensation:

A crossed aldol condensation will be synthetically useful if one of the carbonyl compounds does not have an a -hydrogen atom because in that case mainly one product is obtained.

Example: In die-crossed aldol condensation involving formaldehyde and acetaldehyde, the chief product is P -hydroxy propionaldehyde.

In the reaction between formaldehyde and acetone, mainly 4-hydroxybutan-2-one is formed.

Aldehydes Ketones Notes

The formation of self-condensation products of acetaldehyde and acetone is avoided by the slow addition of these carbonyl compounds containing α-hydrogen to the mixture of the carbonyl compound lacking α-hydrogen i.e., formaldehyde and dilute alkali.

Claisen-Schmidt Reaction

Since aromatic aldehydes have no α-H atom, the crossed aldol condensation between aliphatic aldehydes or ketones having α-H atom and aromatic aldehydes is synthetically useful. But in this case, the product always undergoes dehydration in the presence of a dilute alkali (catalyst). Such reactions are called Claisen-Schmidt reactions.

Example:

Intramolecular Aldol Condensation

When 1,5 or 1,6-dicarbonyl compounds are heated with dilute alkali, an intramolecular aldol condensation reaction occurs to form a 5 or 6-membered cyclic compound.

Example:

Preparation of phorone from acetone by acid-catalysed aldol condensation: When HQ gas is passed through acetone, aldol condensation followed by dehydration occurs to yield mesityl oxide. This reacts further with an excess of acetone in a similar way to form a solid compound known as phorone.

Cannizzaro Reaction

Aldehydes that have no a -H atom undergo self-oxidation- reduction in the presence of 50% aqueous solution of NaOH or KOH. It means that half of the participating aldehyde molecules is reduced to alcohol and the other half is oxidised to carboxylic acid (as sodium or potassium salt).

This disproportionation or self-oxidation-reduction reaction is called the Cannizzaro reaction. Due to the absence of α-H atom, formaldehyde, trimethyl acetaldehyde, benzaldehyde or any other aromatic aldehyde participates in the Cannizzaro reaction.

Example:

When formaldehyde (HCHO) is heated with 50% NaOH solution, between the two participating molecules, one is oxidised to formic acid (salt) and the other is reduced to methyl alcohol.

Reaction Mechanism

The overall reaction is irreversible. The reaction occurs in three steps:

First Step: Nucleophilic attack by the OH^– ion on the carbonyl carbon occurs to give monoanion of a gem-diol.

Second step: The hydride ion originating from the anion, attacks the carbonyl carbon of a second molecule of aldehyde and as a result, one molecule of carboxylic acid and one molecule of alkoxide are obtained.

Third step: Rapid proton transfer from the carboxylic acid to the alkoxide ion occurs to form the salt of the acid and the primary alcohol (the stabler pair).

⇒ \(\mathrm{HCOOH}+\mathrm{CH}_3 \mathrm{O}^{\ominus} \stackrel{\mathrm{H}^{+} \text {transfer }}{\longrightarrow} \mathrm{HCOO}^{\ominus}+\mathrm{CH}_3 \mathrm{OH}\)

Crossed Cannizzaro Reaction

When two different aldehydes having no a-H atom react with each other in the presence of 50% NaOH or KOH, then the reaction is called crossed Cannizzaro reaction. In this reaction, all possible products (4 pairs) are obtained.

However, if one of the participating aldehydes is formaldehyde (HCHO), then it is preferentially converted into sodium formate by oxidation and the other aldehyde is reduced to an alcohol. In fact, between the two participating aldehydes, the one whose carbonyl group is more reactive (i.e., more positively polarised carbonyl carbon is susceptible to nucleophilic attack) is oxidised.

Example:

Intramolecular Cannizzaro Reaction

When ethanediol or glyoxal is heated with 50% NaOH or KOH solution, it undergoes an intramolecular Cannizzaro reaction. Between the two aldehyde groups, one is oxidised and the other is reduced forming sodium or potassium salt of glycolic acid.

Example:

Some Notable Points On Cannizzaro’s Reaction

Chloral (CCl₃CHO) does not undergo Cannizzaro reaction in the presence of alkali, even though it does not contain α-H atom. When it is heated with a cone. NaOH solution, C — C bond cleavage occurs to yield chloroform and sodium formate.

Although 2-methyl propanal contains one α-H atom, yet it undergoes Cannizzaro reaction in the presence of alkali.

When an aldehyde containing α-hydrogen is heated with a concentrated solution of alkali, a brown-coloured resin is obtained due to a polycondensation reaction.

When a mixture of acetaldehyde and excess formaldehyde is treated with a cone. NaOH solution, pentaerythritol, C(CH₂OH)₄, is formed as the final product. The reaction proceeds by repeated crossed Aldol condensation followed by crossed Cannizzaro reaction.

Pentaerythritol reacts with cone. HNO3 to give pentaerythritol tetranitrate (PETN).

When a mixture of propanal and excess methanol IN is treated with a cone. NaOH solution, the repealed crossed aldol condensation followed by crossed Cannizzaro reaction occurs.

Tischenko Reaction

Aldehydes having α-H atom undergo a Cannizzaro reaction in the presence of aluminium ethoxide. However, in this case, an ester is obtained instead of a carboxylic acid and an alcohol. This modified form of the Cannizzaro reaction is called the Tischenko reaction.

Example: Acetaldehyde gives ethyl acetate when treated with aluminium ethoxide.

Some important information regarding Aldol condensation and Cannizzaro reaction:

1. Compounds which participate in aldol condensation are—
   1. Aldehydes have α-H atom;
   2. Ketones having α-H atom;
   3. A mixture of two different aldehydes, both or one of them must contain α-H atom (crossed aldol condensation);
   4. A mixture of two different ketones, both or one of them must contain α-H atom (crossed aldol condensation);
   5. A mixture of an aldehyde and a ketone, both or one of which must have α-H atom (crossed aldol condensation).
2. Aldol condensation occurs In the presence of a dilute aqueous solution of NaOH, KOH, Ba(OH)₂, Na₂CO₃, K₂CO₃, etc.
3. Aldol condensation may also take place In the presence of dilute acid as a catalyst. However, the product Aldol undergoes dehydration immediately.
4. As formaldehyde (HCHO) and trimethylacetaldehyde [(CH₃)₃CCHO) contain no α-H atom, individually they do not undergo aldol condensation.
5. Compounds which undergo Cannizzaro reaction
   1. Aldehydes having no tr li atom;
   2. Mixture of two different aldehydes having no a II atom (crossed Cannizzaro reaction);
   3. 1,2-dialdehydes (e.g., OHC—CHO) or ketoaldehydes (e.g., C₆H₅COCHO) (intramolecular Cannizzaro reaction).
6. Cannizzaro reaction occurs In the presence of a concentrated solution of alkali (NaOH, KOH).

Class 12 Chemistry Unit 12 Aldehydes Ketones And Carboxylic Acids Miscellaneous Reactions Of Carbonyl Compounds

Reaction With Schiff Reagent

When aldehydes are shaken with a cold Schiff reagent, they readily restore the magenta (pink) colour of the reagent. Ketones cannot restore the colour of the Schiff reagent. So aldehydes and ketones can be differentiated by this reagent.

Schiff reagent: The aqueous solution of rosaniline hydrochloride is pink in colour. When SO₂ is passed through this, it becomes colourless. This colourless solution is called Schiff reagent.

To identify the —CHO group, the test of an aldehyde with Schiff reagent is performed in cold conditions and this is because the reagent on heating restores its magenta colour automatically. However, some aldehydes cannot restore the magenta colour of the Schiff reagent.

Reaction With Phosphorus Pantachloride

Both aldehydes and ketones react with phosphorus pentachloride to form gem-dichlorides by the replacement of the carbonyl oxygen atom with two chlorine atoms.

Example:

The gem-dichlorides, on hydrolysis, regenerate the parent carbonyl compounds.

Knoevenagel Reaction

Aldehydes and ketones react with active methylene compounds (e.g., malonic ester, acetoacetic ester, etc.) in the presence of a basic catalyst (e.g., pyridine) to form, unsaturated esters. This is called the Knoevenagel reaction.

Example:

The product of alkaline hydrolysis followed by acidification produces a dicarboxylic acid which, when heated, produces an α, β-unsaturated carboxylic acid.

Reformatsky Reaction

The β-hydroxy esters can be prepared from aldehydes and ketones by this reaction. When aldehydes or ketones are made to react with α-bromo esters in the presence of zinc in an inert solvent (ether, benzene etc.), additional compounds are obtained. These on acidification produce β-hydroxy esters.

Example:

When the product of the Reformatsky reaction is hydrolysed, and then heated, an α, β-unsaturated acid is obtained.

Wittig Reaction

A Wittig reagent may be prepared by allowing an alkyl halide (methyl, 1° or 2°) to interact with triphenylphosphine (Ph₃P), followed by treatment of the resulting phosphonium salt with a strong base like PhLi, BuLi, etc. For example, methylidene-triphenyl phosphorane, Ph₃P⁺—C^–H₂, may be prepared as follows:

The reaction of an aldehyde or ketone with a Wittig reagent [e.g., Ph₃P⁺—C⁺H₂ ) to form an alkene replacement of is known as the Wittig reaction.

Example:

Polymerisation Reaction

Among lower aldehydes, formaldehyde & acetaldehyde and among ketones, acetone form polymer in the presence of acid or alkali.

Polymerisation Reactions Of Formaldehyde

Aqueous solution of formaldehyde on evaporation yields a white crystalline solid known as paraformaldehyde or paraform. When paraformaldehyde is heated, it gives back formaldehyde on decomposition.

When gaseous formaldehyde is allowed to stand at ordinary temperature, it is converted into a solid compound known as metaformaldehyde trioxymethylene or trioxan.

Trioxan is also obtained when a 60% aqueous solution of formaldehyde is distilled in the presence of a few drops of cone. H₂SO₄.

Metaformaldehyde has no reducing property. This supports the cyclic structure of this compound. Formaldehyde is regenerated when meta-formaldehyde is heated.

When an aqueous solution of formaldehyde is allowed to stand in the presence of lime water [Ca(OH)₂] or baryta [Ba(OH)₂], a mixture of sugars having molecular formula C₆H₁₂O₆ is produced. This mixture of sugars is called formose.

Polymerisation Reactions Of Acetaldehyde

When a few drops of cone. H₂SO₄ is added to acetaldehyde at 25°C, and a cyclic trimer (CH₃CHO)₃ known as paraldehyde is obtained. It is a sweet-smelling colourless liquid. Due to the absence of a free aldehyde group, paraldehyde does not exhibit reducing properties. It is used as a hypnotic. Acetaldehyde is regenerated when paraldehyde is refluxed with dilute H₂SO₄.

When a few drops of cone. H₂SO₄ is added to acetaldehyde at 0°C or HCl gas is passed through acetaldehyde at 0°C, and a cyclic tetramer (CH₃CHO)₄ known as metaldehyde is obtained. It is a white solid and has no reducing property due to the absence of a free aldehyde group ( —CHO). It is used as a fuel. Metaldehyde on distillation with dilute H₂SO₄ gives back acetaldehyde.

When acetaldehyde is left to stand in the presence of cone. NaOH solution, an orange-coloured sticky liquid with an obnoxious smell is formed. It is a resin obtained as a result of the polymerisation of acetaldehyde.

Polymerisation Reactions Of Acetone

When acetone is distilled with a cone. H₂SO₄, three molecules of it condense to form an aromatic compound known as mesitylene or 1,3,5-trimethylbenzene.

Ring Substitutions Of Aromatic Aldehydes And Ketones

Since the aldehyde group ( —CHO) and ketonic group (—COR or —COAr) are electron-withdrawing, they are deactivating and meta-directing, Le., electrophilic substitutions in aromatic carbonyl compounds occur preferably at m-position.

Halogenation reaction

In aromatic aldehydes and ketones, side-chain halogenation occurs at a faster rate than nuclear halogenation.

Example:

However, when acetophenone reacts with Br2 in the presence of excess anhydrous aluminium chloride (AlCl₃), m-bromo acetophenone is produced.

Nitration Reaction

Aromatic aldehydes and ketones undergo nitration to give meta-substituted compounds.

Sulphonation Reaction

Aromatic aldehydes and ketones undergo sulphonation to yield meta-substituted compounds.

Example:

Condensation Reactions

Definition: A condensation reaction involves a combination of two or more molecules (same or different) to form compound having complex structure quite different from the starting compound.

In the reaction, one or more small molecules like H₂O, NH₃ etc., are eliminated and a new carbon-carbon or carbon-nitrogen bond is usually formed.

Condensation Reaction Of Acetaldehyde

Acetaldehyde reacts with hydroxylamine to form acetaldoxime. In this case, a new carbon-nitrogen double bond is formed and one molecule of water is eliminated.

Condensation Reactions Of Benzaldehyde

Claisen-Schmidt Reaction

Benzaldehyde, in the presence of dilute alkali, condenses with aliphatic aldehydes or ketones containing α-H atom to form α, β-unsaturated carbonyl compound. This aldol-type condensation is called Claisen-Schmidt condensation or Claisen reaction.

Example: Acetaldehyde and acetone react with benzaldehyde to form cinnamaldehyde and benzylidene acetone respectively.

The intermediate aldol readily eliminates a molecule of water because, in the resulting compound, a carbonyl group is in conjugation with the benzene ring via the newly generated olefinic bond.

Benzoin Condensation

When an aqueous ethanolic solution of benzaldehyde is refluxed with potassium cyanide, an α-hydroxy ketone known as benzoin is obtained. This reaction is named as benzoin condensation after the name of the product. This self-condensation reaction is catalysed by cyanide ion (CN^–).

An α-diketone known as benzil is obtained when benzoin is oxidised with HNO₃. When benzil is heated with ethanolic KOH, benzilic acid (as salt) is obtained. This reaction is known as benzil-benzilic acid rearrangement.

Perkin Reaction

When benzaldehyde is heated with the anhydride of an aliphatic acid (containing at least two H-atoms) in the presence of sodium salt of that acid, condensation occurs and a β-aryl acrylic acid is formed. This reaction is known as the Perkin reaction.

Example: Benzaldehyde reacts with acetic anhydride in the presence of sodium acetate to form cinnamic acid.

Preparation Of Malachite Green

When benzaldehyde is heated at 100°C with dimethylaniline in the presence of cone. H₂SO₄, a condensation reaction occurs to form a triphenylmethane derivative (colourless leuco base*). This colourless product is oxidised by lead dioxide in an acetic acid medium and then excess HCl is added to the mixture when a green organic dye called ‘Malachite green’ is obtained.

Condensation Reaction With Trinitrotoluene

Since trinitrotoluene is sufficiently acidic due to the presence of three electron-withdrawing —NO₂ groups, it undergoes a condensation reaction with benzaldehyde to yield trinitrotoluene.

Comparison between chemical reactions of aldehyde and ketone: similar reactions

Comparison between chemical reactions of aldehyde and ketone: dissimilar reactions

Preparation And Reactions of Aldehydes And Ketones:

Preparation Of Formaldehyde(HCHO):

Reaction Of Formaldehyde:

Preparation Of Acetaldehyde (CH₃CHO):

Reactions Of Acetaldehyde:

Preparation Of Acetone (CH₃COCH₃):

Reactions Of Acetone:

Preparation Of Benzaldehyde (C₆H₅CHO):

Reactions Of Benzaldehyde:

Uses of Aldehydes And Ketones

Uses of Formaldehyde, Acetaldehyde, Acetone and Benzaldehyde:

Identification Of Aldehydes And Ketones

Identification of Formaldehyde, Acetaldehyde, Acetone and Benzaldehyde:

Distinctive Chemical Tests

Acetaldehyde And Acetone:

Ethyl Alcohol And Acetone:

Ethyl Alcohol And Acetaldehyde:

Formaldehyde And Acetaldehyde

Formaldehyde And Acetone:

Ethyl Alcohol And Formaldehyde:

Pentan-2-One And Pentan-3-One:

Acetophenone And Benzophenone:

Acetaldehyde And Benzaldehyde:

Benzaldehyde And Acetophenone:

Transformations

1. Lower aldehyde (RCHO)→Higher aldehyde (RCH₂CHO):

2. Higher aldehyde (RCH₂CHO) → Lower aldehyde (RCHO):

Alternative method:

3. Acetaldehyde (CH₃CHO) → Acetone (CH₃COCH₃):

4. Acetone (CH₃COCH₃) → Acetaldehyde (CH₃CHO):

Alternative method:

5. Acetone (CH₃COCH₃) → n-propyl alcohol (CH₃CH₂CH₂OH):

Alternative method:

6. Formaldehyde (HCHO) Isopropyl alcohol (CH₃CHOHCH₃):

7. Ethyl alcohol (CH₃CH₂OH) → 2-butanol (CH₃CHOHCH₂CH₃):

8. Acetone (CH₃COCH₃)-Formaldehyde (HCHO):

9. Formaldehyde (HCHO) Ethyl alcohol (CH₃CH₂OH) and Acetaldehyde (CH₃CHO):

Alternative method:

10. Acetaldehyde (CH₃CHO) → n-propyl alcohol or 1-propanol (CH₃CH₂CH₂OH):

11. Ethanol (CH₃CH₂OH) → Acetone (CH₃COCH₃):

12. Propyne (CH₃—C = CH) → Acetone (CH₃COCH₃):

13. Acetylene (HC = CH) → Acetone (CH₃COCH₃):

14. Acetaldehyde (CH₃CHO) → Aldol, Crotonaldehyde, n-butyl alcohol and n-butyl chloride:

15. Acetaldehyde (CH₃CHO) Crotonicacid (CH₃—CH=CH—COOH) and n -butyric acid (CH₃CH₂CH₂COOH):

[In the preparation of crotonic acid from 2-butenal, Tollens’ reagent is used. Tollens’ reagent (ammoniacal silver nitrate solution) being a mild oxidising agent oxidises only the —CHO group to the —COOH group; the carbon-carbon double bond remains unaffected. If a strong oxidising agent like K₂Cr₂O₇ /H₂SO₄ is used, both the carbon-carbon double bond and the —CHO group will be attacked.

Again, if the reduction of CH₃CH=CHCHO is carried out first by H₂/Ni, both the carbon-carbon double bond and —CHO group are reduced. But when the reduction of CH₃CH=CHCOOH is carried out by H₂/Ni, only the carbon-carbon double bond is reduced while the —COOH group remains unaffected.]

16. Methane (CH₄) Formaldehyde (HCHO) and vice-versa:

17. Methanol (CH₃OH) Formaldehyde (HCHO) and vice-versa:

18. 1-butanol (CH₃CH₂CH₂CH₂OH) 2-butanone (CH₃COCH₂CH₃) :

Alternative method:

19. Acetone (CH₃COCH₃) Propyne (CH₃C= CH):

20. Formaldehyde (HCHO) n-butane (CH₃CH₂CH₂CH₃) :

Alternative method:

21. Acetaldehyde (CH₃CHO) — butane-2-one (CH₃COCH₂CH₃):

22. Acetone (CH₃COCH₃) Mesityl oxide [(CH₃)₂C=CHCOCH₃]:

23. Propanone (CH₃COCH₃) —’ Propene (CH₃—CH—CH₂):

24. Acetaldehyde (CH₃CHO) —»Malonic acid (HOOC—CH₂—COOH) :

25. Propyne (CH₃—C = CH) — Butanone (CH₃—CH₂—COCH₃) :

26. Benzaldehyde — m-hydroxybenzaldehyde:

27. Benzaldehyde (C₆H₅CHO) —> Mandelic acid (C₆H₅CHOHCOOH):

28. Benzaldehyde (C₆H₅CHO) —*Benzilic acid [Ph₂C(OH)COOH]:

29. Benzaldehyde (C₆H₅CHO) —» Phenylacetic acid (C₆H₅CH₂COOH) :

30. Benzaldehyde (C₆H₅CHO) — Benzophenone (C₆H₅COC₆H₅) :

Alternative method:

31. Benzaldehyde (C₆H₅CHO) Acetophenone (C₆H₅COCH₃):

32. Benzaldehyde (C₆H₅CHO) 3-phenylpropanoid-1-ol (C₆H₅CH₂CH₂CH₂OH) :

33. Benzene (C₆H₆) —*• m-nitroacetophenone:

34. Benzaldehyde (C₆H₅CHO) —*ÿ Cyanobenzene (C₆H₅CN):

35. Benzene (C₆H₅) p-nitrobenzaldehyde:

36. Toluene (C₆H₅CH₃) Cinnamaldehyde (C₆H₅CH=CHCHO):

37. Benzonitrile (C₆H₅CN) — Cinnamic acid (C₆H₅CH=CHCOOH) :

Carboxylic Acid Introduction

The —COOH group is called the carboxyl group. The carboxyl group is made up of a carbonyl group and a hydroxyl (—OH) group. The name has been derived from the union of the two terms ‘carb’ (from carbonyl) and ‘oxyl’ (from hydroxyl). The alkyl groups combine with the — COOH group to form aliphatic carboxylic acids (R-COOH). Similarly, the aryl groups combine with the —COOH group to form aromatic carboxylic acids (Ar —COOH).

For example—1. Aliphatic carboxylic acids (R—COOH): H—COOH (Formic acid), CH₃ — COOH (Acetic acid) etc.

2. Aromatic carboxylic acids (Ar—COOH):

The compounds containing the —COOH group release proton (H⁺) in an aqueous solution and because of this, these compounds are considered acids: R—COOH(carboxylic acid) RCOO^–+ H⁺. The compounds containing the carboxyl (—COOH) group are called carboxylic acids.

Class 12 Chemistry Unit 12 Aldehydes Ketones And Carboxylic Acids Classification Of Aliphatic Carboxylic Acids

Carboxylic acids are classified into several groups depending on the number of carboxyl groups (—COOH) present in a molecule of the organic acid—

Monocarboxylic Acids

Organic acids containing only one carboxyl group in their molecules are called monocarboxylic acids. These are monobasic acids and can be represented by the general formula C_nH_{2n +1}COOH.

Example: When n = 0, the acid is formic acid (HCOOH); when n = 1, the acid is acetic acid (CH₃COOH), when n = 2, the acid is propionic acid (CH₃CH₂COOH), etc.

Aliphatic monocarboxylic acids are generally known as fatty acids as the higher members of this series, viz., palmitic acid (C₁₅H₃₁COOH), stearic acid (C₁₇H₃₅COOH), etc., exist as esters in animal fats and are obtained by their hydrolysis. However, lower members of this series such as formic acid, acetic acid, etc., are not found in fats.

Dicarboxylic Acids

Organic acids containing two carboxyl groups in their molecules are termed dicarboxylic acids. Dicarboxylic acids are dibasic.

Example:

Tricarboxylic Acids

Organic acids containing three carboxyl groups in their molecules are called tricarboxylic acids.

Example:

Polycarboxylic acids: Organic acids containing more than two carboxyl groups are generally called polycarboxylic acids. Therefore, tricarboxylic acids may also be considered polycarboxylic acids.

Basicity of carboxylic acids: The number of carboxylic groups present in the molecule of a carboxylic acid gives the measure of its basicity.

Example: The basicities of monocarboxylic, dicarboxylic and tricarboxylic acids are 1, 2 and 3 respectively.

Class 12 Chemistry Unit 12 Aldehydes Ketones And Carboxylic Acids Nomenclature Of Carboxylic Acids

Common System Of Nomenclature:

The first few members of this homologous series are known well by their common names. The common names of carboxylic acids are derived from the Latin or Greek word related to their sources (plant or animal). There is no generalised rule for their nomenclature. Their names end with ‘ic acid’ only.

Straight-chain monocarboxylic acids are generally referred to as ‘normal (n)’ acids. If there is a (CH₃)₂CH— -group at the end of the carbon chain, the prefix ‘iso’ is added to the common name of the acid.

Example:

Carboxylic acids containing more than two carbon atoms are often named as derivatives of acetic acid.

Example:

In the case of branched or substituted carboxylic acids, the carbon atoms (except the —COOH group) are designated as α, β, γ, δ, etc., starting from the carbon atom next to the —COOH group to indicate the position of the substituent.

Example:

IUPAC System Of Nomenclature

The IUPAC names of carboxylic acids are derived by replacing the terminal ‘e’ from the name of the corresponding alkane with the suffix ‘oic acid’ (i.e., Alkane – e + oic acid = Alkanoic acid). For substituted acids, the numbering of the longest chain is to be done from the side of the —COOH group, i.e., by giving its carbon atom the serial number 1. The positions of the substituents are indicated by writing numerals 2, 3, 4, etc., before the names of the substituents.

Common and IUPAC names of some monocarboxylic acids:

Nomenclature Of Dicarboxylic Acids

Most of the aliphatic dicarboxylic acids are known by their common names. In the IUPAC system, names of the dicarboxylic acids are obtained by adding the suffix ‘dioic acid’ to the name of the parent alkane. Some of the examples are given below.

Common and IUPAC names of some dicarboxylic acids:

Nomenclature Of Tricarboxylic Acids

dicarboxylic acids If an unbranched carbon chain is directly linked to the three carboxyl groups, then the acid is named as a derivative of the parent alkane which does not include the carbon atoms of the carboxyl groups. However, when all three carboxyl groups are not directly linked to the unbranched carbon chain, the two similar carboxyl groups are included in the parent chain while the third carboxyl group is considered as a substituent.

Example:

Nomenclature Of Aromatic Carboxylic Acids

In the common system, the simplest aromatic monocarboxylic acid, C₆H₅COOH is called benzoic acid. The common names of substituted aromatic acids are derived by prefixing the name of the substituent with the name benzoic acid. The positions of substituents are indicated by the prefixes ortho-(o-), meta-(m-) and para-(p-).

In the IUPAC system, benzoic acid is called benzenecarboxylic acid. However, the common name benzoic acid has also been adopted by the IUPAC system. In this case, the positions of the substituents concerning the —COOH group are indicated by numerals 2, 3, 4, … etc. with the carbon attached to the group being numbered as 1.

Example:

Common name: Benzoic acid

IUPAC name: Benzenecarboxyllc acid or, Benzoic acid

Common name: O-aminobenzoic acid or, Anthranllic acid

IUPAC name: 2-aminobenzolc acid

Most of the aromatic polycarboxylic acids are known by their common names. In the IUPAC system, these are called benzene dicarboxylic acid, benzene tricarboxylic acid, etc. The relative positions of the carboxyl groups in these acids are indicated by Arabic numerals 1, 2; 1, 3; 1, 4; 1, 2, 3; 1, 2, 4; etc.

Nomenclature Of Cycloalkane Carboxylic Acids

If a —COOH group remains directly attached to a cycloalkane ring, the acid is named cycloalkane carboxylic acid.

Example:

If more than one —COOH groups are present in the ring, the acids are called cycloalkane dicarboxylic acid, cycloalkane tricarboxylic acid, etc. The relative positions of the —COOH groups are indicated by the numerals l, 2; 1, 3; 1, 4; 1, 2, 3; etc.

Common and IUPAC names of some aromatic and cycloalkane carboxylic acids:

Functional Group Isomerism

The monocarboxylic acids exhibit functional isomerism with monocarboxylic esters, hydroxy aldehydes and hydroxy ketones. Some examples are given below—

Example: The functional isomers of acetic acid (CH₃COOH) are:

The functional isomers of propionic acid (CH₃CH₂COOH) are:

Class 12 Chemistry Unit 12 Aldehydes Ketones And Carboxylic Acids Structure Of Carboxyl Group

The bonds attached to the carbon atom of the carboxyl group lie in the same plane and the angle between any two bonds is approximately 120°. The carboxyl group may be considered as a resonance hybrid of the following three structures:

In structures 1 and 3, the C-atom and two O-atoms have their octet filled up (eight electrons in their respective valence shells) while in structure 2, the C-atom has only six electrons. Therefore, the structures I and UI are relatively more stable than the structure II and consequently, their contributions to the hybrid are much higher than that of the structure 2.

Some Notable Points Regarding The Structh:

Since the hydroxylic oxygen atom of the carboxyl group becomes involved in resonance interaction with the C=O π bond, therefore, there exists a very small amount of positive charge on the central carbon atom. For this reason, the carbonyl carbon is less electrophilic than the carbonyl carbon of aldehydes and ketones.

Due to the considerable contribution of structure 3 in the resonance hybrid, in carboxylic acid—

1. The C=O bond length is slightly longer than that of aldehydes and ketones and
2. The C— O bond length is somewhat shorter than that of alcohols.

From the resonance structure of the carboxyl group, it becomes clear that the central C-atom and the two O-atoms are sp²-hybridised.

Class 12 Chemistry Unit 12 Aldehydes Ketones And Carboxylic Acids Genera Methods Of Preparation Of Carboxylic Acids

By oxidation of primary (1°) alcohols:

Primary alcohols on oxidation with strong oxidising agents e.g., acidified K₂Cr₂O₇ or Na₂Cr₂O₇ or alkaline KMnO4 or dilute HNO₃ produce carboxylic acids containing the same number of C-atoms. This oxidation occurs in two steps.

Example:

By Oxidation Of Aldehydes

Strong oxidising agents e.g., K₂Cr₂O₇ /conc.H₂SO₄ or mild oxidising agents e.g., Tollens’ reagent, Fehling’s solution or bromine-water oxidise aldehydes to carboxylic acids with the same number of carbon atoms.

Example:

By oxidation of secondary (2°) alcohols or ketones

When secondary alcohols or ketones are oxidised with strong oxidising agents, a mixture of more than one carboxylic acid is obtained. This method of preparing carboxylic acids is not used because it is difficult to separate the constituents from the mixture of acids.

Example:

By Oxidation Of Methyl Ketones

When methyl ketones are oxidised with sodium hypohalite (NaOX) or a mixture of halogen and alkali, haloform (CHX₃) and salt of carboxylic acid are obtained. The resulting salt on acidification liberates the carboxylic acid. The carboxylic acid obtained in this method contains one carbon less than that of the parent ketone.

Example:

By Oxidation Of Alkylbenzenes

Primary or secondary alkylbenzenes undergo oxidation with chromic acid (K₂Cr₂O₇/H₂SO₄) or alkaline KMnO₄ to yield benzoic acid. Due to the higher reactivity of the benzylic hydrogens as compared to other hydrogens on the alkyl chain, oxidation occurs at this carbon and thus all alkyl-benzenes give the same benzoic acid.

It means the entire side chain is oxidised to a —COOH group irrespective of the length of the carbon chain. Tertiary alkylbenzenes do not undergo oxidation due to the absence of benzylic hydrogen.

Example:

By Hydrolysis Of Alkyl Cyanides Or Nitriles

When alkyl cyanides are refluxed with dilute HCl or H₂SO₄, they undergo hydrolysis to form carboxylic acids.

