WBCHSE Class 12 Chemistry Chapter 11 Alcohols Phenols And Ethers

Class 12 Chemistry Chapter 11 Alcohols Phenols And Ethers Alcohols Phenols And Ethers Introduction

Substitution of one or more H-atoms of a hydrocarbon molecule by equal number of hydroxyl groups results in the formation of alcohols or phenols. Alcohols are hydroxy derivatives of aliphatic hydrocarbons while phenols are hydroxy derivatives of aromatic hydrocarbons in which, the hydroxyl -OH group is directly attached to the carbon atom of the aromatic ring.

• Compounds of these classes find wide applications in industry as well as in our daily lives. For example, ordinary spirit used for polishing wooden furniture and also as an antiseptic is chiefly ethanol, a compound containing a hydroxyl group.
• But its real importance lies in its ability to act as a beverage in the form of beer, wine, brandy, whisky, etc.
• The sugar (i.e., sucrose) we eat, the cotton (i.e., cellulose) we use for making fabrics, the paper (i.e., cellulose) we use for writing and for manufacturing books, notebooks, newspapers, etc., and the antiseptic Dettol (a phenolic compound) we apply on wounds to prevent the growth of microorganisms, are all made up of compounds containing —OH groups.
• The phenolic compound, hexachlorophene, is a constituent of several mouthwashes, deodorant soaps and medicinal skin cleansers.
• Phenols are also used for the manufacture of dyes, drugs, resins (bakelite), etc. Therefore, life without alcohol and phenols would be very difficult.
• Substitution of a hydrogen atom of an aliphatic or aromatic hydrocarbon by an alkoxy (—OR) or aryloxy (—OAr) group gives another class of compounds known as ethers.

Ethers may also be regarded as a class of compounds obtained by substituting the hydrogen atom of the alcoholic or phenolic —OH group by an alkyl or aryl group.

Classification Of Alcohol, Phenol And Ether

Classification Of Alcohols And Phenols:

Alcohols are divided into two broad categories: Aliphatic alcohols: Alcohols in which the hydroxyl group is attached to an aliphatic carbon chain are called aliphatic alcohols.

Example:

Aromatic alcohols: Alcohols in which the hydroxyl group is attached to the side chain of an aromatic hydrocarbon are called aromatic alcohols.

Aromatic alcohols Example:

Alcohols and phenols may also be classified on the basis of the number of hydroxyl groups present in their molecules. If they contain one, two, three or more —OH group(s), they are called monohydric, dihydric, trihydric or polyhydric alcohols or phenols, respectively.

Aromatic alcohols Example: Alcohols: CH₃CH₂OH [ethanol (monohydric)],

Phenols:

Dihydric alcohols containing two —OH groups attached to the same carbon atom are called gem-diols. These are very unstable and readily eliminate one molecule of water, forming compounds containing group.

However, gem-diols containing electron-withdrawing groups are stable and can be isolated.

Phenols Example:

Chloral hydrate, CCl₂CH(OH)₂ (2,2,2- trichloro-1,1-ethanediol),

Cyclopropane hydrate,  (1,1-dihydroxycyclopropane). The stability of this gem-diol is due to less angle strain as compared to cyclopropanone.

Monohydric alcohols may be further classified on the basis of the hybridisation state of the carbon to which the hydroxyl group is attached—

1. Alcohols containing Csp³—OH bond and
2. Alcohols containing Csp³-OH bond.

Alcohols Containing Csp³—OH Bond:

In this class of alcohols, the —OH group is linked to an sp3 -hybridised carbon atom. These are further classified as:

Primary (1°), secondary (2°) and tertiary (3°) alcohols: If the —OH group of alcohol is attached to a primary, secondary or a tertiary carbon atom, then the corresponding alcohols are called primary or 1°, secondary or 2° and tertiary or 3° alcohols, respectively.

Example:

It is evident from the above structures that the functional groups of primary, secondary and tertiary alcohols are —CH₂OH, and ,respectively.

Allylic alcohols: In these alcohols, the —OH group is attached to an allylic carbon (a sp³ -hybridised carbon which is directly attached to a doubly bonded carbon). Allylic alcohols are further classified as 1°, 2° or 3° depending on the nature of the allylic carbon atom.

Example:

Benzylic alcohols: In these alcohols, the —OH group is linked to a benzylic carbon (a sp³ -hybridised carbon which is directly attached to an aromatic ring). Benzylic alcohols are further classified as 1°, 2° or 3° depending on the nature of the benzylic carbon.

Example:

Alcohols Containing Csp²—OH Bond:

• In these alcohols, the —OH group is attached to a doubly bonded carbon atom, i.e., to a vinylic carbon or an aryl group.
• Vinylic alcohols are obtained when the —OH group is attached to a vinylic carbon atom. They are usually unstable.
• However, phenols are obtained when the —OH group is attached to the aryl group and are quite stable.

Alcohols Containing Csp²—OH Bond Example:

Vinylic alcohol: CH₂=CH—OH Vinyl alcohol (unstable)

Phenol:

Classification Of Ethers: Ethers are classified on the basis of the two alkyl or aryl groups (similar or dissimilar) attached to the oxygen atom. Ethers in which the two groups are the same are called simple or symmetrical ethers, while ethers in which the two groups are different are called mixed or unsymmetrical ethers.

Classification Of Ethers Example:

Simple Ether: C₂H₅ — O —C₂H₅ (Diethyl ether)

C₆H₅ —O —C₆H₅(Diphenyl ether)

Mixed ether: CH₃—O —CH₂CH₃ (Ethyl methyl ether)

C₆H₅ —O —CH₃ (Methyl phenyl ether)

C₆H₅ —O —CH₂C₆H₅ (Benzyl phenyl ether)

Ethers are further classified on the basis of the nature of the two groups attached to the oxygen atom.

Aliphatic ethers: Ethers containing two alkyl groups attached to oxygen are called aliphatic ethers.

Aliphatic ethers Example:

1. CH₃—O — CH₃
   Dimethyl ether
2. CH₃—O—C₂H₅
   Ethyl methyl ether

Aromatic ethers: Ethers containing one or two aryl groups attached to oxygen are called aromatic ethers.

Aromatic ethers Example:

1. C₆H₅—O—CH₃
   Methyl phenyl ether
2. C₆H₅—O—C₆H₅
   Diphenyl ether

Aromatic ethers are of two types. Ethers in which one of the groups is aryl are called phenolic ethers or alkyl aryl ethers and where both the groups are aryl are called diaryl ethers. Methyl phenyl ether is an example of phenolic ether while diphenyl ether is an example of diaryl ether.

Nomenclature Of Alcohol Phenol And Ether

Nomenclature Of Alcohols:

1. Monohydric alcohols: Monohydric alcohols are named by the following systems:
2. Common system: According to this system, monohydric alcohols are called alkyl alcohols.
   1. Common system Example: CH₃OH Methyl alcohol (CH₃)₂CHOH Isopropyl alcohol (CH₃)₃COH Tert-butyl alcohol
3. Carbinol system: In this system methyl alcohol (CH₃OH) is called carbinol while the other alcohols are regarded as derivatives of carbinol, i.e., they are called alkyl or aryl carbinol.
   1. Carbinol system Example:

IUPAC System: In this system, alcohols are called alkanols. The IUPAC names of alcohols are written by replacing ‘e’ of the class suffix ‘ane’ of the corresponding alkane with the class suffix ‘ol’ of alcohol (i.e., alkane-e + ol = alkanol) and the positional number (the lowest possible serial number) of the —OH group is written just before the ‘ol’.

IUPAC System Example:

Dihydric Alcohols

These are named by the following two systems:

Common system: In this system α- or vic-glycols are named by adding the word ‘glycol’ to the name of the alkene from which they are prepared by hydroxylation.

Example:

IUPAC System: In this system, the dihydric alcohols are called alkanediols. The positions of the two —OH groups are indicated by arabic numerals.

IUPAC System Example:

Trihydric Alcohols: There is no general rule for naming these alcohols by a common system.

IUPAC System: According to this system, trihydric alcohols are called alkanetriols. The positions of the three —OH groups are indicated by arabic numerals.

IUPAC System Example:

Common name: Glycerol

IUPAC name: Propane- 1,2,3-triol

Nomenclature Of Phenols

• The simplest hydroxy derivative of benzene is known as phenol, which is the common as well as the accepted IUPAC name. In both the above systems, all the substituted phenols are named as derivatives of phenol. However, some phenols are best known by their trivial names.
• In the common system, the position of the substituent on the benzene ring concerning the —OH group is indicated by prefixes such as ortho (o-), meta (m-) or para (p-), when the substituent is at position— 2, 3 or 4 respectively concerning the —OH group. In the IUPAC system, the position of the substituents concerning the —OH group is indicated by Arabic numerals (the carbon carrying the —OH group is numbered 1).

Nomenclature Of Phenols Example:

In the IUPAC system, di, trl or polyhydric phenols are named as hydroxy derivatives of benzene. However, they are popularly known by their common names.

Nomenclature Of Phenols Example:

Common name: Hydroquinone or Quinol And Phyogallol

IUPAC name: Benzene-2,4-diol And Benzene-1,2,3-triol

Nomenclature Of Ethers

Common System: In the case of simple or symmetrical ethers, the prefix di is used before the name of the alkyl or aryl group. However, in the case of mixed or unsymmetrical ethers, the names of the two alkyl or aryl groups are written in alphabetical order followed by the word ‘ether’.

Common System Example: CH₃—O—CH₃(Dimethyl ether)

CH₃—O—CH₂CH₃

IUPAC System: In this system, ethers are considered alkoxy ( — OR) substituted alkanes. The smaller alkyl group forms the part of the alkoxy group while the larger alkyl group forms the part of the alkane.

The names of ethers are then derived by adding the suffix ‘alkoxy’ to the name of the alkane (Le., alkoxy + alkane = alkoxy alkane) followed by adding the positional number (the lowest possible number) of the alkoxy group.

IUPAC System Example:

Structures Of Functional Groups

Alcohol: In an alcohol molecule, the oxygen atom and the carbon atom attached to oxygen are sp³-hybridised. Two out of the four sp³-orbitals of oxygen overlap separately with the Is -orbital of H -atom and an sp³ -orbital of Catom to form O—H and O—C σ -bonds, respectively. The remaining two sp³ -orbitals contain two lone pair of electrons. Since lone pair-lone pair repulsions are greater than lone pair-bond pair repulsions, C—O—H bond angle is slightly less (108.9°) than the tetrahedral angle (109.5°). The C — O bond length is 142pm.

Phenol: The O—H bond in phenols, like the O—H bond in alcohols, is formed by the overlap of an sp³– orbital of oxygen with the Is -orbital of hydrogen. However, unlike alcohols, the C —O bond is formed by the overlap of an sp³-orbital of oxygen with an sp²– orbital of carbon of the benzene ring.

• The C— O—H bond angle (109°) in phenols is approximately the same as in alcohols (109°). But the C—O bond length in phenols is slightly less (136pm) due to the following reasons:
• The carbon-oxygen bond acquires some double bond character due to resonance interaction of the lone pair of electrons on oxygen with the benzene ring,
• An sp³-orbital of oxygen overlaps with a sp²-orbital of carbon which is relatively shorter in length.

Ether: The O-atom in ether is sp³-hybridised. Two out of four sp³-hybrid orbitals of oxygen overlap separately with two sp³-orbitals of two carbon atoms of alkyl groups to form two C —O cr -bonds. The remaining two sp³– orbitals contain two lone pairs of electrons. Due to steric interaction (repulsive interaction) between the two bulky R groups, C—O—C bond angle in ether is slightly greater than the tetrahedral angle. The C —O —C bond angle in CH₃ —O —CH₃ is 111.7°. The C —O bond length (141pm) in ethers is almost the same as that in alcohols (142pm).

Preparations Of Alcohols And Phenols

General Methods Of Preparation Of Alcohols

From Haloalkanes:

When haloalkanes or alkyl halides are heated with an aqueous solution of sodium or potassium hydroxide or with a suspension of silver oxide in water, they undergo hydrolysis to yield alcohols.

From Haloalkanes Example:

• This process is suitable for 1° alcohol. However, tertiary haloalkanes mainly produce alkenes through dehydrohalogenation and secondary haloalkanes produce a mixture of alkenes and alcohols.
• In an aqueous solution, NaOH or KOH acts both as a nucleophile and a base, producing some alkene (through elimination reactions) along with the desired alcohol. So, a weaker base like moist Ag₂O is used to avoid the formation of alkene.
• Alcohol cannot be used as a solvent in this reaction because in that case alkene is formed as the major product.

From Alkenes

By hydration of alkenes: Alkenes can be hydrated directly or indirectly. In the indirect method, alkenes are allowed to pass through concentrated H₂SO₄ when alkyl hydrogen sulphates (addition compound) are produced. These, when boiled with water, undergo hydrolysis to give alcohols containing the same number of carbon atoms.

By hydration of alkenes Example:

• No primary alcohol except ethanol can be prepared by the above method. This is because the addition of H₂SO₄ to unsymmetrical alkenes takes place according to the Markownikoffs rule. Subsequent hydrolysis of the addition product gives secondary or tertiary alcohols.
• In the direct process, reactive alkenes are subjected to the addition of a water molecule in the presence of mineral acids such as H₂SO₄ as a catalyst to form alcohols.

By hydration of alkenes Example:

Reaction Mechanism

The reaction occurs through the following three steps:

Step-1: The alkene undergoes electrophilic attack by the hydronium ion to form a carbocation intermediate. The addition of a proton occurs on a doubly bonded carbon in such a way that a more stable carbocation is formed. For example, in a molecule of 2-methylpropene, the addition of a proton occurs on C-l instead of C-2, because such addition leads to the formation of a more stable 3° carbocation instead of a less stable 1° carbocation.

Step-2: Nucleophilic attack by water occurs on the 3° carbocation to form a protonated alcohol (the conjugate acid of tert-butyl alcohol).

Step-3: The protonated alcohol loses a proton to form the corresponding alcohol, i.e., tert-butyl alcohol.

• By hydroboration-oxidation of alkenes: Alkenes react with diborane to form trialkylboranes which produce alcohols when oxidised by alkaline H₂O₂ followed by hydrolysis.
• This two-stage process involves the addition of a molecule of water to the reactant alkene molecule. Unsymmetrical alkenes undergo hydration in antiMarkownikoffmanner.
• Therefore, primary alcohols can be prepared from the alkenes of the type R—CH=CH₂ by this process. Generally, primary alcohols cannot be prepared from such alkenes by other processes.
• In this process, the addition of H and OH occurs from the same side of the double bond, i.e., it is a case of syn-addition.

Example:

Reaction Mechanism: When B₂Hg is allowed to react with an alkene, it undergoes splitting and forms BH₃, which adds to the double bond.

• BH₃ is an electrophile (octet of boron is incomplete). The boron atom may be attached to any of the double-bonded carbons.
• However, the addition of boron to the less substituted doubly bonded carbon leads to the formation of a more stable transition state (the positively polarised C-2 is stabilised by the +1 effect of the —CH₃ group).
• The reaction proceeds through this stable transition stage to form propylborane (the addition product) predominantly. The resulting propylborane reacts further with two more molecules of propene to form tripropylborane.

Tripropylborane thus obtained reacts with alkaline H₂O₂ (i.e., with HOO- ion) to form the boronic ester [(CH₃CH₂CH₂O)₃B], which further undergoes alkaline hydrolysis to form 1-propanol (CH₃CH₂CH₂OH).

By oxymercuration-demarcation of alkenes: Alkenes react with mercuric acetate in a mixture of tetrahydrofuran (THF) and water to produce hydroxyalkyl mercury compounds. In this step, —OH (from water) and —H_gOCOCH₃ (from mercuric acetate) are added to the double bond.

This reaction is called oxymercuration. In the next step, the resulting compounds are reduced with sodium borohydride (NaBH₄) to form alcohol. The acetoxymercury (—HgOAc) group is replaced by H. This reaction is called demercuration. The net addition of water to the double bond occurs in accordance with Markownikolf’s rule.

Example:

• This process of hydration of alkenes is more advantageous than the process of acid-catalysed hydration as there is no possibility of rearrangement.
• 3,3-dimethylbut-1-ene (Me₃C—CH =CH₂ ), for example, produces 3,3-dimethyl- 2-butanol by oxymercuration-demercuration while it produces 2,3-dimethyl-2-butanol (Me₂COHCHMe₂)by acid-catalysed hydration.

From carbonyl compounds (aldehydes and ketones): Primary and secondary alcohols having the same number of C-atoms can be prepared by reducing aldehydes and ketones respectively with a suitable reducing agent.

Reagents used: H₂ in the presence of a catalyst such as finely divided platinum, nickel, palladium or ruthenium (catalytic hydrogenation),

Nascent hydrogen is obtained by the action of sodium on ethanol, Complex metal hydrides like lithium aluminium hydride (LiAlH₄) in dry ether, Sodium borohydride in methanol, and Aluminium isopropoxide in isopropanol.

From carbonyl compounds Reagents used Example:

Lithium aluminium hydride also reduces acid chlorides, esters, amides, nitriles, alkyl halides, azides and nitro compounds while sodium borohydride reduces only acid chlorides and alkyl halides.

From Carboxylic Acids: Primary alcohols can be prepared by reducing carboxylic acids in dry ether medium, in this process, the —COOH group is first converted into a —(‘HO group and then into a —CH₂OH group. It is to be noted that diborane does not reduce functional groups such as ester, nitro, halo, etc.

From Carboxylic Acids Example:

From Esters

• When esters are reduced with sodium and ethanol, lithium aluminium hydride (LiAlH4) or H2 In the presence of copper chromite (CuO-CuCr₂O₄) catalyst at high temperatures and pressures (hydrogenolysis), two molecules of alcohols (same or different type) arc obtained.
• The alcohol produced from the acyl (RCO—) part of the ester Is always found to be a primary (1°) alcohol while the alcohol produced from the alkoxy ( —OR’) purt may be u primary (1°), secondary (2°) or tertiary (3°) alcohol.
• The reduction of an ester by Na and ethanol Is known as Bouveault-Blanc reduction (the reduction of aldehydes or ketones by Na/EtOH Is also known by the same name).

From Esters Example:

• Although LiAlH₄ reduces carboxylic acids to produce alcohols In good yield, this reducing agent being an expensive one, Is used for preparing some special chemicals only.
• Commercially, adds are reduced to alcohols by first converting them Into esters and then reducing these esters with H₂ in the presence of a catalyst (hydrogenolysis) at high temperatures and pressures or Nu/ethanol.

From Grignard reagent

Primary, secondary and tertiary alcohols are prepared by the reaction of Grignard reagents with aldehydes and ketones. The reaction actually occurs in two steps:

1. In the first step: a solution of (grinned reagent in dry Is added to a solution of the carbonyl compound in dry ether. As a result of the nucleophilic addition of RMgX (R acts as u nucleophile) to the carbonyl compound, an adduct (addition compound) is obtained.
2. In the second step: this adduct is decomposed (hydrolysed) with dilute HCl or dilute H₂SO₄ or aqueous NH₄Cl solution to form alcohol.

Synthesis of primary or 1° alcohol: The carbon atom bearing the —OH group in a 1° alcohol Is attached to two hydrogen atoms. Therefore, 1° alcohols can be prepared by the reaction of Grignard reagent with formaldehyde.

• It is to be noted that the primary alcohol produced by using Grignard nard reagent contains one carbon atom more than the alkyl group of the Grignard reagent.
• Since Grignard reagents are prepared from alkyl halides which in turn can be prepared from alcohols, an alcohol can be converted into its next higher homologue by using the above reaction (ROH→RCH₂OH).

Primary alcohols containing two carbon atoms more than the Grignard alkyl groups can be prepared by the reaction of Grignard reagents with ethylene oxide or oxirane.

Synthesis of primary or 1° alcohol Example:

Similarly, primary alcohol containing three more carbon atoms than Grignard alkyl groups can be prepared from the Grignard reagent by using trimethylene oxide.

Synthesis of secondary or 2° alcohols: Since the carbon atom bearing the —OH group in a 2° alcohol is attached to only one hydrogen atom, 2° alcohols can be prepared by the reaction of Grignard reagents with any aldehyde other than formaldehyde.

When two alkyl groups are identical (Le., isopropyl alcohol) only one combination of reactants is possible.

Synthesis of secondary or 2° alcohols Example: Isopropyl alcohol [(CH₃)₂CHOH] can be prepared by using acetaldehyde (CH₃CHO) and methylmagnesium iodide (CH₃MgI).

When two alkyl groups are different (for example., butane-2-ol), two combinations of reactants are possible.

Synthesis of secondary or 2° alcohols Example: Butan-2-ol (CH₃CH₂CHOHCH₃) can be prepared by using acetaldehyde (CH₃CHO) and ethylmagnesium bromide (C₂H₅MgBr) or propanal (CH₃CH₂CHO) and methylmagnesium iodide(CH₃MgI).

Secondary alcohols can also be prepared by the reaction of Grignard reagents (2 moles) with ethyl formate (1 mole).

Synthesis of secondary or 2° alcohols Example:

Substituted oxiranes also react with Grignard reagents to yield 2° alcohols.

Example:

Synthesis of tertiary or 3° alcohols: Since the carbon atom bearing —OH group in a 3° alcohol is not attached to any H-atom, 3° alcohols can be prepared by the reaction of Grignard reagents with any ketone.

When the three alkyl groups are identical (e.g., 2-methylpropan-2-ol, Me3COH), only one combination of reactants is possible.

Synthesis of tertiary or 3° alcohols Example:

When two alkyl groups are identical and one is different (for example., 2-methylbutan-2-oI), two combinations of reactants are possible.

Synthesis of tertiary or 3° alcohols Example:

When all three alkyl groups are different (e.g., 3-methyihexan-3-ol), three combinations of reactants are possible.

Synthesis of tertiary or 3° alcohols Example Example:

• it is to be noted that tertiary alcohols easily dehydrate in the presence of acid to produce alkanes. For this reaction, NH₄Cl solution is very often used to hydrolyse the reaction of the Grignard reagent with ketones.
• Tertiary alcohols can also be prepared by the reaction of Grignard reagent (2 mol) with an ester (1 mol) other than formic ester.

Example:

From aliphatic primary amine or 1° amine

Primary alcohols can be prepared by the reaction of aliphatic primary amines with sodium nitrite in an HCl solution

From aliphatic primary amine or 1° amine Example:

General Methods Of Preparation Of Phenols

• Phenol, commonly known as carbolic acid, was first isolated from the middle oil fraction of coal-tar distillation in the early 19th century.
• However, the present demand for phenol can hardly be met from this source. Nowadays, phenol is manufactured synthetically to cater to almost 90% of the total demand.
• In the laboratory, phenols are prepared from benzene derivatives by the following methods.

From benzene sulphonic acid: Benzene is sulphonated with oleum (H₂S₂O₇, i.e., conc.H₂SO₄ + SO₃) and benzene sulphonic acid so formed is converted to sodium phenoxide by heating strongly with molten sodium hydroxide at about 623K. Acidification of the sodium salt gives phenol.

From Diazonium Salts

• This is a convenient method for the preparation of phenols. A diazonium salt is prepared by treating an aromatic primary amine with nitrous acid [NaNO₂ + HCl] at 273-278K.
• The diazonium salt is then steam distilled when it undergoes hydrolysis to produce phenol.
• The use of benzene diazonium chloride is normally avoided because some chlorobenzene is formed along with phenol. In the laboratory, phenol is generally prepared by this method.

From Salicylic Acid: When sodium salt of salicylic acid is heated with soda lime (CaO + NaOH), it undergoes decarboxylation to form sodium phenoxide, which on acidification produces phenol.

From Grignard Reagent: When dry O₂ is passed through a dry ethereal solution of phenyl magnesium bromide, an addition compound is produced, which is hydrolysed with dil. acid to yield phenol.

Manufacture Of Phenol

From aryl halide (Dow process): When chlorobenzene is heated with 6-8% NaOH solution at 623K under 300-atmosphere pressure, it undergoes hydrolysis to form sodium phenoxide. The resulting solution is cooled and acidified with dilute HCl to produce phenol.

From cumene: Cumene (isopropylbenzene) is prepared by Friedel Crafts alkylation of benzene with propene, in the presence ofphosphoric acid.

• When air (O₂) is passed through cumene, it is oxidised to produce cumene hydroperoxide which on acidifying with 10% H₂SO₄ undergoes hydrolytic rearrangement to give phenol and acetone.
• In this process, two valuable organic compounds, phenol and acetone are obtained from two relatively inexpensive organic compounds, benzene and propene. Most of the worldwide production of phenol is based on this method.

Physical Properties Of Alcohols And Phenols

Physical Properties Of Alcohols

Physical State Colour And Odour: At ordinary temperature, lower alcohols are colourless neutral liquids with a burning taste and a distinct smell while higher alcohols are colourless, odourless, neutral waxy solids.

Solubility: Lower alcohols are lighter than water and completely miscible with water.

Solubility Example: Methanol, ethanol, 1-propanol, 2-propanol, tert-butyl alcohol.

• The solubility of alcohols in water is due to the formation of hydrogen bonds between alcohol and water molecules.
• The energy required to break a hydrogen bond between two water molecules or two alcohol molecules is provided by the formation of a hydrogen bond between water and an alcohol molecule.
• The solubility of alcohols in water decreases as the molecular mass increases. In alcohols, the hydrocarbon part i.e., the alkyl group (R—) is non-polar and hydrophobic while the hydroxyl (—OH) group is highly polar and hydrophilic.
• In lower alcohols, the polar and hydrophilic hydroxyl (—OH) group constitutes a large part of the molecule and as its characteristic effect predominates over the effect of a non-polar and hydrophobic alkyl group, these alcohols are soluble in water by forming H-bonds.

• In higher alcohols, the effect of the non-polar and hydrophobic hydrocarbon part (R— ) which constitutes a large part of the molecule predominates over the effect of the polar and hydrophilic hydroxyl (—OH) group.
• A large alkyl group resists the formation of hydrogen bonds with water molecules and so the higher alcohols are insoluble in water.
• Example: n-decyl alcohol [CH₃(CH₂)₈CH₂OH] , n-dodecyl alcohol [CH₃(CH₂)₁₀CH₂OH], n-tetradecyl alcohol [CH₃(CH₂)₁₂CH₂OH], etc., are insoluble in water.
• Amongst the isomeric alcohols, the solubility increases as the branching in the alkyl group increases.
• This is because with increase in branching of the alcohol molecule, the surface area of the non-polar hydrocarbon part decreases, i.e., it becomes progressively less predominant and hence, the solubility increases.

• The solubility of organic compounds in water increases with an increase in the number of —OH groups in the molecule.
• Example: Glycerol (HOCH₂CHOHCH₂OH) and glucose [HOCH₂(CHOH)₄CHO] molecules contain three and five —OH groups, respectively.
• Due to the presence of an increased large number of —OH groups, the molecules of these compounds are involved in extensive H -bonding with water molecules. This accounts for the high solubility of these organic compounds in the polar solvent, water.

Boiling points: The boiling point of any alcohol is generally higher than those of alkane and ether of comparable molecular mass.

• This is due to the fact that the molecules of alcohol remain associated through intermolecular hydrogen bonding. Hence, a considerable amount of thermal energy is required to separate the molecules by breaking numerous H-bonds.
• Hence, the boiling point of alcohol is relatively much higher. On the other hand, alkanes and ethers are incapable of forming H-bonds.
• Instead, van der Waals forces of attraction and weak dipole-dipole interactions are operative in alkanes and ethers, respectively. So, alkanes and ethers exist as discrete molecules and their boiling points are much lower relative to alcohols.

Boiling points Example: The molecular mass of ethyl alcohol, C₂H₅OH (46), dimethyl ether, CH₃OCH₃ (46) and propane, CH₃CH₂CH₃ (44) are almost equal. Yet, the boiling point of ethyl alcohol (78.7°C) is much higher than the respective boiling points of dimethyl ether (-24.9°C) and propane (-42°C).

• Amongst isomeric alcohols, the boiling point decreases as branching increases. This is because the molecules tend to become spherical, thereby resulting in a decreased surface area available for van der Waals interactions.
• Steric hindrance also prevents the formation of stronger hydrogen bonds among molecules of branched alcohols. Therefore, boiling points of isomeric alcohols decrease in the order: of primary > secondary > tertiary.

Boiling points Example: Boiling points of n -butyl alcohol, CH₃(CH₂)₃OH, isobutyl alcohol, (CH₃)₂CHCH₂OH, sec-butyl alcohol, CH₃CH₂CHOHCH₃ and tert-butyl alcohol (CH₃)₃COH are 118,108,100 and 83°C, respectively.

• With an increase in molecular mass, the boiling points show a regular increase.
• Amongst polyhydric alcohols, the degree of molecular association through intermolecular hydrogen bonding increases with an increase in several —OH groups and as a result, viscosity and boiling point gradually increase.

Example:

Physiological Effect: Lower alcohols have adverse effects on the human body Methyl alcohol is poisonous.

• Drinking or breathing it for a prolonged period, or allowing it to remain on the skin for a long time can lead to blindness or death.
• Ethyl alcohol, on the other hand, has an intoxicating effect and hence, is widely used in alcoholic beverages.

Physical Properties Of Phenols

Physical State, colour and odour: At ordinary temperatures, pure phenols are either colourless liquids or solids.

• However, they usually turn reddish brown due to oxidation when exposed to air and light. When pure colourless phenol, for example, is exposed to air and light, it undergoes easy oxidation to form yellow-coloured p-quinone along with some other quinone derivatives.
• Phenol, in turn, combines with p-quinone to form a red addition compound known as phenoquinone.
• Due to the formation of these coloured substances, phenol, on long exposure to air and light, assumes a deep brown colour. On distillation, coloured phenol turns colourless. Phenols have a characteristic smell, known as ‘phenolic odour’.

Boiling points: Like alcohols, phenols remain associated through intermolecular hydrogen bonding.

• Due to this, the boiling points of phenols are much higher than the aromatic hydrocarbons having comparable molecular mass.
• The boiling point of phenol, C₆H₅OH(182°C), for example, is much higher than toluene, C₆H₅CH₃ (b.p. 111°C) having almost the same molecular mass.

In substituted phenols, it is the nature and position of the substituent which influence the boiling point.

Boiling points Example: Due to the formation of intramolecular hydrogen bonds, o-nitrophenol exists as a discrete molecule (monomer) and consequently, the compound possesses a lower boiling point.

• On the other hand, in m – and p nitrophenols, the —OH and —NO₂ groups, unlike the o-isomer, are situated far apart from each other so intramolecular hydrogen bonding is not possible.
• The molecules of these two isomers remain associated through intermolecular hydrogen bonding and therefore, the boiling points of these two isomers are much higher than that of the o-isomer.

Solubility: Phenols, like alcohols, also form hydrogen bonds with water and hence, are expected to be soluble in water.

• But, due to the presence of a large hydrocarbon part (benzene ring, etc.), the solubility of phenols in water is actually much lower than that of alcohols. However, phenol itself and some dihydric and trihydric phenols are fairly soluble in water.
• It is the large hydrocarbon part for which phenols are soluble in organic solvents like ether, alcohol, etc.
• Due to chelation, o nitrophenol is also less soluble in water than its m -and p isomers. Due to lower solubility and higher volatility, o-nitrophenol can be easily separated from its m -and p -isomers by steam distillation.