Example:

Salts of carboxylic acids are obtained when alkyl cyanides are refluxed with dilute NaOH or KOH solution. These salts, on acidification, give the free carboxylic acids.

Example:

The resulting carboxylic acids have the same number of C-atoms as that present in the starting alkyl or aryl cyanides.

Preparation of carboxylic acids from alcohols or alkyl halides:

The resulting carboxylic acid contains one carbon atom more than the parent alcohol or alkyl halide.

From Grignard Reagents

When carbon dioxide gas is passed through a dry ethereal solution of Grignard reagent in cold conditions (0°C) and the resulting addition compound is hydrolysed, a carboxylic acid is obtained. Alternatively, when solid carbon dioxide (dry ice) is made to react with a dry ethereal solution of a Grignard reagent and the resulting addition compound is hydrolysed, a carboxylic acid is obtained.

Example:

The carboxylic acid produced in this method contains one C-atom more than that of the Grignard reagent.

Carboxylic acids can be prepared from the alkyl or aryl halides via the intermediate formation using Grignard reagents. The carboxylic acids thus obtained contain one carbon atom more than that in the alkyl or aryl halides. Here, CO₂ is the source of the —COOH group.

By hydrolysis of 1,1,1-tri haloalkane or haloform:

Alkaline hydrolysis of 1,1,1-tri haloalkanes or haloforms by NaOH or KOH solution followed by acidification of the resulting mixture lead to the formation of carboxylic acids.

Example: Chloroform produces formic acid in this method.

By Hydrolysis Of Esters

Esters, when heated with acid or alkali solution, undergo hydrolysis to yield carboxylic acids. Alkaline hydrolysis requires acidification to get the free acid. Alkaline hydrolysis of esters is called saponification.

Example:

By Hydrolysis Of Amides

Alkaline hydrolysis of amides followed by acidification produces carboxylic acids.

Example:

By Hydrolysis Of Acyl Halides And Acid Anhydrides

Acyl halides undergo slow hydrolysis by water and ready hydrolysis by alkali solution to yield carboxylic acids.

Acid anhydrides undergo ready hydrolysis by yielding carboxylic acids.

Example:

By Carboxylation Of Alkenes

When a mixture of an alkene, carbon monoxide (CO) and steam is heated at about 350-400°C temperature and under high pressure (50-5000 kPa) in the presence of phosphoric acid (H₃PO₄) catalyst, a carboxylic acid is obtained. This reaction is known as the Koch reaction.

Example:

From Sodium Alkoxides

Na-salts of carboxylic acids are obtained when sodium alkoxides are heated in the presence of carbon monoxide under high pressure (50-500 kPa). Sodium salts, on acidification, yield carboxylic acids. This reaction is also known as the Koch reaction.

Example:

From Malonic Acid Or Alkyl Malonic Acids

When malonic acid or alkyl malonic acids are heated at 150-200°C, monocarboxylic acids are obtained.

Example:

When dicarboxylic acids in which the two —COOH groups attached to the same carbon atom are heated at 150-200°C, they undergo decarboxylation to yield monocarboxylic acids.

Class 12 Chemistry Unit 12 Aldehydes Ketones And Carboxylic Acids Physical Properties Of Carboxylic Acids

Physical State And Odour:

The first three members of the monocarboxylic acid family (formic acid, acetic acid and propionic acid) are colourless-smelling liquids. Carboxylic acids containing 4 to 9 carbon atoms are oily liquids with an obnoxious smell but those with 10 or more carbon atoms are colourless solids.

Solubility

The first four members of the monocarboxylic acid family (formic acid, acetic acid, propionic acid and butyric acid) are completely soluble in water because their molecules form intermolecular H-bonds with water molecules.

1. The solubility of carboxylic acids decreases with increase in molecular mass. This is because with an increase in molecular mass, the size of the hydrophobic hydrocarbon part, i.e., the alkyl group, present in the molecule increases and as a consequence, solubility in water decreases.
2. Benzoic acid, the simplest aromatic carboxylic acid, is nearly insoluble in cold water since the non-polar hydrocarbon part outweighs the effect of the polar —COOH part. However, benzoic acid is sufficiently soluble in hot water.
3. Nearly all carboxylic acids are soluble in less polar organic solvents like benzene, ether, alcohol etc.

Boiling Point

The order of boiling points of alkanes, alcohols, ethers, carbonyl compounds (aldehydes and ketones) and carboxylic acids of comparable molecular masses may be given as follows: Alkanes < ethers < carbonyl compounds (aldehydes or ketones) < alcohols < carboxylic acids.

Explanation:

1. Only very weak van der Waals forces of attraction are operative among the molecules of alkane.
2. Besides the weak van der Waals forces of attraction, weak dipole-dipole attractive forces operate among ether molecules.
3. The dipole-dipole attractive forces operating among the molecules of carbonyl compounds are stronger than the similar forces operating among ether molecules and this is because the carbonyl compounds are more polar than ether molecules.
4. Due to the adjacent electron-withdrawing C=O group, the O— H bond in carboxylic acids is more strongly polarised as compared to that in alcohols and because of this, carboxylic acids form stronger intermolecular H-bonds than alcohols.
5. In fact, in both the vapour phase and aprotic solvents, most of the carboxylic acids exist as cyclic dimers in which the two molecules of the acid are held together by two strong hydrogen bonds. Resonance makes these H-bonds especially strong. It is for this reason the boiling points of carboxylic acids are higher than those of alcohols of comparable molecular masses.

The lower members of carboxylic acid exist as dimers in aqueous solution or even in the vapour state. But in the liquid state, they exist as polymers.

The boiling points of carboxylic acids increase with the increase in molecular masses due to an increase in van der Waals forces of attraction involving the non-polar hydrocarbon part.

Melting Point

For the first ten members of the carboxylic acid family—

The melting point of an acid containing an even number of carbon atoms is higher than the next hitter homologue containing an odd number of carbon atoms.

Explanation:

1. Carboxylic acids containing an even number of carbon atoms have carboxyl and terminal methyl groups on the opposite sides of the zig-zag carbon chain and hence, they fit better in the crystal lattice thereby increasing intermolecular forces of attraction resulting in higher melting points.
2. On the other hand, carboxylic acids containing an odd number of carbon atoms have the carboxyl and the terminal methyl groups on the same side of the zig-zag carbon chain.
3. Consequently, such molecules being less symmetrical, fit poorly in the crystal lattice. Therefore, the magnitude of intermolecular forces of attraction becomes relatively less in this case and so, these acids have lower melting points.

Example:

1. Carboxylic acid with an even number of C-atoms (—CH₃ and —COOH groups are on the opposite sides of the carbon chain), melting point =-5° C.
2. Carboxylic acid with an odd number of C-atoms (— CH₃ and —COOH groups are on the same side of the carbon chain), melting point = -35°C.
3. The melting points of aromatic acids are usually higher than those of aliphatic acids of comparable molecular masses. This is because the planar benzene ring in these acids can pack more closely in the crystal lattice than the zig-zag structure of aliphatic acids.

Class 12 Chemistry Unit 12 Aldehydes Ketones And Carboxylic Acids Acidic Property Of Carboxylic Acids

The strength of an acid primarily depends on—

1. The tendency of the acid to release proton (H+) and
2. The stability of the anion (conjugate base) is obtained as a result of proton release.

Resonance structure of the carboxylic acid

1. The tendency of carboxylic acids to release protons: In carboxylic acids, the bonding electrons of the O—H bond present in the —COOH group are displaced more towards the oxygen atom because the carbonyl group present in the —COOH group exerts its electron-withdrawing inductive (-I) effect and due to resonance, the hydroxyl oxygen becomes positively polarised. As a result, the O—H bond of the —COOH group becomes weak and dissociates easily to release proton (H⁺). Hence, carboxylic acids exhibit acidic properties.

2. Stability of the carboxylate anion: Both carboxylic acid and carboxylate ion are resonance hybrids and two resonance structures can be drawn for each of them.

Since one of the two resonance structures of the carboxylic acid involves the separation of charges, the carboxylic acid is not much stabilised by resonance. On the other hand, the two resonance structures of the carboxylate ion are equivalent and there is no separation of charges but, only a negative charge is delocalised.

Therefore, the carboxylate ion is relatively much stabilised by resonance. The difference in stabilisation causes the equilibrium to shift in the direction of increased ionisation, i.e., to the right. The carboxylic acids, therefore, exhibit acidic properties.

Dissociation Constants [Ka] Of Carboxylic Acids

In aqueous solution, the dissociation equilibrium of a carboxylic acid may be shown as follows:

⇒ \(\mathrm{RCOOH}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{RCOO}^{\ominus}+\mathrm{H}_3 \mathrm{O}^{\oplus}\)

According to the law of mass action, the equilibrium constant K_eq can be expressed as:

[K_eq = equilibrium constant, K_a = dissociation constant of the acid]

The strength of a carboxylic acid can be expressed by its dissociation constant, Ka and it is also called the acidity constant. The strength of an acid is also expressed by pKa.

⇒ \(p K_a=-\log K_a\)…….[3]

Explanation:

1. The greater the tendency of an acid (RCOOH) to release proton, the more it will dissociate in aqueous solution to yield a large number of H₃O⁺ ions, i.e., the acid will be stronger. The equation [2], therefore, suggests that the stronger the acid, the more will be its value of Ka. For example, formic acid (K_a = 1.77 x 10^-4) is a stronger acid than acetic acid (K_a = 1.8 x 10^-5).
2. From equation [3], it can be said that the smaller the numerical value of pKa, the stronger the carboxylic acid. For example, formic acid (pK_a = 3.77) is a stronger acid than acetic acid (pKa = 4.76). Inorganic acids are very strong. pKa values of hydrochloric and hydroiodic acids are -7 and -10 respectively. CF₃COOH (pK_a = 0.23) is the strongest acid among the carboxylic acids.

Nature of acids based on pK_a values

Comparison of acidic character of alcohols and carboxylic acids

The ionisation equilibrium of each of the two classes of compounds may be given as follows:

The resonance stability of the carboxylate ion is much greater than that of the carboxylic acid (already discussed) and because of this, the carboxylic acids tend to get converted into carboxylate ions by releasing proton. On the other hand, none of the alcohol (ROH) and the alkoxide ion (ROe) is stabilised by resonance and for this reason, alcohols do not tend to be converted into the alkoxide ions by releasing proton. The carboxylic acids are, therefore, much stronger acids than alcohols.

The weak O— H bond in the — COOH group of a carboxylic acid molecule dissociates easily due to the -I effect of the adjacent group (already discussed). Since there is no electron-attracting group attached to oxygen in an alcohol molecule, the O— H bond is relatively strong and does not dissociate easily to release protons. This explains why the carboxylic acids are much stronger acids than the alcohols.

Comparison Of Acidic Character Of Phenols And Carboxylic Acids

Both phenol and carboxylic acid are stabilised by resonance:

Similarly, their conjugate bases (RCOO^– and C₆H₅O^– )are also stabilised by resonance:

1. The two resonance structures of the carboxylate ion, i.e., structures I and II, are equivalent and contribute equally to the hybrid. In both of these structures, the negative charge is placed on the highly electronegative oxygen atom. Therefore, resonance in this case is much more effective.
2. On the other hand, the resonance structures of the phenoxide ion are not equivalent. In only two structures (3 and 7), the negative charge is placed on the highly electronegative oxygen atom but in the other three structures (4, 5 and 6), the negative charge is placed on the less electronegative carbon atom. Therefore, structures 3 and VII are more stable and more contributing while structures 4, 5 and 6 are less stable and less contributing.
3. Now, in structures I and II, the negative charge on the carboxylate ion is delocalised over two oxygen atoms while in structures III and VII, the negative charge on the oxygen atom remains localised. The electrons of the benzene ring are only delocalised. Since delocalisation of the ring electrons contributes little towards the stability of the phenoxide ion, the carboxylate ion is much more stabilised by resonance than the phenoxide ion.
4. As a consequence, the difference in stability between the carboxylic acid and the carboxylate ion becomes much more than the difference in stability between phenol and the phenoxide ion. For this reason, the carboxylic acids are stronger acids than phenols.

Relative Strength Of Aliphatic Carboxylic Acids

The nature of the substituent present in the molecule of a carboxylic acid influences the acidity of the acid. If the group is electron-repelling (having + I effect), the strength of the acid decreases but if the group is electron-attracting (having -I effect), the strength of the acid increases.

Effect of electron-releasing group: If an electron-releasing group is present in a molecule of carboxylic acid, the electron density on the oxygen atom in the —OH part of the —COOH group increases and consequently, the O —H bond acquires stability.

So, the O—H bond is not cleaved easily, i.e., the generation of a proton (H+) is hindered. Also, the electron-releasing group destabilises the carboxylate ion by intensifying the negative charge. Therefore, the tendency of the carboxylic acid to undergo dissociation decreases, i.e., its acidity decreases.

Since an alkyl group is electron-repelling (+I effect), the presence of an alkyl group as a substituent in the carboxylic acid decreases the strength of the acid. The +1 effect of different alkyl groups increases in the order: CH₃— < CH₃CH₂— < (CH₃)₂CH— < (CH₃)₃C—

Therefore, the strength of carboxylic acids containing the above alkyl groups follows the order:

1. There is no alkyl group in formic acid (HCOOH). Here the I —COOH group is directly attached to one H-atom. Since hydrogen has no electron-repelling property like an alkyl group, the acidity of formic acid is much higher than that of acetic acid (H—COOH > CH₃ — COOH).
2. Effect of electron-withdrawing group: If an electron-attracting group is present as a substituent in a molecule of carboxylic acid, the electron density of the hydroxyl oxygen atom of the —COOH group decreases and consequently, the O — H bond becomes weak. As a result, the cleavage of the O— H bond occurs easily and the release of H+ ions is facilitated. Also, an electron-attracting group stabilises the carboxylate ion by delocalising the negative charge. Therefore, the tendency of the carboxylic acid to undergo dissociation increases, i.e., its acidity increases.

Effect of the nature, number and position of the electron-attracting groups on the acidity of carboxylic acids

Effect of the nature of electron-withdrawing groups: The halogen atoms on the haloalkyl groups have -I effect, i,e., they are electron-withdrawing in nature. Consequently, their presence in the molecule of an acid increases its acidity. Since the electron-withdrawing nature of halogens (-1 effect) follows the order: F > Cl > Br > I, the strength of halogen-substituted acetic acids decreases in the same order:

The strength of the -I effect of some groups which increase the acidity of acids follows the order:

—C₆H₅ < —I < —Br <—Cl < —F < —CN < —NO₂ < —CF₃

Effect of the number of electron-withdrawing groups: The strength of the acid increases with an increase in the number of electron-withdrawing groups in a particular position of the carbon chain. Therefore, the acidic strength of chloroacetic acids decreases in the order:

Effect of the position of electron-withdrawing groups: ‘ Since the electron-withdrawing inductive effect (-I effect) decreases rapidly with distance, the dispersal of the negative charge of the corresponding carboxylate ion becomes less pronounced. Naturally, the acidity of the carboxylic acid decreases as the distance between the -I group and the —COOH group increases.

For example, the acidic strength of chlorobutanol acids decreases in the order:

Relative strengths of branched and unbranched isomeric carboxylic acids: In a series of isomeric carboxylic acids, the acid strength decreases as branching increases.

This is because as branching increases, the greater number of electron¬ releasing alkyl substituents (+1 effect) intensify the ve charge on the —COO^– group of the corresponding conjugate base resulting in the decreased stability and thereby causing a decrease in acidity of the acid. As an illustration let us consider isomeric C-5 -acids (i.e., pentanoic acids).

Order of stability of conjugate base: 1a > 2a > 3a > 4a

Effect of the type of hybridisation of the α-carbon:

When the —COOH group remains directly attached to groups like acetylenic, vinylic or phenyl, the acidity of the corresponding carboxylic acid increases. Although due to the +R effect of the double or triple bond, the acidity is expected to decrease, it increases due to greater electronegativity of sp² -and sp-hybridised C-atoms.

The +R effect of the vinyl group makes the C=O group less active in making hydroxyl oxygen-positive

The acidity increases as the electronegativity of the α-carbon increases or the s-character of its state of hybridisation increases. For example:

The order of decreasing acidity of some carboxylic acids (based on their pKa values) is as follows:

1. F₃CCOOH >Cl₃CCOOH >Cl₂CHCOOH >
2. O₂NCH₂COOH >NCCH₂COOH >FCH₂COOH >
3. ClCH₂COOH >BrCH₂COOH >HCOOH >
4. ClCH₂CH₂COOH >C₆H₅COOH >C₆H₅CH₂COOH >
5. CH₃COOH >CH₃CH₂COOH

Peracids (RCOOOH) are weaker than their corresponding parent acids (RCOOH). Peracetic acid (CH₃COOOH, pKa = 8.2), for example, is a weaker acid than acetic acid (CH₃COOH, pK_a = 4.76 ). This is because the conjugate base of acetic acid is stabilised by resonance while the conjugate base of peracetic acid is not (the C—O group is not in proper conjugation with the negative charge).

Benzoic acid (pK_a – 4.19) is a stronger acid than acrylic acid (pKa = 4.25). For having a greater number of sp2 -carbon atoms, the electron-donating effect of the phenyl group through the +R effect is less than that of the vinyl group. Moreover, delocalisation destroys the aromatic character of the benzene ring, this also decreases the electron-donating power of the phenyl group.

Strength of Aromatic Carboxylic Acids

In the benzoic acid (C₆H₅COOH) molecule, the —COOH group is attached to an sp²-carbon atom of the ring. Since the +R effect of the phenyl group is greater than its -I effect (+R>-I), benzoic acid is a weaker acid than formic acid (H —COOH). Again, in acetic acid (CH₃COOH), the —COOH group is attached to a —CH₃ group having a +1 effect. Since the electron¬ releasing +1 effect of the methyl group is greater than the net electron-releasing effect [-I + (+R)] of the phenyl group, benzoic acid is a stronger acid than acetic acid.

When a phenyl group is attached to a carboxyl group through one saturated carbon atom, the phenyl group exerts its weak electron-withdrawing inductive effect (-I effect). For this reason, phenylacetic acid is a stronger acid than acetic acid.

Relative Strength Of Aromatic Carboxylic Acids

The acidic strength of substituted benzoic acids depends on the nature and position of the substituent present in the aromatic ring.

Electron-withdrawing substituents (—NO₂, —Cl) tend to increase while electron-donating substituents ( — CH₃, —OCH₃) tend to decrease the acid strength of substituted benzoic acids relative to benzoic acid.

Acid-strengthening effect of the -R group (electron withdrawing substituent) and the acid-weakening effect of the +R group (electron-donating substituent) is more pronounced at para- than at meta-position. Because a -R group makes the carbon attached to the —COOH group positive while a +R group makes that carbon negative.

Irrespective of the nature (electron-releasing or electron-attracting) of the substituent, the ortho-substituted benzoic acids are nearly always stronger than benzoic acid. This is called ortho-effect.

Explanation of ortho-effect: In the benzoic acid molecule, the benzene ring and the —COOH group exist in the same plane. As a consequence, the C=0 group becomes involved in resonance interaction with the ring. But in ortho-substituted benzoic acids, due to steric interaction between the two groups, the —COOH group comes out of the plane of the ring and as a result, the resonance interaction between the ring and the group is inhibited. Consequently, the hydroxyl oxygen atom becomes more electron deficient and as a result, the acidic strength of the corresponding acid increases.

Relative Acid Strength Of Nitrobenzoic Acids

The nitro (— NO₂) group has a powerful electron-withdrawing resonance effect (-R effect) as well as an electron-withdrawing inductive effect (-I effect) and because of this, all the nitrobenzoic acids are stronger acids than benzoic acid.

The para-isomer is relatively stronger than the meta-isomer because the para-nitro group makes the carbon attached to the —COOH group partially positive. The ortho isomer is the strongest one because in this case—

1. The ortho-effect operates,
2. The -I effect of the — NO₂ group is most powerful (because distance from the —COOH group is minimal) and
3. The corresponding carboxylate ion is stabilised by intramolecular electrostatic interaction between the carboxyl and the nitro groups.

Relative Acid Strength Of Toluic Acids

O-toluic acid is the strongest acid due to its ortho-effect.

Due to the +I effect of the —CH₃ group, both m- and p-toluic acids are weaker acids than benzoic acid. Since the hyperconjugative effect operates at para-position but not at m-position, m-toluic acid is a stronger acid than p-toluic acid.

Relative Acid Strength Of Chlorobenzoic Acids

In chlorobenzene acid, the +R effect of Cl-atom is not very effective as the considerable p-orbital overlap between two portals (2p-orbital of C and 3p-orbital of Cl) of dissimilar size does not take place. Thus, the relative strength of chlorobenzene acid is controlled mainly by the -I effect of Cl-atom. Hence, the strength of those acids decreases with an increase in the distance between the —COOH group and the Cl-atom.

Relative Acid Strength Of Hydroxybenzoic Acids

The relative acid strength of hydroxybenzoic acids as compared to benzoic acid follows the order:

O-hydroxybenzoic acid is the strongest acid because the corresponding carboxylate ion is considerably stabilised by intramolecular H-bonding.

Since at meta-position, the —OH group cannot exert its +R effect but can only exert its -I effect, m-hydroxybenzoic acid is a stronger acid than benzoic acid. Due to the strong +R effect of the —OH group, the electron density of the carboxyl group of p-hydroxybenzoic acid increases and hence, it is a weaker acid than benzoic acid.

Relative Strength Of Methoxybenzoic Acids

Since at meta-position, the —OCH₃ group cannot exert its +R effect but can only exert its -I effect, m-methoxybenzoic acid Is a stronger acid than benzole acid. In p-methoxy benzoic acid, the —OCH₃ group exerts a strong +R effect (the effect is very weak due to the long distance from the —COOH group) and hence, it is a weaker acid than benzoic acid. O-isomer is slightly stronger than benzoic acid.

Class 12 Chemistry Unit 12 Aldehydes Ketones And Carboxylic Acids Chemical Properties Of Carboxylic Acids

The aliphatic and aromatic carboxylic acids are made up of an alkyl (R—) or an aryl (Ar— ) group and a carboxyl group (—COOH) (except formic acid). Again, the carboxyl group is made up of a carbonyl group and a hydroxyl group (—OH).

Naturally, the reactions of the carboxyl group may be additive, i.e., the amount of the total of the separate reactions of carbonyl and hydroxyl groups. But in reality, it is observed that the mutual influence of these two groups brings about some changes in the chemical properties of carboxylic acids. For example, the group of carbonyl compounds reacts with hydroxyl amine to yield oxime, with hydrazine to yield hydrazone, etc.

However, the carbonyl group of the carboxylic acids does not react with these ammonia derivatives. Again, the hydrogen atom present in the —OH group of carboxylic acids is more acidic than the hydrogen atom present in the —OH group of alcohols. So, when sodium bicarbonate (NaHCO₃) is added to alcohol, CO₂ gas is not liberated, while CO₂ gas is evolved when sodium bicarbonate is added to the solution of any carboxylic acid.

Chemical reactions of carboxylic acids can be divided into the following groups:

1. Replacement of H-atom of the carboxyl group,
2. Replacement of the —OH group,
3. Reactions of the carbonyl group,
4. Reactions of the carboxyl group and
5. Replacement of H-atom of the alkyl or aryl group.

Substitution Of Carboxylic H-Atom

The molecules of aliphatic and aromatic carboxylic acids (RCOOH) dissociate partially in aqueous solution to produce RCOO^– and H⁺ ions, i.e., they behave as weak acids.

Action On Litmus

Aqueous solutions of carboxylic acids turn blue litmus red.

Reaction With Metals

Metals in the(e.g., Na,k, Mg, Zn, etc.) occupying higher places in the electrochemical series react with carboxylic acids to form their respective salts liberating H₂ gas.

Example:

Reaction With Alkalis And Bases

Carboxylic acids with alkalis and bases form salts and water.

Example:

Reaction With Carbonates And Bicarbonates

Carboxylic acids decompose metallic carbonates and bicarbonates to liberate carbon dioxide gas.

Example:

⇒ \(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{NaHCO}_3 \rightarrow \mathrm{CH}_3 \mathrm{COONa}+\mathrm{CO}_2 \uparrow+\mathrm{H}_2 \mathrm{O}\)

⇒ \(2 \mathrm{CH}_3 \mathrm{COOH}+\mathrm{Na}_2 \mathrm{CO}_3 \rightarrow 2 \mathrm{CH}_3 \mathrm{COONa}+\mathrm{CO}_2 \uparrow+\mathrm{H}_2 \mathrm{O}\)

⇒ \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}+\mathrm{NaHCO}_3 \rightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{COONa}+\mathrm{CO}_2 \uparrow+\mathrm{H}_2 \mathrm{O}\)

Reaction With Ammonia

Ammonium salts are obtained when carboxylic acids react with ammonia.

Example:

The ammonium salts, when heated, produced acid amides.

Reaction With Diazomethane

When carboxylic acids are treated with an ethereal solution of diazomethane, corresponding methyl esters are obtained.

Example:

Substitution Of Hydroxyl Group

When the —OH group of carboxylic acids is substituted by alkoxy (—OR), amino ( — NH₂), chloro (—Cl) and acyloxy (—OCOR) groups, esters, amides, acid chlorides and acid anhydrides are produced respectively.

Formation Of Esters

In the presence of a suitable catalyst (cone. H₂SO₄ or dry HCl gas, alcohols react with carboxylic acids to form esters. This reversible reaction for the formation of ester is called esterification.

Example:

The order of reactivity of carboxylic acids towards ester formation reaction is:

R₃CCOOH < R₂CHCOOH < RCH₂COOH < CH₃COOH and that of alcohols Is ternary alcohol < secondary alcohol < primary alcohol < methyl alcohol. These observations may be explained by the fact that the rate of esterification involving a tetrahedral Intermediate In the rate determination step is sensitive to steric hindrance.

Reaction Mechanism

Formation Of Acid Amides

Ammonium salts produced in the reaction of NH₃ with carboxylic acids, when heated strongly, yield added amides.

Acid amides are also obtained when chlorides are reacted with NH₃. The resulting MCI combines with excess NH₃ to form ammonium chloride (NH₄Cl).

Example:

Formation Of Acid Chlorides

When carboxylic acids are made to react with phosphorus pentachloride (PCl₅), phosphorus trichloride (PCl₃) or thionyl chloride (SOCl₂), acid chlorides are obtained.

Example:

Among all the chlorinating agents, SOCl₂ is used preferably because all the by-products (SO₂, HCl) are gaseous. Hence separation of the acid chloride (product) from the reaction mixture is comparatively easier.

Amdt-Eistert synthesis: Any carboxylic acid converted into its next higher homologue by this reaction. Carboxylic acid is first converted into its acid chloride by reacting with PCl₅ or SOCl₂. The acid chloride is treated with diazomethane to yield a diazo ketone. When it is heated with Ag₂O (catalyst) and the resulting ketene is hydrolysed, the higher homologue is produced.

Formation Of Acid Anhydrides

When carboxylic acids are heated in the presence of phosphorus pentoxide, one molecule of water is eliminated from two molecules of carboxylic acid to form acid anhydrides.

Example:

Carboxylic acids also react with acid chlorides in the presence of pyridine as a base to form acid anhydrides.

Example:

Reactions Of Carbonyl Group

Reduction Reaction:

When carboxylic acids are reduced with a strong reducing agent, e.g., LiAlH₄ or H₂ in the presence of a copper chromite (CuCr₂O₄) catalyst, primary alcohols are produced.

Example:

It is to be noted that carboxylic acids cannot be reduced with sodium borohydride (NaBH₄) or Na/C₂H₅OH.

Reactions Of Carboxyl Group

Decarboxylation Or Removal Of The Carboxyl Group:

Decarboxylation: Na-salts of carboxylic acids (except formic acid) on being heated with soda lime produce alkanes containing one carbon less than parent acids.

Example:

Kolbe’s electrolysis decarboxylation: When concentrated aqueous solution of sodium or potassium salt of a carboxylic acid is electrolysed using Pt-electrodes, an alkane is liberated at the anode.

Example:

Ketones are produced when vapours of carboxylic acids are passed over MnO heated at 300°C temperature.

Example:

When Ca-salts of carboxylic acids (except formic acid) are subjected to dry distillation, ketones are obtained.

Example:

Hunsdiecker reaction: When the silver salts of carboxylic acids dissolved in carbon tetrachloride are treated with bromine, alkyl or aryl bromides containing one carbon atom less than the parent acids are obtained with the evolution of CO₂ gas.

Example:

Schmidt reaction: Carboxylic acids react with hydrazoic acid (HN₃) in the presence of a cone. H₂SO₄ to yield primary amines containing one carbon atom less than the parent acids.

Example:

Complete Reduction Of The Carboxyl Group

Alkanes are obtained when carboxylic acids are heated at 200-250°C with hydroiodic acid in the presence of red phosphorus. In this reaction, the —COOH group is converted into a —CH₃ group.

Example:

Substitution Of H-Atom Of Alkyl And Aryl Groups

Hell-Volhard-Zelinsky Reaction

Carboxylic acids react with Cl2 or Br2 in the presence of small quantities of red phosphorus to form exclusively chloro or a -bromo acids. This halogenation reaction is called the Hell-Volhard-Zelinsky reaction or simply, the HVZ reaction.

Example:

If the excess of halogen is used, more than one α-H -atoms are replaced by halogen atoms.

Preparation of α-hydroxy acids: When α-halo acids are treated with KOH solution, α-hydroxy acids are obtained.

Example:

Preparation of α-amino acids: α-amino acids are obtained when α-halo acids are treated with ethanolic ammonia.

Example:

Preparation of α-cyano acids and 1,1-dicarboxylic acids: α-halo acids react with ethanolic KCN to give α-cyano acids which on hydrolysis give 1,1-dicarboxylic acids.

Example:

Ring Substitution In Aromatic Acids

Due to the -I and -R effects of the —COOH group, the electrophilic substitution reactions in benzoic acid occur at meta-position and at a rate slower than benzene. The carboxylic acids, however, do not undergo Friedei-Crafts reaction because the —COOH group becomes strongly deactivating by co-ordinating with the Lewis acid AlCl₃.

Example:

Class 12 Chemistry Unit 12 Aldehydes Ketones And Carboxylic Acids Identification Of Carboxylic Acid

Litmus Test:

Aqueous or alcoholic solution of a carboxylic acid turns blue litmus red.

Sodium Bicarbonate Test:

When a saturated solution of NaHCO₃ is added to an aqueous or alcoholic solution of carboxylic acid, CO₂ gas is liberated in the form of bubbles.

Esterification Test:

When a carboxylic acid is treated with anhydrous ethanol in the presence of a small amount of concentrated H₂SO₄, an ester having a characteristic sweet odour is obtained.