Physiological Effect: Phenols are poisonous and corrosive substances. On account of their toxic effect on microorganisms, phenols also have antiseptic properties.

Chemical Reactions Of Alcohols

Most of the chemical reactions of alcohols are due to the hydroxyl (—OH) functional group. The chemical reactions of alcohols can be divided into the following three categories on the basis of bond cleavage.

Reactions involving cleavage of O—H bond [C—O+H], i.e., in which alcohols act as nucleophiles and removal of hydrogen occurs as a proton.

Reactions involving cleavage of C—O bond [C+OH], i.e., in which protonated alcohols act as electrophiles.

Reactions in which both the hydroxyl and the alkyl group participate, i.e., reactions in which both C—O and C—H bond cleavages occur.

Reactions Involving The Cleavage Of The O-H Bond

Reaction with active metals: Acidic nature

Alcohols react with active (highly electropositive) metals like Na, K, Mg and A1 to form alkoxides with the evolution of hydrogen gas. The hydrogen atom of the hydroxyl (—OH) group is replaced by the metal atom.

⇒ \(2 \mathrm{ROH}+2 \mathrm{M} \rightarrow 2 \mathrm{ROM}+\mathrm{H}_2 \uparrow \quad[\mathrm{M}=\mathrm{Na}, \mathrm{K}, \mathrm{Mg}, \mathrm{Al}]\)

Example:

Acidic character of alcohols: The reaction of alcohols with electropositive metal indicates that alcohols are slightly acidic in nature. In fact, alcohols behave as Bronsted acids, i.e., they can donate a proton to a strong base (: B^– ).

An alkoxide ion reacts with water to form the corresponding alcohol.

Since a strong base reacts with a strong acid to form a weak base and a weak acid, water is a stronger acid them alcohol and alkoxide is a stronger base than hydroxide ion.

Cause of acidic character of alcohols: The acidic nature of alcohols is due to the presence of a polar O^δ-—H^δ+ bond. Oxygen being more electronegative than hydrogen withdraws electrons of the O—H bond towards itself. As a result, the O—H bond becomes weak and alcohols exhibit acidic character by donating its proton to strong bases. In fact, alcohols are very weak acids (K_a = 1 x 10^-16-10^-18, i.e., pK_a = 16-18) and are neutral to litmus.

Alcohols are weaker acids than water: The alkyl (—R) group by its electron-donating inductive (+1) effect increases the electron density of oxygen (R→O —H) and as a result, the polarity of the O—H bond decreases. Hence, alcohols are weaker acids than water
(K_w = 1×10^-14)

Reason for decreasing acidity from 1° to 3° alcohol: As the number of electron-releasing alkyl ( —R) groups increases, the polarity of the O—H bond gradually decreases. Therefore, the decreasing order of acidity of primary, secondary and tertiary alcohols containing one, two and three alkyl groups respectively is:

• Due to this order of acidity, the rate of reaction of alcohols with sodium decreases gradually from 1° to 3° alcohol.
• Tert-butyl alcohol, being a very weak acid, reacts with less electropositive sodium very slowly and therefore, relatively more electropositive potassium is normally used to prepare its alkoxide.
• Comparison of acidity of alcohols with some common compounds: Although alcohols are less acidic than water, they are relatively more acidic than terminal alkynes, hydrogen, ammonia and alkanes.
• Their acidity order (including alcohols) and the basicity order of their conjugate bases are as follows:

Acidity: H₂O > ROH > RC=CH > H₂ > NH₃ > RH

Basicity: R^– > NH^–₂ > H^– > RC=C^– > RO^– > OH

• From this orderit is clear that, RC=C^–N^–a, N^–aH^–, N^–aNH₂^– and R^δ-—M^δ+gX react with alcohols (an acid-base reaction) to form the corresponding alkoxide along with RC=CH, H₂, NH₃ and RH.
• Due to the presence of active hydrogen, alcohols react with Grignard reagents to form alkanes corresponding to the alkyl groups of the Grignard reagents. For example:

The procedure for the determination of the number of active hydrogen atoms present in any compound using methylmagnesium iodide is known as Zerewitinoff active hydrogen determination.

∴ At STP, 22.4L of CH4= 1 active H -atom

With the help of the Zerewitinoff procedure, it is possible to distinguish between mono, di and trihydric alcohols.

Due to the presence of a lone pair of electrons on the oxygen atom, alcohols also act as Bronsted bases. They can accept a proton from strong acids to form protonated alcohols (conjugate acids).

Reaction With Carboxylic Acids (Esterification): Alcohols react with carboxylic acids in the presence of cone. H₂SO₄ or dry HCl gas as a catalyst to form ester.

• This reaction is called esterification. The reaction is reversible and in this reaction, the —OH group of the carboxylic acid is combined with the H atom of the alcoholic —OH group to form a molecule of water. Besides acting as a catalyst, cone.
• H₂SO₄ also absorbs water produced in the reaction to facilitate the forward process. The yield of ester may be increased by removing water from the reaction medium.

Example:

• As the size of the alkyl (R—) group in acids or that of the alkyl (R;—) group in alcohols increases, the rate of the above esterification reaction decreases due to steric hindrance in the tetrahedral transition state of the rate-determining step of the reaction.
• The reactivity of alcohols follows the order: CH₃OH > CH₃CH₂OH > (CH₃)₂CHOH > (CH₃)₃COH and that of carboxylic acids follows the order: HCOOH> CH₃COOH> CH₃CH₂COOH > (CH₃)₂CHCOOH > (CH₃)₃CCOOH.
• When benzoic acid reacts with methanol that has been labelled with 180, the labelled oxygen appears in the ester and not in water.
• This labelling experiment proves that the OH part of the carboxyl group and the H -atom of the alcoholic —OH group are removed as water.

Reaction With Acid Chlorides And Anhydrides: When alcohols are treated with acid chlorides or anhydrides in the presence of pyridine (base), the H -atom of the —OH group is replaced by the acyl (RCO—) group to form esters. This reaction is called acylation.

• If the acid chloride and the anhydride used are acetyl chloride and acetic anhydride respectively, i.e., if the H atom of the —OH group is replaced by an acetyl (CH₃CO—) group, the reaction is usually called acetylation. However, if benzoyl chloride is used as an acid chloride, the reaction is called benzoylatlon.

Example:

The number of —OH groups present in any compound is Thionyl the same as the number of acetyl (CH₃CO— ) groups present in the acetyl derivative of that compound.

Example: Ethylene glycol contains two —OH groups. So it forms a diacetyl derivative i.e., a diacetate.

HOCH2CH₂OH→CH₃COOCH₂CH₂OCOCH₃

• The presence of —OH group in a compound can be proved and its number can be determined by acetylation. Problem: The molecular mass of a polyhydric alcohol is 180.
• When it is heated with a mixture of acetic anhydride and sodium acetate, a compound is obtained whose molecular mass is 390. Calculate the number of hydroxyl ( —OH) groups in the alcohol. Solution:
• The change of functional group due to acetylation of the hydroxyl group is: —OH-+ —OCOCH. Therefore, 17 59 due to acetylation of one —OH group, the molecular mass of the compound increases by (59- 17) = 42 units.
• Here, the increase in molecular mass of the compound due to acetylation =390-180 = 210. /. The number of —OH groups in the original compound

⇒ \(-\underset{17}{\mathrm{OH}} \rightarrow-\underset{59}{\mathrm{OCOCH}_3}\)

Reactions involving cleavage of the C—OH bond

Reaction with halogen acids

Alcohols react with halogen acids to form haloalkanes (alkyl halides) and water.

R—OH + HX→R—X + H₂O

Reaction with phosphorus halides

When alcohols are treated with PCl₅, PCl₃, PBr₃ (P + Br₂) or Pl₃(P + I₂), respective haloalkanes are obtained.

Reaction with thionyl chloride: When alcohols are refluxed with thionyl chloride in the presence of pyridine (organic base), chloroalkanes are formed.

Example:

Reaction with ammonia: When a mixture of the vapours of alcohol and ammonia is passed over heated alumina (Al₂O₃ ) at 360°C, a mixture of primary, secondary and tertiary amines is obtained.

Reaction with ammonia Examples:

Reactions Involving Both Alkyl (R-)And -on Groups

Dehydration Of Alcohols: When alcohol containing at least one β-hydrogen (hydrogen attached to carbon next to the carbonyl carbon) is heated with dehydrating agents, it undergoes intramolecular dehydration (elimination of a molecule of water) to form an alkene.

• This dehydration can be carried out either with protonic acids like cone. H₂SO₄ or H₃PO₄ or with Lewis acids such as anhydrous zinc chloride or alumina as catalyst.
• Depending on the nature of the alcohol, sulphuric add of different concentrations Is used and the reaction is made to occur at different temperatures.
• The ease of dehydration of alcohols follows the order: 3° alcohol > 2° alcohol > 1° alcohol. The conditions of the reaction (for example, the concentration of the acid used, temperature, etc.) become milder gradually, from 1° to 3° alcohol.

Dehydration Of Alcohols Example:

Reaction Mechanism

It is an El (elimination unimolecular) reaction which occurs through the following three steps. The second step is the rate-determining step of the reaction.

Step-1: Alcohol undergoes protonation.

Step-2: The protonated alcohol loses a molecule of water to form a carbocation.

Step-3: Elimination of a proton from the carbocation to form an alkene.

Since dehydration of alcohols to alkenes occurs through the formation of intermediate carbocations and the stability of carbocations follows the order: 3° > 2° > 1°, therefore, the ease of dehydration of alcohols follows the same order, i.e., 3° > 2° > 1°.

The dehydration of alcohols always takes place in accordance with the Saytzeff rule, i.e., the more substituted alkene is always the major product.

Example:

Since carbocations are very susceptible to rearrangement and rearrangement of carbocations occurs more readily than proton elimination, dehydration of alcohols often leads to the formation of alkenes from the rearranged carbocations.

Example:

If excess of alcohol is present, intermolecular dehydration occurs to form ether.

Example:

Oxidation Of Alcohol: Oxidation of alcohols involves cleavage of the O—H and the C—H bonds and the formation of the C=O bond.

Since these reactions involve the loss of a molecule of hydrogen, these are also called dehydrogenation reactions. The oxidation of alcohols can be carried out by oxidising agents like acidified dichromate (Na₂Cr₂O₇/H₂SO₄ or K₂Cr₂O₇/H₂SO₄), acidic or alkaline KMnO₄ or dilute nitric acid. The ease of oxidation and nature of the products, however, depends on the type of alcohol used in oxidation.

Primary or 1° alcohols are oxidised first to aldehydes and then to carboxylic acids, both containing the same number of carbon atoms as the starting alcohol.

In the first stage, the H -atom of the —OH group and one H -atom attached to the carbonyl carbon are eliminated, i.e., a —CH₂OH group is converted into a —CHO group. In the second stage, one oxygen atom is introduced between the carbonyl carbon and the hydrogen attached to it, i.e., a —CHO group is converted into a —COOH group.

Controlled oxidation of 1° alcohols to aldehydes can be carried out by using some special reagents like pyridinium chlorochromate (PCC, CroC₃H₅NHCl) or, C₅H₅N⁺HCrO₃Cl^–) in dry dichloromethane (CH₃Cl₃) solvent, Collin’s reagent (CrO_{3 .}2C₅H5N), pyridinium dichromate [PDC, (C₅H₅NH)₃Cr₃O₃^2-] or Dess Martin’s reagent.

Example:

Weak oxidising agents: Weak oxidising agents convert 1° alcohols to aldehydes. Some weak oxidising agents are—

1. CrO₃/2C₅H₅N (Collin’s reagent)
2. CrO₃/C5₅H₅N/HCl (PCC, Corey’s reagent)
3. (CH₃)₂SO /(COCl)₂/(C₂H₅)₃N [Swern oxidation]
4. (C₅H₅ NH)₂Cr₂O₂– (PDC)
5. 

(Dess Martin’s Reagent)

Strong oxidising agents: Strong oxidising agents convert most 1° alcohols to carboxylic acids. Some strong oxidising agents are—

1. Na₂Cr₂O₇/H₂SO₄, K₂Cr₂O₂/H₂SO₄
2. KMnO₄/acid, KMnO₄/base (Baeyer’s reagent)
3. CrO₃/CH₃COOH
4. CrO₃/H₂SO₄/ Acetone (Jones’ reagent)
5. Dilute HNO₃

Secondary or 2° alcohols are easily oxidised to ketone containing the same number of carbon atoms. Ketones resist further oxidation. However, under drastic conditions (prolonged treatment with a strong oxidising agent), they are oxidised to carboxylic acids with a lesser number of C-atoms. Symmetrical ketones on oxidation give two carboxylic acids. But in the case of unsymmetrical ketones, C—C bond fission occurs in such a manner that the smaller alkyl group remains attached to the carbonyl group (Popoff’s rule) and the two carboxylic acids produced by such type of fission are the major products.

Both weak and strong oxidising agents convert 2° alcohols to ketones.

Tertiary or 3° alcohols (RgCOH) have no hydrogen atom attached to the carbinol carbon and so, elimination of two H -atoms is not possible. Therefore, oxidation of these alcohols is possible only by C —C bond cleavage. So, they are generally resistant to oxidation under normal conditions. They cannot be oxidised even by using strong oxidising agents in neutral or alkaline medium. However, they are oxidised by very strong oxidising agents in acidic medium (i.e., K₂Cr2O₇/H₂SO₄ or KMnO₄/H₂SO₄). In such an oxidation, a tertiary alcohol first gets dehydrated to yield an alkene which on further oxidation produces a ketone with a lesser number of carbon atom(s) than the parent alcohol. The ketone thus produced is subsequently oxidised to carboxylic acid containing a lesser number of carbon atoms.

Example:

Oppenauer oxidation: When secondary alcohols are heated with aluminium tert-butoxide in the presence of excess acetone, they undergo oxidation to give the corresponding ketones.

Example:

Primary alcohols can be oxidised to aldehydes by using p-benzoquinone instead of acetone.

Example:

Oppenauer oxidation is, in fact, the reverse of Meerwein- Ponndorf-Verley (MPV) reduction.

Dehydrogenation of alcohols

When alcohol vapours are passed over heated copper at 300°C, different classes of alcohol yield different products. Primary and secondary alcohols undergo dehydrogenation to yield aldehydes and ketones respectively. Since it is a dehydrogenation reaction (no oxidising agent is used), the resulting carbonyl compound (aldehyde or ketone) is not further oxidised to carboxylic acid. Due to the absence of hydrogen on the carbonyl carbon atom, tertiary alcohols (R₃COH) do not undergo dehydrogenation but they undergo dehydration to give alkenes.

Chemical Reactions Of Phenols

Due to the presence of —OH group in phenols, they undergo some reactions similar to alcohols. For example, phenols act as weak nucleophiles in reactions involving oxygen-hydrogen bond [C—O+H] cleavage like alcohols.

Phenols usually do not undergo reactions involving cleavage of the C—OH bond, i.e., phenols do not act as electrophiles as the lone pair of oxygen is involved in resonance with the benzene ring. Phenols do not undergo elimination and oxidation reactions like alcohols. However, they undergo a number of electrophilic substitution reactions of the benzene ring. The reactions of of phenols can be divided into three types: Reactions involving the phenolic —OH group reactions involving the benzene ring and some special reactions.

Reactions Involving Phenolic —OH group

Acidic character of phenols

Reaction with active metals: Phenols react with active metals like sodium and potassium to form sodium or potassium phenoxides (salts) and hydrogen gas.

Reaction with alkalies: Phenols (pK_a = 8-10) are stronger acids than alcohols (pK_a = 16-18). They turn blue litmus red and react with aqueous alkalies to form phenoxides or phenates.

Alcohols, however, do not turn blue litmus red or react with NaOH solution. Therefore, these tests can be used to distinguish phenols from alcohols. However, phenols are weaker acids than carboxylic acids (pK_a = 5) or carbonic acids (pK_a = 7). Hence, phenols cannot decompose carbonates and bicarbonates evolving CO₂ gas, i.e., they are not soluble in sodium carbonate or bicarbonate solution. Instead, phenols can be precipitated from aqueous solutions of phenoxides by passing CO₂ gas through them.

Phenols containing electron-withdrawing groups at ortho and para-positions (i.e., 2,4-dinitrophenol, 2,4,6- trinitrophenol, etc.) are soluble in sodium bicarbonate solution.

Explanation of stronger acidic character of phenols than alcohols: Acidic property of any compound In an aqueous medium Is measured by the ease with which it can donate a proton to water. The greater the tendency to give up a proton, the higher will be the acidic strength. In an aqueous solution, phenol dissociates to produce phenoxide ions.

Although phenol is a weak acid, It is a much stronger acid than alcohol. The acidic property of phenol can be explained in terms of resonance. Phenol is, in fact, n resonance hybrid of the following five resonance structures or canonical forms or simply canonicals:

Due to the contribution of three (2,3 and 4) out of five resonance structures, the oxygen atom in phenol becomes partially positively polarised. As a consequence, the oxygen atom attracts the O—H bonding electrons more towards itself with greater force, thereby favouring the dissociation of the O—H bond to liberate proton (H⁺). But in case of alcohols (ROH), similar resonance is not possible. So, the alcoholic O —H bond Is much stronger than the phenolic 0 —H bond and the tendency of the hydrogen atom to be liberated as a proton (H⁺) is much less in this case. Therefore, phenols are more acidic than alcohols.

Alternate explanation: This relative acidic property may also be explained by considering phenol-phenoxide Ion and alcohol-alkoxide ion equilibrium systems. like phenol, phenoxide Ion can be represented as a resonance hybrid of the following five resonance structures.

A close observation of the resonating structures shows that the structures of phenoxide Ion do not Involve any separation of charge (the negative charge Is only delocalised) while the three resonance structures (2,3 and 4) of phenol involve separation of charge. Since monopolar (carrying either five or -ve charge) resonance structures are more stable than bipolar (currying both +ve and -ve charges) resonance structures, phenoxide ion Is more resonance-stabilised than phenol (phenoxide ion is not more stable than phenol). As a result, the difference In stability between phenol and phenoxide Ion decreases, which favours the equilibrium. This explains why phenols are acidic in nature. Both the alcohol and the alkoxide ion, on the other hand, can be represented satisfactorily by single (localised) structures.

In alkoxide ion, the negative charge is centred on the oxygen atom. There is no resonance stabilisation in either of die species. The driving force, Le., the differential stabilisation, that promotes dissociation is thus absent here. The equilibrium is, therefore, not favoured. It is further disfavoured due to the destabilisation of the anion caused by the +R effect of the alkyl group. Therefore, alcohols are much less acidic than phenols.

Effect of substituents on the acidity of phenols: The acidity of phenols is due to the greater resonance stabilisation of the phenoxide ion with respect to phenol. A substituent present in the ring may increase or decrease the acidity of phenols by regulating the charge delocalisation.

An electron-withdrawing group (EWG) stabilises the phenoxide ion by delocalising the negative charge effectively and thereby increasing the acidity of phenol. An electron-donating group (EDG) destabilises the phenoxide ion by localising the negative charge and thereby decreases the acidity of phenol.

[EWG withdraws electrons, disperses the negative charge, stabilises the phenoxide ion with respect to phenol and hence, increases the acid strength.]

[EDG donates electrons, intensifies the negative charge, destabilises the phenoxide ion with respect to phenol and hence, decreases the acid strength.]

Comparison of acidic strength of substituted phenols:

p-nitrophenol is more acidic than o-nitrophenol. Due to intramolecular H-bonding, o-nitrophenol is more stable than phenoxide ion and does not release proton easily.

Reaction With Acid Chlorides And Anhydrides

When phenols are allowed to react with acid chlorides in presence of a base, such as a pyridine or with acid anhydrides in the presence of a few drops of cone. H₂SO₄ (or sodium acetate) as catalyst, the H-atom of the phenolic —OH group is replaced by an acyl (RCO—) group. This reaction is called acylation. If acetyl chloride CH₃COCl is used as the acid chloride and acetic anhydride (CH₃COOCOCH₃) is used as the acid anhydride, the reaction is usually termed as acetylation.

Example:

Only benzoyl chloride is very slowly decomposed by water or dilute NaOH and because of this, compounds like phenol containing an active hydrogen atom can be benzoylated in the presence of dilute aqueous sodium hydroxide. This method of benzoylation is known as the Schotten-Baumann reaction.

It is to be noted that, unlike alcohols, phenols do not react with carboxylic acids to produce esters. This is because phenols are weaker nucleophiles than alcohols.

Reaction with Grignard reagents

Phenols react with Grignard reagents to form alkanes.

Example:

Reaction with ammonia

When phenol is heated at 300°C with ammonia under high pressure or in the presence of anhydrous zinc chloride, aniline is obtained. Since phenol is very less reactive towards nucleophilic substitutions, a drastic condition is required for the reaction.

Reaction With Zinc Dust

Phenols on distillation with zinc dust get reduced to the corresponding aromatic hydrocarbon. The —OH group can be removed from the benzene ring.

Reaction With Ferric Chloride

When a few drops of neutral FeCl₃ solution is added to phenol, the mixture becomes violet due to the formation of the violet-coloured complexion [Fe(OC₆H₅)₆]₃. Phenols can be identified by this reaction.

In fact, all compounds (aliphatic or aromatic) containing the enol group, give characteristic colour with neutral FeCl3 solution. Depending on the nature of the phenol used in this reaction, the colour may vary from violet to blue, green or even red.

Reactions involving benzene ring

The —OH group in phenol strongly activates the ring towards electrophilic substitution, i.e., phenol undergoes electrophilic substitution reaction at a rate faster than benzene and in many cases, it is not possible to avoid polysubstituted. As the —OH group by its +R effect makes the ortho- and para-carbons more electron-rich as compared to meta-carbon, it is an ortho-, para-directing group.

Halogenation

Although the halogenation of benzene takes place in the presence of a Lewis acid (e.g., FeBr3 ), halogenation of phenol, due to much higher reactivity of the ring, takes place in the absence of a Lewis acid.

Reaction with bromine: When bromine-water is added to phenol, the three H -atoms of the ring are substituted by the three bromine atoms to give a white precipitate of 2, 4, 6 -tribromophenol. However, when a less polar or non-polar solvent (e.g., CS₂, CHCl₃, CCl₄, etc.) is used instead of water, a controlled substitution occurs at low temperature to give a mixture of o- and p-bromophenol. The para-isomer is obtained as the major product.

In aqueous solution, phenol ionises to form phenoxide ion  which actually reacts with phenol. The phenoxide ring is highly activated owing to powerful electron release by the negatively charged oxygen through its +1 and stronger +R effects and so polybromination occurs.

However, in non-polar solvents, phenol remains undissociated. The ringing phenol is less activated because of -I effect and the relatively weak +R effect of the —OH group and as a consequence, monobromination occurs.

Reaction With Chloride: In a similar reaction with chlorine-water, phenol gives a white precipitate of 2,4,6- trichlorophenol. A mixture of o- and p-chlorophenol is obtained when the reaction is carried out in less polar or non-polar solvent.

Nitration

Phenol reacts with cold and dilute nitric acid to yield a mixture of o- and p-nitrophenol.In this reaction, a mixture of cone. HNO₃ and cone. H₂SO₄ cannot be used because cone. HNO₃ brings about simultaneous oxidation and polynitration of the highly activated benzene ring, Ortho-nitrophenol, being steam volatile, is separated from the para isomer by steam distillation.

In the above method, the yield of p-nitrophenol is actually very low ( ~ 13%) due to partial oxidation of the ring by HNO₃. p-Nitrophenol may be prepared in good yield by nitrosation followed by oxidation.

2,4,6-trinitrophenol (picric acid) is usually not prepared by nitrating phenol with a mixture of cone. HNO₃ and cone. H₂SO₄ because of the poor yield, since most of the phenol is oxidised by cone. HNO₃. Nowadays picric acid is prepared commercially by first sulphonating phenol and then nitrating the product.

Due to sulphonation, the reactivity of the ring decreases and so, in the next step, phenol sulphonic acids are not destroyed by oxidation with HNO₃. In the second step, the ring undergoes desulphonative nitration (replacement of —SO₃H group by —NO₂ group) is one position.

Sulphonation

When phenol is treated with concentrated H₂SO₄, it undergoes sulphonation to give a mixture of O- and p-phenol sulphonic acids. If the reaction is carried out at a low temperature, the kinetically controlled orth-isomer predominates and if the reaction is carried out at a high temperature, the thermodynamically controlled para-isomer is the major product. Further, when the ortho-isomer is heated, it is converted into the para-isomer.

Friedel Crafts Reaction

When phenol is treated with alkyl halides in the presence of anhydrous AlCl₃, it undergoes Friedel Crafts alkylation to form a mixture of o- and p-alkylphenol. The p-isomer is usually the major product.

Coupling Reaction

When a cold solution of benzene diazonium chloride is added to a cold alkaline solution of phenol, a coupling reaction occurs and an orange-coloured azo-compound, phydroxyazobenzene is produced.

Due to steric hindrance, coupling preferably occurs at the p- position with respect to the —OH group. But if this position is blocked, o-coupling occurs. If all the o-and p-positions are blocked, then no coupling reaction occurs. It is an electrophilic substitution reaction in which C₆H₅N₂⁺ acts as an electrophile.

Kolbe-Schmitt Reaction

When dry sodium phenoxide or phenate is heated at 120-140°C with carbon dioxide under high pressure (4-7atm), the sodium salt of o-hydroxybenzoic acid, Le., sodium salicylate, is obtained almost exclusively.

Salicylic acid is produced when the aqueous solution of sodium salicylate is acidified with dilute HO. This reaction is called the Kolbe-Schmitt reaction. A carboxyl (—COOH) group can be directly introduced into the benzene ring by this reaction.

If potassium phenoxide Is used instead of sodium phenoxide, then p-hydroxybenzoic acid is obtained predominantly. It is an electrophilic substitution reaction in which CO₂ acts as an electrophile. The six-membered cyclic transition state involving large K+ ion is not stable and so product is not obtained when KOH is used. In this reaction, phenoxide ion is used instead of phenol because the ring of phenoxide ion is relatively more activated and so, it reacts easily with the weak electrophile CO₂.

Reaction Mechanism

Aspirin (acetylsalicylic acid or 2-acetoxybenzoic acid) is prepared by acetylation of salicylic acid with acetic anhydride. It is widely used for relieving pain (analgesic) and to bring down body temperature (febrifuge or antipyretic). Due to its anti-blood clotting action (anticoagulant property), aspirin finds use in the prevention of heart attack.

Salol (phenyl salicylate) and methyl salicylate (oil of wintergreen) can also be prepared from salicylic acid.

Salol is used as an internal antiseptic and methyl salicylate is used in pain-relieving ointments and in perfumery.

Reimer-Tiemann Reaction

When a mixture of phenol, chloroform and an aqueous solution of sodium hydroxide is heated at 65-70°C and the resulting solution is acidified with dilute HCl, o-hydroxybenzaldehyde (salicylaldehyde) is formed as the major product. A small amount of p-hydroxybenzaldehyde is also formed. This reaction is called the Reimer-Tiemann reaction. An aldehyde (—CHO) group can be directly introduced into the benzene ring with the help of this reaction.

It is an electrophilic substitution reaction in which dichlorocarbene (: CCl₂) acts as an electrophile.

Reaction Mechanism

If carbon tetrachloride (CCl₄) is used instead of chloroform, salicylic acid is obtained as the major product.

Gattermann Reaction

When phenol is treated with a mixture of HCN(g) and HCl(g) in presence of any. AlCl₃  the imine so produced is decomposed with water, p-hydroxybenzaldehyde is mainly obtained. This reaction is known as the Gattermann reaction.

Lederer-Manasse Reaction

When phenol is treated with formaldehyde in the presence of acid or base as a catalyst, H -atoms of ortho- and para-positions are substituted by —CH₂OH group forming ortho- and mainly para-hydroxybenzyl alcohols. Any other aliphatic or aromatic aldehyde may also be used instead of formaldehyde. This reaction is called the Lederer-Manasse reaction.

However, the reaction does not stop at this stage but proceeds with the introduction of more than one —CH₂OH group into the benzene ring, forming bis-hydroxymethyl phenol and tris-hydroxymethyl phenol. If the reaction is carried out for a long time, these phenolic alcohols participate in condensation reactions forming a cross-linked polymer of higher molecular mass. This polymer is known as phenol-formaldehyde resin or Bakelite. It is a thermosetting plastic used for making radios, electrical goods, washing machines, etc.

Reaction With Phthalic Anhydride

Phenol reacts with phthalic anhydride in the presence of conc.H₂SO₄ to form the acid-base indicator phenolphthalein.

Some Special Reactions Of Phenol

Reduction

Hydrogenation: When a mixture of hydrogen gas and vapours of phenol is passed over a nickel catalyst heated at 160°C, phenol is reduced to cyclohexanol by consuming 3 molecules of hydrogen.

Birch Reduction: When phenol is allowed to react with sodium dissolved in liquid NH₃, the aromatic ring is partially reduced to an enol which readily tautomerises to yield 3- cyclohexenone. The first step of this conversion is Birch reduction.

Oxidation

Phenol is a reactive aromatic compound with an electron-rich ring system capable of donating electrons easily. So, it is readily oxidised by oxidising agents. Alkaline potassium permanganate oxidises it to form tartaric acid, oxalic acid and CO₂. This oxidation causes complete destruction of the benzene ring. On the other hand, phenol on being oxidised by chromyl chloride or chromic acid and alkaline potassium persulphate gives p-benzoquinone and p-dihydroxy benzene or Quinol respectively. The latter reaction is known as Elbs persulphate oxidation.

When colourless phenol is exposed to air and light, it is easily oxidised to form p-benzoquinone along with some other quinone derivatives, p-benzoquinone combines with phenol to form a red-coloured addition compound known as phenoquinone. Due to the presence of this coloured substance, phenol becomes purple-red or pink-red. However, on long exposure to air and light, phenol assumes a deep brown colour.

Libermann Reaction

When a mixture of phenol and a few crystals of NaNO₂ is warmed with conc.H₂SO₄, the mixture turns dark green or greenish-blue. On dilution with water, the colour changes to red. When caustic soda solution is added to the mixture, the red colour of the solution again changes to blue.

Identification And Distinction

Identification of alcohols: Iodoform test When an alcohol containing CH₃CH(OH)— group is heated with an aqueous solution of NaOH in the presence of iodine, a yellow precipitate of iodoform is produced. This test is known as iodoform test for alcohol.

1. If R = H, the alcohol is CH₃CH₂OH (ethanol).

Example: Ethanol is the only primary alcohol which responds to the iodoform test.

2. If R = an alkyl group, the alcohol is an aliphatic secondary alcohol.

Example: All aliphatic 2° alcohols containing CH₃CH(OH)—group, for example, CH₃CH(OH)CH₃ (propan-2-ol), CH₃CH₂CH(OH)CH₃ (2-butanol). respond to the iodoform test.