⇒ \(\mathrm{RCOOH}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} \stackrel{\text { conc. } \mathrm{H}_2 \mathrm{SO}_4}{\rightleftharpoons} \mathrm{RCOOC}_2 \mathrm{H}_5+\mathrm{H}_2 \mathrm{O}\)

Preparation And Reactions Of Carboxylic Acids

Preparation Of Aliphatic Monocarboxylic Acids:

Reaction Of Aliphatic Monocarboxylic Acids:

Preparation Of Formic Acid (HCOOH):

Reaction Of Formic Acid:

Preparation Of Acetic Acid:

Reaction Of Acetic Acid:

Preparation Of Oxalic Acid (HOOC—COOH):

Reaction Of Oxalic Acid (HOOC—COOH):

Preparation Of Benzoic Acid (C₆H₅COOH):

Reaction Of Benzoic Acid (C₆H₅COOH):

Identification Of Carboxylic Acids

Identification Of Formic, Acetic, Oxalic And Benzoic Acid:

Uses Of Carboxylic Acids

Uses Of Formic, Acetic, Oxalic And Benzoic Acid:

Distinctive Chemical Tests

Formic Acid And Acetic Acid:

Acetic Acid And Acetone

Ethanol And Acetic Acid:

Formaldehyde And Formic Acid:

Oxalic Acid And Acetic Acid:

Phenol And Acetic Acid:

Benzoic Acid And Salicylic Acid:

Benzoic Acid And Phenol:

Transformations

1. Higher acid (acetic acid) from lower acid (formic acid):

Alternative method: acetic acid to propionic acid:

2. Lower acid (formic acid) from higher acid (acetic acid):

3. Ethanoic acid from propanoic acid:

4. Isopropyl alcohol from propionic acid:

5. N-Propylamine from acetic acid:

6. Ethylamine from acetic acid:

7. Methylamine from acetic acid:

8. Acetone from acetic acid:

9. Acetaldehyde from acetic acid:

10. Pyruvic acid from acetic acid:

11. Propionic acid from sodium ethoxide:

12. N-butane from propionic acid:

13. α-aminopropionic acid (alanine) from propionic acid:

14. α-hydroxyacetic acid from acetic acid:

15. Malonic acid from acetic acid:

16. Tert-butyl alcohol from acetic acid:

17. Propionic acid from acetylene:

Alternative method:

18. Acrylic add from propionic acid:

19. Methanol from acetic add:

20. Lactic acid from propionic acid:

21. Lactic acid from acetic acid:

22. Acetic acid from sodium formate:

23. Acetylene from acetic acid:

24. Acetic acid from acetylene:

25. Isopropyl alcohol from acetic acid:

26. Oxalic acid from formic acid and vice-versa:

27. Crotonic acid from acetic acid:

28. Methylamine and trichloroacetic acid from acetic acid (In one step):

29. Ethane from acetic acid:

30. Propane from acetic acid:

31. Ethyl bromide from propanoic acid:

32. Propanoic acid from propene:

33. Propenoic acid from ethanol:

34. Propenoic acid from propanoic acid:

35. M-hydroxybenzoic acid from benzoic acid:

36. m-bromoaniline from benzoic acid:

Class 12 Chemistry Unit 12 Aldehydes Ketones And Carboxylic Acids Very Short Answer Type

Question 1. why aliphatic aldehydes do not show position isomerism?
Answer:

In aliphatic aldehydes, the —CHO group always exists at the end of the carbon chain and for this reason, they do not exhibit position isomerism.

Question 2. Write the name of a straight-chain ketone which does not form a bisulphite addition compound.
Answer:

Diethyl ketone (CH₃CH₂COCH₂CH₃).

Question 3. Why in the preparation of acetaldehyde by the oxidation process, Na₂Cr₂O₇ is used instead of K₂Cr₂O₂?
Answer:

In the presence of ethanol, Na₂Cr₂O₇ easily dissolves in aqueous solution. However, the solubility of K₂Cr₂O₇ is much less in aqueous ethanol. For this reason, in the preparation of acetaldehyde by oxidation of ethanol, Na₂Cr₂O₇ instead of K₂Cr₂O₇ is used.

Question 4. What is formalin? Mention one of its uses.
Answer:

The 40% aqueous solution of formaldehyde is called formalin. It contains 8% methanol and 52% water. Formalin is widely used for preserving biological specimens.

Question 5. Is it possible to prepare formaldehyde by the Rosenmund reduction process? Explain your answer.
Answer:

It is not possible to prepare formaldehyde (HCHO) by the Rosenmund reduction process. The reason is that formyl chloride (HCOCl), the necessary reactant for preparing formaldehyde by this process, is an unstable compound at ordinary temperature.

Question 6. Write the name of an aldehyde which does not reduce Fehllng’s solution.
Answer:

Benzaldehyde (or any other aromatic aldehyde) does not reduce Fehling’s solution.

Question 7. Name an unstable derivative of formic acid.
Answer:

Formyl chloride (HCOCI). It dissociates readily to form CO and HCl.

Question 8. Which is the strongest monobasic fatty acid?
Answer: Formic acid (HCOOH).

Question 9. Write the structure and IUPAC name of the optically active fatty acid with the lowest molecular mass.
Answer:

Question 10. What happens when an excess of CI₂ gas is passed through boiling acetic acid?
Answer:

Trichloroacetic acid is obtained when excess Cl₂ gas is passed through boiling acetic acid.

⇒ \(\mathrm{CH}_3 \mathrm{COOH} \text { (boiling) } \stackrel{\text { excess } \mathrm{Cl}_2 \text { gas }}{\longrightarrow} \mathrm{CCl}_3 \mathrm{COOH}\)

Question 11. Is it possible to separate formic acid from its aqueous solution by fractional distillation?
Answer:

No. This is because the boiling points of formic acid (100.5°C) and water (100°C) are very close.

Question 12. Predict whether the solubility of carboxylic acids in water increases or decreases with an increase in molecular mass. Give reason.
Answer:

The solubility of carboxylic acids decreases with increase in molecular mass. This is because, with an increase in molecular mass, the size of the hydrophobic hydrocarbon part gradually increases.

Question 13. Arrange carboxylic acids, carbonyl compounds, ethers, alkanes and alcohols in order of increasing boiling point.
Answer:

The boiling point increases in the order: of alkanes <ethers < carbonyl compounds < alcohols < carboxylic acids.

Question 14. Which out of carboxylic acid (RCOOH) and carboxylate ion (RCOO^–) is more stabilised by resonance?
Answer:

The carboxylate ion (RCOO^–) is more stabilised than carboxylic acid (RCOOH) by resonance.

Question 15. Arrange in order of decreasing acidic strength: ClCH₂COOH, CI₃CCOOH and Cl₂CHCOOH.
Answer: CI₃CCOOH > Cl₂CHCOOH > ClCH₂COOH

Question 16. Mention the number of water of crystallisation In lead formate crystal.
Answer:

No water of crystallisation is present in lead formate [(HCOO)2Pb] crystals.

Question 17. Convert in one step: methanol → acetic acid.
Answer:

⇒ \(\mathrm{CH}_3 \mathrm{OH}+\mathrm{CO} \stackrel{\mathrm{I}_2 \text {-Rh catalyst }}{\longrightarrow} \mathrm{CH}_3 \mathrm{COOH}\)

Question 18. Why (CH₃)₃CCOOH does not take part In the HVZ reaction?
Answer:

There is no α-H atom in the (CH₃)₃CCOOH molecule.

Question 19. Arrange RC≡CH, RCOOH, ROH, H₂O and ArOH in order of decreasing acidic strength.
Answer:

The decreasing order of acidic strength is:

RCOOH > ArOH > H₂O > ROH > RC ≡ CH

Question 20. Which alkene produces acetone on ozonolysis?
Answer: 2,3-dimethyl but-2-ene

Question 21. Lower aldehydes and ketones are soluble in water—why?
Answer: They form hydrogen bonds with water molecules;

Question 22. What is the role of barium sulphate in the catalyst used in Rosenmund reduction?
Answer: Poison-catalyst

Question 23. Which reagent is used to differentiate aldehydes and ketones?
Answer: Tollens’ reagent

Question 24. Write the chemical name of Rochelle salt.
Answer: Alkaline sodium potassium tartrate

Question 25. Write a suitable reagent to oxidise an unsaturated alcohol into an unsaturated ketone.
Answer: Aluminium tertiary butoxide (acetone solvent)

Question 26. Write the structure and the IUPAC name of the compound obtained when propyne is passed through a hot H₂SO₄ solution in the presence of mercuric sulphate.
Answer: Propanone (CH₃COCH₃)

Question 27. Between which two atoms a new covalent bond is formed during the formation of a bisulphite addition compound?
Answer: Carbon and sulphur atom

Question 28. Arrange HCHO, CH3COCH3 and CH3CHO in order of increasing reactivity towards HCN.
Answer: HCHO > CH₃CHO > CH₃COCH₃

Question 29. What is the chemical name of urotropine? Mention one use of it.
Answer:

Hexamethylene tetramine; is used as antibacterial medicine in the treatment of urinary infection.

Question 30. Write the name and structure of a primary alcohol which responds to a haloform reaction.
Answer: Ethyl alcohol (C₂H₅OH)

Question 31. Write the name of a reagent which can be used to convert acetaldehyde into ethyl acetate in one step.
Answer: Aluminium ethoxide

Question 32. Write the name and structure of a simple aldehyde which cannot be prepared by Rosenmund reduction.
Answer: Formaldehyde (HCHO)

Question 33. Write the name and structure of an aldehyde which does not reduce Fehling’s solution.
Answer: Benzaldehyde

Question 34. Which compound undergoes oxidation and which undergoes a reduction in the crossed Cannizzaro reaction between benzaldehyde and formaldehyde?
Answer: Formaldehyde is oxidised and benzaldehyde is reduced

Question 35. Name the green-coloured dye that can be prepared from benzaldehyde
Answer: Malachite Green

Question 35. Which reagent can be used to convert carboxylic acids directly into primary alcohols?
Answer: LiAlH₄

Question 36. Write the structure of the acid having molecular formula C₂H₂O₄.
Answer: (COOH)₂

Question 37. What is obtained when cane sugar is oxidised with a cone.HNO₃?
Answer: Oxalic acid

Question 38. Write the structure of two functional isomers of acetic acid.
Answer: Methyl formate (HCOOCH₃); 2-hydroxyethanal (OH— CH₂—CHO)

Question 39. Which class of compound is obtained when a saturated carboxylic acid is dehydrated?
Answer: Acid anhydride

Question 40. What is the equivalent weight of anhydrous oxalic acid?
Answer: 45

Question 41. What is vinegar?
Answer: Dilute solution (4-10%) of acetic acid

Question 42. Which reagent can be used to distinguish between phenol and acetic acid?
Answer: Sodium bicarbonate

Question 43. The value of which dissociation constant (1st or 2nd) of oxalic acid is higher?
Answer: Ka1 > Ka2

Question 44. Which disease can be treated with the vapours of benzoic acid?
Answer: To resist the infection of a bronchial tube

Question 45. Which one is a stronger acid?

Answer:

Question 46. Why benzoic acid does not undergo Friedel-Crafts reaction?
Answer: The electron-attracting —COOH group deactivates the benzene ring

Question 47. What is obtained when benzoic acid is heated with P₂O₅?
Answer: Benzoic anhydride

Question 48. Which salt of benzoic acid is used as a preservative?
Answer: Sodium benzoate (C₆H₅COONa)

Class 12 Chemistry Unit 12 Aldehydes Ketones And Carboxylic Acids Short Answer Type

Question 1. In the preparation of acetaldehyde by oxidation of ethanol, why is a mixture of ethanol and dichromate slowly added to boiling sulphuric acid instead of heating the mixture of ethanol, and Na₂Cr₂O₇ and H₂SO₄?
Answer:

When a mixture of ethanol, Na₂Cr₂O₇ and H₂SO₄ is heated, ethanol is first oxidised to yield acetaldehyde which undergoes further oxidation in the presence of an excess oxidising agent to yield acetic acid. So, to avoid the formation of acetic acid, the mixture of dichromate and ethanol is added slowly to boiling H₂SO₄. In this process, acetaldehyde (b.p. 21°C) thus produced, instead of remaining in contact with an excess oxidising agent, is distilled out as it is formed and so, further oxidation to acetic acid is avoided. Ethanol having a higher boiling point (b.p. 78°C) is left behind in the solution.

Question 2. why are the boiling points of aldehydes higher than those expected from their molecular masses?
Answer:

The carbonyl group present in an aldehyde molecule is highly polar becomes evident from its resonance structures. Due to the presence of polar carbonyl groups, the aldehyde molecules are held together by strong dipole-dipole attractive forces. Therefore, a greater amount of thermal energy is required to separate these molecules by making them free from such attractive forces. For this reason, the boiling points of the aldehydes are higher than those expected from their molecular masses.

Question 3. How acetaldehyde is prepared directly from ethylene?
Answer:

See the Wacker process for the preparation of acetaldehyde.

Question 4. Starting from an alkene, how will you prepare an aldehyde containing one carbon atom more than that of the alkene?
Answer:

When an alkene is heated with CO and H₂ at high temperature in the presence of octacarbonyl dicobalt catalyst under high pressure, an aldehyde having one carbon atom more than that of the alkene is obtained (Oxo process).

Question 5. Although aldehydes are more susceptible to oxidation than alcohol, propanal may be prepared from 1-propanol by using acidified K₂Cr₂O₇ as an oxidising agent. Explain.
Answer:

During the preparation of an aldehyde by the oxidation of an alcohol, the oxidising agent used (K₂Cr₂O₇/H₂SO₄) is kept at the minimum possible concentration. Further, the product aldehyde (having a lower boiling point than the parent alcohol) is distilled out from the reaction mixture as soon as it is formed, thereby preventing further oxidation to carboxylic acid.

Question 6. Mention one chemical test to distinguish between acetaldehyde and other aldehydes.
Answer:

Acetaldehyde is the only aldehyde which contains the ketomethyl group (CH₃CO — ). So, it responds to the iodoform test, i.e., it yields yellow crystals of iodoform having a characteristic smell when heated with I₂/NaOH solution. Other aldehydes do not respond to this test.

Question 7. To prepare ketone from acid chloride, dialkyl cadmium, Instead of Grlgnard reagent, is preferably used. Why?
Answer:

As a nucleophilic reagent, Grignard reagent is much more reactive than dialkyl cadmium, because the electropositivity of Mg is more than that of Cd. Dialkyl cadmium reacts with acid chloride to form ketone but cannot produce tertiary alcohol by further reaction with the resulting ketone.

On the other hand, a more reactive Grignard reagent at first reacts with acid chloride to form ketone which in turn reacts with Grignard reagent to form tertiary alcohol. For this reason, dialkyl cadmium, instead of Grignard reagent, is preferred in the preparation of ketone from acid chloride.

Question 8. Which two alkynes on hydration give the same ketone?
Answer:

Both 1-butyne and 2-butyne on hydration yield the same ketone, CH₃COCH₂CH₃.

Hydration in the unsymmetrical alkyne 1-butyne occurs according to Markownikoff’s rule.

Question 9. Mention a suitable method for the preparation of an unsaturated ketone from an unsaturated 2° alcohol.
Answer:

Unsaturated 2° alcohols can be oxidised to unsaturated ketones by the oppenauer oxidation process. In this process, the 2° alcohol is made to react with aluminium tertiary butoxide in acetone. Under this condition, the secondary alcohol is oxidised to ketone but the double or the triple bond present in the alcohol molecule is not attacked by the oxidising agent.

Question 10. Identify the dichloroalkane which on alkaline hydrolysis produces a ketone having molecular formula C₃H₆O.
Answer:

The ketone having molecular formula C₃H₆O is acetone (CH₃COCH₃). 2, 2-dichloropropane on alkaline hydrolysis yields this ketone.

Question 11. Ketones have higher boiling points than their isomeric aldehydes. Explain.
Answer:

The carbonyl carbon of an aldehyde is attached to only one electron-releasing (+I effect) alkyl group while the carbonyl carbon of a ketone is attached to two electron-releasing alkyl groups.

So, ketones are more polar than their isomeric aldehydes. As a consequence, the magnitude of dipole-dipole attractive forces operating among the molecules of a ketone are greater than those operating among the molecules of an aldehyde. This accounts for the slightly higher boiling point of a ketone as compared to its isomeric aldehyde.

Question 12. The values of dipole moments of aldehydes and ketones are higher than those of alcohols, even though a polar C —O bond is present in both compounds. Explain with reason.
Answer:

In the molecules of aldehydes and ketones, there is a weak H-bond between the C and the σ-atoms and bond between them. Since oxygen is more electronegative than carbon, the electron pair of the weak π-bond is shifted more towards the oxygen atom. The molecules of aldehydes and ketones can be represented by the following resonance hybrids:

Hence, the positive and negative charge densities on carbon and oxygen atoms of the C=O group of aldehydes and ketones are much greater than the values of charge densities on the carbon and oxygen atoms of the C —O bond of alcohols. So, the dipole moments of aldehydes and ketones (2.3-2.8D) are higher than those of alcohols (1.6-1.8D).

Question 13. Arrange the following compounds in increasing order of their tendency to participate in nucleophilic addition reactions CH₃CHO, CH₃COCH₃, HCHO, C₂H₅COCH₃.
Answer:

The tendency of carbonyl compounds to undergo nucleophilic addition reaction decreases with a decrease in electron deficiency of the carbonyl carbon atom.

Again, with an increase in the number and size of the alkyl groups attached to the carbonyl carbon atom, the tendency of carbonyl compounds to undergo nucleophilic addition reaction decreases due to the steric effect.

Therefore, based on these two effects, it may easily be predicted that the tendency of nucleophilic addition reaction decreases in the following order:

(In the case of compound J, the electron deficiency on the carbonyl carbon is maximum and the steric effect Is minimum. In the case of compound IV, the electron deficiency on the carbonyl carbon is minimal and the steric effect Is maximum.)

Question 14. Arrange the following compounds In Increasing order of their reactivity towards HCN: formaldehyde, acetone, di-tert-butyl ketone, and methyl tert-butyl ketone.
Answer:

The reactivity of carbonyl compounds towards a nucleophilic reagent (e.g., HCN) increases with an increase in electron deficiency at the carbonyl carbon and with a decrease in steric effect. Therefore, the order of reactivity of the given compounds towards HCN is as follows—

Question 15. Pure HCN does not react with aldehydes and ketones—Why? What type of catalyst will cause the reaction to occur? Explain.
Answer:

HCN is a covalent compound and behaves as a very weak acid. Hence, the concentration of CN^– ion (nucleophile) produced by the dissociation of this weak acid is very low to bring about the nucleophilic attack on the carbonyl carbon atom of aldehydes and ketones.

The reaction occurs in the presence of a basic catalyst. In an alkaline medium, due to the reaction between OH^– ion and HCN, the concentration of CN^– ion increases appreciably and as a consequence, HCN can form additional compounds with ‘aldehydes and ketones.

Question 16. Aldehyde and ketones have higher boiling points than alkanes of comparable molecular masses but lower boiling points than alcohols and carboxylic acids of comparable molecular masses. Explain with reason.
Answer:

The non-polar alkane molecules are held together only by weak van dar Waak force of attraction. The polar molecules of aldehydes and ketones, on the other hand, are held together by relatively stronger dipole-dipole attractive forces. Again, due to the presence of the —OH group in alcohols and carboxylic acids, their molecules remain associated through the formation of intermolecular H-bonds.

Since the strength of intermolecular attractive forces follows the order: van der Waals forces of attraction < dipole-dipole attractive forces < H-bonding, the boiling points of aldehydes and ketones are higher than those of alcohols and carboxylic acids of comparable molecular masses.

Question 17. Hydrazone or oxime of aldehydes and ketones cannot be prepared in a strongly acidic medium—Why?
Answer:

Hydrazine and hydroxylamine react with aldehydes or ketones to form hydrazones and oximes respectively. In these reactions, hydrazine and hydroxylamine act as nucleophiles.

But in a strongly acidic medium, these reagents undergo protonation to form their conjugate acids. These conjugate acids can no longer function as nucleophiles. No reaction of aldehydes or ketones occurs with these reagents in a strong acidic medium.

Question 18. In the reaction with aldehydes and ketones, which nitrogen atom of phenylhydrazine takes part in the formation of a bond with the carbonyl C-atom and why?
Answer:

In the reaction with aldehydes and ketones, phenylhydrazine acts as a nucleophile. Hence, out of two N-atoms present in phenyl hydrazine, the one having greater availability, of lone pair will take part in the formation of a bond with the carbonyl C-atom.

The lone pair of electrons on the N-atom adjacent to the phenyl group becomes involved in resonance interaction with the ring. As a result, this lone pair of electrons is scarcely available for bond formation. So, the N-atom lying apart from the phenyl group forms a bond with the carbonyl carbon.

Question 19. In a semicarbazide molecule, all three N-atoms contain a lone pair of electrons. But in the reaction with aldehydes and ketones, only a particular N-atom participates in bond formation with the carbonyl carbon. What is the reason?
Answer:

In the reaction with aldehydes and ketones, semicarbazide acts as a nucleophilic reagent. Out of the three N-atoms, the lone pair of electrons of the two N-atoms adjacent to the carbonyl group becomes involved in resonance interaction with the carbonyl group.

So, these two lone pairs of electrons are not easily available. However, the lone pair of electrons on the third N-atom, not being involved in resonance, is easily available. Therefore, this N-atom lying further from the carbonyl group forms a bond with the carbonyl carbon of aldehydes and ketones.

Question 20. Explain why during the reactions of ammonia derivatives (e.g., NH₂NH₂, C₆H₅NHNH₂, NH₂OH etc.) with aldehydes and ketones, the pH of the medium is to be controlled.
Answer:

The reactions of ammonia derivatives with aldehydes and ketones are carried out in a weak acidic medium (pH ≈ 3.5). In the presence of an acidic catalyst, the ) group of the aldehydes and ketones becomes protonated to produce the cation in which the extent of electron deficiency on carbonyl carbon increases.

As a result, the carbonyl carbon undergoes nucleophilic attack readily and the rate of the reaction increases. Experimental results show that the rate of the reaction becomes maximum when the pH of the reaction medium lies in the vicinity of 3.5.

If the pH of the reaction medium is very low, i.e., the concentration of H⁺ ion is very high, then the ammonia derivatives undergo protonation to form cations which cannot act as nucleophiles and naturally, reactions with carbonyl compounds do not take place.

The extent of electron deficiency  on the carbonyl carbon is low (reactions with nucleophilic reagents do not take place easily) The extent of electron deficiency on carbonyl carbon is much higher than  (reactions with nucleophilic  reagents take place easily)

⇒ \(\mathrm{Z}-\ddot{\mathrm{NH}}_2 \text { (nucleophile) }+\stackrel{\oplus}{\mathrm{H}} \rightleftharpoons \mathrm{Z}-\stackrel{\oplus}{\mathrm{N}} \mathrm{H}_3 \text { (not a nucleophile) }\)

Question 21. Unlike ordinary aldehydes, chloral forms stable hydrate—Why?
Answer:

Since the carbonyl carbon of chloral is attached to the strong electron attracting —CCl₃ group, it is highly electron deficient. Due to this, H₂O, despite being a very weak nucleophile, reacts very easily with chloral to yield the hydrate.

Moreover, the hydrate thus obtained is stabilised by intramolecular hydrogen bonding involving Cl atoms as well as by the electron-withdrawing inductive effect of the —CCl₃ group that prevents loss of H₂O to regenerate chloral. For these reasons, chloral forms a stable hydrate.

Question 22. How can aldehydes and ketones be purified by using sodium bisulphite?
Answer:

Impure aldehydes or ketones, when treated with a saturated sodium bisulphite solution, form a crystalline precipitate of bisulphite addition compounds. The resultant addition compounds are separated by filtration and then hydrolysed with dilute HCl or NaOH solution when pure aldehydes or ketones are regenerated.

Question 23. When an aldehyde reacts with HCN, a mixture of two isomeric compounds is obtained. Explain why it is not possible to separate the two isomers from that mixture by fractional distillation?
Answer:

The cyanohydrin obtained as a result of the reaction between an aldehyde and HCN is, in fact, an equimolecular mixture of a pair of enantiomers (optically active isomers).

All the physical (except their behaviour towards plane polarised light) and chemical (except their behaviour towards a chiral reagent) properties of enantiomers are identical. Thus, the boiling points of these two enantiomers are the same and cannot be separated by fractional distillation.

Question 24. Halogen acids (HX) form additional compounds with alkenes but not with carbonyl compounds—Why?
Answer:

Carbon-carbon double bonds in alkenes act as nucleophiles. On the other hand, halogen acids ( H^δ+ —X^δ-) act as electrophiles and easily react with alkenes to produce alkyl halides. However, the carbonyl group present in carbonyl compounds acts as electrophile. This electrophile is unable to react with another electrophile (in this case HX). Thus, carbonyl compounds do not react with halogen acids.

Question 25. HCN forms additional compounds with carbonyl compounds but not with alkenes —Why?
Answer:

The carbonyl group of aldehydes and ketones acts as an electrophile. Again, in the presence of a basic catalyst, HCN acts as a nucleophile.

⇒ \(\mathrm{H}-\mathrm{CN}+\mathrm{OH}^{\ominus} \rightleftharpoons \mathrm{H}_2 \mathrm{O}+\mathrm{CN}^{\ominus} \text { (nucleophile) }\)

So, HCN easily reacts with carbonyl compounds in the presence of a basic catalyst to form an additional compound cyanohydrin. On the other hand, the carbon-carbon double bond of alkenes acts as a nucleophile. This nucleophile is unable to react with another nucleophile (here CN^–) and hence, HCN does not react with alkenes to form additional compounds.

Question 26. Mention three processes for converting the group into )CH₂ group.
Answer:

1. Clemmensen reduction (Zn-Hg + cone. HCl) .
2. Wolff-Kishner reduction [NH₂NH₂, KOH or C₂H₅ONa, ethylene glycol, 180°C ]
3. Reduction by red P and HI, 150°C.

Question 27. Formaldehyde and benzaldehyde respond to Cannizzaro reaction but acetaldehyde does not— Why?
Answer:

Aldehydes having no a-H atom undergo the Cannizzaro reaction in the presence of a strong alkali (50% NaOH solution), while the aldehydes having α-H atom do not take part in the Cannizzaro reaction. Formaldehyde and benzaldehyde do not contain α-H atoms while acetaldehyde contains three α-H atoms. So, formaldehyde and benzaldehyde undergo a Cannizzaro reaction while acetaldehyde does not.

Question 28. Acetaldehyde participates in aldol condensation but trimethylacetaldehyde does not. Explain.
Answer:

Aldehydes having α-H-atom participate in aldol condensation reactions. As acetaldehyde contains α-H atom, it participates in the aldol condensation reaction. On die other hand, trimethylacetaldehyde having no α-H-atom does not take part in the aldol condensation reaction.

Question 29. Name the reagent with the help of which most of the aldehydes participate in a reaction similar to the Cannizzaro reaction. Write the equation.
Answer:

Most of the aldehydes (the presence or absence of a-H is not important) undergo a special reaction (Tischenko reaction) very similar to the Cannizzaro reaction. But here, the resulting alcohol and acid do not exist separately. Instead, they unite together to form an ester. For example:

Question 30. Convert acetone into an aromatic compound.
Answer:

Acetone on distillation with concentrated H₂SO₄ produces mesitylene, an aromatic compound.

Question 31. What products will be produced if a mixture of HCHO and DCDO is treated with a 50% NaOH solution?
Answer:

In this case, due to normal and crossed Cannizzaro reaction, the products expected to be formed are CH₃OH, CD₃OH, CH₂DOH, CHD₂OH, HCOONa and DCOONa.

Question 32. How will you convert formaldehyde into methanol without using any reducing agent?
Answer:

When formaldehyde is treated with 50% NaOH solution, it undergoes a disproportionation reaction to yield a mixture of methanol and sodium formate. Therefore, without using any reducing agent formaldehyde can be converted into methanol through the Cannizzaro reaction.

⇒ \(\mathrm{HCHO}+\mathrm{HCHO} \stackrel{50 \% \mathrm{NaOH}}{\longrightarrow} \mathrm{CH}_3 \mathrm{OH} \text { (methanol) }+\mathrm{HCOONa}\)

Question 33. Write the name (in the IUPAC system) of an aldehyde (aliphatic) other than formaldehyde which undergoes the Cannizzaro reaction.
Answer:

Aldehydes having no α-H atom undergo the Cannizzaro reaction. Trimethylacetaldehyde [(CH₃)₃CCHO] is such an example whose IUPAC name is 2,2-dimethyl propanal.

Question 34. Give two examples of reactions in which formaldehyde participates but acetaldehyde does not.
Answer:

Formaldehyde containing no a-H atom undergoes a Cannizzaro reaction but acetaldehyde containing α-H atom does not undergo a Cannizzaro reaction.

⇒ \(\mathrm{HCHO}+\mathrm{HCHO} \stackrel{50 \% \mathrm{NaOH}}{\longrightarrow} \mathrm{HCOONa}+\mathrm{CH}_3 \mathrm{OH}\)

Formaldehyde reacts with ammonia to form urotropine instead of an additional compound. On the other hand, acetaldehyde reacts with ammonia to form an additional compound known as acetaldehyde-ammonia.

⇒ \(6 \mathrm{HCHO}+4 \mathrm{NH}_3 \longrightarrow\left(\mathrm{CH}_2\right)_6 \mathrm{~N}_4 \text { (urotropine) }+6 \mathrm{H}_2 \mathrm{O}\)

Question 35. Which isomer of butyl alcohol will give a positive iodoform test (give structure)?
Answer:

Alcohols that on oxidation produce compounds having (CH₃CO—) group, i.e., alcohols with [CH₃CH(OH)— ] group give positive iodoform test. Thus, an isomer of butyl alcohol that responds to the iodoform test is:

⇒ \(\mathrm{CH}_3-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}_2-\mathrm{CH}_3 \text { (Butan-2-ol) }\)

Question 36. Differentiate between acetaldehyde and benzaldehyde through an experiment with visible change.
Answer:

When acetaldehyde is heated with I2 in the presence of NaOH solution, yellow crystals of iodoform are precipitated. A similar reaction does not take place in the case of benzaldehyde.

Question 37. What happens when acetylene gas Is passed through hot glacial acetic acid in the presence of HgSO₄ and the product obtained is distilled?
Answer:

Acetylene reacts with glacial acetic acid in the presence of HgSO₄ to form ethylidene diacetate. During distillation, it decomposes to yield acetaldehyde and acetic anhydride. The low boiling (21°C) acetaldehyde comes out as distillate.