3. If R = an aryl group, the alcohol is an aromatic secondary alcohol.

Example: Any aromatic alcohol containing CH₃CH(OH)— group responds to the iodoform test, for example, C₆H₅CH(OH)CH₃ (1-phenylethanol).

Therefore, the iodoform test can be used to distinguish between ethanol and methanol, 1-propanol and 2-propanol and 1-phenoxyethanol and 2-phenylethanol.

Distinction between 1°, 2° and 3° alcohols

Lucas test

The Lucas reagent is a mixture of concentrated HCl and anhydrous ZnCl₂. This reagent is added to alcohol, under test and shaken well. Alcohols are converted into alkyl chlorides, which being insoluble, results in the appearance of turbidity in the reaction mixture. This test is applicable only to the case of liquid alcohols and invalid for the alcohols with six or more carbon atoms, as these higher alcohols are insoluble In Lucas reagent.

1° Alcohol: At room temperature, primary alcohol does not react with the Lucas reagent. Turbidity appears only when the reaction mixture is heated or left to stand for a few hours.

⇒ \(\left.\mathrm{R}-\mathrm{CH}_2 \mathrm{OH}\left(1^{\circ}\right) \frac{\text { anhy. } \mathrm{ZnCl}_2 / \mathrm{HCl}}{(\text { room temperature }}\right) \text { Alkyl chloride is not formed. }\)

Exception: Benzyl alcohol (C₆H₅CH₂OH) and allyl alcohol (CH₂=CHCH₂OH) react readily with Lucas reagent, even though they are 1° alcohols.

2° Alcohol: In this case, turbidity appears within 5 minutes as the reaction occurs slowly.

3° Alcohol: In this case, turbidity appears immediately as the reaction occurs readily.

On shaking a sample of alcohol with Lucas reagent if the turbidity appears immediately, it is a tertiary alcohol. If the turbidity appears within five minutes, it is a secondary alcohol and if no turbidity appears it is a primary alcohol.

Reaction Mechanism

The reaction involved in the Lucas test proceeds by the S_N1 mechanism and therefore, an intermediate carbocation is formed in the first step (rate-determining step) of the reaction. Since a 3° carbocation is very stable, a 3° alcohol reacts readily with the Lucas reagent to form the corresponding alkyl chloride. A 2° alcohol reacts relatively slowly with Lucas reagent because a 2° carbocation is relatively less stable. A 1° alcohol does not react with Lucas reagent at ordinary temperature because a 1° carbocation is highly unstable.

Since benzyl and allyl alcohols react with ZnCl₂ to form resonance-stabilised carbocations, they respond to this test readily, even though they are 1° alcohols.

Oxidation method

In this method, the alcohol to be identified is subjected to oxidation by K₂Cr₂O₇/H₂SO₄ or Na₂Cr₂O₂/H₂SO₂ and the colour of the resulting solution is observed. The resulting compounds are then identified by some special tests.

1° Alcohol: An aldehyde or a carboxylic acid with the same number of carbon atoms is produced and colour of the solution changes from orange to bluish-green.

The aldehyde produced is detected by Tollen’s reagent and the carboxylic acid by esterification reaction.

2° Alcohol: A ketone with the same number of carbon atoms is produced and the colour of the solution changes from orange to bluish-green.

The ketone on treatment with Brady’s reagent gives orange or red precipitate.

3° Alcohol: These are not oxidised under ordinary conditions.

Orange-coloured dichromate [Cr(6)] oxidises 1° or 2° alcohols and itself is reduced to chromic salt [Cr(3)].

Example:

Victor Meyer’s Test

The alcohol to be identified is first converted into a nitro compound which is at first treated with HNO₂ and then with NaOH solution. Different colours are obtained depending on the type of alcohol. Thus the alcohols are identified from the colour.

1° Alcohol:

Pseudonitrol does not dissolve in NaOH solution. The solution looks blue due to its own blue colour.

3° Alcohol:

Distinction between alcohols and phenols

Alcohols can be distinguished from phenols by—

Litmus Test: Phenols turn blue litmus red due to acidic nature but alcohols do not.

Ferric Chloride Test: Phenols react with a neutral ferric chloride solution to produce blue, violet or green colouration while alcohols do not.

Bromine-Water Test: Phenols react with bromine-water to form a white precipitate of the corresponding bromo derivatives but alcohols do not.

Coupling Reaction: Phenols react with diazonium salts in a weakly alkaline solution to form yellow or orange-coloured azo dyes but alcohols do not.

Libermann’s test: When a few drops of phenol is heated with a few crystals of NaNO₂ and 1 ml of cone. H₂SO₄, a green or blue colouration develops. When the mixture is diluted with water, the colour of the aqueous solution turns red. When excess NaOH solution is added to that solution, thered colour changes to blue or green. Alcohols do not respond to this test.

Interconversion Among Different Classes Of Alcohols

1. 1° to 2° alcohol having the same number of C- atoms:

Example:

2. 2° to 1° alcohol having the same number of C-atoms:

Example:

3. 1° to 3° alcohol having the same number of C-atoms:

Example:

4. 3° to 1° alcohol having the same number of C-atoms:

Example:

5. Higher alcohols to lower alcohols (‘step down’):

Example:

6. Lower alcohols to higher alcohols (‘step up’):

Example:

Commercially Important Alcohols And Phenols

Methyl Alcohol Or Methano

Manufacture of methanol Methanol is also known as wood spirit because it was first prepared by destructive distillation of wood. It is prepared commercially by the following two methods.

From water gas: When a mixture of purified water gas and half its volume of H₂ is compressed under 200-300 atmosphere pressure is passed over copper oxide, zinc oxide and chromium oxide catalyst heated at 300-400°C, methanol is obtained in good yield (99% ).

Water gas (equal volumes of CO and H₂) is produced by passing steam over red hot coke.

By oxidation of methane: When a 9:1 (by volume) mixture of methane and oxygen is passed through a heated copper tube at 200-250°C under 100-atmosphere pressure, methane undergoes oxidation to give methanol. In this oxidation, copper tube acts as a catalyst.

Properties Of Methanol

Methanol is a colourless, neutral, inflammable liquid having characteristic smell. Its boiling and melting points are 64.5°C and -97.8°C, respectively. It is soluble in most of the organic solvents and miscible with water in all proportions. It is highly poisonous. When taken internally, it causes blindness or even death. Therefore, methanol is not used as an alcoholic drink. It gives all the general reactions of alcohol.

Uses Of Methanol

1. In industry, methanol finds extensive use as a cheap solvent for paints, varnishes, celluloid, fat, etc.
2. It is used for the manufacture of formaldehyde which is widely used as a very important raw material in the plastic industry and as a preservative for biological specimens.
3. It is used substitute of petrol.
4. In cold countries, it is used as an anti-freezing agent in automobile radiators.
5. It is used for denaturing ethyl alcohol i.e., to make it unfit for drinking purposes. The denatured ethanol is commonly known as methylated spirit.

Ethyl Alcohol Or Ethanol

Manufacture Of Ethanol

In many countries like the USA, ethanol (C₂H₅OH) is prepared mainly by the hydration of ethylene. However, in India, ethanol is mainly prepared by the fermentation of carbohydrates. The slow process of biochemical decomposition of complex organic compounds into simpler ones in the presence of suitable microorganisms [e.g., yeast, Acetobacter aceti, etc.] which secrete the biochemical catalysts (or enzymes) is called fermentation. Ethanol is manufactured from the following two types of carbohydrates:

From starchy materials such as potato, rice, barley, maize, etc.: In this method, the enzyme diastase, produced from malt (germinated cereal grains) hydrolyses starch into maltose. Then maltose undergoes hydrolysis to produce glucose in the presence of the enzyme maltose secreted by yeast. Another enzyme zymase produced by yeast causes the fermentation of glucose resulting in the formation of ethanol (also known as grain alcohol) and carbon dioxide.

From Molasses: In this method, the enzyme invertase secreted by yeast hydrolyses sucrose to give glucose and fructose.

Both glucose and fructose on fermentation, under the influence of another enzyme zymase secreted by yeast, produce ethanol and carbon dioxide.

Rectified Spirit

When diluted aqueous solution of ethanol obtained by fermentation of carbohydrates is subjected to fractional distillation, a mixture containing 95.6% ethanol and 4.4% water by mass is obtained. This mixture is called rectified spirit.

Preparation of absolute alcohol from rectified spirit: The rectified spirit is mixed with quick lime (calcium oxide) and the mixture is kept for 1-2 days. Quick lime absorbs water present in the rectified spirit forming calcium hydroxide [CaO + H₂O→Ca(OH)₂]. It is then distilled when 99.5% ethanol is obtained as distillate. Generally, it is called absolute alcohol.

Manufacture of absolute alcohol by azeotropic distillation method: A mixture of rectified spirit and a suitable amount of benzene is subjected to distillation. Three fractions are collected as distillate

1. The first fraction obtained at 64.8°C is a ternary azeotrope (constant boiling mixture) consisting of water (7.4%), ethanol (18.5%) and benzene (74.1%). All the water present is eliminated in this way.
2. The second fraction obtained at 68.2°C is a binary azeotrope consisting of the remaining benzene (67.6%) and ethanol (32.4%).
3. Absolute alcohol is then obtained at 78.3°C as the last fraction.

Preparation of super dry alcohol (>99.95%) from absolute alcohol: Absolute alcohol is refluxed with magnesium in the presence of a trace of iodine. Ethanol reacts completely with magnesium to form magnesium ethoxide. Magnesium ethoxide thus produced undergoes hydrolysis by water present in absolute alcohol, to give insoluble magnesium hydroxide and ethanol. In this way, water present in absolute alcohol is completely removed. The mixture on distillation gives super dry alcohol (>99.95%) and is collected as distillate.

⇒ \(\mathrm{Mg}+2 \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} \rightarrow \mathrm{Mg}\left(\mathrm{OC}_2 \mathrm{H}_5\right)_2+\mathrm{H}_2 \uparrow\)

⇒ \(\mathrm{Mg}\left(\mathrm{OC}_2 \mathrm{H}_5\right)_2+2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Mg}(\mathrm{OH})_2 \downarrow+2 \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\)

Properties Of Ethanol

Ethanol is a colourless, neutral, sweet-smelling, volatile inflammable liquid. Its boiling point is 78.3°C and its melting point is -114°C. It is lighter than water (specific gravity 0.789 ) and miscible with water in all proportions due to intermolecular hydrogen bonding.

Therefore, water is very effective in extinguishing fires caused by ethanol. Many organic compounds are soluble in ethanol. It acts as a stimulant If taken internally in a low dose. However, consumption in excess results in intoxication and even loss of consciousness.

Uses Of Ethanol

1. Ethanol is a very important solvent. In industry, it is widely used as a solvent for resin, soap, varnish, paints and dyes, rayon, perfumes, shellac, artificial rubber, synthetic fibres and medicines.
2. It finds extensive use in the preparation of ether, ethyl halide, ethyl ester, haloform, acetic acid, acetaldehyde, etc.
3. It is used as a chief ingredient in preparing alcoholic drinks like beer, whisky, brandy, rum, etc.
4. It is used in hospitals as an antiseptic in the form of a rectified spirit.
5. In cold countries, it is used as an antifreeze spirit. 0 In cold countries, it is used as an antifreeze in automobile radiators.
6. It finds use in the production of power alcohol. In countries having a shortage of petrol, a mixture of petrol ( — 80%), ethanol ( -20%) and benzene or tetracene is used as fuel to cut down its use. As alcohol is immiscible with petrol, a third compound {e.g., benzene or tetracene) is added to function as a cosolvent. This petrol mixed with alcohol is known as power alcohol (as it is used as a source of power).
7. Since the mobility of ethanol is high and the freezing point is low, it is used in spirit levels and low-temperature thermometers.

Methylated or denatured spirit: In order to prevent the use of ethanol for beverage purposes, the rectified spirit is denatured or made unfit for drinking by adding highly poisonous methanol (up to 10%) and a small amount of pyridine, caoutchouc or naphtha having an obnoxious smell and bad taste. Rectified spirit mixed with methanol is known as methylated spirit or denatured spirit. It is quite cheap and can be used particularly in industries.

Phenol

Manufacture Of Phenol

Phenol is manufactured either from cumene or from the middle oil fraction obtained from destructive distillation of coal tar.

Uses Of Phenol

1. Phenol is widely used in the manufacture of bakelite (used for making electrical goods) and nylon in the plastic industry.
2. It finds extensive use in the preparation of medicines like salicylic acid, salol, aspirin, methyl salicylate phenacetin, etc., explosives like picric acid, acid-base indicators like phenolphthalein and solvents for rubber and nitrocellulose like cyclohexanol.
3. Due to its germicidal property, phenol is used in different soaps, lotions and germicidal liquids.
4. It is also used in the preservation of wood.

Preparations Of Ethers

From Alcohols

Dehydration of alcohols

Acid-catalysed dehydration: Alcohols undergo dehydration in the presence of protonic acids like cones. H₂SO₄ or H₃PO₄ to produce alkenes or ethers. However, the product which is actually obtained depends on the reaction conditions. When alcohol, for example, is heated with excess cone. H₂SO₄ at 165-170°C, alkene is obtained as the major product while with excess alcohol at 140°C, ether is obtained as the major product. Therefore, excess alcohol is heated with acid at a relatively low temperature, in the preparation of ethers.

Example:

Reaction Mechanism

The dehydration of primary alcohols occurs by the SN2 mechanism. The reaction actually takes place through the following three steps.

1st Step: Protonation of the 1° alcohol to form a protonated alcohol or oxonium ion.

2nd Step: S_N2 attack on the protonated alcohol by a second molecule of alcohol to form a protonated ether.

3rd Step: Loss of a proton to form an ether.

• The yield of ether actually depends on the nature of alcohol (1°, 2° or 3° ) used.
• Primary alcohols mainly lead to the formation of ether by S_N2 reaction along with a small amount of alkene. Therefore, these are the most suitable alcohols for the preparation of ethers.
• Secondary alcohols react by the E1/S_N1 mechanism and alkenes are obtained as the major product.

Tertiary alcohols react only by the El mechanism and alkenes are obtained nearly exclusively.

Example:

Thus, the tendency of dehydration of alcohols leading to the formation of ethers follows the order:

primary (1°) > secondary (2°) > tertiary (3°)

Limitations: This method is suitable for the preparation of symmetrical ethers. The preparation of an unsymmetrical ether (R—O—R’) requires the use of two different alcohols and because of this, a mixture of three different ethers is obtained. Naturally, the yield of the unsymmetrical ether becomes very poor. Moreover, as the boiling points of the ethers in the mixture are very close to each other, separation of the constituents becomes very difficult. Hence, unsymmetrical ethers cannot be prepared by this method.

Although acid-catalysed dehydration of alcohols is not a suitable method for the preparation of unsymmetrical ethers, tert-butyl ethyl ether can easily be prepared in good yield from the corresponding alcohols by this method. Since a tertiary (3°) carbocation is relatively more stable than a primary (1°) carbocation, tert-butyl alcohol leads to the formation of the corresponding 3° carbocation at a faster rate than ethyl alcohol. The resulting ferf-butyl cation then reacts with ethanol to give tert-butyl ethyl ether nearly exclusively.

Catalytic dehydration (Manufacture of ethers): Alcohols can be dehydrated to ethers by passing alcohol vapours over heated catalysts like alumina (Al₂O₃) or thoria (ThO₂). Diethyl ether is manufactured by this process.

Example:

By The Reaction Of Diazomethane With Alcohol

Methyl ethers are formed by treating alcohols with diazomethane (CH₂N₂) in presence of catalysts like tetrafluoroboric acid (HBF₄) or aluminium alkoxide [Al(OR)₃].