Question 38. How will you carry out the following conversion: acetylene into dichloro acetaldehyde?
Answer:

When acetylene gas is passed through a hypochlorous acid solution, dichloroacetaldehyde is obtained.

Question 39. Under which condition acetone undergoes an iodoform reaction but ethanol does not?
Answer:

When acetone is treated with I₂ in the presence of a weak base like ammonium hydroxide, iodoform is obtained but under this condition, ethanol does not form iodoform.

Question 40. What is the chemical compound present in Brady’s reagent? For which purpose is it used?
Answer:

The chemical compound present in Brady’s reagent is 2,4-dinitrophenylhydrazine. This reagent is used in the identification of carbonyl compounds (aldehydes or ketones). When a carbonyl compound is treated with Brady’s reagent, 2,4-dinitrophenylhydrazone is precipitated as yellow or orange crystals.

Question 41. Complete the following reactions and mention the names of the products:

Answer:

Question 42. What is bakelite? Mention its preparation and use.
Answer:

Formaldehyde undergoes a condensation reaction with phenol in the presence of an alkaline catalyst (NaOH or NH₄OH ) to produce a kind of polymer called phenol-formaldehyde resin or bakelite.

⇒ \(\text { Phenol }+ \text { Formaldehyde } \stackrel{\text { catalyst }}{\longrightarrow} \text { Phenol-formaldehyde resin }\)

Uses: Being an electrical insulator, bakelite is used in making electrical goods. Bakelite is a solid substance at ordinary temperature but it becomes soft when heated. Thus, articles of different shapes are made by pouring molten bakelite into moulds.

Question 43. Write the structural formulas and IUPAC names of the isomeric aldehydes and ketones having the molecular formula C₅H₁₀O.
Answer:

Question 44. What happens when propanal is made to react with excess formaldehyde in the presence of sodium hydroxide solution?
Answer:

The reaction occurs in three steps. In the first two steps, two aldol condensation reactions occur successively and in the third step, a crossed Cannizzaro reaction takes place.

Question 45. In which of the following cases Clemmensen reduction and Wolff-Kishner reduction reaction should be employed and why?

1. BrCH₂CH₂CH₂COCH₃→BrCH₂CH₂CH₂CH₂CH₃
2. HOCH₂CH₂COCH₃→HOCH₂CH₂CH₂CH₃

Answer:

Clemmensen reduction is carried out in an acidic medium, but Wolff-Kishner reduction is carried out in an alkaline medium. If the reduction of halo ketone is carried out in an alkaline medium, then along with reduction, dehydrobromination (loss of HBr) will also take place. So, BrCH₂CH₂CHCOCH₃ is reduced by Clemmensen reduction instead of Wolff-Kishner reduction.

⇒ \(\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COCH}_3 \underset{\text { conc. } \mathrm{HCl}}{\stackrel{\mathrm{Zn} / \mathrm{Hg}}{\longrightarrow}} \mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3\)

If the hydroxy ketone is reduced in an acidic medium, then dehydration along with reduction will take place. So, HOCH₂CH₂CH₂COCH₃ is reduced by Wolff-Kishner reduction instead of Clemmensen reduction process.

⇒ \(\mathrm{HOCH}_2 \mathrm{CH}_2 \mathrm{COCH}_3 \frac{\text { (1) } \mathrm{NH}_2 \mathrm{NH}_2}{\left(\text { 2) } \mathrm{C}_2 \mathrm{H}_5 \mathrm{ONa}, 180^{\circ} \mathrm{C}\right.} \mathrm{HOCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3\)

Question 46. Convert:

1. CO and H₂ → HCHO
2. Acetaldehyde → pentaerythritol.

Answer:

Pentaerythritol is obtained when CaO dust and CH₃CHO are added to paraformaldehyde suspended in water.

Question 47. Which of the given compounds will react with NaOH: CH₃COCH₃, CH₃CH(OH)CH₃, CH₃CH₂CHO, CBr₃COCH₃?
Answer:

CH₃COCH₃ (acetone), CH₃CH₂CHO (propanal) and CBr₃COCH₃ (tribromoacetone) are the three compounds that will react with NaOH. CH₃CH(OH)CH₃ (propane-2-ol) does not react with NaOH.

Question 48. One mole of an organic compound reacts with 0.5 mole of oxygen to form an organic acid. Mention a reaction to identify these types of compounds.
Answer:

The organic compound is an aldehyde as one oxygen atom (1/2 molecule O₂) is required to oxidise 1 molecule of an aldehyde to yield 1 molecule of a carboxylic acid.

Identification of aldehydes: Aldehydes, when warmed with Tollens’ reagent, produce metallic silver which appears as a silver mirror on the walls of the reaction vessel.

Question 49. A carboxylic acid, when heated with HI in the Br presence of red phosphorus, yields propane. What will happen if the sodium salt of that acid is heated with soda lime and subjected to electrolysis?
Answer:

When a carboxylic acid is heated with red phosphorus in the presence of hydroiodic acid (HI), it undergoes reduction to yield an alkane containing the same number of carbon atoms as that of the parent carboxylic acid. Therefore, the carboxylic acid that forms propane on reduction with red phosphorus and HI is propanoic acid (CH₃CH₂COOH).

Na-salt of propanoic acid on being heated with soda lime gives ethane.

When a saturated aqueous solution of sodium propanoate is subjected to electrolysis, butane is obtained.

Question 50. What is the role of red phosphorus in the HVZ reaction?
Answer.

When carboxylic acids react with chlorine (Cl₂) or bromine (Br₂) in the presence of red phosphorus, α-chloro or α -bromo acids are obtained.

⇒ \(\mathrm{RCH}_2 \mathrm{COOH} \stackrel{\mathrm{Red} P+\mathrm{Br}_2}{\longrightarrow} \mathrm{R}-\mathrm{CH}(\mathrm{Br})-\mathrm{COOH}\)

Phosphorus tribromide, produced in the reaction between red phosphorus and bromine, converts the carboxylic acid into an acid bromide. The acid bromide forms α-bromo acid bromide very rapidly through enol formation, α -bromo acid bromide then reacts with the carboxylic acid to form α-bromo acid and acid bromide.

⇒ \(2 \mathrm{P}+3 \mathrm{Br}_2 \longrightarrow 2 \mathrm{PBr}_3\)

Question 51. In the presence of red phosphorus, chlorine reacts with acetic acid to form chloroacetic acid but formic acid does not respond to a similar reaction. Explain.
Answer:

When acetic acid (or any other carboxylic acid) is treated with chlorine in the presence of red phosphorus, one or more α-H-atoms are replaced by Cl-atoms.

⇒ \(\stackrel{\alpha}{\mathrm{C}} \mathrm{H}_3 \mathrm{COOH} \stackrel{\text { Red } \mathrm{P}+\mathrm{Cl}_2}{\longrightarrow} \mathrm{Cl}-\stackrel{\alpha}{\mathrm{C}} \mathrm{H}_2-\mathrm{COOH}\)

As there is no a -carbon atom in the formic acid (H—COOH) molecule, a similar reaction does not take place.

Question 52. Give the equation for the preparation of carboxylic acids using a haloform reaction.
Answer:

The carbonyl compounds containing ketomethyl (CH₃CO—) group and the alcohols containing —CH(OH)CH₃ group react with halogen (Cl₂, Br₂ or I₂) in the presence of strong alkali (e.g., NaOH) to form haloform (CHCl₃, CHBr₃ or CHI₃) and salts of carboxylic acids. The reaction mixture on acidification yields carboxylic acid.

Question 53. What is the source of CO₂ produced in the reaction of carboxylic acid with sodium bicarbonate?
Answer:

The bicarbonate ion combines with the proton (H+) released due to the dissociation of carboxylic acid to form unstable carbonic acid. It dissociates readily to form CO₂ and water. Therefore, the source of CO₂ evolved is sodium bicarbonate.

Question 54. Name two monobasic and dibasic carboxylic acids with their structural formulae.
Answer:

Monobasic carboxylic acid: HCOOH (formic acid) and CH₃COOH (acetic acid).

Diabasic carboxylic acid: COOH—COOH (oxalic acid) and HOOCCH₂CH₂COOH (succinic acid)

Question 55. Explain why formic acid reduces Tollens’ reagent.
Answer:

The group present in the formic acid molecule is directly attached to a H-atom and a hydroxyl group, i.e., formic acid contains both an aldehyde (—CHO) and a carboxyl (—COOH) group. So, it displays the properties of both aldehydes and the carboxylic acids. For this reason, formic acid reduces Tollens’ reagent.

Question 56. Three test tubes contain samples of ethyl alcohol, phenol and acetic acid. How would you identify them?
Answer:

These three liquids are tested with moist blue litmus paper. The liquid which does not turn blue litmus red must be ethyl alcohol. The liquids in the other two test tubes (phenol and acetic acid) turn blue litmus red. @ Now, sodium bicarbonate solution is added to each of these two test tubes. The liquid which gives effervescence of C02 is identified to be acetic acid. Hence, the liquid in the other test tube is phenol.

Question 57. Explain why carboxylic acids fail to exhibit the characteristic properties of the carbonyl group?
Answer:

In R—COOH, an unshared pair of electrons on the O-atom of the group participate in resonance with the electrons of the group. Consequently, the extent of the positive character of the carbonyl carbon in the carboxyl group is much less compared to the carbonyl carbon of aldehydes or ketones. Moreover, the carbonyl group of carboxylic acid is attached to a leaving group (—OH). For these reactions, carboxylic acids fail to exhibit the characteristic reactions (e.g., formation of oxime, hydrazone, etc.) of the carbonyl group.

Question 58. The boiling points of carboxylic acids are higher than those of alcohols of comparable molecular masses. Explain with reason.
Answer:

In the liquid state, the molecules of both alcohols and carboxylic acids exist in the associated state through the formation of intermolecular H-bonds. Since the molecules of carboxylic acids exist as resonance hybrids, the amount of partial positive charge on the O-atom of the group and that of partial positive charge on the H-atom of the —OH group are sufficiently high. So, the H-bonds formed between the molecules of carboxylic acids are very strong and a greater amount of thermal energy is required to break these bonds.

On the other hand, the molecules of alcohols do not exist as resonance hybrids. So, the intermolecular H-bonds in this case are relatively much weaker and hence less amount of thermal energy is required for the cleavage of these bonds. For this reason, the boiling points of alcohols are less than that of the carboxylic acids of comparable molecular masses.

Question 59. What is grey acetate of lime? How can aqueous acetic acid solution be obtained from this?
Answer:

When the vapours evolved by boiling pyroligneous acid in a copper vessel are passed through milk of lime [Ca(OH)₂], the acetic acid present in the vapours reacts with milk of lime to form calcium acetate. The resulting solution of calcium acetate is evaporated by heating when dry calcium acetate is obtained. It is called grey acetate of lime. When it is distilled with concentrated H₂SO₄, 40-50% acetic acid solution is obtained.

Question 60. Why is acetic acid more acidic than ethanol?
Answer:

In acetic acid, due to the -I and -R effect of the group, the O— H bond becomes weaker and hence the release of the proton occurs easily. In ethanol, there is no such electron-attracting group and so the release of a proton by the cleavage of the O— H bond occurs with much difficulty. Again, the conjugate base of acetic acid is stabilised by resonance while a conjugate base of ethanol is not. Hence, acetic acid is more acidic than ethanol.

Question 61. Convert acetic acid directly into methyl amine.
Answer:

When acetic acid is heated with hydrazoic acid (HN₃) in the presence of a cone. H₂SO₄ and methylamine are obtained.

⇒ \(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{HN}_3 \stackrel{\text { conc. } \mathrm{H}_2 \mathrm{SO}_4}{\longrightarrow} \mathrm{CH}_3 \mathrm{NH}_2+\mathrm{CO}_2+\mathrm{N}_2\)

Question 62. The molecular weight of acetic acid is 120 in the vapour state—explain.
Answer:

Acetic acid exists as a cyclic dimer in the vapour state. As a result, its molecular weight becomes almost double (2 × 60 = 120). Consequently, its apparent vapour density also becomes double (120+2 = 60), i.e., the experimentally observed vapour density of acetic acid is abnormally high.

Dimer of acetic acid

Question 63. Name an organic compound that displays both acidic and reducing properties. Give an example of each.
Answer:

Formic acid displays both acidic & reducing properties. Acidic property: The aqueous solution of formic acid turns blue litmus red and CO₂ is evolved in the form of bubbles when sodium bicarbonate is added to this solution.

⇒ \(\mathrm{HCOOH}+\mathrm{NaHCO}_3 \longrightarrow \mathrm{HCOONa}+\mathrm{CO}_2 \uparrow+\mathrm{H}_2 \mathrm{O}\)

Reducing property: Formic acid reduces Tollens’ reagent when a grey precipitate of metallic silver is obtained.

⇒ \(\mathrm{HCOOH}+\mathrm{Ag}_2 \mathrm{O} \stackrel{\text { Tollens’ reagent }}{\longrightarrow} 2 \mathrm{Ag} \downarrow+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}\)

Question 64. Write the IUPAC names and structural formulas of the following compounds:

1. Acetaldehyde (acro line),
2. Crotonaldehyde,
3. A-phenyl propionaldehyde,
4. Isobutyraldehyde,
5. Glyoxal,
6. Methyl isopropyl ketone,
7. Methyl isobutyl ketone,
8. Ethyl sec-butyl ketone,
9. Di-tert-butyl I ketone,
10. Diacetyl.

Answer:

Question 65. Write the IUPAC names of the following compounds:

1. CH₂(OH)CHCICHO
2. C₆H₅CH₂CH₂CHO
3. CH₃CH(NH₂)CH(OH)COCH₃
4. (CH₃)₂CHCHOHCHOHCHO
5. CH₃CH(OCH₃)COCH(OCH₂CH₃)CH₃
6. CH₃COCH₂CH(CH₃)CH₂Cl
7. (CH₃)₂CHCH₂CH₂COCH₂Cl
8. CH₃CH₂COCH₂CHO

Answer:

Question 66. What happens when :

1. Acetone is heated with hydrazine and sodium ethoxide is added to it.
2. (CH₃)₃CCHO+HCHO→(50%NaOh solution)
3. The product obtained on the hydration of ethyne is allowed to react with dil. alkali.
4. Cl2 gas is passed through acetone in an alkaline solution.
5. Acetone is refluxed in the presence of baryta.
6. Acetone is distilled with cone. H₂SO₄.
7. Vapours of EtOH are passed separately over—(a) heated alumina, (b) heated copper powder.
8. The product obtained in the reaction of CH₃MgI with CH₃CHO at low temperature is hydrolysed.

Answer:

1. When acetone is heated with hydrazine, acetone hydrazine is obtained. It reacts with sodium ethoxide to yield propane. This is called Wolff-Kishner reduction.

2. This is a crossed Cannizzaro reaction. In this case, formaldehyde (HCHO) containing the more reactive carbonyl group is oxidised to yield sodium formate and trimethylacetaldehyde with relatively less reactive carbonyl group gets reduced to give neopentyl alcohol.

3. When ethyne gas is passed through dil. H₂SO₄ solution in the presence of Hg²⁺ ion (catalyst), acetaldehyde is obtained that reacts with dilute alkali to form aldol.

4. When Cl₂ gas is passed through an alkaline solution of acetone, chloroform and sodium acetate are formed.

5. Acetone on distillation with cone. H₂SO₄ produces mesitylene (1,3,5-trimethyl benzene).

6. When vapours of ethyl alcohol are passed over heated alumina at 350°C, ethylene is obtained.

⇒ \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} \underset{\text { dehydration }}{\stackrel{\mathrm{Al}_2 \mathrm{O}_3, \Delta}{\longrightarrow}} \mathrm{CH}_2=\mathrm{CH}_2 \text { (Ethylene) }+\mathrm{H}_2 \mathrm{O}\)

7. When vapours of ethyl alcohol are passed over heated copper powder at 300°C, acetaldehyde is obtained.

⇒ \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} \frac{\mathrm{Cu}, 300^{\circ} \mathrm{C}}{\text { dehydrogenation }} \mathrm{CH}_3 \mathrm{CHO}+\mathrm{H}_2\)

8. In this reaction, propan-2-ol is produced.

Question 67. Write the IUPAC names of the following compounds:

1. HOOC—CH₂CH(CH₃)COOH
2. HOOCC(CH₃)₂COOH
3. CH₃C(CH₃)=CHCOOH
4. HOOCCH(OH)CH(OH)COOH
5. OHC—COOH
6. CH₂=CH —CH=CH —COOH
7. CH₃CH₂CH(CHO)CH₂COOH
8. CH₃COCH₂CH=CH—COOH

Answer:

 

Question 68. Write the structural formulas and IUPAC names of

1. N-valeric acid,
2. Oxalic acid,
3. Malonic acid,
4. Succinic acid,
5. Lactic acid,
6. Glycine.
7. Alanine,
8. Pyruvic acid,
9. Isobutyric acid.

Answer:

Question 69. Identify the compounds which will respond to the iodoform test: CH₃CH₂COCH₃, CH₃CH₂COCH₂I, CH₃CH₂COCH₂CH₂I, CH₃COOCH₂CH₃.
Answer:

CH₃CH₂COCH₂I and CH₃CH₂COCH₃ will respond to the iodoform test.

Question 70. Would it be possible to distinguish between formaldehyde and formic acid by Tollens’ reagent? Justify your answer. (Equation is not required)
Answer:

Both formaldehyde and formic acid contain the H—C— group. Hence, both of them can reduce Tollen’s reagent to metallic Ag. Thus, they cannot be distinguished by Tollen’s reagent.

Question 71. An organic compound (A) of formula C₄H₆O₃ produces two organic compounds (B) and (C) on reaction with methanol. (C) on refluxing with methanol in the presence of the catalytic amount of cone. H₂SO₄ gives (B). When (A) is allowed to react with excess conc.NH₄OH it furnishes (D) and (E). (D) is obtained on heating (E). Write down the structures of (A), (B), (C), (D) and (E) and explain the reactions.

Answer:

1. A : Acetic anhydride,
2. 
3. B: Methyl ethanoate,
4. 
5. C: Acetic acid, CH₃COOH
6. D: Acetamide, CH₃CONH₂
7. E: Ammonium acetate, CH₃COONH₄

Question 72.

1. Identify the ester and aldol from the following compounds:

CH₃CH₂OCH₂CH₃> CH₃CHOHCH₂CHO, CH₃COOCOCH₃, CH₃COOCH₂CH₃, CH₃CHOHCHOHCH₃

How the aldol and the ester identified by you can be prepared starting from the same compound in a single step in each case?

Ester: CH₃COOCH₂CH₃

Aldol: CH₃CHOHCH₂CHO

2. How can you prepare HCOONa and CH₃OH simultaneously from HCHO using only one reagent?

Answer:

⇒ \(\mathrm{HCHO} \underset{\text { Cannizzaro reaction }}{\stackrel{50 \% \mathrm{NaOH}}{\longrightarrow}} \mathrm{HCOONa}+\mathrm{CH}_3 \mathrm{OH}\)

Question 73. Mention the reagent for the following conversion in a single step:

Answer:

Question 74. Which two of the following four compounds will produce the same product (organic) on treatment with excess Br₂ /water at room temperature? Write down the structure of the organic product.

Answer:

will produce the same product (organic) on treatment with excess Br2 /water at room temperature. The product is—

Question 75. Consider the following compounds and answer the question that follows:

1. Which will produce benzoin on refluxing with alcoholic KCN? Write the structure of benzoin.
2. Which will produce a hydrocarbon on heating with soda lime? Write the structure of the hydrocarbon.

Answer:

will produce benzoin on refluxing with alcoholic KCN. The structure of benzoin is—

will produce a hydrocarbon (benzene) on hearing with soda lime. The structure of the hydrocarbon is

Question 76. Indicate the reagents for the following transformations:

Answer:

Question 77. Which of the following will respond to Cannizzaro’s reaction—

1. CH₃CHO
2. (CH₃)₂CHCHO
3. (CH₃)₃CCHO
4. 

Answer: 3. Aldehydes containing no a-H atom undergo the Cannizzaro reaction.

Question 78.

1. Identify A, B, C, D, E and F in the following reactions:

2. Write the reagents required in the following reactions:

Or, How would you convert?

Answer:

A: RCOCl,

B: RCH₂COOH,

C: (CH₃)₂C(OH)—CH₂COCH₃,

D: CH₃CH(I)CO₂H

Question 79. Which of the following compounds is obtained when calcium acetate is dry distilled—

1. Formic acid
2. Formaldehyde
3. Acetone
4. Butanone

Answer: 3. Acetone

Question 80. Give examples of the following reactions: Identify A, B, C and D in the following reactions:

Or,

1. An organic compound A, of molecular formula C₆H₁₄O₂ on acid-hydrolysis, produces one molecule of ethanol and two molecules of ethanol from one molecule of A. Identify A. How can A be prepared?
2. How would you convert? CH₃CHO→CH₃CH=CHCHO
3.  Write the appropriate reagents for the following two conversions:

4. Distinguish between formic acid and acetaldehyde by a suitable chemical test.

Answer:

Or,

On addition to a saturated NaHCO₃ solution, formic acid evolves CO₂ in the form of bubbles but acetaldehyde does not.

⇒ \(\mathrm{HCOOH}+\mathrm{NaHCO}_3 \rightarrow \mathrm{HCOONa}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \uparrow\)

Question 81. Which of the following compounds is formed when acetophenone is treated with bromine in an acidic medium —

Answer: 3

Question 82. An organic compound A (molecular formula C₂H₄O) on reaction with excess formaldehyde in the presence of Ca(OH)₂ provides B (C₅H₁₂O₄). A on reaction with Al(OEt)₃ gives C (molecular formula C₄H₈O₂ ) Both A and C on treatment with LiAlH₄ in dry ether furnish the same compound D (molecular formula C₂H₆O ). A reaction with AgNO₃ (ammoniacal) gives E (molecular formula C₂H₇NO₂). The reaction of A with D in the presence of dry HCl provides F (molecular formula C₆H₁₄O₂). Write down the structures of A to Falong with the arrowhead equations for the above reactions.

Or, 1. How would you convert?

2. Identify A, B, D, E, F, and G In the following reactions:

Answer:

Or,

1. 
2. B: CH₃OH
3. D: CHCl₃
4. E: CH₃COCHO
5. 
6. G: (HCOO)₂Ca

Question83. Which of the following compounds will take part in nucleophilic addition reaction most readily—

1. CH₃COCH₃
2. CH₃CHO
3. C₆H₅CHO
4. C₆H₅COC₆H₅

Answer: 2

Question 84.

1. Benzoic acid in reaction with SOCl₂ gives (A). (A) on reduction with Pd-BaSO₄, H₂ in the presence of quinoline affords (B). (B) reacts with NH₂OH. HCl in the presence of CHgCOONa in aqueous ethanol to furnish (C). (C) on reaction with PCl₅ give (D). Write the structures of (A), (B), (C) and (D). What is the role of CH₃COONa in the conversion of (B) to (C).
2. Identify A, B, C and D in the following steps of the reaction of acetaldehyde with dilute aqueous solution of NaOH:

⇒ \(\mathrm{CH}_3 \mathrm{CHO}+\stackrel{\ominus}{\mathrm{O}} \mathrm{H} \rightleftharpoons A+B\)

⇒ \(\mathrm{CH}_3 \mathrm{CHO}+A \rightleftharpoons C\)

⇒ \(C+B \rightleftharpoons D+\stackrel{\ominus}{\mathbf{O}} \mathrm{H}\)

1. An organic compound produces acetic acid and ethanol on acid hydrolysis. Write the structural
   formula of the compound. How can you prepare the compound from acetaldehyde In one step?
2. Write the structures of A to D In the following reactions

How would you convert?

Answer:

Normally hydroxylamine hydrochloride (NH₂OH.HCl) does not react with carbonyl compounds. But, in the presence of CH₃COONa, it releases free NH₂OH which readily reacts with carbonyl compounds. Apart from this, CHjCOONa retains the pH of the reaction medium unchanged by forming CH₃COOH-CH₃COONa buffer. For these reasons, CH₃COONa is used.

Or,

Question 85. Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions: ethanal, propanal, propanone, and butanone.
Answer: Ethanal > propanal > propanone > butanone

Question 86.

1. Illustrate the following name reactions giving suitable examples in each case—
   1. Clemmensen reduction,
   2. Hell-Volhard-Zelinsky reaction
2. How are the following conversions carried out?
   1. Ethyl cyanide to ethanoic acid
   2. butan-l-ol to butanoic acid.

Answer:

Question 87. Write the structure of 2-hydroxybenzoic acid.
Answer:

5.

1. Write the products formed when CH₃CHO reacts with the following reagents:

1. HCN
2. H₂N-OH
3. CH₃CHO in the presence of dilute NaOH.

2. Give simple chemical tests to distinguish between the following pairs of compounds:

1. Benzoic acid and Phenol
2. Propanal and Propanone.

Answer:

Propanal gives a positive test with Fehling’s solution and gives a red precipitate of cuprous oxide. Propanone does not respond to this test.

Question 88.

1. Account for the following:

1. Cl—CH₂COOH is a stronger acid than CH₃COOH.
2. Carboxylic acids do not give reactions of the carbonyl group,

2. Out of CH₃CH2—CO—CH₂   —CH₃ and CH₃CH₂—CH₂ —CO—CH₃, which gives iodoform test?

Answer:

1. Due to the presence of —Cl as an electron-withdrawing group in the molecule of Cl —CH₂COOH, its acidity increases. This is because, Cl attracts the electron cloud towards itself, making the proton release easy.
2. The carbonyl group in carboxylic acids is involved in resonance, leading to the decrease in the double bond character Hence, the —COOH group in carboxylic acids does not give a reaction of the carbonyl group although it has a group.

Due to the presence ofgroup CH₃CH₂—CH₂—CO—CH₃ will give an iodoform test.

Question 89. Draw the structure of 3-methylpentanal.
Answer:

Question 90. An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br₂ and KOH forms compound ‘C’ of molecular formula C₆H₇N. Write the structures and IUPAC names of compounds A, B and C.
Answer:

Question 91. Describe the following giving chemical equations: Friedel-Crafts reaction
Answer:

Friedel-Crafts reaction involves alkylation or acylation of an aromatic ring with alkyl or acyl halide using a strong Lewis add as a catalyst.

Question 92. How will you bring about the following conversions?

1. Benzoic acid to Benzaldehyde
2. Benzene to m-nitroacetophenone
3. Ethanol to 3-hydroxy butanal

or,

1. Describe the following reactions:

1. Acetylation
2. Aldol condensation.

2. Write the main product in the following equations:

Answer:

The process of introduction of an acetyl group (CH₃CO —) into a compound, usually alcohols, phenols and amines is called acetylation.

Question 93. Write the IUPAC name of the following: CH₃CH₂CHO
Answer: Propane-1-al

Question 94.

1. Draw the structure of the following:

1. P-methyl benzaldehyde
2. 4-methylpent-3-ene-2-one

2. Give the chemical tests to distinguish between the following pairs of compounds:

1. Benzoic acid and Ethyl benzoate
2. Benzaldehyde and Acetophenone
3. Phenol and Benzoic acid

or,

1. Draw the structure of the following derivatives:

1. Propanone oxime
2. Semicarbazone of CH₃CHO

2. How will you convert ethanal into the following compounds? Give the chemical equations involved.

Answer:

1. 
2. (CH₃)₂C=CHCOCH₃

Benzoic acid does not respond to the iodoform test while ethyl benzoate does. On heating with excess NaOH, ethyl benzoate gives ethyl alcohol, which on heating with iodine in the presence of NaOH gives a yellow precipitate of iodoform.

Question 95. Write the structures of A, B, C, D and E in the following reactions:

Or,

1. Write the chemical equation for the reaction involved in the Cannizzaro reaction.
2. Draw the structure of semicarbazone of ethanal.
3. Why pKa of F— CH₂—COOH is lower than that of Cl—CH₂—COOH.
4. Write the product in the following reaction:

5. How can you distinguish between propanal and propanone?

Answer:

1. H₃C—CH=N— NHCONH₂
2. The strength of α-halocarboxylic acids increases with an increase in the electronegativity of the halogen.
3. H₃C—CH=CH—CH₂—CHO

Question 96. An organic compound ‘X’ having molecular formula C₄H₈O gives orange-red ppt with 2,4-DNP reagent. It does not reduce Tollens’ reagent but gives a yellow ppt of iodoform on heating with NaOI. Compound ‘X’ on reduction with LiAlH4 gives compound ‘ Y’ which undergoes dehydration reaction on heating with cone. H₂SO₄ to form but-2-one. Identify the compounds X and Y.
Answer:

Question 97. Give reasons:

1. The α-hydrogen atoms of aldehydes and ketones are acidic in nature.
2. Oxidation of aldehydes is easier than ketones.
3. H₂C=CH—COOH is more acidic than CH₃CH₂—COOH.

Answer:

1. The α-hydrogen atoms of aldehydes and ketones are acidic in nature due to—

1. The strong electron-withdrawing (-I) effect of the carbonyl group.
2. The resonance stabilisation of the conjugate bases of aldehydes and ketones.

2. Oxidation of aldehydes involves the cleavage of the C —H bond whereas oxidation of ketones involves the cleavage of the C—C bond, which is much stronger than the C—H bond. So, oxidation of aldehydes is easier than ketones.

Question 98. Write structures of compounds A, B and C in each of the following reactions:

Answer:

1. A: C₆H₅MgBr B: C₆H₅COOH C: C₆H₅COCl
2. A: CH₃CHO B: CH₃CH(OH)CH₂CHO C: CH₃CH=CHCHO

Question 99. Do the following conversions in not more than two steps:

1. Benzoic acid to Benzaldehyde
2. Ethyl benzene to Benzoic acid
3. Propanone to Propene.

Answer:

Question 100.

1. Write the product(s) in the following reactions:

2. Give a simple chemical test to distinguish between the following pairs of compounds:

1. Butanal and Butan-2-one
2. Benzoic add and Phenol.

Answer:

Butanal does not respond to the iodoform test whereas butan-2-one gives yellow-coloured iodoform.

Question 101.

1. Write the reactions involved in the following:

1. Etard reaction
2. Stephen reduction

2. How will you convert the following in not more than two steps:

1. Benzoic acid to Benzaldehyde
2. Acetophenone to Benzoic acid
3. Ethanoic acid to 2-hydroxyethannoic acid

Answer:

Question 102. How do you convert the following?

1. Ethanal to Propanone
2. Toluene to Benzoic acid

Answer:

Question 103. Account for the following:

1. Aromatic carboxylic acids do not undergo FriedelCrafts reaction.
2. pKa value of 4-nitrobenzoic acid is lower than that of benzoic acid.

Answer:

The —COOH group attached to the benzene ring is electron-withdrawing in nature and thus deactivates the ring. Secondly, the catalyst used in the reaction, i.e., AlCl₃ is a Lewis acid which has the tendency to form a bond with the carboxyl group. That is why aromatic carboxylic acids do not undergo FriedelCrafts reaction.

Question 104. A, B and C are three non-cyclic functional isomers of a carbonyl compound with molecular formula C4H8oi Isomers A and C give positive Tollens’ test whereas isomer B does not give Tollens’ test, but gives position Iodoform test. Isomers A and B on redaction with Zn(Hg)/conc. HQ give the same product D.

1. Write the structures of A, B, C and D.
2. Out of A, B and C isomers, which one is least reactive towards the addition of HCN?

Answer:

The possible functional isomers of carbonyl compounds with molecular formula C₄H₈O are—

(CH₃)₂CHCHO, CH₃CH₂CH₂CHO, CH₃CH₂COCH₃

Isomers A and C give positive Tollens’ test. Hence they must be aldehydes. On the other hand, B does not give Tollen’s test, so it is a ketone. Again it gives a positive iodoform test, hence it should contain a —COCH₃ group.

Thus B is CH₃CH₂COCH₃ .

A and B on reduction with Zn-Hg/conc. HCl, give same product D. On reduction with Zn-Hg/conc. HCl, CH₃CH₂CH₂CHO and CH₃CH₂COCH₃ are reduced into CH₃CH₂CH₂CH₃.

1. Hence A: CH₃CH₂CH₂CHO, B: CH₃CH₂COCH₃, C: (CH₃)₂CHCHO, D: CH₃CH₂CH₂CH₃
2. The isomer B is least reactive towards the addition of HCN. This is because aldehydes are more reactive towards nucleophilic addition reactions than ketones due to less inductive and steric effects.

Question 105. Give names of the reagents to bring about the following transformations:

1. Hexan-1-ol to hexanal
2. Cyclohexanol to cyclohexanone
3. P-fluoro toluene to p-fluoro benzaldehyde
4. Ethanenitrile to ethanal
5. Allyl alcohol to propenal
6. But-2-ene to ethane

Answer:

Question 106. Arrange the following compounds in the increasing order of their boiling points: CH₃CH₂CH₂CHO, CH₃CH₂CH₂CH₂OH, H₅C₂— O—C₂H₅, CH₃CH₂CH₂CH₂CH₃
Answer:

The given compounds (having molecular masses in the range of 72 to 74) belong to aldehyde, alcohol, ether and hydrocarbon series, respectively. Out of these compounds, only butan-l-ol undergoes extensive intermolecular hydrogen bonding therefore its boiling point is the highest. Butanal (carbonyl compound) is more polar than diethyl ether.

So intermolecular dipole-dipole attraction is stronger in the former. Thus boiling point of butanal is greater than that of diethyl ether, n-Pentane molecules have only weak van der Waals forces of attraction so its boiling point is the lowest. Hence, the order of increasing boiling point is given by,

CH₃CH₂CH₂CH₃ < C₂H₅—O—C₂H₅ < CH₃CH₂CH₂CHO < CH₃CH₂CH₂CH₂OH

Question 107. Would you expect benzaldehyde to be more reactive or less reactive in nucleophilic addition reactions than propanal? Explain your answer.
Answer:

The +ve charge on the C-atom of the carbonyl group in propanal is reduced slightly as the ethyl group has a weak +I effect. However, the +ve charge on the similar C-atom of benzaldehyde is reduced by a large amount due to the strong +R effect of the phenyl group. Thus the carbonyl carbon of benzaldehyde is less electrophilic than that of propanal and hence, benzaldehyde is less reactive in nucleophilic addition reactions.

Question 108. An organic compound (A) with molecular formula C₈H₈O forms an orange-red precipitate with 2, 4- DNP reagent and gives a yellow precipitate on heating with iodine in the presence of sodium hydroxide. It neither reduces Tollens’ or Fehlings’ reagent nor does it decolourise bromine water or Baeyer’s reagent. On drastic oxidation with chromic acid, it gives a carboxylic acid (B) having molecular formula C₇H₆O₂. Identify the compounds (A) and (B) and explain the reactions involved.
Answer:

Compound A(C₈H₈O) forms a 2, 4-DNP derivative, so it is an aldehyde or a ketone. It does not reduce Tollens’ or Fehling’s solution, so it must be a ketone. It responds to the iodoform test, so it is a methyl ketone. The molecular formula of ‘A’ indicates a high degree of unsaturation, but it does not decolourise bromine water or Baeyer’s reagent. This shows that ‘A’ contains an aromatic ring.

Now compound B(C₇H₆O₂)is the oxidation product of the methyl ketone A(C₈H₈O). Since B is a carboxylic acid, It must be benzoic acid (C₆H₅COOH) and compound A should, therefore, be a monosubstituted aromatic methyl ketone.

The molecular formula of A suggests that it is phenyl methyl ketone (C₆H₅—CO—CH₃).

Question 109. Write chemical reactions to affect the following transformations:

1. Butan-l-ol to butanoic acid
2. Benzyl alcohol to phenylethanoid acid
3. 3- nltrobromobenzene to 3-nitrobenzoic acid
4. 4- methyl acetophenone to benzene-1,4-dicarboxylic acid
5. Cyclohexene to hexane-1,6-dioic acid
6. Butanal to butanoic acid.

Answer:

Question 110. Write the structures of the following compounds:

1. α-methoxypropionaldehyde
2. 3-hydroxy butanal
3. 2-hydroxy cyclopentane carbaldehyde
4. 4-oxo pentanal
5. Di-sec. butyl ketone
6. 4-fluoroaceto phenone.

Answer:

Question 111. Write the structures of products of the following reactions:

Answer:

Question 112. Arrange the following compounds in increasing order of their boiling points. CH₃CHO, CH₃CH₂OH, CH₃OCH₃, CH₃CH₂CH₃
Answer:

The molecular masses of these compounds are in the range of 44 to 46. Since only ethanol molecules are associated due to extensive intermolecular hydrogen bonding, therefore, the b.p. of ethanol would be the highest. Ethanal is more polar than methoxymethane. Therefore the intermolecular dipole-dipole attraction is stronger in the former and hence, ethanal has higher b.pt. than methoxymethane. Propane molecules have only weak van der Waals forces of attraction. Hence, the increasing order of boiling points of the given compounds is— CH₃CH₂CH₃ < CH₃— O—CH₃ < CH₃CHO < CH₃CH₂OH.

Question 113. Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions.

1. Ethanal, propanal, propanone, butanone.
2. Benzaldehyde, p-tolu aldehyde, p-nitrobenzaldehyde, acetophenone.

Hint: Consider the steric effect and electronic effect.

Answer:

1. Ongoing from ethanal — propanal-propanone —butanone, the +1 effect of the alkyl groups (attached to the carbonyl carbon) increases. As a result, electron density on the carbonyl carbon (i.e., the carbon atom of the carbonyl group) progressively increases. Furthermore, an increase in the size as well as the number of alkyl groups on the carbonyl carbon causes increasing steric crowding around it Hence, attack by the nucleophile becomes more and more difficult as one passes from ethanal to butanone. Thus the reactivity increases in the reverse order i.e., butanone < propanone < propanal < ethanal.
2. Out of the given compounds, only acetophenone is the aromatic ketone, while all others are aromatic aldehydes, therefore, it is the least reactive. In p-tolu aldehyde, the methyl group at the para position causes an increase in electron density on the carbon atom of the carbonyl group by hyperconjugation effect thereby, making it less reactive than benzaldehyde.

In p-nitrobenzaldehyde, however, the NO₂ group at the p -position withdraws electrons, both by -I and -R effects, causing a decrease in electron density on the carbon atom of the carbonyl group. Therefore attack by nucleophiles becomes favourable indicating that it is most reactive.

Thus the reactivity of the given compounds increases in the order: of acetophenone < p -tolualdehyde < benzaldehyde < p -nitrobenzaldehyde.

Question 114. Predict the products of the following reactions:

Answer:

Question 115. Give the IUPAC names of the following compounds:

Answer:

1. 3-phenyl propanoic acid
2. 3-methyl but-2-enoic acid
3. 2-methyl cyclopentane carboxylic acid
4. 2,4,6- trinitro benzoic acid

Question 116. Show how each of the following compounds can be converted to benzoic acid.

1. Ethylbenzene
2. Acetophenone
3. Bromobenzene
4. Phenylethene (Styrene)

Answer:

Question 117. Which acid of each pair shown here would you expect to be stronger?

1. CH₃CO₂H or CH₂FCO₂H
2. CH₂FCO₂H or CH₂ClCO₂H
3. CH₂FCH₂CH₂CO₂H or CH₃CHFCH₂CO₂H
4. 

Answer:

Thus FCH₂COOH is a stronger acid than CH₃COOH.

Due to the much stronger -I effect of F over Cl, the release of a proton from FCH₂COOH is more facile than that from ClCH₂COOH. For the same reason, the stability of FCH₂COO^– is greater than that of ClCH₂COO^–. Both these factors are responsible for the greater acid strength of FCH₂COOH over ClCH₂COOH.

In halogen-substituted acids, the influence of acid strengthening -I effect of the halogen atom decreases as its distance from the carboxyl group increases. Thus 3-fluorobutanoic acid is stronger than 4-fluorobenzoic acid.

Question 118. Name the following compounds according to the IUPAC system of nomenclature:

1. CH₃CH(CH₃)CH₂CH₂CHO
2. CH₃CH₂COCH(C₂H₅)CH₂CH₂Cl
3. CH₃CH=CHCHO
4. CH₃COCH₂COCH₃
5. CH₃CH(CH₃)CH₂C(CH₃)₂COCH₃
6. (CH₃)₃CCH₂COOH
7. OHCC₆H₄CHO-p

Answer:

1. 4-methylpentanal
2. 6-chloro-4-ethylhexan-3-one
3. But-2-enal
4. Pentane-2,4-dione
5. 3,3,5-trimethylhexan-2- one
6. 3, 3-dimethyl butanoic acid
7. Benzene-1,4- dicarbaldehyde

Question 119. Draw the structures of the following compound:

1. 3-methyl butanal
2. P-nitropropiophenone
3. P-methyl-benzaldehyde
4. 4-methylpent-3-en-2-one
5. 4-chloro-pentan-2-one
6. 3-bromo-4-phenyl pentanoic acid
7. P, p -dihydroxy benzophenone
8. Hex-2-en-4-ynoic acid.

Answer:

Question 120. Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.

1. CH₃CO(CH₂)₄CH₃
2. CH₃CH₂CHBrCH₂CH(CH₃)CHO
3. CH₃(CH₂)₅CHO
4. Ph —CH=CH —CHO
5. 
6. PhCOPh

Answer:

1. The common names are given in the brackets.
2. Heptan- 2-one (Methyl n -pentyl ketone)
3. 4-bromo-2-methyl hexanal (γ-bromo-α-methyl caproaldehyde)
4. Heptanal
5. 3- phenyl prop-2-enal (Cinnamaldehyde)
6. Cyclopentane-carbaldehyde
7. Diphenylmethanone (Benzophenone)

Question 121. Draw structures of the following derivatives:

1. The 2, 4-dinitrophenylhydrazone of benzaldehyde
2. Cyclopropane oxime
3. Acetaldehydedimethylacetal
4. The semicarbazone of cyclobutanone
5. The ethylene ketal of hexane-3-one
6. The methyl hemiacetal of formaldehyde.

Answer:

Question 122. Predict the products formed when cyclohexanecurbaldchydc reacts with the following reagents.

1. PhMgllr und then H3O⁺
2. Tollens’ reagent
3. Scmlcarbuzide and weak acid
4. Excess ethanol and acid
5. Zinc amalgam and dilute hydrochloric acid

Answer:

Question 123. How will you convert ethanal into the following compounds?

1. Butane-1, 3-dlol
2. But-2-enal
3. But-2-enoic acid

Answer:

Question 124. Write structural formulae and names of four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde acts as a nucleophile and which is an electrophile.
Answer:

The four possible products are—

Question 125. An organic compound with the molecular formula C₉H₁₀O forms a 2,4-DNP derivative reduces Tollens’ reagent and undergoes the Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound.
Answer:

The compound (C₉H₁₀O) forms 2, 4-DNP derivative, reduces Tollens’ reagent and undergoes Cannizzaro reaction. So it is an aldehyde having no a-H atom. The compound on vigorous oxidation gives 1, 2-benzene dicarboxylic acid which shows that there are two carbon residues in the ortho positions of a benzene ring. One carbon residue must be an aldehyde group (in accordance with the Cannizzaro reaction). Thus, the compound must be 2-ethylbenzaIdehyde.

Question 126. An organic compound (A) (molecular formula C8H16°2 ) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-one. Write equations for the reactions involved.
Answer:

Compound (A) on hydrolysis gives an alcohol (C) and a carboxylic acid (B). So A must be an ester. The alcohol (C) on oxidation gives the carboxylic acid (B). This shows that both (B) and (C) contain the same number of carbon atoms, i.e. each contains 4 carbon atoms. Again, the alcohol (C) on dehydration gives 2-butene. So (C) is butan-l-ol, and its oxidation product (B) is butanoic acid. Accordingly, the ester (A) is butylbutanoate.

Question 127. Give simple chemical tests to distinguish between the following pairs of compounds.

1. Propanal and Propanone
2. Acetophenone and Benzophenone
3. Phenol and Benzoic acid
4. Benzoic acid and Ethyl benzoate
5. Pentan-2-one and pentan-3-one
6. Benzaldehyde and Acetophenone
7. Ethanal and propanal

Answer: Similar to the destination between acetaldehyde and acetone.

Question 128. How will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom.

1. Methyl benzoate
2. M-nitrobenzoic acid
3. P-nitrobenzoic acid
4. Phenylacetic acid
5. P-nitrobenzaldehyde.

Answer:

Question 129. Why is there a large difference in the boiling points of butanal and butan-l-ol?
Answer:

Butan-1-oil has a higher boiling point as the molecules undergo extensive association through intermolecular H-bonding. Molecules of butanal are held together only by weak dipole-dipole interaction.

Question 130. Write a test to differentiate between pentan-2-one and pentan-3-one.
Answer:

Pentan-2-one is a methyl ketone. So it undergoes haloform reaction on treatment with I₂/NaOH to give yellow ppt. of iodoform. Pentan-3-one fails to do so.

Question 131. Give the IUPAC names of the following compounds

Answer:

1. 3-phenyl prop-2-enal,
2. Cyclohexanecarbaldehyde,
3. 3-oxopentanal,
4. But-2-enal.

Question 132. Give the structure of the following compounds.

1. 4-nitropropiophenone
2. 2-hydroxycyclopentanecarbaldehyde
3. Phenyl acetaldehyde

Answer:

Question 133. Write the IUPAC names of the following structures.

Answer:

1. Ethane-1,2-dial or Ethanedial,
2. Benzene-1,4- dicarbaldehyde,
3. 3-bromobenzenecarbaldehyde or 3-bromobenzaldehyde.

Question 134. Benzaldehyde can be obtained from benzal chloride. Write reactions for obtaining benzal chloride and then benzaldehyde from it.
Answer:

Question 135. Name the electrophile produced in the reaction of benzene with benzoyl chloride in the presence of anhydrous AlCl₃. Name the reaction also.
Answer:

This is the Friedel-Crafts acylation reaction.

Question 136. Oxidation of ketones involves carbon-carbon bond cleavage. Name the products formed on oxidation of 2, 5-dimethylhcxan-3-one.
Answer:

Question 137. Arrange the following in decreasing order of their acidic strength and give a reason for your answer. CH₃CH₂OH , CH₃COOH, CICH₂COOH, FCH₂COOH, C₆H₅CH₂COOH
Answer:

Carboxylic acids are stronger acids than alcohols because carboxylate ions (RCOO^–) are stabilised by resonance but alkoxide (RO^–)ions have no such resonance stabilisation. So CH₃CH₂OH is the weakest acid. Again the strength of carboxylic acid increases by the introduction of the electron-withdrawing group in the carbon chain. Now, the electron-withdrawing -I effect decreases in the sequence:

FCH₂COOH > CICH₂COOH > C₆H₅CH₂COOH CH₃COOH > CH₃CH₂OH

Question 137. What product will be formed in the reaction of propanal with 2-methyl propanal in the presence of NaOH? What products will be formed? Write the name of the reaction.
Answer:

Both propanal (CH₃CH₂CHO) and 2-methyl propanal [(CH₃)₂CHCHO] contain α- H atoms. So a mixture of 4 different aldol condensation products (two normal products and two crossed products) will be formed when a mixture of the given compounds is treated with NaOH.

Question 138. Compound’A’was prepared by oxidation of the compound with alkaline KMnO₄. Compound ‘A’ on reduction with lithium aluminium hydride gets converted back to compound ‘ B’. When compound ‘A’ is heated with compound B in the presence of H₂SO₄ it produces the fruity smell of compound ‘C’. To which family do the compounds ‘ A ‘ B ’ and ‘ C’ belong?
Answer:

Compound on oxidation gives compound ‘A Again compound ‘A’ on reduction gives ‘B’. So both ‘ A ‘ and ‘ B’ contain the same number of C-atoms. Compounds ‘A’ and ‘B’ react together in the presence of H₂SO₄ to give a compound. ‘ C ‘ has a fruity smell. Thus, ‘ C ’ is an ester, ‘A ’ is a carboxylic acid and ‘ B ‘ is an alcohol.

Question 139. Arrange the following in decreasing order of their acidic strength. Give an explanation for the arrangement. C₆H₅COOH, FCH₂COOH, NO₂CH₂COOH
Answer:

The strength of a carboxylic acid increases with the presence of an electron-withdrawing group in the molecule. Since the -I effect of — NO₂ is greater than that of -F, so acid strength of NO₂CH₂COOH is greater than that of FCH₂COOH. NOW in C₆H₅—COOH, the — C₆H₅ group has a weak -I effect and also a +R effect. Thus it is the weakest acid. Hence, acid strength decreases in the sequence:

O₂NCH₂COOH > FCH₂COOH > C₆H₅COOH

Question 140. Alkenesand carbonyl compoundsboth contain a π bond but alkenes show electrophilic addition reactions whereas carbonyl compounds show nucleophilic addition reactions. Explain.
Answer:

In alkenes, the π-electron cloud of the bond is almost symmetrical because the C-atoms involved in such bonding have roughly the same electronegativity. Hence, no separation of charge takes place involving the olefinic carbons. Thus, the approach of any nucleophile will be strongly repelled by this symmetrical electron cloud.

So alkenes do not undergo nucleophilic attack in the first step of a reaction. But such an electron cloud welcomes the approach of any electrophile in the first step of a reaction. So alkenes undergo electrophilic addition reactions. In contrast, the π-electron cloud of the bond is unsymmetrical and it shifts more towards the electronegative O-atom. Thus the carbon atom of the carbonyl group acquires a partial +ve charge and hence, is readily attacked by any approaching nucleophile. So carbonyl compounds undergo nucleophilic addition reactions.

Question 141. Carboxylic acids contain carbonyl groups but do not show nucleophilic addition reactions like aldehydes or ketones. Why?
Answer:

The carbon atom of a carbonyl group has an appreciably large amount of + ve charge because such groups exist as resonance hybrids of the following structures. So they undergo nucleophilic addition reactions readily. On the other hand, the +ve charge on the C-atom of a carboxyl group is reduced remarkably because the C=O group is involved in resonance with the —OH group, as shown below. Hence, carboxylic acids do not undergo nucleophilic addition reactions.

Question 142. Identify the compounds A, B and C in the following reaction.

Answer:

Question 143. Why are carboxylic acids more acidic than alcohols or phenols although all of them have a hydrogen atom attached to an oxygen atom ( —O — H )?
Answer:

Both carboxylate ion and phenoxide ion are stabilised by resonance. In carboxylate ion, the -ve charge is delocalised over two oxygen atoms (having high electronegativity). In phenoxide ion, the -ve charge is delocalised over only one oxygen atom and three carbon atoms (having very low electronegativity). Thus carboxylate ion is more stable than phenoxide ion. So carboxylic acids are stronger than phenol in their acid strengths. Again alkoxide ions (RO^–) are not stabilised by resonance so alcohols are least acidic. A sequence of acid strength: RCOOH > C₆H₅OH > R— OH.

Question 144. Complete the following reaction sequence.

Answer:

Question 145. Ethylbenzene is generally prepared by acetylation of benzene followed by reduction and not by direct alkylation. Think of a possible reason.
Answer:

During Friedel-Craft’s ethylation, the first product of the reaction is ethylbenzene. Since the ethyl group is electron donating, it causes an increase in electron density in the ring, thereby making the ring system more reactive for further alkylation. Thus ethylbenzene competes with benzene and hence, would preferentially undergo further alkylation to give first di-and then polyethylbenzene.

It is because of the possibility of such polyalkylation, that ethylbenzene is not prepared by Friedel-Crafts ethylation of benzene. Instead, Friedel-Crafts acylation is used. The first-formed acetyl benzene is less reactive than benzene (because of the electron-withdrawing nature of the — COCH₃ group), and hence, di- or polyacylation does not occur. Acetylbenzene, on Clemmensen reduction, gives ethylbenzene.

Question 146. Can the Gatterman-Koch reaction be considered similar to Friedel-Crafts acylation? Discuss.
Answer:

In Friedel-Crafts acylation, the reaction proceeds via the formation of an acyl cation (R—C⁺=0) electrophile (formed by a reaction between acyl chloride and AICI₃). In the Gattermann-Koch reaction, the active electrophile is a formyl cation (H—C⁺=O) which is formed by the reaction of CO and HCl in the presence of AlCl₃. Thus Gattermann-Koch formylation can be considered as a reaction similar to that of Friedel-Crafts acylation reaction.

Question 147. What will be the product obtained when itis treated with 50% NaOH solution?

Answer:

Question 148. Explain why a white substance is found to be present at the surface of the stopper of the bottle filled with benzaldehyde.
Answer:

In contact with air, benzaldehyde is slowly oxidized to form perbenzoic acid and benzoic acid, these are deposited as a white crystalline solid at the surface of the stopper of the bottle containing benzaldehyde.

Question 149. Benzaldehyde cannot be prepared by the alkaline hydrolysis of benzalchloride. Explain.
Answer:

The resultant benzaldehyde undergoes a Cannizzaro reaction in the presence of NaOH to form benzyl alcohol and sodium benzoate.

Question 150. Arrange p-chlorobenzoic acid, benzoic acid and p-toluic acid in order of decreasing acidic strength and explain the order.
Answer:

p-chlorobenzoic acid> benzoic acid > p-toluic acid; the presence of an electron-attracting group in the ring increases acidity and the presence of an electron-releasing group decreases acidity.

Class 12 Chemistry Unit 12 Aldehydes Ketones And Carboxylic Acids Long Answers Type

Question 1. Iodoform can be prepared by reacting acetone with hypoiodide but not with iodine. How can this observation be explained?
Answer:

Acetone reacts with hypoiodide (e.g., NaOI) to form first triiodoacetone and OH-ion. This triiodoacetone reacts with alkali to form iodoform.

Acetone does not react with iodine alone and So, iodoform is not obtained in this case.

Question 2. The amount of ketone obtained on oxidation of secondary alcoholism or than the amount of aldehyde obtained on oxidation of primary alcohol. Justify whether the statement is correct or wrong.
Answer:

The aldehyde obtained on oxidation of primary alcohol is further oxidised to give carboxylic acid in the presence of the oxidising agent. This is because the tendency of aldehyde to undergo oxidation is greater than that of alcohol.

On the other hand, ketones obtained on oxidation of secondary alcohols, do not undergo oxidation easily in the presence of the oxidising agent. For this reason, the amount of ketone obtained on oxidation of secondary alcohol is more than the amount of aldehyde obtained on oxidation of primary alcohol.

Question 3. What happens when isobutyraldehyde is made to react with excess formaldehyde in the presence of a cone. NaOH solution?
Answer:

The reaction occurs in two steps. A crossed aldol condensation followed by a crossed Cannizzro reaction occurs to form sodium formate and diol.

Question 4. What happens when acetaldehyde and CaO powder are added to paraformaldehyde suspended in water?
Answer:

Calcium oxide reacts with water to form Ca(OH)₂ which acts as a catalyst. Acetaldehyde reacts with formaldehyde (generated from paraformaldehyde) in the presence of Ca(OH)₂.

Since the acetaldehyde molecule contains three a-H atoms, three crossed aldol condensation reactions occur successively with formaldehyde in the first three steps. In the fourth step, a crossed Cannizzaro reaction occurs.

Question 5. A hydrocarbon containing two carbon atoms decolourises bromine water and gets hydrolysed by H₂SO₄ in the presence of HgSO₄ to form a compound which on heating with bleaching powder produces chloroform. Identify the hydrocarbon and give equations in favour of your statement.
Answer:

The hydrocarbon with two carbon atoms which decolourises bromine-water may be ethylene (CH₂=CH₂) or acetylene (CH = CH). Since the given hydrocarbon is hydrolysed by H₂SO₄ in the presence of HgSO₄ therefore, it must be acetylene. Considering the given hydrocarbon as acetylene, the chemical reactions as mentioned in the question, may easily be explained. The reactions are as follows:

⇒ \(\mathrm{CH} \equiv \mathrm{CH}+2 \mathrm{Br}_2 \rightarrow \mathrm{CHBr}_2-\mathrm{CHBr}_2\)

Question 6. Write the names and structures of two organic compounds of two different classes having molecular formula C₃H₈O. Mention one reaction which is applicable to both of them. Mention the other two reactions each of which is applicable to one but not to the other.
Answer:

The two compounds belonging to different classes are CH₃CH₂CHO (propanal) and CH₃COCH₃ (acetone). Both of them reacts with 2,4-DNP to form the orange-coloured precipitate of the corresponding 2,4-dinitrophenylhydrazone.

Again, propanal reduces Tollens’ reagent when metallic silver is precipitated. Acetone does not reduce Tollens’ reagent.

On the other hand, acetone reacts with I₂ /NaOH to form yellow crystalline precipitate of iodoform. Propanal does not react with I₂ /NaOH to form yellow precipitate of iodoform.

Question 7. An organic compound (A) having molecular formula C₄H₉Cl reacts with hot alcoholic caustic potash solution to produce two isomeric alkenes (B) and (C). When the mixture of (B) and (C) is subjected to ozonolysis, three compounds are obtained :

1. HCHO,
2. CH₃CHO
3. CH₃CH₂CHO. Determine the structural formulas of (A), (B) and (C).

Answer:

The general formula of the organic compound ‘A’ (C₄H₉Cl) is C_nH_2n+1Cl. Therefore, it may be concluded that it is an alkyl chloride. The compound on dehydrochlorination by alcoholic KOH solution produces two isomeric alkenes ‘B’ and ‘C’. Hence, the chlorine atom is not attached to the terminal carbon atom of the alkyl chloride ‘A’. In that case, the probable structure of ‘A’ is (1) or (2).

The compound (1) on dehydrochlorination will produce two alkenes while the compound (2) on dehydrochlorination will produce only one alkene. Hence, the actual structural formula of the compound ‘A’ is (I).

The two isomeric alkenes obtained on dehydro-chlorination are, therefore, but-l-ene (B) and but-2-ene (C). From the structural formulas of ‘B’ and ‘C, it is easily understandable that their ozonolysis will produce (HCHO + CH₃CH₂CHO) and CH₃CHO respectively. This result is in conformity with the given data.

Therefore, the compounds ‘A’ ‘B’ and ‘C are respectively CH₃CHClCH₂CH₃ (2-chlorobutane), CH₂=CHCH₂CH₃ (but-l-ene) and CH₃CH=CHCH₃ (but-2-ene).

Question 8. An unknown compound contains 69.77% carbon, 11.63% hydrogen and rest of oxygen. The molecular weight of the compound is 86. The compound responds to the iodoform test and forms a bisulphite addition compound but cannot reduce Fehling’s solution. What may be the probable structural formula of the compound?
Answer:

The amount of oxygen in the compound

= 100 -(69.77 + 11.63) = 18.6%

Dividing the percentage compositions by respective atomic weights, we get,

⇒ \(\mathrm{C}: \mathrm{H}: \mathrm{O}=\frac{69.77}{12}: \frac{11.63}{1}: \frac{18.6}{16}=5: 10: 1\)

∴ The empirical formula of the compound: C₅H₁₀O and its molecular formula (C₅H₁₀O)n.

∴ n(5 × 12 + 10 × 1 + 16) = 86 , or n = 1 ;

∴ Molecular formula: C₅H₁₀O=C₅H_2×5O=C_nH_2nO

The molecular formula of the compound is in conformity with the general formula of aldehydes and ketones. Therefore, the compound ‘A’ may be an aldehyde or a ketone. But according to the given data the compound is not an aldehyde because it cannot reduce Fehling’s solution. Hence, it is a ketone. Again, as it responds to iodoform test, it is a methyl ketone. Therefore, the compound is pentan-2-one or 3- methylbutan-2-one. Their structural formulas are given below:

Question 9. A compound (A) having molecular formula C₅H₁₀O forms phenylhydrazone. It does not form iodoform and does not reduce Tollens’ reagent (A) on reduction produces n-pentane. Write the structural formula of (A) and explain the reactions.
Answer:

The given compound ‘A’ forms phenylhydrazone. Therefore, it must be an aldehyde or a ketone. The compound does not reduce Tollens’ reagent. Hence, it is not an aldehyde but a ketone. Again, as the compound does not form iodoform, so it cannot be a methyl ketone. Therefore, the only probable structural formula of the compound is: CH₃CH₂COCH₂CH₃ (pentan-3-one).

Pentan-3-one on reduction produces npentane (CH₃CH₂CH₂CH₂CH₃) which is in conformity with the given data. Now, on the basis of the above mentioned structure of ‘A’ the reactions may be given as follows:

Question 10. Which out of propanoic and propenoic acids is a stronger acid and why?
Answer:

Propenoic acid is a stronger acid than propanoic acid. Their structures are—

In propenoic acid, the —COOH group is attached to an electron-attracting (-I) vinyl group (—CH=CH₂) while in propanoic acid, the —COOH group is attached to an electron-releasing (+I) ethyl group (CH₂CH₃). Due to this, the O—H bond in propenoic acid is weaker than the O—H bond in propanoic acid and dissociates easily to release H⁺. Since the release of H⁺ occurs easily from propenoic acid, therefore, it is a stronger acid than propanoic acid.

Question 11. Adipic acid (C₆H_xO_y is the homologue of oxalic acid. What are the values of x and y?
Answer:

Oxalic acid is the first member of the dicarboxylic acid family. Adipic acid is also a dicarboxylic acid, i.e., it is a homologue of oxalic acid. Therefore, the molecular formula of adipic acid is (CH₂)_n(COOH)₂ or C_2n+1H_2n+1O₄. Given, the formula of adipic acid is C₆H_xO_y.