Example:

Methyl phenyl ether or anisole (C₆H₅OCH₃) can be prepared by treating phenol (C₆H₅OH) with diazomethane. In this case, the presence of a catalyst (Lewis acid) is not essential because phenol is more acidic than alcohol. Note that, for methylation by diazomethane, the substrate must contain acidic hydrogen. The acidity of alcohols increases in the presence of Al(OR)₃ or HBF₄.

By The Reaction Of Alcohols With Alkenes

Alcohols add to reactive alkenes (e.g., alkyl-substituted alkenes) in the presence of acids as catalysts to form ethers.

Example:

The main disadvantage of this method is that the carbocation formed initially sometimes rearranges itself to form the more stable carbocation, which then undergoes nucleophilic attack by alcohol to form an unexpectedly rearranged ether.

Example:

From Alkyl Halides

By Williamson Synthesis

When alkyl halides are heated with sodium or potassium alkoxide (RO^–Na⁺ or RO^–K⁺), ethers are obtained. This method is known as Williamson synthesis. It is a versatile method for the synthesis of both symmetrical and unsymmetrical ethers. Since a mixture of ethers is not obtained, it is highly suitable for preparing unsymmetrical ether. The reaction, which occurs by the S_N2 mechanism, involves nucleophilic displacement of halide ion from alkyl halide, by alkoxide ion.

Preparation Of Simple Or Symmetrical Ethers:

Example: When a mixed solution of sodium ethoxide and ethyl iodide in alcohol is heated, diethyl ether (simple or symmetrical) is obtained along with sodium iodide.

Preparation Of Mixed Or Unsymmetrical Ethers:

Example: When a mixed alcoholic solution of sodium ethoxide and methyl iodide is heated, ethyl methyl ether (mixed or unsymmetrical) is obtained along with sodium iodide.

Preparation Of Alkyl Aryl Ethers (Phenolic Ethers): These ethers can be easily prepared by treating sodium phenoxides with suitable alkyl halides.

Example:

Limitations: Although Williamson synthesis is a very effective procedure for the synthesis of both symmetrical and unsymmetrical ethers, a proper choice of reactants is necessary for the synthesis of unsymmetrical ethers.

1. Since Williamson synthesis involves S_N2 reaction, it is very susceptible to steric hindrance. Due to the least steric hindrance, the primary (methyl) halides are most reactive in this reaction.
2. Due to considerable steric hindrance, the S_N2 attack on secondary alkyl halides is inhibited and as a result, alkenes are obtained as the major product by the E2 mechanism. Due to severe steric hindrance, tertiary alkyl halides are completely unreactive towards S_N2 rection and alkenes are obtained exclusively by the E2 mechanism.
3. The alkoxide may be either primary, secondary or tertiary. Therefore, primary alkyl halide (or methyl halide) should be used in this synthesis.
4. In this reaction, it is better to use alkyl bromides and iodides because I^– and Br^– are good leaving groups. Thus, for preparing an ether containing an isopropyl or a tert-butyl group, these groups must be present in the alkoxide but not in the alkyl halide.

Example: tert-butyl ethyl ether can be prepared by treating ethyl bromide with potassium tert-butoxide but not by treating tert-butyl bromide with potassium ethoxide, because in that case, 2-methylpropene is obtained as the only product.

Since the backside attack by the nucleophile does not take place in aryl halides due to repulsive interaction involving n electron cloud, they are unreactive towards S_N2 reaction. Therefore, aryl halides cannot be used for the preparation of alkyl aryl ethers or diaryl ethers.

Example:

By The Reaction Of Alkyl Halides With Dry Silver Oxide

Ethers can be prepared by heating a mixture of alkyl halide and dry silver oxide (Ag₂O).

Example:

An epoxide is a three-membered cyclic ether, also called an oxirane. Epoxides are very important synthetic intermediates used for converting alkenes to a variety of other functional groups. An alkene can be converted to an epoxide by a peroxyacid (also called peracids) which are highly selective oxidising agents. For example:

The reaction takes place in one step and it is stereospecific i.e., the epoxide retains whatever stereochemistry is present in the alkene.

Physical Properties Of Ethers

Physical State, Colour And Odour

At ordinary temperature, ethers of low molecular masses are colourless gases or colourless volatile liquids having a characteristic ‘ethereal smell’ For example, dimethyl ether and ethyl methyl ether are colourless gases while diethyl ether is a colourless volatile liquid (b.p.34.5°C ) at ordinary temperature. Vapours of ether are highly inflammable.

Dipolar Nature Of Ethers

The oxygen atom of ether (ROR) is sp³-hybridised. So, in this molecule, the two unshared electron pairs and the two C—O bonds are arranged tetrahedrally.

Polar ether molecule μ= 1.13D-1.18D

Due to the greater electronegativity of oxygen atoms than carbon, the two C—O bonds in ether are slightly polar. The C—O—C bond angle in ethers is nearly 110° and thus, the two C—O bond moments do not cancel each other. In fact, a resultant of two C—O bond moments operates and it adds to the resultant moment of two unshared pairs of electrons. Therefore, a net moment operates and the molecule becomes polar in nature.

Boiling Points Of Ethers

1. Although ethers are polar in nature, because of steric hindrance between alkyl groups, ether molecules do not come close to each other and only weak dipole-dipole attractions exist between them along with van der Waals forces of attraction.

Hence, the boiling points of lower ethers such as dimethyl ether (CH₃OCH₃) and ethyl methyl ether (CH₃OC₂H₅) are only slightly higher than those of the n-alkanes having comparable molecular masses.

Example: The boiling point of dimethyl ether, CH₃OCH₃ (Mol. mass = 46) is -23.7°C while that of propane, CH₃CH₂CH₃ (Mol. mass = 44 ) is -42.1°C. Similarly, the boiling point of ethyl methyl ether, CH₃OC₂H₅ (Mol. mass=60) is 10.8°C while that of butane, CH₃CH₂CH₂CH₃ (Mol. mass =58) is -0.5°C.

2. As the hydrocarbon part increases beyond four carbons, the boiling points of ethers are slightly lower than n-alkanes of comparable molecular masses. This is due to the fact that the van der Waals forces of attraction between relatively less symmetrical ether molecules are slightly lower than those in little more symmetrical alkanes of comparable molecular masses.

Example: The boiling point of diethyl ether, C₂H₅OC₂H₅(Mol. mass =74) is 34.6°C while that of pentane, CH₃(CH₂)₃CH₃ (Mol. mass =72) is 36.1°C.

3. Ethers do not form hydrogen bonds because the hydrogen atoms of ether are attached not to oxygen but to carbon. On the other hand, the alcohol molecules remain associated through intermolecular hydrogen bonding. Therefore, more thermal energy is required to separate these associated molecules. Therefore, the boiling points of ethers are much lower than alcohols of comparable molecular masses.

Example: The boiling point of dimethyl ether (CH₃OCH₃ ) is -23.6°C while that of isomeric ethyl alcohol (CH₃CH₂OH) is 78.3°C.

Solubility Of Ethers

1. Ethers containing up to three carbon atoms are soluble in water like alcohols of comparable molecular masses due to the formation of hydrogen bonds with water molecules.

Example: Like ethanol (C₂H₅OH), dimethyl ether is completely miscible with water. Similarly, the solubilities of diethyl ether and n-butyl alcohol are approximately the same (7.5 and 9.0 g per 100 g H₂O, respectively).

H-bonding between ether and water

2. As the molecular mass increases, the solubility of ethers in water decreases because of the gradual increase in the volume of the hydrophobic hydrocarbon part. Ethers are soluble in common organic solvents like alcohol, benzene, chloroform, acetone, etc.

3. When diethyl ether is shaken with water in a test tube and then the mixture is allowed to stand, two separate layers are formed. Ether forms the upper layer while water forms the lower layer. This observation suggests that diethyl ether or in general, ethers are almost insoluble in water and are lighter than water.

Solvent Property Of Ethers

1. In a molecule of ether, there is no acidic hydrogen essential for H-bond formation. So, ether is called an aprotic solvent. Many organic compounds dissolve in ether but ionic compounds do not. So, ether can be used as a solvent for preferential extraction of organic compounds from the mixed aqueous solution containing ionic compounds. Ethers are relatively inert by nature. So, they are frequently used as solvents. Some important ethers used as solvents are as follows:

Example:

2. Due to the absence of any hydroxyl group (—OH ) in ether molecules, they are normally unreactive towards strong bases. Due to poor leaving group characteristics of alkoxide ion (RO“) (a very strong base) ethers resist nucleophilic attack by bases. The non-bonding electron pairs of an ether effectively solvate cations. These special features of ether make it a suitable solvent in case of many reactions.

Chemical Reactions Of Ethers

Ethers are chemically almost as inert as alkanes. They are stable to bases, dilute acids, oxidising agents and reducing agents. This is due to the fact that the functional group of ethers ( —O —) does not contain any active site in their molecules just as the hydroxyl (-OH) group of alcohols and phenols, even though the oxygen atom in both the functional groups contains two lone pair of electrons. However, under suitable conditions, they undergo some reactions mainly because of the unshared electron pairs on the oxygen atom or cleavage of the C—O bond.

Reactions Of The Ethereal Oxygen

Due to the presence of two lone pairs of electrons on the oxygen atom, ethers behave as Lewis bases and hence, undergo some reactions in which coordinate bond formation occurs.

Formation Of Oxonium Salt

Ethers dissolve in cold concentrated mineral acids such as cone. H₂SO₄ or cone. HCl due to the formation of stable oxonium salts.

Example:

Since ethers behave as Lewis bases, they do not react with alkalies and consequently do not dissolve in alkalies.

When a solution of ether in a concentrated mineral acid is diluted with water, the oxonium salt undergoes decomposition (hydrolysis) to regenerate the original ether. For example, when ether dissolved in concentrated H₂SO₄ is shaken with water and kept for some time, two separate layers of ether and water are formed.

This is due to the fact that water is a stronger base than ether. So, when an excess of water is present in the mixture, the water molecules accept H⁺ ions. As a consequence, the ether molecule becomes free from acid and the acid molecule dissolves in water with the formation of hydroxonium hydrogen sulphate.

Unlike alkanes or alkyl halides, ethers dissolve in concentrated mineral acids forming salt, and on dilution of this solution with water, ether is again liberated. On the basis of this principle, ether can be separated from its mixture with alkane or alkyl halide.

Formation Of Etherates (Coordination Complexes)

Being Lewis bases, ethers form coordination complexes known as etherates with Lewis acids such as BF₃, AlCl₃, Grignard reagents (RMgX), etc.

Example:

Grignard reagents dissolve in ether due to the formation of such complexes and hence, Grignard reagents are usually prepared in ethers.

Reactions Involving Cleavage Of C—O Bond

By Halogen Acids

1. When ethers are heated with hydroiodic acid (HI) or hydrobromic acid (HBr) at 100°C, they undergo C—O bond cleavage to form an alcohol and an alkyl halide. HI, and HBr are very suitable reagents for the cleavage of ethers because they are relatively stronger acids and the respective anions (Bre or Ie) are also stronger nucleophiles.

Example:

2. If an excess of halogen acid is used, the alcohol thus formed reacts further with the halogen acid to form the corresponding alkyl halide.

Reaction Mechanism

The cleavage of ethers with an excess of halogen acids takes place by the following mechanism involving two stages:

First stage: Diethyl ether reacts with HI to form ethyl iodide and ethanol through the following two steps:

Step 1: A molecule of ether takes up a proton from halogen acid and is converted into an oxonium ion.

Step 2: Iodide ion (I^–) is a good nucleophile. It displaces the alcohol molecule from the oxonium ion by the S_N2 mechanism.

Second stage: The resulting ethanol reacts with another molecule of HI to form ethyl iodide through the following two steps:

Step 1 and 2:

Reactivity Of Halogen Acids: The cleavage of ethers by halogen acids actually involves an S_N2 attack by the halide ion (X^–) on the protonated ether molecule. Thus, how fast and easily the cleavage will take place depends on the strength of the nucleophilic attack, Le., the nucleophilicity of the halide ion. Since the nucleophilicity of the halide ions follows the sequence: I^– > Br^– > Cl^–, the reactivity of the halogen acids follows the same sequence, Le., HI > HBr > HCl.

Site Of Cleavage: In the case of unsymmetrical ethers having two different alkyl groups, the alcohol and the alkyl halide expected to be formed, i.e., the site of cleavage, depends on the nature of the alkyl group.

1. If the ether contains a methyl group or a 1 ° alkyl group, the cleavage occurs by the S_N2 mechanism and the nucleophilic attack occurs on the smaller alkyl group because of lesser steric hindrance. Therefore, the alkyl halide is always formed from the smaller alkyl group.

Example: When isopropyl methyl ether is treated with HI, the smaller methyl groups form the alkyl halide.

2. When the ether contains a 3° alkyl group, the cleavage occurs by the S_N1 mechanism and a 3° alkyl halide is produced through the formation of a stable 3° carbocation.

Example: tert-butyl methyl ether reacts with HI to from tert-butyl iodide and methyl alcohol.

3. The cleavage of alkyl aryl ethers or phenolic ethers always leads to the formation of phenols and alkyl halides. The nucleophilic attack (S_N2) by the halide ion on the benzene ring does not occur from the back side of the leaving group. This is because of the repulsive interaction caused by the ring π-electrons. Again, an S_N1 reaction does not take place because phenyl cation (C₆H⁺₅) very unstable. Therefore, cleavage of Ar—O bond does not take place and a phenol is always obtained.

Example: Anisole (C₆H₅OCH₃) reacts with HI to form phenol and methyl iodide.

4. The cleavage of benzyl alkyl ethers occurs by the SNI mechanism and benzyl halide is produced through the formation of resonance-stabilised benzyl cation.

Example: Benzyl methyl ether reacts with HI to form benzyl iodide and methyl alcohol.

Application Of The Cleavage Reaction Involving HI: The cleavage of ethers by hydroiodic acid (HI ) forms the basis of the Zelsel method for the estimation of alkoxy groups like methoxy (—OCH₃) and ethoxy (—OC₂H₅) group, in a compound.

When a known mass of the given ether is heated with HI, alkyl iodide is formed. The vapours of volatile alkyl iodide (CH₃I or C₂H₅I ) are absorbed in alcoholic AgNO₃ solution when a yellow precipitate of Agl is obtained. It is filtered, washed, dried and finally weighed. From the amount of silver iodide, a number of alkoxy groups can be estimated.

By Sulphuric Acid

1. When ethers are heated with dil. H₂SO₄ under high pressure, undergoes hydrolysis to give alcohols.

⇒ \(\mathrm{R}-\mathrm{O}-\mathrm{R} \text { (Ether) }+\mathrm{H}_2 \mathrm{O} \underset{\text { pressure }}{\stackrel{\text { dil. } \mathrm{H}_2 \mathrm{SO}_4, \Delta}{\longrightarrow}} 2 \mathrm{R}-\mathrm{OH} \text { (Alcohol) }\)

Example:

2. When heated with cone.H₂SO₄, ethers form alkyl hydrogen sulphates.

Example:

3. Ethers containing 2° or 3° alkyl groups form alkenes when heated with cone. H₂SO₄.

Example:

By Phosphorus Pentachloride

When ethers are heated with phosphorus pentachloride, the cleavage of the carbon-oxygen bond leads to the formation of alkyl chlorides and phosphorus oxychloride (POCl₃).

Example:

In this reaction, HCl is not produced (An alcohol can be distinguished from an ether by this reaction).

By Acid Chlorides And Anhydrides

When ethers are heated with acid chlorides in the presence of anhydrous ZnCl₂ or AlCl₃, alkyl chlorides and esters are produced. However, when heated with acid anhydrides, only esters are formed.