By comparing these two formulas, we get n + 2 = 6, 2n + 2 = x and y = 4.

∴ n = 6-2 = 4 and x = 2n + 2 = 2×4 + 2 = 10

i.e., x = 10 and y =4

Question 12. An organic compound A (C₅H₁₂O) reacts with sodium to liberate H₂ gas. A on oxidation produces, B (C₅H₁₀O). B reacts with I₂/KOH to form a yellow precipitate. When A is heated with a cone. H₂SO₄ at 170°C, C(C₅H₁₀) is obtained. C, when oxidised by hot KMnO₄ solution, produces D (C₃H₄O) and E (C₂H₄O₂). When the calcium salt of E is heated, D is obtained. Identify A, B, C and E giving proper reasons.
Answer:

i.e., A = C₅H₁₂O≡C₅H_2×5+2 O≡C_nH_2n+2O.

The molecular formula of ‘A’ is in conformity with that of alcohols or ethers. Since it liberates H₂ when reacts with sodium, therefore, it is an alcohol. Again, ‘B’ (C₅H₁₂O) obtained on oxidation of A reacts with I₂/KOH to form a yellow crystalline precipitate of iodoform. Therefore ‘B’ is a methyl ketone [CH₃COCH₂CH₂CH₃ or CH₃COCH(CH₃)₂]. Naturally, the structural formula of the alcohol ‘A’ is CH₃CH(OH)CH₂CH₂CH₃ or CH₃CH(OH)CH(CH₃)₂.

Now, the probable structural formulas of ‘C’ (C₅H₁₀) obtained on dehydration of ‘A’ is— CH₃—CH=CHCH₂CH₃ or CH₃CH=C(CH₃)₂. Again, when ‘C is oxidised with KMnO₄, ‘D’ (C3H6O) and’E’ (C2H₄O2) are obtained. The molecular formulas of ‘D’ and ‘E’ suggest that ‘D’ is a carbonyl compound (CH₃CH₂CHO or CH₃COCH₃) and ‘E’ is a carboxylic acid (CH₃COOH). That is the alkene ‘C’ on oxidation produces a carbonyl compound and a carboxylic acid (CH₃COOH). From this it becomes clear that the structural formula of ‘C’ cannot be CH₃CH=CHCH₂CH₃, rather its structural formula will be CH₃CH=C(CH₃)₂.

When the calcium salt of (CH₃COOH) is heated, acetone (CH₃COCH₃) is obtained, i.e., ‘D ’ is obtained which conforms with the given data.

Thus, the structural formula of the secondary alcohol ‘A ‘ is CH₃CH(OH)CH(CH₃)₂. Therefore, the structural formulas oi’A’, ‘B’, ‘C, ‘D’ and ‘F are respectively—

According to the given data, the reactions are as follows:

Question 13. Unbranched carboxylic acids are stronger acids than the isomeric branched carboxylic acids—why?
Answer:

The more the conjugate base (the carboxylate ion, RCOO^– ) of the carboxylic acid is stabilised by solvation in an aqueous medium, the more the acid becomes stronger; Again, the more the alkyl group is branched, the more the solvation of the conjugate base is hindered due to steric effect and hence more it becomes less stabilised. For this reason, unbranched carboxylic acids are stronger acids than isomeric branched carboxylic acids.

Question 14. Carboxylic acids release CO₂ from sodium bicarbonate or sodium carbonate solution but phenol does not. Why?
Answer:

Sodium bicarbonate and sodium carbonate are respectively the monosodium and disodium salts of carbonic acid. The carboxylic acids are stronger acids than carbonic acid. So, they react with sodium bicarbonate or sodium carbonate to form unstable carbonic acid which readily dissociates to liberate CO₂.

On the other hand, phenol (C₆H₅OH) is a weaker acid than carbonic acid and so, it cannot react with NaHCO₃ or Na₂CO₃ to form H₂CO₃, Le., to liberate CO₂.

Question 15. How will you prepare (CH₃)₃C —COOH from (CH₃)₃CBr? Can this conversion be carried out by the cyanide process?
Answer:

The carboxylic acid can be prepared by converting (CH₃)₃CBr into the corresponding Grignard reagent followed by treating the Grignard reagent with carbon dioxide.

This conversion cannot be carried out by the cyanide process because the SN₂ reaction involving an attack by CN^– ion from the backside on the 3° alkyl halide does not take place due to steric hindrance. An elimination reaction occurs and as a result, an alkene is obtained.

 

Question 16. The two carbon-oxygen bond lengths in formic acid are different but both the carbon-oxygen bond lengths in sodium formate have the same value. Explain.
Answer:

Formic acid can be represented as a resonance hybrid of the following two resonance structures (1a and 1b):

In between the two non-equivalent resonance structures, (1a) is relatively more stable than (1b) because (1b) involves the separation of charges. Therefore, the contribution of (1a) to the resonance hybrid is greater than that of (1b).

Due to this, the C—OH bond had a greater single bond character while the C=O bond has a greater double bond character and hence the former carbon-oxygen bond is longer than the latter carbon-oxygen bond. That is, in the formic acid molecules, the two carbon-oxygen bond lengths are different.

On the other hand, formate ion in sodium formate can also be represented as a hybrid of 2a and 2b:

Since these two structures are equivalent, the two carbon-oxygen bond lengths in sodium formate have the same value.

Question 17. Considering the first dissociation, which out of fumaric and maleic acids is more acidic and why?
Answer:

In maleic acid, the two — COOH groups are situated on the same side of the double bond and so the maleate ion (the conjugate base) is stabilised by intramolecular hydrogen bonding. On the other hand, in fumaric acid, the two —COOH groups are held farther apart so, intramolecular hydrogen bonding is not possible in fumarate ion (the conjugate base). Due to the greater stability of the monoanion, maleic acid has a greater tendency to ionize than fumaric acid, i.e., for the first ionization it is a stronger acid than fumaric acid.

Question 18. An organic liquid compound made up of carbon, hydrogen and oxygen is soluble in water. It liberates CO₂ from sodium carbonate and reduces both KMnO₄ and Tollens’ reagent. Identify the compound and give equations of the reactions taking place.
Answer:

Since the given compound reacts with sodium carbonate to liberate CO₂, therefore, it is a carboxylic acid. Again, the compound reduces both KMnO₄ and Tollens’ reagent, i.e., it is a reducing agent. Among the liquid organic compounds, formic acid is the only acid which has reducing properties and is soluble in water. So, the given compound composed of carbon, hydrogen and oxygen is formic acid.

Question 19. Although the number of canonical structures In the resonance hybrid of phenoxide ion (C₆H₅O^–) is more than the number of canonical structures in the resonance hybrid of carboxylate ion (RCOO^–), phenol is a weaker acid than carboxylic acid. Explain with reason.
Answer:

Although the number of canonical structures in the resonance hybrid of phenoxide ion is more, the negative charge is dispersed over three carbon atoms (having low electronegativity) and one oxygen atom (having high electronegativity).

On the other hand, in spite of having only two canonical structures in the resonance hybrid of carboxylate ion, the negative charge of the ion is distributed over two different oxygen atoms having high electronegativity. So, the carboxylate ion is more stabilised by resonance than the phenoxide ion. Hence, the carboxylic acid loses proton more easily than phenol to be converted into the stable conjugate base, Le., RCOO^–.

Question 20. A liquid reacts with sodium bicarbonate to liberate carbon dioxide gas in the form of bubbles. When the liquid is heated with a cone. H₂SO₄ in a test tube and the mouth of it is held in the burner, the liberated gas burns with a blue flame. The liquid gives a grey precipitate when treated with ammoniacal silver nitrate solution. What will be your conclusion about the identity of that liquid?
Answer:

Since the given liquid reacts with sodium bicarbonate to liberate a gas in the form of bubbles, therefore, it is an acidic compound. Again, when the liquid is heated with a cone. H₂SO₄, a gas is liberated which burns with a blue flame. Hence, the gas that evolved is possibly carbon monoxide (CO). The liquid reacts with silver nitrate (ammoniacal) to yield a grey precipitate, i.e., the liquid has a reducing property. Thus, these reactions lead to the conclusion that the given liquid is formic acid (HCOOH).

⇒ \(\mathrm{HCOOH}+\mathrm{NaHCO}_3 \longrightarrow \mathrm{HCOONa}+\mathrm{CO}_2 \uparrow+\mathrm{H}_2 \mathrm{O}\)

⇒ \(\mathrm{HCOOH}+\mathrm{H}_2 \mathrm{SO}_4 \text { (conc.) } \stackrel{\Delta}{\longrightarrow} \mathrm{CO} \uparrow+\left[\mathrm{H}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{O}\right]\)

⇒ \(\mathrm{HCOOH}+\mathrm{Ag}_2 \mathrm{O} \longrightarrow 2 \mathrm{Ag} \downarrow+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}\)

Question 21. Why is the value of the first dissociation of oxalic acid higher than that of its second dissociation?
Answer:

In oxalic acid, the two —COOH groups are linked with one another by a single bond. For the first dissociation of the acid, the electron-attracting -I effect of any one —COOH group facilitates the release of a proton from the other —COOH group. Moreover, the anion (conjugate base) produced attains stability by the -I effect of the undissociated —COOH group.

In the case of second dissociation, the electron-repelling +I effect of the — COO^– group makes the O—H bond stronger, and proton release from the second — COOH group becomes difficult. Moreover, the stability of the anion (conjugate base) decreases by the +1 effect of the — COO^– group.

From the above discussion, it is easily understood that the first proton is released from the molecule of oxalic acid quite easily while the loss of the second proton is hindered. Hence, the first dissociation constant (K_a1) of oxalic acid is higher than its second dissociation constant (K_a2).

Question 22. 22.5 mL 0.2 (N) NaOil solution Is required to neutralise 0.333g of a monocarboxylic acid. Identify the acid. What will happen if the sodium suit of that acid is heated with soda lime?
Answer:

22.5mL 0.2(N) NaOH neutralises 0.333g of monocarboxylic acid.

∴ 100mL 1(N) NaOH neutralises = \(\frac{0.333 \times 1000}{22.5 \times 0.2}=74 \mathrm{~g}\) of monocarboxylic acid.

∴ The molecular mass of the monocarboxylic acid = its equivalent weight = 74.

Let, the molecular formula of the acid is C_nH_2n+1, COOH.

∴ Its molecular mass =12n + (2n + I) + 12 + (2×16) + 1

= 14n + 46

∴ 14n + 46 = 74 or, n = 2

∴ The carboxylic acid is C₂H₅COOH (propanoic acid).

When the sodium salt of propanoic acid is heated with soda lime, ethane (C₂H₆) is produced.

⇒ \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COONa}+\mathrm{NaOH} \underset{\Delta}{\stackrel{\mathrm{CaO}}{\longrightarrow}} \mathrm{CH}_3 \mathrm{CH}_3+\mathrm{Na}_2 \mathrm{CO}_3\)

Question 23. The molecular formulas of two organic acids A and B are C₂H₄O₂ and C₂H₂O₄ respectively. Identify A and B and determine their basicity and equivalent weights.
Answer:

Acid A: CH3 —COOH (Acetic acid). It is a monocarboxylic acid. Therefore, its basicity = 1 and equivalent weight

⇒ \(=\frac{\text { Molecular mass }}{\text { Basicity }}=\frac{60}{1}=60\)

Acid B: (COOH)₂ (Oxalic acid)

It is a dicarboxylic acid. Therefore, Its basicity = 2 and equivalent weight

⇒ \(=\frac{\text { Molecular mass }}{\text { Basicity }}=\frac{90}{2}=45 .\)

Question 24. The empirical formula of two organic compounds (A) and (B) is CH2O and their vapour densities are 15 and 30 respectively. (A) reduces Tollens’ reagent but does not liberate CO2 gas from sodium bicarbonate solution. On the other hand, (B) does not reduce Tollens’ reagent but liberates CO2 gas from sodium bicarbonate solution. Identify the compounds (A) and (B) and write the reactions involved.
Answer:

The molecular mass of ‘A’= 2×vapour density

= 2 x 15 = 30

The molecular mass of ‘B’= 2×vapour density

= 2 x 30 = 60

Given, that the compound ‘A’ has a reducing property but has no acidic property (because it cannot liberate CO₂ from NaHCO₃ ). So, this compound consisting of carbon, hydrogen and oxygen is a compound of aldehyde series.

Now, the general formula of aldehyde is RCHO or C_nH_2n+1CHO.

= 12n + (2n + 1) + 12 + 1 + 16= 14n + 30

∴ In case of compound ‘ A ‘ , 14n + 30 = 30 ; or n = 0

i.e., the molecular formula of ‘A’ is C₀H_2×0+1CHO or HCHO.

Hence, the compound is formaldehyde.

Again, according to the question, the compound ‘B’ has no reducing property but has acidic property (because it liberates CO₂ gas from NaHCO₃ solution), i.e., the compound ‘B’ is a carboxylic acid. But it is not formic acid because it does not reduce Tollens’ reagent.

Now, the general formula of carboxylic acids is: RCOOH or C_nH_2n+1COOH

The molecular mass of C₂H_2n+1COOH

= 12n + (2n + 1) + 12 + 2 × 16 + 1= 14n + 46

∴ In the case of compound ‘B’, 14n + 46 = 60; or, n = 1

That is, the molecular formula of the compound ‘B’ is:

C₁H_2×1+1 COOH or, CH₃COOH

Therefore, the compound ‘B’ is acetic acid.

Reactions of the compound ‘A’:

Reactions of the compound ‘B’:

Question 25. Write the names of two isomers of the dicarboxylic acid having the general formula C₂H₆O₄. Show their structural formulas.
Answer:

The molecule of a dicarboxylic acid contains two carboxyl groups. So, the names and structural formulas of the two dicarboxylic acids are—

Question 26. 30g of a monobasic carboxylic acid (A) requires a 1000cm³ 0.5 (TV) NaOH solution for neutralisation. If it is made to undergo the following reactions successively, (A) is again obtained. Identify (A), (B),(C) and (D).

Answer:

1000 cm³ 0.5 (N) NaOH solution neutralises 30g of the monobasic acid.

∴ 1000 cm³ 1 (N) NaOH solution neutralises 30/0.5 = 60g of the monobasic acid.

i.e., 1 g-equivalent alkali neutralises 60g of the monobasic acid. Therefore, the molecular mass of the acid = equivalent weight = 60.

The general formula of a monocarboxylic acid =C₂H_2n+1COOH.

In case of the given acid, 12n + (2n+1) + 12 + (2 × 16) + 1 = 60

or, 14n+ 46 = 60

or, n = 1

Therefore, the carboxylic acid in question is CH₃COOH (acetic acid). When the reactions taking place are written successively, we get—

Question 27. When excess chlorine gas Is passed through an organic compound A (C₇H₈) in the presence of sunlight, compound B (C₇H₅Cl₃) is obtained. The compound B, when hydrolysed by lime water under pressure, gives another compound C (C₇H₆O₂). The compound C is also obtained by oxidation of A with K₂Cr₂O₇ and H₂SO₄> Identify A, B and C and explain the reactions involved.
Answer:

The molecular formula of the organic compound A is C₇H₈. So, a benzene ring is present in the compound, i.e., the compound is the aromatic hydrocarbon toluene. Its structural formula is C₆H₅CH₃.

When an excess of Cl₂ gas is passed through boiling toluene in the presence of sunlight, the H-atom of the methyl side-chain is replaced by the Cl atom one after another ultimately forming benzotrichloride. So, the compound B is benzotrichloride (C₆H₅CCl₃). Benzotrichloride, on hydrolysis by lime water under pressure, produces benzoic acid (molecular formula C₇H₆O₂ )• Hence, the compound C is benzoic acid.

Again, when toluene (A) is oxidised by K₂Cr₂O₇ and cone. H₂SO₄, benzoic acid (C) is obtained.

Therefore, A: Toluene (C₆H₅CH₃>, B: Benzotrichloride (C₆H₅CCI₃) and C: benzoic acid (C₆H₅COOH ).

Question 28. The molecular formula of 2 isomeric compounds A and B is C₆H₁₀. When A is oxidised with K₂Cr₂O₇ / H₂SO₄, benzoic acid (C₆H₅COOH) is obtained. B, on similar oxidation, produces benzene-1, 3- dicarboxylic acid (isophthalic acid). Identify compounds A and B.
Answer:

Oxidation of A by K₂Cr₂O₇/H₂SO₄ produces benzoic acid which contains only one —COOH group. Therefore, there is only one side-chain in the benzene ring of the compound A and that side-chain contains two C-atoms. That is, the compound A is ethyl benzene (C₆H₅CH₂CH₃). On similar oxidation, compound B produces isophthalic acid which contains two —COOH groups. Therefore, the benzene ring of compound B contains two side chains and each of them contains one C-atom. In isophthalic acid, the two —COOH groups exist meta to each other. Therefore, in B the two side-chains are oriented meta to each other. Hence, compound B is m-xylene or 1,3- dimethylbenzene.

Question 29. What is meant by the following terms? Give an example of the reaction in each case:

1. Cyanohydrin
2. Acetal
3. Semicarbazone
4. Aldol
5. Hemiacetal
6. Oxime
7. Ketal
8. Imine
9. 2, 4-DNP-derivative
10. Schiff’s base

Answer:

Cyanohydrin: Compounds in which cyano and hydroxyl groups are on the same carbon atom are called cyanohydrins. These are the addition products of HCN to aldehydes and ketones in a weakly basic medium.

Acetal: Compounds in which two alkoxy groups are on the same carbon atom are called acetals. These are formed by the action of two molecules of a monohydric alcohol with one molecule of an aldehyde in the presence of dry HCl gas.

Semicarbazone: These are the derivatives of aldehydes or ketones and are produced by the action of semicarbazide on aldehydes or ketones.

Hemiacetal: Compounds in which one alkoxy group and one hydroxyl group are on the same carbon atom are called hemiacetals. These are produced by the reaction between an aldehyde and a monohydric alcohol (in 1: 1 mole ratio) in the presence of dry HCl.

Ketal: Compounds in which two alkoxy groups are on the same carbon atom are called ketals. These are formed when a ketone reacts with a dihydric alcohol in the presence of dry HC1 gas or p-toluene sulphonic acid as a catalyst.

Imine: These are the compounds which contain a group. Aldehydes or ketones react with ammonia to form such compounds.

 

Question 30. Which of the following compounds would undergo aldol condensation, which is the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction.

1. Methanal
2. 2-methylpentanal
3. Benzaldehyde
4. Benzophenone
5. Cyclohexanone
6. 1-phenyl propanone
7. Phenyl-acetaldehyde
8. 2,2-dimethyl butanal

Answer:

1. Methanal undergoes Cannizzaro reaction:

⇒ \(\mathrm{HCHO}+\mathrm{HCHO} \stackrel{50 \% \mathrm{NaOH}}{\longrightarrow} \mathrm{CH}_3 \mathrm{OH}+\mathrm{HCOONa}\)

2. 2-methyl pentanal undergoes aldol condensation:

3. Benzaldchyde undergoes Cannizzaro reaction:

4. Benzophenone(Ph—CO — Ph) will not undergo any of these reactions.

5. Cyclohexanone undergoes aldol condensation:

6. 1-phenyl propanone undergoes aldol condensation:

7. Phenylacetaldehyde undergoes aldol condensation:

8. 2,2-dimethyl butanal undergoes Cannizzaro reaction:

 

Question 31. Arrange the following compounds In Increasing order of their property as indicated:

1. Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tort-butyl ketone (reactivity towards HCN)
2. CH₃CH₂CH(Br)COOH, CH₃CH(Br)CH₂COOH, (CH₃)₂CHCOOH, CH₃CH₂CH₂COOH (acid strength)
3. Benzoic acid, 4-nitrobenzoic acid, 3,4-dinitro benzoic acid, 4-methoxy benzoic acid (acid strength)

Answer:

Electron density and steric effect on carbonyl-carbon increases as we move from left to right. The reactivity of carbonyl compounds in nucleophilic addition reactions decreases as the electron density and the steric effect on the carbonyl carbon increases. Thus, the reactivity of the given compounds towards HCN addition reaction increases in the order: Di-tert-butyl ketone < Methyl tertbutyl ketone < Acetone < Acetaldehyde.

The presence of electron-withdrawing Br-atom on the α or β-carbon of butyric acid causes an increase in the acid strength. Again the acid strength decreases as the distance of the electron-withdrawing group from the carboxyl group increases. Furthermore, isobutyric acid is weaker than n-butyric acid because the acid strength weakening +I effect of (CH₃)₂CH —group is greater than that of CH₃CH₂CH₂ —group. Thus, the strength of the given acids increases in the order:

(CH₃)₂CHCOOH < CH₃CH₂CH₂COOH < CH₃CH(Br)CH₂COOH < CH₃CH₂CH(Br)COOH

4-methoxybenzoic acid is weaker than benzoic acid because the electron-donating +R effect of the — OCH₃ group increases the electron density on the C-1 of the former. On the other hand, 4-nitrobenzoic acid is stronger than benzoic acid because the electron-withdrawing -R effect of the — NO₂ group decreases the electron density of C-1 of the former. Again the presence of an additional electron withdrawing — NO₂ group in 3,4-dinitrobenzoic acid makes it the strongest acid. So acid strength follows the sequence:

4-methoxybenzoic acid < Benzoic acid < 4-nitrobenzoic acid < 3,4-dinitrobenzoic acid.

Question 32. How will you about the following conversions in not more than two steps?

1. Propanone to Propene
2. Benzoic acid to Benzaldehyde
3. Ethanol to 3-hydroxy butanal
4. Benzene to m-nitroacetophenone
5. Benzaldehyde to Benzophenone
6. Bromo-benzene to 1-phenyl ethanol
7. Benzaldehyde to 3-phenyl-propane-1-ol
8. Benzaldehyde to a -hydroxy-phenyl-acetic acid
9. Benzoic acid to m-nitrobenzyl alcohol

Answer:

Question 33. Describe: Acetylation
Answer:

Acetylation: Replacement of an H-atom from the —OH group of an alcoholic or phenolic compound or from the — NH₂ group of the amino compound by an acetyl (—COCH₃) group is known as an acetylation reaction. The reaction is carried out in the presence of a basic catalyst such as pyridine. Alcohols and phenols give esters, while amines give amides.

Question 34. Complete each synthesis by giving missing starting material, reagent or products.

Answer:

NaBH₄ reduces aldehyde and keto groups but not 2- methyl-3-phenyl prop-2-enal ester groups:

Question 35. Give a plausible explanation for each of the following:

1. Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not.
2. There are two — NH₂ groups in semicarbazide. However, only one is involved in the formation of semicarbazones.
3. During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, water or the ester should be removed as soon as it is formed.

Answer:

In 2,2,6-trimethylcyclohexane, the carbonyl group is highly crowded due to the presence of 3 methyl groups at a -position. Thus, nucleophilic attack by CN^– at the carbonyl carbon does not occur. In cyclohexanone, such steric hindrance is absent, so CN^– ion attacks readily at the carbonyl carbon, thus forming cyclohexanone cyanohydrin with HCN.

The formation of the ester by the reaction of a carboxylic acid and alcohol in the presence of an acid catalyst is a reversible reaction.

To shift the equilibrium in the forward direction, one of the products (water or ester) is to be removed when it is formed.

Question 36. An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce the reagent but forms an additional compound with sodium hydrogen sulphite and gives a positive iodoform test. On vigorous oxidation, it gives ethanoic and propanoic acid. Write the possible structure of the compound.
Answer:

Percentage of oxygen = 100- (69.77 + 11.63) = 18.6

⇒ \(\mathrm{C}: \mathrm{H}: \mathrm{O}=\frac{69.77}{12}: \frac{11.63}{1}: \frac{18.6}{16}=5.81: 11.63: 1.16\)

= 5:10:1

∴ Empirical formula = C₅H₁₀O ,

molecular formula = (C₅H₁₀O)_n

∴ n(12 × 5 + 10 + 16) = 86 , or, n = 1

So, the molecular formula of the compound =C₅H₁₀O

The compound forms a bisulphite addition compound but does not reduce Tollens’ reagent. However, it gives a +ve iodoform test. So it is a methyl ketone. On oxidation, it gives a mixture of ethanoic acid and propanoic acid. Hence, the compound is pentan-2-one (CH₃COCH₂CH₂CH₃).

Question 37. An alkene ‘A’ ( C₅H₁₀ ) on ozonolysis gives a mixture of two compounds ‘B’ and ‘C’. Compound ‘B’ gives positive Fehling’s test and also forms iodoform on treatment with I₂ and NaOH. Compound ‘C’ does not give Fehling’s test but forms iodoform. Identify the compounds A, B and C. Write the reaction for ozonolysis and formation of iodoform from B and C.
Answer:

Ozonolysis of alkene A (M.F. C₅H₁₀) gives two compounds B and C, both of which form iodoform on treatment with I₂/NaOH. So both the compounds must contain —COCHg. Now, ‘B’ reduces Fehling’s solution but ‘C’ does not. Thus the only possibility is that ‘B’ is CH₃—CHO and ’ C ’ is CH₃— CO—CH₃.

Hence, the compound ‘ A’ is

Reactions:

Question 38. An aromatic compound ‘A’ (C₈H₈O) gives a positive 2, 4-DNP test. It gives a yellow precipitate of compound ‘B’ on treatment with iodine and sodium hydroxide solution. Compound ‘ A’ does not give Tollens’ or Fehling’s test. On drastic oxidation with potassium permanganate, it forms a carboxylic acid’ C’ (Molecular formula C₇H₆O₂ ), which is also formed along with the yellow compound in the above reaction. Identify A, B and C and write all the reactions involved.
Answer:

Compound ‘A’ (Mol. formula C₈H₈O) gives +ve 2, 4- DNP test but does not reduce Tollens’ reagent or Fehling’s solution. So it contains a keto-carbonyl group. It forms a yellow precipitate of compound ‘B’ on treatment with I₂/NaOH. This shows that compound ‘B’ is iodoform (CHI₃). Thus compound ‘A ‘ contains a keto-methyl (— CO— CH₃) group. Since it is an aromatic compound, so it must be C₆Hg—CO—CH₃. This is supported by the observation that on drastic oxidation it gives a carboxylic acid, C (C₇H₆O₂) i.e., C₆H₅COOH.

Reactions:

Question 39. Write down functional isomers of a carbonyl compound with molecular formula C₃H₆O. Which isomer will react faster with HCN and why? Explain the mechanism of the reaction also. Will the reaction lead to the completion with the conversion of the whole reactant into the product at the reaction condition? If a strong acid is added to the reaction mixture what will be the effect on the concentration of the product and why?
Answer:

The functional isomers of the carbonyl compound with molecular formula C₃H₆O are:

The carbon atom of the carbonyl group of compound ‘A’ has more +ve charge than that of compound ‘B’, because the former contains only one electron-donating alkyl (CH₃CH₂— ) group, while the latter contains two alkyl groups. The carbonyl carbon of ‘ B ’ is more crowded than that of ‘A ‘ because the former is attached to two alkyl groups. So on the basis of both electronic and steric factors, compound ‘A ’ will react faster with HCN.

Mechanism: The reaction is carried out in the presence of a base.

The reaction will not lead to completion with the conversion of the whole reactant into the product as it is a reversible reaction.

The addition of a strong acid to the reaction mixture inhibits the reaction because the die formation of CN from HCN is prevented.

Question 40. When liquid ‘A’ is treated with a freshly prepared ammoniacal silver nitrate solution, it gives a bright silver mirror. The liquid forms a white crystalline solid on treatment with sodium hydrogen sulphite. Liquid ‘B’ also forms a white crystalline solid with sodium hydrogen sulphite but it does not give a test with ammoniacal silver nitrate. Which of the two liquids is aldehyde? Write the chemical equations of these reactions also.
Answer:

Liquid ‘A’ is an aldehyde because it reduces Tollens’ reagent and forms a bisulphite addition compound. Liquid 1 fl ‘ is a methyl ketone because it does not reduce Tollens’ reagent but forms a bisulphite addition compound.

Reactions:

Class 12 Chemistry Unit 12 MCQ’S

Question 1. Which of the following has maximum volatility—

Answer: 3. Due to the formation of intramolecular H-bond, volatility is maximum for

Each of the other three compounds can form an intermolecular H-bond. So they are relatively less volatile.

Question 2. Which of the following compounds is not formed in the iodoform reaction of acetone —

1. CH₃COCH₂I
2. ICH₂COCH₂I
3. CH₃COCHI₂
4. CH₃COCI₃

Answer: 2. The steps of the iodoform reaction of acetone are as follows—

Aldehydes Ketones Notes

Question 3. Identify the method by which Me₃CCO₂H can be prepared—

1. Treating 1 mol of MeCOMe with 2 mol of MeMgl
2. Treating 1 mol of MeCO₂ Me with 3mol MeMgl
3. Treating 1 mol of MeCHO with 3mol of MeMgl
4. Treating 1 mol of dry ice with 1 mol of Me₃CMgI

Answer: 4. Dry ice (O—C—O) gives trimethyl acetic acid in reaction with Me₃CMgI

Question 4. Upon treatment with I₂ and aqueous NaOH, which of the following compounds will form iodoform—

1. CH₃CH₂CH₂CH₂CHO
2. CH₃CH₂COCH₂CH₃
3. CH₃CH₂CH₂CH₂CH₂OH
4. CH₃CH₂CH₂CH(OH)CH₃

Answer: 4. Alcohols having group produce iodoform on reaction with I₂ and aq. NaOH.

Question 5. Bromination of PhCOMe in acetic acid medium produces mainly—

Answer: 4

Due to the -R effect of the neighbouring carbonyl group, the H-atoms of the —CH₃ group are more acidic compared to the aromatic hydrogens. Hence the bromination proceeds through the abstraction of the H-atom of the — CH₃ group. Further, the reaction proceeds through the formation of enol. As enol I is more reactive than enol II (formed after monobromination), the reaction does not continue to the di- and tribromination stage and stops after monobromination.

Aldehydes Ketones Notes

Question 6. The reaction of benzene with Me₃CCOCl in the presence of anhydrous AlCl₃ gives-

Answer: 2

Initially, Me₃CCOCl forms Me₃CC⁺O ion when it reacts with anhydrous AlCl₃. This carbocation is transformed into a more stable tert-butyl carbocation (Me₃C⁺) by the removal of CO. Due to the formation of stable Me₃C⁺ ion, we get PhCMe3 as the product by FriedelCrafts reaction.

Question 7. In the following species, the one which is likely to be the intermediate during benzoin condensation of benzaldehyde is—

Answer: 3

Aldehydes Ketones Notes

Question 8. In the following reaction, the product E is—

Answer: 3

Question 9. The reagents to carry out the following conversion are —

1. HgSO₄/dil.H₂SO₄
2. BH₃; H₂O₂/NaOH
3. OsO₄; HIO₄
4. NaNH₂/CH₃I; HgSO₄/dil.H₂SO₄

Answer: 4

Aldehydes Ketones Notes

Question 10. Among the following compounds, the one(s) that give(s) effervescence with aq. NaHCO₃ solution is (are)—

1. (CH₃CO)₂O

2. CH₃COOH

3. PhOH

4. CH₃COCHO

1. 1 and 2
2. 1 and 3
3. Only 2
4. 1 and 4

Answer: 3. Only 2

NaHCO₃ is the monosodium salt of carbonic acid. Acetic acid is a stronger acid than carbonic acid. CH₃COOH reacts with aq. NaHCO₃ produces unstable carbonic acid. This carbonic acid is decomposed to give effervescent CO₂.

⇒ \(\mathrm{CH}_3 \mathrm{COOH}+\mathrm{NaHCO}_3 \longrightarrow \mathrm{CH}_3 \mathrm{COONa}+\mathrm{H}_2 \mathrm{CO}_3\)

⇒ \(\mathrm{H}_2 \mathrm{CO}_3 \rightarrow \mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \uparrow\)

Question 11. The intermediate / in the following Wittig reaction is—

Answer: 1

Question 12.

The product of the above reaction is—

Aldehydes Ketones Notes

Answer: 3

Question 13. In the following reaction:

1. RCHO
2. R₂CHOEt
3. R₃CH
4. RCH(OEt)₂

Answer: 1. RCHO

Aldehydes Ketones Notes

Question 14. Which of the following will be dehydrated most readily in an alkaline medium—

Answer: 2

Aldehydes Ketones Notes

Question 15. Amongst the following compounds, the one(s) which readily react with ethanolic KCN—

1. Ethyl chloride
2. Chlorobenzene
3. Benzaldehyde
4. Salicylic acid

Answer: 1 and 3

Aldehydes Ketones Notes

Question 16. The major products obtained during ozonolysis of 2,3-dimethyl-1-butene and subsequent reductions with Zn and H2O are—

1. Methanoic acid and 2-methyl-2-butanone
2. Methanal and 3-methyl-2-butanone
3. Methanol and 2,2-dimethyl-3-butanone
4. Methanoic acid and 2-methyl-3-butanone

Answer: 2

Question 17. Which of the following reactions will not result in the formation of carbon-carbon bonds—

1. Cannizzaro reaction
2. Wurtz reaction
3. Reimer-Tiemann reaction
4. Friedel-Crafts acylation

Answer: 1

In the Cannizzaro reaction, mutual oxidation-reduction occurs between two aldehydes (same or different). So, no C— C bond is formed.

Question 18. Amongst the following compounds, the one which would not respond to the iodoform test is—

1. CH₃CH(OH)CH₂CH₃
2. ICH₂COCH₂CH₃
3. CH₃COOH
4. CH₃CHO

Answer: 3. CH₃COOH

Aldehydes Ketones Notes

In the case of acetic acid, the most acidic proton is attached to an oxygen atom. Thus deprotonation of hydrogen does not occur. Hence acetic acid will not respond to iodoform rest.

Question 19. Ozonolysis of an alkene produces only one dicarbonyl compound. The structure of the alkene is—

Answer: 2

Question 20. For the reaction below:

the structure of the product Q is—

Answer: 2

Aldehydes Ketones Notes

Question 20. The reaction sequence given below gives product R.

The structure of the product R is —

Answer: 4

Question 21. The correct order of acid strengths of benzoic acid (A), hydroxybenzoic acid (Y) and p-nitrobenzoic acid (Z) is—

1. Y>Z>X
2. Z>Y>X
3. Z> X>Y
4. Y>X>Z

Answer: 3. Z> X>Y

In the case of a conjugated base of compound Z, the -ve charge of the —COO® group can be delocalised through resonance. Again the extent of delocalisation of the -ve charge is enhanced by the -R and -I effect of the —NO₂ group, which is situated at the p -position thearomatic ring. So, the order of acid strength is Z > X > Y.

Question 22. In the IUPAC system, PhCH₂CH₂COOH is named as—

1. 3-phenyl propanoic acid
2. Benzylacetic acid
3. Carboxyethylbenzene
4. 2-phenyl propanoic acid

Answer: 1. 3-phenyl propanoic acid

Question 23. The ease of hydrolysis in the compounds CH₃COCl(1), CH₃CO—O—COCH₃(2), CH₃COOC₂Hs(3) and CH₃CONH₂(4) is of the order—

1. 1> 2 > 3 >4
2. 4> 3 > 2 > 2
3. 1> 2 > 4 > 3
4. 2 > 1 > 4 > 3

Answer: 1. 1> 2 > 3 >4

Aldehydes Ketones Notes

With the increase of +ve charge on the carbonyl carbon, the probability of attack by H₂O or OH^– ion on that carbon atom will be increased.

In—Cl has a very strong -I effect but less +R effect. So, the +ve charge on carbonyl carbon is maximum.

For +R the effect of oxygen(—C—O—C—) affects two carbonyl groups. So, the overall amount of +R effect on any one of the carbonyl groups is much less. In the case of and  — NH₂ groups have a sufficient amount of +R effect. However, the electronegativity of oxygen is greater than that of nitrogen. So, the extent of the +R effect in CH₃CONH₂ is greater than that of CH₃COOEt.

Thus, the amount of +ve charge on the carbonyl carbon of CH₃CONH₂ is lesser than that of CH₃COOEt. Hence, the order of the amount of +ve charge on carbonyl carbon is—

So, in order of ease of hydrolysis: 1 > 2 > 3 > 4.

Question 24. The correct order of reactivity for the addition reaction of the following carbonyl compounds with ethyl magnesium iodide is—

1. 1 > 3 > 2 > 4
2. 4 > 3 > 2 > 1
3. 1 > 2 >4 > 3
4. 3 > 2 > 1 > 4

Answer: 1. 1 > 3 > 2 > 4

The reactivity of the carbonyl group is—

Inversely proportional to the steric effect of carbonyl carbon atom and

Directly proportional to the partial positive charge on the carbonyl carbon atom. Hence, with an increase of the +ve charge over carbonyl C-atom, the tendency of the nucleophilic reagent to attack it gets enhanced. Again the tendency of the nucleophilic reagent to attack the carbonyl C-atom decreases with an increase in the number of —CH₃ groups around it (steric effect).

So, the correct order of activity is 1 > 3 > 2 > 4.

Question 25. Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the presence of—

1. A vinyl group
2. An isopropyl group
3. An acetylenic triple bond
4. Two ethylenic double bonds

Answer: 1. A vinyl group

Aldehydes Ketones Notes

On ozonolysis, organic compounds containing vinyl groups can produce formaldehyde as one of the products.

Question 26. Trichloroacetaldehyde was subjected to a Cannizzaro reaction by using NaOH. The mixture of the products contains sodium trichloroacetate and another compound. The other compound is—

1. Trichloroethane
2. 2, 2, 2-trichloro propanol
3. Chloroform
4. 2, 2, 2-trichloroethane

Answer: 4. 2, 2, 2-trichloroethane

Question 27. Iodoform can be prepared from all except—

1. Isopropyl alcohol
2. 3-methyl-2-butanone
3. Isobutyl alcohol
4. Ethyl methyl ketone

Answer: 3. Isobutyl alcohol

Isobutyl alcohol does not contain orgroup. So it cannot produce iodoform upon treatment with I₂ and NaOH.

Question 28. In the given transformation, which of the following is the most appropriate reagent—

1. Zn-Hg/HCl
2. Na, Liq.NH₃
3. NaBH₄
4. NH₂NH₂,O^–H

Answer: 4. NH₂NH₂, O^–H

Aldehydes Ketones Notes

—The OH group is an acid-sensitive group. Thus Clemmension reduction cannot be used here. NaBH₄ reduces the Carbonyl group, which cannot be reduced to the —CH₂ — group by metallic sodium in the presence of. NH₃.

Question 29. An organic compound A upon reacting with NH₃ gives B. On heating, B gives C. C in the presence of KOH reacts with Br₂ to give CH₃CH₂NH₂. A is—

1. CH₃COOH
2. CH₃CH₂CH₂COOH
3. CH₃CH(CH₃)COOH
4. CH₃CH₂COOH

Answer: 4. CH₃CH₂COOH

Question 30. In the reaction, CH₃COOH \(\stackrel{\mathrm{LiAlH}_4}{\longrightarrow} A \stackrel{\mathrm{PCl}_5}{\longrightarrow} \boldsymbol{B} \stackrel{\text { alc. } \mathrm{KOH}}{\longrightarrow} C\) product C is-

1. Acetyl chloride
2. Acetaldehyde
3. Acetylene
4. Ethylene

Answer: 4. Ethylene

Aldehydes Ketones Notes

Question 31. Toluene \(\stackrel{\mathrm{KMnO}_4}{\longrightarrow} A \stackrel{\mathrm{SOCl}_2}{\longrightarrow} B \stackrel{\mathrm{Pd} / \mathrm{H}_2}{\mathrm{BaSO}_4} C.\) C is-

1. C₆H₅CH₂OH
2. C₆H₅COOH
3. C₆H₅CHO
4. C₆H₅CH₃

Answer: 3. C₆H₅CHO

Question 32. The major product obtained in the following reaction is—

Answer: 4

DIBAL-H is a reducing agent, which reduces both esters and carboxylic acids to aldehydes.

Aldehydes Ketones Notes

Question 33. The correct sequence of reagents for the following conversion will be —

1. CH₃MgBr, [Ag(NH₃)2]⁺OH^–, H⁺/CH₃OH
2. [Ag(NH₃)2]⁺OH^– , CH₃MgBr, H⁺/CH₃OH
3. [Ag(NH₃)2]⁺OH-, H⁺/CH₃OH, CH₃MgBr
4. CH₃MgBr, H⁺/CH₃OH, [Ag(NH₃)2]⁺OH^–

Answer: 3. [Ag(NH₃)2]⁺OH^–, H⁺/CH₃OH, CH₃MgBr

Question 34. The sodium salt of an organic acid ‘X produces effervescence with cone. H₂SO₄. ‘X reacts with the acidified aqueous CaCl₂ solution to give a white precipitate which decolourises the acidic solution of KMnO₄ ‘X’ is—

1. CH₃COONa
2. Na₂C₂O₄
3. C₆H₅COONa
4. HCOONa

Answer: 2. Na₂C₂O₄

Aldehydes Ketones Notes

Question 35. In a set of reactions, m-bromobenzoic acid gives a product S. Identify the product S—

Answer: 3

Question 36. Match the compounds given in List A with List B and select the suitable option using the code given below —

1. 1-D, 2-A, 3-C,4-B
2. 1-D, 2-B, 3-C, 4-A
3. 1-B, 2-C, 3-4, 4-1
4. 1-B, 2-A, 3-D, 4-C

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

Aldehydes Ketones Notes

Question 37. The order of reactivity of phenylmagnesium bromide (PhMgBr) with the following compounds—

1. 3 > 2 > 1
2. 2 > 1 > 3
3. 1 > 3 > 2
4. 1 > 2 > 3

Answer: 4. 1 > 2 > 3

The greater the number of alkyl or aryl groups attached to the carbonyl carbon, the greater will be the steric hindrance. Again, a phenyl group is more bulky than a methyl group. The reactivity of the carbonyl carbon atom decreases with an increase in steric hindrance. Therefore the order of reactivity will be 1 > 2 > 3.

Question 38. CH₃CHO and C₆H₅CHO cannot be distinguished chemically by—

1. Tollens’ reagent
2. Benedict test
3. Fehling’s solution test
4. Iodoform test

Answer: 1. Tollens’ reagent

Aliphatic aldehydes produce red ppt. of Cu₂O in the presence of Fehling’s solution but aromatic aldehydes produce no such ppt. Aliphatic aldehydes produce red ppt. of Cu₂O in the presence of Benedict’s solution but aromatic aldehydes produce no such ppt. Due to the presence of the keto-methyl group, CH₃CHO responds to the iodoform test but PhCHO does not respond to this test. Both CH₃CHO and PhCHO give ppt. of Ag in the presence of Tollens’ reagent.

Question 39. Consider the following reaction:
The product ‘A’ is-

1. C₆H₅OH
2. C₆H₅COCH₃
3. C₆H₅Cl
4. C₆H₅CHO

Answer: 4. C₆H₅CHO

Rosenmund reduction gives aldehyde from acyl chloride.

Question 40. Which of the following compounds will give a yellow precipitate with iodine and alkali—

1. Methyl acetate
2. Acetamide
3. 2-hydroxy propane
4. Acetophenone

Answer: 3 and 4

Both 2-hydroxypropane and acetophenone will give yellow ppt. on reaction with iodine and alkali. Methyl acetate and acetamide do not participate in this reaction because of the presence of less reactive ar-H in —COCH₃ and —CONH₂ groups. These or-H-atoms are less reactive due to the +R effect of — OCH₃ and — NH₂ groups.

Question 41. RCHO + NH₂NH₂→RCH=N—NH₂, what sort of reaction is it—

1. Free radical addition-elimination reaction
2. Electrophilic substitution-elimination reaction
3. Nucleophilic addition-elimination reaction
4. Electrophilic addition-elimination reaction

Answer: 3. Nucleophilic addition-elimination reaction

Question 42. Reaction by which benzaldehyde cannot be prepared—

Aldehydes Ketones Notes

Answer: 1.

In the presence of Zn-Hg and cone. HCl, reduction reaction occurs for aldehyde and ketone but carboxylic acid groups remain unaffected.

Question 43. The structure of the compound whose IUPAC name is 3-ethyl-2-hydroxy-4-methylhex-3-en-5-yonic acid is—

Answer: 3

Question 44. The order of stability of the following tautomeric compounds is —

Aldehydes Ketones Notes

1. 2 > 3 > 1
2. 1 > 2 > 3
3. 3 > 2 > 1
4. 2 > 1 > 3

Answer: 3. 3 > 2 > 1

The stability of enol 3 is highest due to intramolecular Hbonding and resonance. Again dike to form 2 is more stable than enol 1 (no resonance stability).

Question 45. Which is most reactive towards nucleophilic addition—

Answer: 4

— The NO₂ group has -I as well as strong -R effect. So, the electron-withdrawing capacity of this group is much higher. It makes the ring more electron-deficient and more prone to nucleophilic addition.

Question 46. Reaction of a carbonyl compound with one of the following reagents Involves nucleophilic addition followed by elimination of water. Hie reagent is—

1. A Grignard reagent
2. Hydrazine in the presence of a feeble acidic solution
3. Hydrocyanic acid
4. Sodium hydrogen sulphite

Answer: 2. Hydrazine in the presence of a feeble acidic solution

Question 47. Which one of the following esters gets hydrolysed most easily under alkaline conditions—

Aldehydes Ketones Notes

Answer: 1

— NO₂ group has a strong -R effect. For this reason, the electrophilic character of the carbonyl carbon of ester will be maximum. So, ester will be hydrolysed most easily under alkaline conditions.

Question 48. The correct structure of the product A formed In the reaction is—

Answer: 3

Selective reduction occurs on the C=C bond of α, β unsaturated carbonyl compound by using H₂/Pd-C.

Question 49. The correct order of strengths of the carboxylic acids is—

Aldehydes Ketones Notes

1. 2 > 1 > 3
2. 1 > 2 > 3
3. 2 > 3 > 1
4. 3 > 2 > 1

Answer: 3. 2 > 3 > 1

The acidity of II and III is higher than that of 1 as oxygen is present in the rings of both 2 and 3. In the case of II oxygen is more adjacent to the —COOH group than that of ni. So, the order of acid strength is 2 > 3 > 1.

Question 50. Which among the given molecules can exhibit tautomerism—

1. Both 2 and 3
2. 3 only
3. Both 1 and 3
4. Both 1 and 2

Answer: 2. 3 only

α-hydrogen of bridged carbon never participates in tautomerism. So compound III only participates in tautomerism.

Question 51. The correct statement regarding a carbonyl compound with a hydrogen atom on its alpha-carbon is—

1. A carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as keto-enol tautomerism
2. A carbonyl compound with a hydrogen atom on its alpha-carbon never equilibrates with its corresponding enol
3. A carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as aldehyde ketone equilibration
4. A carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as carbonylation

Answer: 1. A carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as keto-enol tautomerism

Aldehydes Ketones Notes

Question 52. Which of the following reagents would distinguish ciscyclopenta-l,2-diol from the trans isomer—

1. Aluminium isopropoxide
2. Acetone
3. Ozone
4. MnO₂

Answer: 2. Acetone

Cis-isomer forms ketal when it reacts with acetone, but trans-isomer does not react with acetone.

Aldehydes Ketones Notes

Question 53. Predict the correct intermediate and product in the following reaction—

Answer: 3

Question 54. Of the following, which is the product formed when cyclohexanone undergoes aldol condensation followed by heating—

Aldehydes Ketones Notes

Answer: 1

Question 55. Consider the reaction:

Identify A, X, Y and Z—

1. A: methoxymethane, X: ethanol, Y: ethanoic acid, Z: semicarbazide
2. A: ethanal, X: ethanol, Y: but-2-enal, Z: semicarbazone
3. A: ethanol, X: acetaldehyde, Y: butanone, Z: hydrazone
4. A: methoxymethane, X: ethanoic acid, Y: acetate ione, Z: hydrazine

Aldehydes Ketones Notes

Answer: 2. A: ethanal, X: ethanol, Y: but-2-enal, Z: semicarbazone

Question 56. The IUPAC name of the compound is—

1. 5-formylhex-2-en-3-one
2. 5-methyl-4-oxohex-2-en-5-al
3. 3-keto-2-methylhex-5-enal
4. 3-keto-2-methylhex-4-enal

Aldehydes Ketones Notes

Answer: 4. 3-keto-2-methylhex-4-enal

Question 57. Carboxylic acids have higher boiling points than aldehydes, ketones and even alcohols of comparable molecular mass. It is due to their—

1. Formation of intermolecular H-bonding
2. Formation of intramolecular H-bonding
3. More extensive association of carboxylic acid via van der Waals force of attraction
4. Formation of carboxylate ion

Answer: 1. Formation of intermolecular H-bonding

The formation of intermolecular H-bonding makes the carboxylic acids to boil at higher temperatures.

Question 58. Maximum decarboxylation occurs in—

1. CH₃COOH
2. C₆H₅COOH
3. C₆H₅CH₂COOH
4. CH₃COCH₂COOH

Answer: 4. CH₃COCH₂COOH

CH₃COCH₂COOH is a β-keto acid. The extent of decarboxylation is maximum in a carboxylic acid containing an electron-withdrawing group such as or —COOH at the β-carbon atom with respect to the —COOH group.

Question 59. If phthalic acid is treated with NH3 and then it is first heated weakly and then strongly, the final product formed is—

Answer: 4

Aldehydes Ketones Notes

Question 60. In a set of reactions, acetic acid yielded a product S.

The structure of S would be —

Aldehydes Ketones Notes

Answer: 1

Question 61. Salicylic acid can be easily prepared by reaction between—

1. Phenol and CO₂
2. Benzoic acid and H₂O₂
3. Benzene diazonium chloride and CO₂
4. Phenol and formic acid

Answer: 1

Aldehydes Ketones Notes

Question 62. Arrange the following compounds in Increasing order of reactivity towards nucleophilic addition reaction —

1. C₆H₅COCH₃

2. CH₃CO — C₂H₅

3. C₆H₅CHO

4. Cl — CH₂ — CHO

1. 4 > 3 > 2 > 1
2. 4 > 2 > 3 > 1
3. 1 > 2 > 3 > 4
4. 3 > 4 > 2 > 1

Answer: 1. 4 > 3 > 2 > 1

Two electron-donating alkyl groups in ketones make the carbonyl carbon less electron deficient in comparison to aldehydes. Therefore, ketones are less reactive than aldehydes towards nucleophilic addition reactions. Aromatic aldehydes and ketones are less reactive than corresponding aliphatic aldehydes and ketones due to the +R effect of a benzene ring.

Since Cl is more electronegative than carbon, it increases the reactivity of carbonyl carbon. So, the order of reactivity is,

Cl — CH₂ — CHO>C₆H₅CHO>CH₃CO₂H5>C₆H₅COCH₃

Question 63. The acidic strength of the given compounds follows the order —

Aldehydes Ketones Notes

1. 2 > 3 > 1
2. 3 > 2 > 1
3. 2 > 1 > 3
4. 1 > 2 > 3

Answer: 4. 1 > 2 > 3

The order of electron-withdrawing capability of the groups attached to the —COOH group in the given compound is—

CH₃ — CH=CH— >CH₃ — O — CH=CH —>CH₃—CH₂—

Hence, the order of acid strength is 1 > 2 > 3

Question 64. Benzaldehyde can be prepared from—

Aldehydes Ketones Notes

Answer: 1

Question 65. Product is—

Aldehydes Ketones Notes

Answer: 1

Question 66.

The number of stereoisomers formed in the given reaction is—

1. 2
2. 4
3. 8
4. 6

Answer: 1.2

Aldehydes Ketones Notes

As there is one chiral carbon in the product, the total number of stereoisomers = 2.

Question 67. What are the suitable reagents for the following conversion —

1. Br₂/FeBr₃, KMnO₄ , HNO₃/H₂SO₄
2. KMnO₄, Br₂/FeBr₃, HNO₃
3. HNO₃, Br₂/FeBr₃, KMnO₄
4. HNO₃, KMnO₄, Br₂/FeBr₃

Answer: 1. Br₂/FeBr₃, KMnO₄ , HNO₃/H₂SO₄

Aldehydes Ketones Notes

Question 68. Give the correct sequence of reagents used for the following conversion—

1. DIBAL-H, NaBH₄, H₃O⁺/Δ
2. H₃O⁺/Δ, NaBH₄, DIBAL-H
3. NaBH₄, DIBAL-H, H₃O⁺/Δ
4. DIBAL-H, H₃O⁺/Δ, NaBH₄

Answer: 3. NaBH₄, DIBAL-H, H₃O⁺/Δ

Question 69. The final product of the given reaction is—

Aldehydes Ketones Notes

Answer: 1

Question 70. The final product of the given reaction is—

Answer: 1

Aldehydes Ketones Notes

Question 71. The addition of water to alkynes occurs in an acidic medium and in the presence of Hg²⁺ ions as a catalyst. Which of the following products will be formed on addition of water to but-1-yne under these conditions—

Answer: 2

Question 72. Which of the following compounds is most reactive towards nucleophilic addition reactions—

Answer: 1

Explanation: The C-atom of the carbonyl group of CH₃CHO has the largest amount of +ve charge and minimum steric hindrance. So CH₃CH = O is most reactive towards nucleophilic addition reaction.

Question 73. The correct order of increasing acidic strength is —

1. Phenol < ethanol < chloroacetic acid < acetic acid
2. Ethanol < phenol < chloroacetic acid < acetic acid
3. Ethanol < phenol < acetic acid < chloroacetic acid
4. Chloroacetic acid < acetic acid < phenol < ethanol

Answer: 3. Ethanol < phenol < acetic acid < chloroacetic acid

Explanation: Carboxylic acids are stronger than phenols, which are again stronger than alcohols (acidic strength). Further, the strength of carboxylic acids increases due to the presence of electron-withdrawing groups such as halogen. Thus the increasing order of acid strength is: ethanol < phenol < acetic acid < chloroacetic acid.

Aldehydes Ketones Notes

Question 74. The compound can be prepared by the reaction of—

1. Phenol and benzoic acid in the presence of NaOH
2. Phenol and benzoyl chloride in the presence of pyridine.
3. Phenol and benzoyl chloride in the presence of ZnCl₂
4. Phenol and benzaldehyde in the presence of palladium

Answer: 2. Phenol and benzoyl chloride in the presence of pyridine.

Explanation: Esters are prepared by the reaction of alcohols or phenols with acid chlorides in the presence of pyridine.

Question 75. The reagent which does not react with both, acetone and benzaldehyde—

1. Sodium hydrogen sulphite
2. Phenyl hydrazine
3. Fehling’s solution
4. Grignard reagent

Answer: 3. Fehling’s solution

Aromatic aldehydes (e.g., benzaldehyde) do not reduce Fehling’s solution.

Question 76. Cannizzaro’s reaction is not given by—

Answer: 4.

CH₃CHO (contains n-H atoms) does not undergo Cannizzaro reaction.

Aldehydes Ketones Notes

Question 77. Which product is formed when the compound is treated with concentrated aqueous KOH solution—

Answer: 2

Explanation: In the Cannizzaro reaction, 2 molecules of aldehyde participate; one molecule is oxidised to the corresponding acid (as salt) and the other molecule is reduced to the corresponding alcohol.

Question 78.

The structure of ‘A’ and type of isomerism in the above reaction are respectively—

1. Prop-1-en-2-ol, metamerism
2. Prop-1-en-1-ol, tautomerism
3. Prop-2-en-2-ol, geometrical isomerism
4. Prop-1-en-2-ol, tautomerism

Answer: 4. Prop-1-en-2-ol, tautomerism

Aldehydes Ketones Notes

Question 79. Compounds A and C in the following reactions are—

1. Identical
2. Positional isomers
3. Functional isomers
4. Optical isomers

Answer: 2. Positional isomers

Question 80. Which is the most suitable reagent for the following conversion—

1. Tollens’ reagent
2. Benzoyl peroxide
3. I₂ and NaOH solution
4. Sn and NaOH solution

Answer: 3. I₂ and NaOH solution

Explanation: Methyl ketones react with I₂/NaOH to form CHI₃ and carboxylic acid (as salt); any unsaturation present in the molecule is not affected by the reagent.

Aldehydes Ketones Notes

Question 81. Which of the following compounds will give butanone on oxidation with alkaline KMnO₄ solution—

1. Butan-1-ol
2. Butan-2-ol
3. Both of these
4. None of these

Answer: 2. Butan-2-ol

Question 82. In Clemmensen reduction carbonyl compound is treated with—

1. Zinc amalgam + HCl
2. Sodium amalgam + HCl
3. Zinc amalgam + nitric acid
4. Sodium amalgam + HNO₃

Answer: 1. Zinc amalgam + HCl

Question 83. Which of the following compounds do not undergo aldol condensation—

Answer: 2 and 4

Explanation: Aldehydes or ketones devoid of α-H atom do not undergo aldol condensation.

Aldehydes Ketones Notes

Question 84. Treatment of compound Ph —COO —Ph with NaOH solution yields—

1. Phenol
2. Sodium phenoxide
3. Sodium benzoate
4. Benzophenone

Answer: 2 and 3

Question 85. Which of the following conversions can be carried out by Clemmensen reduction—

1. Benzaldehyde into benzyl alcohol
2. Cyclohexanone into cyclohexane
3. Benzoyl chloride into benzaldehyde
4. Benzophenone into diphenylmethane

Answer: 2 and 4

Explanation: In Clemmensen reduction (aldehydic or ketonic) groups are converted to groups.

Question 86. Through which of the following reactions number of carbon atoms can be increased in the chain—

1. Grignard reaction
2. Cannizzaro’s reaction
3. Aldol condensation
4. HVZ reaction

Answer: 1 and 3

Explanation: In Aldol condensation and Grignard reaction new C— C bond is formed.

Aldehydes Ketones Notes

Question 87. Benzophenone can be obtained by—

1. Benzoyl chloride + benzene + AlCl₃
2. Benzoyl chloride + diphenyl cadmium
3. Benzoyl chloride + phenyl magnesium chloride
4. Benzene + carbon monoxide + ZnCl₂

Answer: 1 and 2

Explanation: In Friedel-Crafts reaction, benzene reacts with acid chloride to form ketone. Dialkyl or diaryl cadmium converts acid chlorides to ketones, but ketones do not react further with dialkyl or diaryl ketone.

Question 88. Which of the following is the correct representation for (A) intermediate of nucleophilic addition reaction to the given carbonyl compound (A)—

Answer: 1 and 2

Explanation: Carbonyl compounds (with planar structure) undergo nucleophilic addition to form additional compounds with a tetrahedral structure.