Example:

By Carbon Monoxide

Ethers react with carbon monoxide at 125-180°C under 500 atm pressure in the presence of boron trifluoride and a small amount of water to yield esters.

Example:

By boron tribromide (BBr₃)

Cleavage of ethers by BBr₃ occurs in two stages.

First stage: \(\mathrm{ROR}^{\prime}+\mathrm{BBr}_3 \rightarrow \mathrm{R}-\mathrm{O}-\mathrm{BBr}_2+\mathrm{R}^{\prime} \mathrm{Br}{\prime}\)

Second Stage:

The smaller alkyl group always forms the alkyl bromide.

Example:

Reactions Involving The Alkyl Groups

Formation Of Peroxides

Ethers having at least one α-H atom undergo slow oxidation in the presence of sunlight and air (O₂) to form hydroperoxide and dialkyl peroxide. It is a free radical reaction and the carbon atom attached directly to the oxygen atom is oxidised in this case.

Example:

Tert-butyl methyl ether [(CH₃)₃C— O—CH₃] does not undergo auto-oxidation as there is no a-H in tert-butyl group and a free radical involving methyl group (Me₃C—O—CH₂) cannot be formed easily as it is not very stable.

Once a large container of ether has been opened, it comes in contact with atmospheric oxygen and the auto-oxidation process starts. After several months, a large amount of peroxide is produced in the container. The peroxides are explosive oily liquids. Distillation or evaporation concentrates the peroxides present in an old sample of ether (peroxides are less volatile) and an explosion may occur. Such an explosion may be avoided by conducting the distillation not up to dryness. Detection of peroxide in a sample of either:

If on shaking an old sample of ether with an aqueous solution of KI mixed with starch becomes blue in colour, then it must be concluded that the sample of ether contains peroxide. In this case, the peroxide oxidises iodide to iodine which then combines with starch to form a blue inclusion complex.

If on shaking an old sample of ether with a freshly prepared solution of ferrous ammonium sulphate (Mohr salt) followed by the addition of an aqueous solution of ammonium thiocyanate a red colour appears, then it must be concluded that the ether sample contains peroxide. In this case, Fe²⁺ ions are oxidised by peroxide to Fe³⁺ ions which then react with thiocyanate to form a red complex ion.

⇒ \(\left[\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3+2 \mathrm{NH}_4 \mathrm{SCN} \rightarrow 2 \mathrm{Fe}(\mathrm{SCN}) \mathrm{SO}_4+\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4\right]\)

Removal Of Peroxides From Ethers: Ether, used as an anaesthetic and for various other purposes, must be completely free from peroxide. Peroxides present In an ether sample can be removed by shaking the ether with KI or FeSO₄ solution or by distilling with cone. H₂SO₄ or LiAlH₄.

Preservation of ethers: When ethers come in contact with air in the presence of sunlight, they undergo oxidation to form peroxide which is highly explosive in nature. Therefore, to avoid contact with air and light, others are stored in dark-coloured bottles with a narrow necks wrapped with black paper in sealed condition. Sometimes a small amount of cuprous oxide, ethanol or hydroquinone is mixed with ether to prevent the formation of peroxides.

Halogenation Reaction

1. Ethers undergo substitution with chloride or bromide at low temperatures and in the dark. One or two a-H atoms are replaced by halogen-producing mono and dihaloethers.

Example:

In the presence of light, all the H-atoms of ether are substituted by chlorine-producing perchlorodiethyl ether.

Example:

Electrophilic Substitution Reactions Of Aromatic Ethers

The alkoxy (-OR) group of aryl alkyl ethers increases the electron density of the ring by its +R effect and at the same time decreases the electron density of the ring by -I effect. However, as +R effect is stronger than -I effect, the group activates the ring towards electrophilic substitution reactions.

Resonance in aryl alkyl ether

From the hybrid structures, it becomes clear that the electron density at each of the two ortho- and para-positions is much higher as compared to that at the two meta-positions. Therefore, the electrophilic substitution occurs mainly at the o-and p-positions.

Anisole is less reactive towards electrophilic aromatic substitution than phenol. In a neutral or alkaline medium, phenol ionises to give phenoxide ion C₆H₅— O^–. However, anisole does not ionise under such conditions.

Since the oxygen atom in the phenoxide ion with full-fledged negative charge is more electron-releasing (stronger +R effect) as compared to the group in anisole (C₆H₅OCH₃), anisole is less reactive than phenol towards electrophilic substitution reactions.

Nitration Reaction

When an anisole is treated with mixed acid (cone. HNO₃ + cone. H₂SO₄ ), o-and p-nitro anisole is obtained.

Bromination Reaction

Anisole reacts with bromine in acetic acid to form a mixture of o- and p-bromoanlsole.

Friedel Crafts Alkylation Reaction

Anisole reacts with methyl chloride in the presence of anhy. AlCl₃ to form a mixture of o-and p-methylanisole.

Friedel-Crafts Acylation Reaction

When anisole is treated with acetyl chloride in the presence of anhydrous AlCl₃, 2 and 4-methoxyacetophenone are obtained.

Noted that aromatic ethers are much less reactive than phenols towards electrophilic aromatic substitution reactions. In fact, they are unreactive towards coupling, Kolbe and Reimer-Tiemann reactions.

Ethers Uses Of Ethers

1. Due to almost inert nature and good dissolving power, ethers are widely used as solvents both in laboratories as well as in industry. For example, diethyl ether finds extensive use as a solvent for resins, oils, fats, alkaloids, cellulose, esters, etc.
2. Diethyl ether is widely used in the extraction of many organic compounds from natural sources.
3. In many reactions {e.g., Wurtz reaction, Grignard reaction, reduction with Lithium aluminium hydride etc.), diethyl ether is used as a solvent.
4. Diethyl ether has been widely used as an inhalation anaesthetic agent in surgery as it produces unconsciousness without affecting lungs or heart. However, it has been replaced by better anaesthetic agents such as ethrane (ClFCHCF₂OCHF₂) because of its slow action and lengthy recovery period. Isoflurane (C₃H₂ClF5O) is also used as a general anaesthetic.
5. Diphenyl ether (C₆H₅—O—C₆H₅ ) is used as a heat transfer medium as its boiling point (258°C) is much higher.
6. A mixture of ether and alcohol is used as a substitute for petrol, under the trade name natalite.
7. A mixture of ether and solid carbon dioxide is used as refrigerant and can generate a very low temperature (-110°C).
8. Some naturally occurring phenolic ethers are used in perfumes and as flavouring agents, for example, vanilline, eugenol, isoeugenol, anethole, etc.

Distinction Between Alcohols And Ethers

1. Alcohols are readily soluble in water due to intermolecular hydrogen bonding while ethers, except dimethyl ether, are insoluble in water.
2. Alcohols containing acidic or active hydrogen react with sodium to liberate H₂ gas while ethers are unreactive towards sodium. Sodium is used to dry them.
3. Alcohols react readily with PCl₅ to form chloroalkanes with the liberation of HCl gas while ethers react with PCl₅ at high temperature to form chloroalkanes without liberation of HCl gas.
4. Alcohols are readily oxidised to aldehydes and ketones while ethers are resistant to oxidation.
5. Air has no effect on alcohols while ethers form peroxides on reacting with air in presence of light.

Distinctive Tests Between Diethyl Ether, Ethanol And Methanol:

Some Transformations Related To Ethers

1. Diethyl ether from ethanol:

2. Ethanol and butane from diethyl ether:

3. Diethyl ether from methanol:

4. Ethyl propanoate from diethyl ether:

5. Ethylene from diethyl ether:

6. Diethyl ether from ethylene:

7. 2-ethoxypropane from propene:

8. Diethyl ether from acetylene:

9. Ethyl phenyl ether from ethanol:

10. Allyl phenyl ether from propene:

 

Class 12 Chemistry Chapter 11 Alcohols Phenols And Ethers Very Short Answer Type Questions

Question 1. Which part of an alcohol (R —OH) is hydrophobic and which part is hydrophilic and why?
Answer: The non-polar alkyl (R— ) group is the hydrophobic part of an alcohol as it has no affinity for the polar water molecule while the hydroxy (—OH) group is the hydrophilic part because it forms hydrogen bond with water molecule.

Question 2. Which is the only 1° alcohol that responds to the iodoform test?
Answer: Ethyl alcohol (CH₃CH₂OH)

Question 3. Arrange 1°, 2° and 3° alcohols in order of decreasing acidic strength.
Answer: Primary (1°) > Secondary (2°) > Tertiary (3°)

Question 4. Arrange CH₃OH, H₂O and C₆H₅OH in order of decreasing acidic strength.
Answer: C₆H₅OH > CH₃OH > H₂O

Question 5. Arrange(CH₃)₃COH, CH₃CH₂OH and (CH₃)₂CHOH in order of increasing reactivity towards metallic Na.
Answer: (CH₃)₃COH < (CH₃)₂CHOH < CH₃CH₂OH

Question 6. Name a reagent by which the three classes of alcohols can be distinguished.
Answer: Lucas reagent (anhy. ZnCl₂ +conc. HCl)

Question 7. Which alcohol is used as an important beverage?
Answer: Ethyl alcohol (CH₃CH₂OH).

Question 8. Which class of alcohols does not react with Lucas reagent at ordinary temperature?
Answer: Primary (1° ) alcohol.

Question 9. Predict the product expected to be formed when vapours of tert-butyl alcohol are passed over heated copper at 300°C.
Answer: (CH₃)₂C=CH₂ (2 -methylpropene)

Question 10. Mention the process by which anti-Markownikoff hydration of propene can be effected.
Answer: Hydroboration-oxidation.

Question 11. Which class of alcohols readily undergoes dehydration?
Answer: Tertiary (3°) alcohol.

Question 12. Write the name and structure of the optically active alcohol having the lowest molecular mass.
Answer: Butan-2-ol (CH₃CH₂CHOHCH₃).

Question 13. Mention the carbonyl compounds which react with Grignard reagents to form 1°, 2° and 3° alcohols.
Answer:

The Grignard reagents (RMgX) are to be treated with formaldehyde (HCHO), an aldehyde (RCHO) other than formaldehyde and ketones (RCOR) to prepare primary or 1°, secondary or 2° and tertiary or 3° alcohols, respectively.

Question 14. Which one out of phenol, p-cresol and 2,4-dinitrophenol is soluble in an aqueous solution of sodium bicarbonate?
Answer: 2,4-dinitrophenol (a stronger acid than carbonic acid).

Question 15. Ethanol responds to the iodoform test while methanol does not—why?
Answer:

There is a CH₃CH(OH) -group in ethanol [CH₃CH₂(OH)], which takes part in the iodoform reaction but not in methanol (CH₃OH).

Question 16. Mention a suitable reagent for preparing heptanal from 1-heptanol.
Answer: Pyridinium chlorochromate(CrO₃-C₅H₅N.HCl).

Question 17. Mention two chemical tests which can be used to distinguish between phenol and alcohol.
Answer:

1. Litmus test
2. Ferric chloride test (phenol responds to both the tests).

Question 18. Which out of CH₃CH₂OH, C₆H₅OH and CH₃OH does not react with acetic acid in the presence of cone. H2SO₄ to form ester?
Answer: C₆H₅OH , because it acts as a weak nucleophile

Question 19. Which out of CH₃CH₂OH, CH₃CHOHCH₃ and (CH₃)₃COH gives a red colour in Victor Meyer’s test?
Answer: CH₃CH₂OH, because it is a 1° alcohol

Question 20. Which one of the following compounds does not react with Br₂ -water to form a white precipitate?

Answer:

Question 21. What are the electrophiles involved in Reimer- Tiemann and Kolbe-Schmitt reactions?
Answer: Dichlorocarbene (:CCl₂) and CO₂, respectively.

Question 22. Write the name and structure of a compound which is used as a drug for reducing fever (antipyretic).
Answer: Aspirin

Question 23. Which one out of proof spirit, rectified spirit and methylated spirit causes blindness if taken internally?
Answer: Methylated spirit.

Question 24. Arrange secondary alcohol, primary alcohol, phenol and tertiary alcohol in decreasing order of tendency towards bond cleavage.
Answer: Tertiary (3°) alcohol > secondary (2°) alcohol > primary (1°) alcohol > phenol

Question 25. Account for the difference in behaviour of phenol with Br₂/H₂O and Br₂/CS₂.
Answer:

Phenol (PhOH) ionizes in water to give some PhO^– which is the actual reacting species. Due to the presence of a full-fledged negative charge on the oxygen atom, the ring is highly activated and as a consequence, tribromination occurs to yield 2,4,6- tribromophenol. The less reactive PhOH does not ionize in non-polar solvent CS₂ and so, monobromination occurs to yield a mixture of o-and p-bromophenol.

Question 26. Arrange dimethyl, di-tert-butyl and diethyl ethers in the order of increasing C —O — C bond angle.
Answer: Me— O— Me < Et— O— Et < Me₃C—O—CMe₃

Question 27. Give examples of three reagents by which ethanol can be distinguished from diethyl ether.
Answer:

1. PCl₅
2. I₂/NaOH and
3. Na

Question 28. Which one out of HC1, HBr and HI is not an effective reagent for ether cleavage?
Answer:

HCl; because it is relatively a weaker acid and the anion (Cl^–) is relatively a weaker nucleophile.

Question 29. What type of alkyl halide should be used in Williamson synthesis?
Answer: Methyl or 1° alkyl halide.

Question 30. Mention the mechanism of the reaction involved in C—O bond cleavage of benzyl methyl ether by HI.
Answer: The cleavage occurs through the S_N1 mechanism.

Question 31. When an ether C₅H₁₂O is hydrolysed with dilute H₂SO₄, two alcohols are obtained which respond to iodoform test. Identify the ether.
Answer: (CH₃)₂CHOCH₂CH₃ (Ethyl isopropyl ether).

Question 32. Di-ferf-butyl ether cannot from peroxide in the presence of light and air. Explain
Answer: Me₃C—O—CMe₃ contains no a-H atom and for this reason it does not form peroxide in the presence of light and air.

Question 33. Arrange the following compounds in the order of increasing boiling point: 1-pentanol, n-pentane, pentanal, ethoxyethane.
Answer: N-pentane < ethoxyethane < pentanal < 1-pentanol

Question 34. Arrange the following ethers in the order of decreasing C—O—C bond angle: CH₃ —O—CH₃, (CH₃)₃C—O—C(CH₃)₃, (CH₃)₂CH—O—C₂H₅
Answer: (CH₃)₃C—O—C(CH₃)₃ > (CH₃)₂CH—O—C₂H₅>CH₃— O—CH₃

Question 35. Give an example of a stable gem-diol.
Answer: Chloral hydrate [CCl₃CH(OH)₂]

Question 36. Which is the only 1° alcohol that responds to the iodoform test?
Answer: Ethyl alcohol (C₂H₅OH)

Question 37. Which class of alcohols does not react with the Lucas reagent at ordinary temperature?
Answer: Primary or 1° alcohol

Question 38. Which class of alcohols undergoes dehydration very rapidly in the presence of concentrated H₂SO₄?
Answer: Tertiary or 3° alcohol

Question 39. Mention one reaction by which the presence of the —OH group in an organic compound can be detected and its number can be determined.
Answer: Acetylation reaction

Question 40. Which is the only primary alcohol that can be prepared by the hydration of alkene?
Answer: Ethanol

Question 41. Which class of alcohols does not undergo dehydrogenation when passed through a copper catalyst heated at 300°C?
Answer: Tertiary or 3° alcohol

Question 42. Which class of alcohols is unable to change the orange colour of Na₂Cr₂O₇/H₂SO₄ solution?
Answer: Tertiary or 3° alcohol

Question 43. Ethanol responds to iodoform test but methanol does not —why?
Answer: A CH₃CHOH- group is present in ethanol but not in methanol

Question 44. What is the maximum number of water molecules with which a molecule of alcohol forms H-bond?
Answer: Three

Question 45. Which is the alcohol having molecular formula C₄H₁₀O, produces a red colouration in Victor Meyer’s test?
Answer: 1-Butanol or 2- methyl-1-propanol

Question 46. Name two reagents which can be used to distinguish between phenol and cyclohexanol.
Answer: Neutral FeCl₃ solution and Lucas reagent can be used

Question 47. Which one of alcohol and phenol is a weak nucleophile and why?
Answer:

The unshared pair of electrons on oxygen in phenol is delocalized with the ring n electrons, therefore, phenol is a weaker nucleophile than alcohol.