Aldehydes Ketones Notes

Question 89. Which reagent converts carbonyl compounds into hydrocarbons—

1. H₂/Pt
2. LiAlH₄
3. K,Cr₂O₇/H2SO₄
4. Zn-Hg/HCl

Answer: 4. Zn-Hg/HCl

Question 90. Acetylene reacts with hypochlorous acid to produce—

1. CI₂CHCHO
2. CICH₂COOH
3. CH₃COCI
4. CICH₂CHO

Answer: 1. CI₂CHCHO

Question 91. Dihydroxy acetone reacts with HIO₄ to form—

1. HCHO
2. HCOOH
3. HCHO and HCOOH
4. HCHO and CO₂

Answer: 4. HCHO and CO₂

Question 92. Which one of the following does not reduce Fehling’s solution —

1. Benzaldehyde
2. Formic acid
3. Glucose
4. Fructose

Answer: 1. Benzaldehyde

Aldehydes Ketones Notes

Question 93. Which of the following will undergo nucleophilic addition reaction most easily—

1. CH₃CH₂CH₂COCH₃
2. (CH₃)₂C=O
3. CH₃CH₂CHO
4. CH₃CHO

Answer: 4. CH₃CHO

Question 94. In the Hoffmann degradation reaction, the carbonyl carbon of the amide comes out as—

1. Co₂
2. CO₃^2-
3. CO
4. HCO^–₃

Answer: 2. CO₃^2-

Question 95. When RCONH₂ reacts with Br₂/KPH, RNH₂ is obtained. intermediate formed during the reaction is-

1. R — NH — Br
2. H — CO — NBr₂
3. R — N=C=O
4. All of these

Answer: 3. R — N=C=O

Aldehydes Ketones Notes

Question 95. Which of the following acids does not form anhydride when heated with P₂O₅ —

1. HCOOH
2. CH₃COOH
3. CH₃CH₂COOH
4. C₆H₅COOH

Answer: 1. HCOOH

Question 96. Condensation of two moles of ethyl acetate in the presence of sodium ethoxide leads to the formation of—

1. Ethyl butyrate
2. Acetoacetic ester
3. Methyl acetoacetate
4. Ethyl propionate

Answer: 2. Acetoacetic ester

Question 97. Which compound does not undergo benzoin condensation reaction—

Answer: 1

Question 98. Ethylbenzene (excess) Product. The product is-

1. PhCH₂CHO
2. PhCOCH₃
3. PhCHO
4. PhCOOH

Answer: 2. PhCOCH₃

Aldehydes Ketones Notes

Question 99. The slow step of the Cannizzaro reaction is—

1. Attack by OH^– on the carboxyl group
2. Transfer of hydride ion to the carbonyl group
3. Abstraction of a proton from the carboxyl group
4. Deprotonation of PhCH₂OH

Answer: 2. Transfer of hydride ion to the carbonyl group

Question 100. On heating with cone. NaOH, phenyl glyoxal (C₆H₅COCHO) produces—

1. C₆H₅COONa and CH₃OH
2. C₆H₅CH2OH and HCOONa
3. C₆H₅CHOHCOONa
4. C₆H₅COONa and HCOONa

Answer: 3. C₆H₅CHOHCOONa

Question 101. The product (R) in the following reaction is—

1. (CH₃)₂C(OH)CH₂COCH₃
2. (CH₃)₂C=CHCOCH₃
3. (CH₃)₂CHCH₂CHOHCH₃
4. (CH₃)₂C=CHCHOHCH₃

Answer: 4. (CH₃)₂C=CHCHOHCH₃

Question 102. The crossed Cannizzaro reaction is actually a type of—

1. Redox reaction
2. Disproportion reaction
3. Both 1 and 2
4. Oxidation

Answer: 1. Redox reaction

Aldehydes Ketones Notes

Question 103. The enol formed when acetone reacts with D₂O is—

Answer: 1

Question 104. The suitable reagent for the following conversion is—

1. Zn-Hg, HCl
2. NH₂NH₂, OH^–
3. H₂/Ni
4. NaBH₄

Answer: 2. NH₂NH₂, OH^–

Question 105. The correct order of reactivity of PhCOPh(P), CH₃CHO(Q) and CH₃COCH₃(P) towards PhMgBr —

1. P > Q > R
2. P > R > Q
3. P < R < Q
4. P < Q < R

Answer: 3. P < R < Q

Aldehydes Ketones Notes

Question 106. Which will undergo dehydration most readily—

Answer: 1

Question 107. Which one of the following reacts with water to form a stable compound—

1. CH₃Cl
2. CCl₄
3. CCI₃CHO
4. CH₂CICH₂CI

Answer: 3. CCI₃CHO

Question 108. Reagent used to convert 2-pentanone to butanoic acid—

1. Sodium hypoiodite
2. O₂
3. Acidic KMnO₄
4. Alkaline KMnO₄

Answer: 1. Sodium hypoiodite

Question 109. Which will undergo decarboxylation most readily at 100-150°C-

1. CH₂=CHCH₂COOH
2. O₂NCH₂COOH
3. NC—CH₂COOH
4. CH₃COCH₂COOH

Answer: 4. CH₃COCH₂COOH

Aldehydes Ketones Notes

Question 110. Treatment of C₆H₅CH(OH)CN with red P/HI produces—

1. C₆H₅CH₂CN
2. C₆H₅CH₂COOH
3. C₆H₅CH₂CH₂NH₂
4. C₆H₅CH(OH)CH₂NH₂

Answer: 2. C₆H₅CH₂COOH

Question 111.Z is—

1. CH₃CHO
2. CH₃CH₂NHOH
3. CH₃CH₂COOH
4. CH₃COOH

Answer: 1. CH₃CHO

Question 112. Which one of the following will react with benzaldehyde to form 1-phenyl ethanol—

1. Methyl bromide
2. Ethyl iodide and magnesium
3. Methyl bromide and aluminium bromide
4. Methyl iodide and magnesium

Answer: 4. Methyl iodide and magnesium

Question 113. X reacts with SeO₂ to form glyoxal. X is—

1. CH₃COCH₃
2. CH₃CHO
3. CH₂=CH-CHO
4. CH₃COOH

Answer: 2. CH₃CHO

Question 114.. X and Y are—

1. CH₃COOH, CH₃COOC₂H₅
2. CH₃COOC₂H₅, CH₃COCH₂COOC₂H₅
3. CH₃COOH, CH₃COCH₃
4. None of these

Answer: 2. CH₃COOC₂H₅, CH₃COCH₂COOC₂H₅

Aldehydes Ketones Notes

Question 115. An ester on hydrolysis produces the acid P and the alcohol Q. The acid reduces Fehling’s solution and Q can be oxidised to the acid P. Hence, the ester is—

1. Methyl formate
2. Ethyl formate
3. Methyl acetate
4. Ethyl acetate

Answer: 1. Methyl formate

Question 116. The medium of the reaction between acetaldehyde and hydroxyl amine should be—

1. Very much alkaline
2. Very much acidic
3. Moderately acidic
4. None of these is correct

Answer: 3. Moderately acidic

Question 117. The major product(s) obtained when pentane-2-one is oxidised with HN03 is(are)—

1. N-butyric acid and formic acid
2. Isobutyric acid and acetic acid
3. Pentanoic acid
4. Ethanoic acid and propanoic acid

Answer: 4. Ethanoic acid and propanoic acid

Question 118. Which will not be obtained when a mixture of calcium formate and calcium acetate is distilled—

1. Acetone
2. Propanal
3. Ethanal
4. Methanal

Answer: 2. Propanal

Question 119. Which two reagents react to yield acetophenone—

1. Benzene and acetone
2. Phenol and sodium acetate
3. Phenol and AcOH
4. Benzene and AcCl

Answer: 4. Benzene and AcCl

Aldehydes Ketones Notes

Question 120. The cyanohydrin of which compound can be used to prepare lactic acid—

1. HCHO
2. CH₃COCH₃
3. CH₃CHO
4. CH₃CH₂CHO

Answer: 3. CH₃CHO

Question 121. The reductive ozonolysis of benzene gives—

1. Acetone
2. Maleic anhydride
3. Phthalic acid
4. Glyoxal

Answer: 4. Glyoxal

Question 122. The product obtained in the following reaction is— CH₃CH₂CH₂CH=PPh₃+ 2 -butanone

1. 3-methyl-3-heptene
2. 4-methyl-3-heptene
3. 5-methyl-3-heptene
4. 1-methyl-5-heptene

Answer: 1. 3-methyl-3-heptene

Question 123. Which dicarboxylic acid contains the most acidic hydrogen—

1. Maleic acid
2. Fumaric acid
3. Succinic acid
4. Malonic acid

Answer: 1. Maleic acid

Question 124. The correct decreasing order of acidic strength is— ClCH₂CH₂CH₂COOH(1), CH₃CHClCH₂COOH(2), CH₃CH₂CHClCOOH(3)

1. 1 > 2 > 3
2. 3 > 2 > 1
3. 1 > 3 > 2
4. 3 > 1 > 2

Answer: 2. 3 > 2 > 1

Aldehydes Ketones Notes

38. Ketones react with Mg-Hg in the presence of H₂O to form—

1. Pinacolones
2. Pinacols
3. Alcohols
4. None of these

Answer: 2. Pinacols

Question 125. A dihaloalkane, on alkaline hydrolysis, produces a ketone having molecular formula C3HgO. The dihaloalkane is—

1. 2, 2-dichloro propane
2. 1, 1-dichloro propane
3. 1,2-dichloro propane
4. 1,3-dichloro propane

Answer: 1. 2, 2-dichloro propane

Question 126. Acetaldehyde reacts with ammonia to form—

1. Ethylamine
2. Hexamethylenetetramine
3. Acetic Acid
4. Acetaldehyde ammonia

Answer: 4. Acetaldehyde ammonia

Question 127. Acetaldehyde reacts with excess of ethanol in the presence of HCl to form—

1. C₂H₅OCH₂OC₂H₅
2. ketal
3. CH₃CH(OC₂H5)₂
4. CH₃CH(OH)₂

Answer: 3. CH₃CH(OC₂H5)₂

Aldehydes Ketones Notes

Question 128. The compound obtained when acetone reacts with trichloromethane in the presence of KOH is—

1. Chloropicrin
2. Chloritone
3. CCl₄
4. Trichloroacetone

Answer: 2. Chloritone

Question 129. Which will not undergo benzoin condensation—

Answer: 4

Question 130. Which one of the following is paraldehyde—

1. (HCHO)_n
2. (CH₃CHO)₃
3. (HCHO)₃
4. (CH₃CHO)₄

Answer: 2. (CH₃CHO)₃

Question 131. Chloritone belongs to the class of—

1. General aldehyde
2. P-chloroketone
3. P-chloroester
4. Trichloro alcohol

Answer: 4. Trichloro alcohol

Question 132. Which has an α-C atom but not an α-H atom—

1. Propionaldehyde
2. Furfural
3. Isobutyraldehyde
4. Formaldehyde

Answer: 2. Furfural

Aldehydes Ketones Notes

Question 133. The suitable reagent to be used to prepare acetone from acetyl chloride is—

1. HI
2. Diethyl cadmium
3. Dimethyl cadmium
4. Methyl magnesium bromide

Answer: 3. Dimethyl cadmium

Question 134.

1. LiAlH₄ /ether
2. H₂/Pd-BaSO₄
3. SnCl₂/HCl; H₂O,Δ
4. NaBH₄ /ether/H₃O⁺

Answer: 3. SnCl₂/HCl; H₂O,A

Question 135. Acetic anhydride reacts with an excess of NH₃ to form—

1. 2CH₃COONH₄
2. 2CH₃CONH₂
3. CH₃CONH₂ + CH₃COONH₄
4. 2CH₃COOH

Answer: 3. CH₃CONH₂ + CH₃COONH₄

Aldehydes Ketones Notes

Question 136. Which one of the following compounds is used in baking powder—

1. Citric acid
2. Lactic acid
3. Tartaric acid
4. Malonic acid

Answer: 3. Tartaric acid

Question 137. Which one of the following compounds reacts with NH₂OH to form two oximes—

1. CH₃COCH₃
2. CH₃CH₂COCH₃
3. CH₃CH₂COCH₂CH₃
4. 

Answer: 2. CH₃CH₂COCH₃

Question 138. The compound formed on passing EtOH vapours over heated Cu at 300°C is treated with NaOH soln. to yield—

1. Aldol
2. β-hydroxy butyraldehyde
3. Both 1 and 2
4. None of these

Answer: 3. Both 1 and 2

Question 139. Which of the given undergoes Cannizzaro reaction, reduces Schiff’s reagent but not Fehling’s reagent—

1. CH₃CHO
2. HCHO
3. C₆H₅CHO
4. C₆H₅CH₂CHO

Answer: 3. C₆H₅CHO

Aldehydes Ketones Notes

Question 140. The reagent used to distinguish between HCHO and HCOOH is—

1. Tollens’ reagent
2. NaHCO₃
3. Fehling’s reagent
4. Benedict’s solution

Answer: 2. NaHCO₃

Question 141. Which one of the following pairs will undergo aldol condensation to give a compound which on dehydration produces methyl vinyl ketone—

1. HCHO and CH₃COCH₃
2. HCHO and CH₃CHO
3. Two molecules of CH₃COCH₃
4. Two molecules of CH₃CHO

Answer: 1. HCHO and CH₃COCH₃

Question 142. Which one of the following reagents does not react with aldehydes and ketones to form a solid derivative—

1. NaHSO₃
2. Phenylhydrazine
3. Semicarbazide hydrochloride
4. Hydrogen sodium phosphate

Answer: 4. Hydrogen sodium phosphate

Aldehydes Ketones Notes

Question 143. Which one is an example of a condensation reaction—

1. HCHO → paraformaldehyde
2. CH₃CHO → paraldehyde
3. CH₃COCH₃ → mesityl oxide
4. CH₂=CH₂ → polyethylene

Answer: 3. CH₃COCH₃ → mesityl oxide

Question 144. Benzaldehyde reacts with which of the following aldehyde to form cinnamaldehyde in Claisen condensation—

1. Formaldehyde
2. Acetaldehyde
3. Crotonaldehyde
4. Propionaldehyde

Answer: 2. Acetaldehyde

Question 145. Decreasing order of boiling points of CH₃CONH₂(1), CH₃COCl(2), CH₃COOH(3) and (CH₃CO)₂O(4) —

1. 1 > 4 > 3 > 2
2. 1 > 2 > 3 > 4
3. 4 > 3 > 2 > 1
4. None of these

Answer: 1. 1> 4 > 3 > 2

Question 146. Which converts carboxylic acids directly into alcohols—

1. LiAlH₄
2. Na + C₂H₅OH
3. NaBH₄
4. All of these

Answer: 1. LiAlH₄

Question 147. Acetaldehyde can be obtained from which of the following reactions—

Aldehydes Ketones Notes

Answer: 1,2,3, and 4

Question 148. CH₃CHO and PhCHO can be distinguished by—

1. Tollens’ reagent
2. Fehlings solution
3. Bendict’s solution
4. H₂N—OH

Answer: 2 and 3

Question 149. A new C—Cbond formation is possible in—

1. Aldol condensation
2. Friedel-Craft’s alkylation
3. Clemmensen reduction
4. Reimer-Hemann reaction

Answer: 1,2 and 3

Question 150. Which of the following are correct about HCOOH—

1. It is a stronger acid than CH₃COOH
2. It forms formyl chloride with PCl₅
3. It gives CO and H₃O on heating with a cone. H₂SO₄
4. It reduces Tollens’ reagent

Answer: 1,3 and 4

Aldehydes Ketones Notes

Question 151. can be distinguished by—

1. I₂ + NaOH
2. NaSO₃H
3. NaCN/HCl
4. 2, 4-DNP

Answer: 1 and 2

Question 152. Which of the following reagents react in the same manner with HCHO, CH₃CHO and CH₃COCH₃—

1. HCN
2. NH₂OH
3. Schiff reagent
4. NH₃

Answer: Both 1 and 2

Question 153. .Here ‘X’ Is—

1. CH₂N₂
2. CH₃OH/H⁺
3. MeCOOH
4. Me₂SO₄

Answer: 1,2 and 3

Question 154. Fehling’s solution gives a red precipitate with—

1. Aromatic aldehyde
2. Aliphatic aldehyde
3. Ketone
4. α-hydroxy ketone

Answer: 2 and 4

Question 155. Which would be decarboxylated readily when heated—

Aldehydes Ketones Notes

Answer: 3 and 4

Class 12 Chemistry Unit 12 Aldehydes Ketones

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

Question 2.

Aldehydes Ketones Notes

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

Question 3.

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

Question 4.

Aldehydes Ketones Notes

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

Class 12 Chemistry Unit 12 Aldehydes Ketones Notes

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

1. Both A and R are true, R is the correct explanation of A.
2. Both A and R are true, R is not a correct explanation of A.
3. A is true but R is false.
4. A is false but R is true.

Question 1. Assertion (A): Formaldehyde is a planar molecule.

Reason (R): It contains sp² a hybridised carbon atom.

Answer: 1. Both A and R are true, R is the correct explanation of A.

Aldehydes Ketones Notes

Question 2. Assertion (A): Compounds having — CHO group are easily oxidised to respective carboxylic acid.

Reason (R): Carboxylic acids can be reduced to alcohol by treatment with LiAlH4.

Answer: 5

C=O group has an electron-withdrawing nature and hence, C—Hbond in aldehyde is weak. Thus —CHO group can easily be oxidised to the —COOH group.

Question 3. Assertion (A): The a -hydrogen atom in carbonyl compounds is less acidic.

Reason (R): The anion formed after the loss of hydrogen atoms is resonance stabilised.

Answer: 4

The α-H atom in carbonyl compounds is acidic.

Aldehydes Ketones Notes

Question 4. Assertion (A): Aromatic aldehydes and formaldehyde undergo the Cannizzaro reaction.

Reason (R): Aromatic aldehydes are almost as reactive as formaldehyde.

Answer: 3

Aromatic aldehydes as well as formaldehyde do not contain any a-H atom

Question 5. Assertion (A): Aldehydes and ketones, both react with Tollens’ reagent to form a silver mirror.

Aldehydes Ketones Notes

Reason (R): Both aldehydes and ketones contain a carbonyl group.

Answer: 4

Aldehydes reduce Tollens’ reagent but ketones do not.

Class 12 Chemistry Unit 12 Aldehydes Ketones And Carboxylic Acids Fill in the blanks

Question 1. The aliphatic aldehydes do not exhibit ____ isomerism.
Answer: Positional

Aldehydes Ketones Notes

Question 2. Fehling A is the aqueous solution of ____ Fehling B is the alkaline solution of_
Answer: CuSO4, Rochelle salt

Question 3. The trimer of acetaldehyde is called ____
Answer: Paraldehyde

Question 4. Aldehydes react with alcohols in the presence of dry hydrogen chloride to form ____.
Answer: Acetal [CH₃CH(OC₂H₅)₂)

Question 5. The aldehyde and ketones containing ____ group respond to the haloform reaction.
Answer: Ketomethyl (—COCH₃)

Question 6. Aldehydes undergo ____ in the Cannizzaro reaction.
Answer: Disproportionation or self-oxidation-reduction

Question 7. ____ is obtained when an aldol condensation reaction between benzaldehyde and acetaldehyde is carried out in the presence of a dilute NaOH solution.
Answer: Cinnamaldehyde

Aldehydes Ketones Notes

Question 8. Polymerisation of ____ results in the formation of forms.
Answer: Formaldehyde

Question 9. Diethyl cadmium reacts with ethanoyl chloride to form a compound which on hydrolysis produces ____
Answer: 2-Butanone (CH₃COCH₂CH₃)

Question 10. Benzaldehyde reacts with ____ to form benzal chloride.
Answer: PCl₅

Question 11. When a mixture of potassium acetate and arsenic oxide is heated, poisonous is ____ obtained.
Answer: Cacodyl oxide

Aldehydes Ketones Notes

Question 12. _____ is liberated when acetic acid reacts with sodium bicarbonate.
Answer: CO₂

Question 13. LiAlH₄ reduces propanoic acid to produce —-
Answer: Propane-1-ol

Question 14. ____ is a dibasic organic acid.
Answer: Oxalic acid

Question 15. Chloroacetic acid is a ____ acid than acetic acid
Answer: Stronger

Question 16. In the HVZ reaction, the H-atom of ____ of carboxylic acid is substituted by ____ atom.
Answer: α-carbon, halogen

Question 17. The reaction PhCOOAg + Br₂→PhBr is known as ____ reaction
Answer: Hunsdiecker

Question 18. Benzoic acid is _____ more soluble in water than acetic acid.
Answer: Less

Aldehydes Ketones Notes

Class 12 Chemistry Unit 12 Aldehydes Ketones And Carboxylic Acids Warm Up Exercise

Question 1. Give examples of two simple and two mixed ketones.
Answer:

Simple Ketone: CH₃CH₂COCH₂CH₃, CH₃COCH₃

Mixed Ketone: CH₃COCH₂CH₃,CH₃COCH(CH₃)₂

Question 2. Write structures of the following compounds:

1. α -methoxypropionaldehyde
2. 3-hydroxy butanal
3. 4-oxopentanal
4. Di-sec-butyl ketone
5. 2-hydroxy-cyclopentanecarbaldehyde
6. p,p’ -dihydroxy benzophenone
7. p -nitropropiophenone

Answer:

Aldehydes Ketones Notes

Question 3. Write the IUPAC names of the following compounds:

1. CH₃CH(CH₃)CH₂CH₂CHO
2. CH₃COCH₂COCH₃
3. CH₃CH₂COCH(C₂H₅)CH₂CH₂Cl
4. PhCOPh
5. CH₃CH=CHCHO
6. C₆H₅ — CH=C(Cl) — CHO
7. CH₃CH(CH₃)CH₂C(CH₃)₂COCH₃
8. 

Answer:

1. 4-methylpentanal
2. Pentane-2, 4-dione
3. 1-chloro-3-ethylhexan-4-one
4. Diphenylmethanone
5. But-2-enaI
6. 2-chloro-3-phenylpropenal
7. 3, 3, 5-trimethylhexan-2-one
8. Cyclopentanecarbaldehyde

Question 4. How many ketones isomeric with 2,2-dimethylpropanal are possible?
Answer:

Three isomeric ketones of 2, 2-dimethylpropanal exist. They are— CH₃COCH₂CH2CH₃, CH₃COCH(CH₃)₂, CH₃CH₂COCH₂CH₃

Question 5. Write IUPAC names and structures of the carbonyl compounds having molecular formula C₄H₈O.
Answer:

CH₃CH₂CH₂CHO(Butanal), (CH₃)₂CHCHO(2-methylpropanal) ,CH₃COCH₂CH3(Butan-2-one)

Question 6. Give examples of two cyclic and two acyclic functional group isomers having molecular formula C₃H₆O.
Answer: Cyclic isomer:

Acyclic isomer: CH₃CH₂CHO, CH₃COCH₃

Aldehydes Ketones Notes

Question 7. Grignard reagents cannot be used for the preparation of ketones from acid chlorides. Explain.
Answer:

Question 8. Write the appropriate reagents for carrying out the following transformations:

1. Hexanal from hexan-1-ol,
2. Propenal from allyl alcohol,
3. P -fluoro benzaldehyde from p -fluoro toluene,
4. Ethanal from but-2-ene,
5. Ethanal from ethanenitrile,
6. Cyclohexanone from cyclohexanol.

Answer:

1. C₆H₅NH⁺CrO₃Cl^–(PCC)
2. Active MnO₂/CH₂Cl₂
3. CrO₃/(CH₃CO)₂O followed by H₃O⁺
4. O₃/CCl₄ followed by Zn/H₂O
5. DIBAL-H
6. K2₂Cr₂O₇/H₂SO₄

Question 9. Predict the product in each of the following reactions:

Aldehydes Ketones Notes

Answer:

Question 10. What is PCC? Mention its uses.
Answer:

PCC is a 1:1: 1 mixture of chromium trioxide (CrO₃), pyridine (C₅H₅N), and HCl. It is dissolved in CH₂Cl₂. It oxidizes primary and secondary alcohols to corresponding aldehydes.

Question 11. Arrange the following compounds in increasing order of boiling points and explain the order: CH₃CHO, CH₃CH₂OH, CH₃OCH₃, CH₃CH₂CH₃.
Answer:

Order of increasing boiling point: CH₃CH₂CH₃ < CH₃OCH₃ < CH₃CHO < CH₃CH₂OH

Question 12. Ethanal is more soluble in water than hexanal. Explain.
Answer:

Aldehyde with lower atomic mass (up to 4 carbon) can form H-bonding with H₂O molecules and dissolve easily in it. Thus ethanal dissolves easily in water. But for the aldehydes with higher atomic mass, the H-bonding cannot take place effectively due to the presence of a builder hydrocarbon chain. Thus hexanal being bulkier in size, does not dissolve in water.

Aldehydes Ketones Notes

Question 13. p -hydroxybenzaldehyde is a solid at ordinary temperature, even though o-hydroxybenzaldehyde is a liquid. Explain.
Answer:

Due to intramolecular H-bonding, orthohydroxybenzaldehyde exists as a discreet molecule. Whereas, due to intermolecular H-bonding para hydroxybenzaldehyde exists as associated molecules. For this reason, the ortho-isomer is a liquid at room temperature, whereas the para-isomer is a solid.

Question 14. Although there are double bonds in both alkenes and carbonyl compounds, they exhibit different types of addition reactions. Explain with reasons.
Answer:

C=C bond in alkene is non-polar. The electrophiles are attracted by the n-electron clouds of the C=C bond resulting in an electrophilic addition reaction. Thus C=C bond gives an electrophilic addition product. However, in the case of carbonyl compounds, due to the higher electronegativity O-atom in the C=O bond, +ve and -ve partial charges are generated on the C and O-atoms respectively. Therefore, nucleophilic reaction takes place with nucleophiles attacking the C-atom of the C=O bond resulting in nucleophilic addition product.

Question 15. Carry out the following transformations in not more than two steps :

1. Propene from propanone,
2. 3-hydroxy butanal from ethanol,
3. Benzophenone from benzaldehyde,
4. 3-phenylpropanoid-l-ol from benzaldehyde,
5. α-hydroxyphenyl acetic acid from benzaldehyde

Answer:

Question 16. Arrange the following compounds in order of increasing reactivity towards nucleophilic addition reaction:

1. Ethanal, propanal, propanone, butanone
2. Benzaldehyde, p-tolu aldehyde, p-nitrobenzaldehyde, acetophenone

Answer:

Due to the higher electronegativity of the O-atom, the cr-electron density of the C— O bond shifts towards the O-atom in all cases of aldehydes, ketones, and carboxylic acids. However, due to the hybrid structure of the carbonyl group dipole moment of aldehydes and ketones is higher than that of alcohols.

Aldehydes Ketones Notes

Question 17. Which of the following compounds participate in the Cannizzaro reaction, in the aldol condensation reaction, and in none of these two reactions:

1. Methanol,
2. 2-methylpentanal,
3. Benzaldehyde,
4. Benzophenone,
5. Cyclohexanone,
6. 1-phenylpropanone,
7. phenylacetaldehyde,
8. Butane-1-ol,
9. 2,2-dimethylbutanal.

Answer:

Aldehydes or ketones hating α-H can participate in aldol condensation. Such compounds are 2-methylpentanal, cyclohexanone, 1-phenylpropanone, and phenyl acetaldehyde.

Aldehydes having no α-H, participate in the Cannizzaro reaction. Such compounds are methanal, benzaldehyde, and 2,2 dimethylbutanal. Butan-1-ol (alcohol) and benzophenone (no α-H ) do not participate in either of the reactions.

Question 18. Arrange as directed:

1. Acetaldehyde, acetone, di-tertbutyl ketone, methyl tert-butyl ketone (in order of increasing reactivity towards HCN)
2. CH₃COCH₃, CH₃COCH₂Cl, CH₃CHO, ClCH₂CHO, HCHO (in order of increasing degree of hydration).

Answer:

Reactivity towards HCN:

(CH₃)₃C—CO—CCH₃)₃ < (CH₃>3C—CO—CH₃ < CH₃COCH₃ < CH₃CHO

Degree of hydration:

CH₃COCH₃ < CH₃COCH₂CI < CH₃CHO CH₂CICHO<HCHO

Question 19. What is the role played by Rochelle salt in Fehling’s solution?
Answer:

The tartrate ion coming from the Rochelle salt reacts with the Cu²⁺ forming a soluble salt. In the absence of Rochelle salt, Cu²⁺ reacts with NaOH and precipitates as CU(OH)₂.

Aldehydes Ketones Notes

Question 20. It is necessary to control the pH of the solution during reactions of aldehydes and ketones with ammonia derivatives—why?
Answer:

At lower pH, the ammonia derivatives get protonated and hence nucleophilicity of the ammonia derivatives decreases. Again at higher pH, the protonation of the carbonyl O-atom does not take place and hence the reactivity of the carbonyl group decreases. Hence an optimum pH should be maintained.

Question 21. Write structures of two isomeric oximes are expected to be formed when acetophenone is allowed to react with hydroxylamine hydrochloride in the presence of sodium
acetate.
Answer:

Question 22. Give examples of two reactions in which aliphatic and aromatic aldehydes behave differently.
Answer:

Aromatic aldehydes undergo perkin reaction and benzoin condensation whereas aliphatic aldehydes cannot participate in such reactions.

Question 23.

1. Give all the products expected to be formed when CH₃CHO and CH₃CH₂CHO are treated with dilute alkali. Write their IUPAC names.
2. What happens when HCl gas is passed through acetone till it becomes saturated with the gas and the mixture is allowed to stand for some time.

Answer:

Aldehydes Ketones Notes

Question 24. Predict the product obtained in each of the following reactions:

Answer:

Question 25. Mention two methods by which a carbonyl group can be converted into a methylene group.
Answer: Clemmensen and Wolff-Kishner reduction.

Question 26. Write the IUPAC names of the given compounds:

Aldehydes Ketones Notes

Answer:

1. 2-methyl cyclopentane carboxylic acid
2. 2,4,6-trinitro benzoic acid
3. 3-methyl but-2-enoic acid
4. 3-phenyl propanoic acid

Question 26. Write the structures of—

1. 3-bromo-A-phenyl pentanoic acid,
2. Hex-4-enoic acid,
3. Pyruvic acid,
4. Isobutyric acid,
5. Benzene-1,2-dicarboxylic acid.

Answer:

Question 27. Arrange as directed

Aldehydes Ketones Notes

3. Benzoic acid, 4-nitrobenzoic acid, 3, 4-dinitrobenzoic acid, 4-methoxybenzoic acid (in the order of decreasing acidic strength).

4. CH₃CH(Br)CH2CO2H C2H5CH(Br)COOH, (CH3)2CHCO2H, CH₃(CH2)2CO2H (in the order of increasing acidic strength).

Answer:

1. Order of increasing pKa value:

2. Order of decreasing acidic strength:

2. Order of increasing acidic strength:

Aldehydes Ketones Notes

(CH₃)₂CHCOOH < CH₃CH₂CH₂COOH < CH₃CH(Br)CH₂COOH < CH₃CH₂CH(Br)COOH

Question 28. Which one is stronger in each of the following pairs of acids?

1. CH₃CO₂H and FCH₂CO₂H,
2. FCH₂CO₂H and ClCH₂CO₂H,
3. FCH2(CH₂)₂CO₂H and CH₃CHFCH₂CO₂H,
4. 

Answer:

The presence of an electron-withdrawing group increases the acidity of a compound whereas the presence of an electron donating group decreases the acidity. The stronger the electron-withdrawing group, the higher the acidity of the compound. The stronger acids among the pairs are:

1. CH₂FCOOH
2. CH₂FCOOH
3. CH₃CHFCH₂COOH
4. 

Question 29. Convert:

1. Butan-1-ol → butanoic acid;
2. Benzyl alcohol → phenylethanoid acid;
3. 3-nitrobromobenzene → 3-nitrobenzoic acid;
4. Butanal → butanoic acid;
5. 4-methylacetophenone → benzene-1,4-dicarboxylic acid.

Answer:

Aldehydes Ketones Notes

Question 30. State, with equations, what happens when-

1. Bromine is added to the silver salt of propanoic acid dissolved in CCl₄.
2. Hydrazoic acid is made to react with benzoic acid in the presence of a cone. H₂SO₄.
3. Ammonium acetate is strongly heated,
4. A concentrated aqueous solution of potassium succinate is electrolyzed,
5. Acetic acid is heated with HI at 200-250°C in the presence of red phosphorus,
6. Sodium acetate is heated with soda lime.

Answer:

Question 31. Convert:

1. CH₃COOH → HOOCCH₂COOH
2. CH₃COOH → (CH₃CO)₂O

Answer:

Aldehydes Ketones Notes

Question 32. How can it be proved that a group of carboxylic acids and a group of carbonyl compounds behave differently?
Answer:

In aldehydes and ketones, no potential leaving group is bonded to the carbonyl carbon. However, in carboxylic acids, the carbonyl carbon is bonded to the —OH group, which is a potential leaving group. Thus, aldehydes and ketones can undergo only addition reactions whereas, carboxylic acids can undergo both addition and nucleophilic substitution reactions.

Again the carbonyl group in carboxylic acids undergoes resonance with the —OH group. Therefore, it shows less electrophilicity compared to the aldehydes and ketones.