Question 48. Which out of 3-methylhexan-3-ol and 5-methylhexan-1-ol reacts with metallic sodium at a faster rate?
Answer: 5-methylhexan-1-ol reacts with metallic Na because it is a primary alcohol

Question 49. Glycerol is a viscous liquid—why?
Answer:

Glycerol (HOCH₂CHOHCH₂OH) contains three —OH groups. So, it remains associated through extensive hydrogen bonding and so, it exists as a viscous liquid.

Question 50. Write the name and structure of an antipyretic.
Answer: Aspirin or acetylsalicylic acid

Question 51. Explain why the —OH group in phenol is o-, p-directing.
Answer:

The -OH group by its +R effect increases the electron density at ortho-and para-positions

Question 52. Which class of alcohol is obtained when formaldehyde is allowed to react with Grignard reagents?
Answer: Primary alcohol

Question 53. Name the by-product obtained in the preparation of phenol from cumene.
Answer: Acetone (CH₃COCH₃)

Question 54. Which of the isomeric nitrophenols is more volatile and why?
Answer:

Due to the formation of intramolecular hydrogen bond, o-nitrophenol exists as single molecule. So, its boiling point is the lowest and it is the most volatile.

Question 55. Predict the product expected, when a mixture of phenol, phthalic anhydride and concentrated H₂SO₄ is heated.
Answer:

Question 56. Arrange 1°, 2° and 3° alcohols in the order of increasing reactivity towards metallic sodium.
Answer: Phenolphthalein (an acid-base indicator)

Question 57. Which one between 2-chloroethanol and ethanol is more acidic and why?
Answer: Due to the -I effect of chlorine, 2-chloroethanol is more acidic than ethanol

Question 58. Mention the factors responsible for solubility of alcohols in water.
Answer:

Hydrophobic character of R-group, a steric effect due to the R-group and hydrophilic character of the —OH group (formation of H-bond).

Question 59. Mention a reagent for converting ethanol to ethanoic acid.
Answer: Na₂Cr₂O₇/H₂SO₄ or K₂Cr₂O₇/H₂SO₄

Question 60. Explain why the reactivity of the three classes of alcohols towards Lucas reagent is different.
Answer:

The stabilities of the three carbocations obtained from three classes (1°, 2° and 3°) of alcohols are different and for this reason, the rates of reactions with Lucas reagent are different

Question 61. Mention a reagent for converting ethanol to ethanal.
Answer: PCC (Pyridinium chlorochromate).

Question 62. How can a small amount of water present in ether be removed?
Answer: It is allowed to stand with sodium for one day and then distilled

Question 63. Ether possesses dipole moment. Predict the shape of ether from this information.
Answer: The ether molecule is an angular

Question 64. Give an example of (1) aprotic solvent and (2) protic solvent.
Answer: Ether is an aprotic solvent and ethanol is a protic solvent

Question 65. Write the names and structures of two metamers of diethyl ether.
Answer: 1-Methoxypropane (CH₃OCH₂CH₂CH₃) and 2-methoxy propane [CH₃OCH(CH₃)₂)]

Question 66. What are the two explosive substances obtained on auto-oxidation of ether?
Answer: 1-Ethoxyethyl hydroperoxide and diethyl peroxide

Question 67. Write the names of three ethers which cannot be prepared by Williamson synthesis.
Answer: Me₃C—O—CMe₃, Ph—O— Ph and Me₃CCH₂—O—CH₂CMe₃

Question 68. Which physical experiment can be used to distinguish between ethanol and diethyl ether?
Answer: Solubility in water

Question 69. Predict the products expected to be formed when anisole is treated with HI.
Answer: Phenol and methyl iodide

Question 70. Which alkyl group of an unsymmetrical dialkyl ether can be converted into an alkyl halide when the ether is treated with HX? (X = I or Br)
Answer: The smaller alkyl group

Question 71. An old ether sample should not be distilled to dryness why?
Answer: To avoid explosions caused by peroxides

Question 72. What are ethrane and isoflurane? Mention their uses.
Answer: Ethrane(ClFCHCF₂OCHF₂) and isoflurane(F₃CCHClOCHF₂) are two ethers used as anaesthetic agents

Question 73. Write the IUPAC name of the optically active ether having molecular formula C₅H₁₂O.
Answer: 2-Methoxybutane [CH₃CH₂CH(CH₃)OCH₃]

Question 74. Name the solvent/reagent in which diethyl ether does not dissolve.
Answer: cone. NaOH solution

Question 75. Which is the most effective halogen acid used for ether cleavage?
Answer: HI

Question 76. What type of reaction is involved during conversion of ether to peroxide?
Answer: Free-radical reaction

Question 77. Explain why di-tert-butyl ether does not form peroxide in the presence of light and air.
Answer: There is no α – H in tert-butyl group

Question 78. Write the structure of one ether which is isomeric with butan-2-ol.
Answer: CH₃CH₂—O—CH₂CH₃ (ethoxyethane)

Question 79. Explain why di-tert-butyl ether cannot be prepared by Williamson synthesis.
Answer: Me₃C—X does not undergo S_N2 reaction due to steric reason

Question 80. O=C=O is nonpolar, but R—O— R is polar—why?
Answer: Due to its linear shape, CO₂ is nonpolar. However, R—O— R is polar as it is angular in shape.

Question 81. Which reagent can be used to distinguish between secondary and tertiary alcohols.

Answer: Lucas reagent

Question 82. How will you distinguish the compounds in each pair by a chemical reaction?

Answer: Phenol (C₆H₅OH) reacts with a neutral FeCl₃ solution to produce a violet colouration but anisole
(PhOCH₃) does not.

Class 12 Chemistry Chapter 11 Alcohols Phenols And Ethers Short Answer Type Question And Answer

Question 1. The boiling point of ethanol (78.3°C) is much higher than that of butane (-0.5°C) even though the molecular mass of butane is higher than that of ethanol. Explain.
Answer:

Ethanol molecules remain associated through intermolecular hydrogen bonding while very weak van der Waals forces of attraction operates among the non-polar butane molecules. Thus, greater amount of thermal energy is required to separate alcohol molecules as compared to butane molecules. Therefore the boiling point of ethanol (C₂H₅OH) is much higher than that of butane (CH₃CH₂CH₂CH₃).

Question 2. Although methanol and ethanol are organic solvents, they are very good solvents for a number of Ionic compounds—why?
Answer:

Methanol (CH₃OH) and ethanol (CH₃CH₂OH), being able to form hydrogen bonds, can solvate anions of an ionic compound. Due to their relatively high values of dielectric constant they are capable of reducing the attractive forces between cations and anions of an ionic compound. Therefore, methanol and ethanol are good solvents for ionic compounds.

Question 3. Which one among the isomeric alcohols having formula C4H10O is most soluble in water and why?
Answer:

The isomeric alcohols having molecular formula C4H10°

1. N-butyl alcohol [CH₃(CH₂)₃OH] ,
2. Secbutyl alcohol [CH₃CH(OH)CH₂CH₃] ,
3. Isobutyl alcohol [(CH₃)₂CHCH₂OH]
4. Tert-butyl alcohol [(CH₃)₃COH].

Among these, tert-butyl alcohol has maximum solubility in water because its spherical hydrophobic hydrocarbon part having a small surface area, is relatively less dominant.

Question 4. Alcohols act both as acids and bases. Explain.
Answer:

The highly electronegative oxygen atom in alcohol can easily accommodate the O—H bonding electrons. Therefore, alcohols behave as acids by donating protons to strong bases. For example:

Due to the presence of unshared pair of electrons on oxygen, alcohols can also attract protons from strong acids and thereby, act as Bronsted bases. For example:

Question 5. Arrange 1°, 2° and 3° alcohols in the order of decreasing acidic and basic strength and explain.
Answer:

Ongoing from 1° (RCH₂OH) to 3° alcohol (R₃COH), the number of electron-releasing alkyl (R-) groups attached to the carbinol carbon (the carbon attached to the —OH group) increases. Therefore, the electron density on oxygen progressively increases and as a result, the tendency of displacement of O —H bonding electrons progressively decreases, i.e., the polarity of the O—H bond progressively decreases.

Therefore, the acidic strength of alcohols decreases in the order: 1°>2°>3°. Since the electron density on oxygen increases gradually with increase in the number of alkyl groups, the tendency of oxygen to donate electrons progressively increases from 1° to 3° alcohol, i.e., their basic strength decreases in the order: 3° > 2° > 1°.

Question 6. Explain why alcohols cannot be used as solvents in I reactions involving Grignard reagent (RMgX) or lithium aluminium hydride (LiAlH₄).
Answer:

The carbanion (R^–) generated from Grignard reagent and the hydride ion (H^–) generated from LiAlH₄ are very strong bases. So, they react readily with alcohols (an acid-base reaction) to liberate RH (hydrocarbon) and H₂ and these reagents are destroyed. Hence, alcohols cannot be used as solvents in the reactions involving RMgX or LiAlH₄.

Question 7. The concentration of ethanol cannot be increased further by distilling rectified spirit—why?
Answer:

Rectified spirit is a constant boding mixture or azeotropic mixture of 95.6% ethanol and 4.4% water by mass. This means that both alcohol and water bod at the same temperature. Thus, on distillation, the composition of the mixture, which bods at 78.15°C, remains unchanged. Therefore, the concentration of ethanol cannot be increased by distilling rectified spirit.

Question 8. What are the products expected to be obtained in the reaction of ethanol with cone. H2SO₄ under the following thermal conditions?

1. 0°C
2. Room temperature
3. 140°C
4. 170°C

Answer:

1. CH₃CH₂OH₂HSO₄ (ethyl oxonium hydrogen sulphate)
2. CH₃CH₂OSO₃H (ethyl hydrogen sulphate)
3. CH₃CH₂OCH₂CH₃ (diethyl ether)
4. CH₂=CH₂(ethylene)

Question 9. Ethanol does not react with NaBr to form ethyl bromide while ethyl bromide is formed if the reaction is carried out in presence of H₂SO₄. Explain.
Answer:

Although Br^– is a moderately strong nucleophile, it displace the very poor leaving group OH^– (a strong base) from a molecule of alcohol by an S_N2 reaction. Hence, ethanol does not react with NaBr to form ethyl bromide. However, in the presence of H₂SO₄, the —OH group of ethanol gets protonated and is converted into — OH^–. Since H₂O (very weak base) is a very good leaving group, it is easily displaced by Br^– ion. Thus, ethanol reacts with NaBr in the presence of H₂SO₄ to form ethyl bromide.

Question 10. Sodium can be used to remove traces of water from benzene but not from alcohol —why?
Answer:

Due to the presence of acidic or active hydrogen in ethanol (C₂H₅OH), it readily reacts with Na to form sodium ethoxide and hydrogen. So, Na cannot be used to remove traces of water from alcohol.

⇒ \(2 \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+2 \mathrm{Na} \rightarrow 2 \mathrm{C}_2 \mathrm{H}_5 \mathrm{ONa}+\mathrm{H}_2 \uparrow\)

On the other hand, benzene having no active or acidic hydrogen, is inert to sodium. So, traces of water can be removed from benzene by using sodium.

Question 11. A polyhydric alcohol of molecular mass 168 gives an acetyl derivative having molecular mass 294. How many hydroxyl groups are present in that alcohol?
Answer:

⇒ \(-\mathrm{OH}(17) \stackrel{\text { Acetylation }}{\longrightarrow}-\mathrm{OCOCH}_3(59)\)

The increase in molecular mass due to acetylation of one —OH group = (59- 17) = 42. Here, the observed increase in molecular mass of the alcohol = (294- 168) = 126.

∴ Number of —OH group present = \(\frac{126}{42}=3\)

Question 12. No primary alcohol except ethanol can be prepared by hydration of an alkene—why? How can primary alcohols be prepared from alkenes?
Answer:

The preparation of a primary alcohol is possible only by the hydration of a terminal alkene (RCH=CH₂). All terminal alkenes except ethylene are unsymmetrical. So, addition of concentrated H₂SO₄ takes place according to the Markownikoff rule and hydrolysis of the resulting alkyl hydrogen sulphate leads to the formation of 2° (or 3°) alcohol, predominantly.

Since ethylene (CH₂ = CH₂) is symmetrical, it produces only the primary alcohol, ethanol (C₂H₅OH ) on hydration. Therefore, no other primary alcohol except ethanol, can be prepared by the hydration of an alkene.

Primary alcohols can be prepared by an indirect hydration of terminal alkenes using the hydroboration-oxidation method.

Question 13. Write structure of the major product(s) in each of the following reactions:

1. Monobromlnation of 3-methylphenol
2. Denitration of 3-methylphenol
3. Mononitration of phenyl ethanoate

Answer:

Question 14. Which of the following compounds exhibit chelation:

1. O-cresol,
2. Methyl salicylate (oil of wintergreen),
3. O-cyanophenol,
4. O-fluorophenol,
5. O-iodophenol

Answer:

Chelation occurs in (2) where the H-atom of the —OH group forms H-bond with the O-atom of the C=O and in (4) since F is very electronegative and is approachable to the H-atom of the —OH group to form H-bond.

Question 15. Unlike diethyl ether, ethanol is completely miscible in water. Explain.
Answer:

In ethanol molecule, the H-atom is bonded to the highly electronegative oxygen atom, which also contains an unshared pair of electrons. So, ethanol molecules acting both as H-bond acceptors and H-bond donors, dissolve in water by forming stronger H-bonds with water molecules.

On the other hand, in an ether molecule, the oxygen atom is not attached to any hydrogen atom. So, it acts only as a hydrogen bond acceptor and forms Hbonds with water molecules to a very small extent. So, diethyl ether is sparingly soluble but immiscible in water.

In (1), H-bond cannot take place due to the lack of an electrosegative atom. In (3), although the N too far away from OR In (4), I is not sufficiently electronegative and also larger in size.

Question 16. Why is strict control of temperature essential during the preparation of diethyl ether from ethanol than the preparation of dimethyl ether from methanol?
Answer:

• Intermolecular dehydration of methanol (CH₃OH) results in the formation of dimethyl ether (CH₃OCH₃). Since it is an alcohol with only one carbon atom, obviously there is no possibility of its intramolecular dehydration.
• Ethanol (CH₃CH₂OH) is an alcohol consisting of two carbon atoms and so, it may undergo both intermolecular and intramolecular dehydration leading to the formation of diethyl ether (C₂H₅OC₂H₅) and ethylene (CH₂=CH₂), respectively.
• Since the nature of dehydration depends on the reaction temperature, control of temperature is more essential in the preparation of diethyl ether from ethanol than in the preparation of dimethyl ether from methanol.

Question 17. What do you mean by protic and aprotic solvents? Name one each of protic and aprotic solvents.
Answer:

The solvents in which acidic proton (H-atom which is capable of forming H-bond) is present are called protic solvents while the solvents in which acidic proton is absent are termed as aprotic solvents. For example, ethanol (CH₃CH₂OH) is a protic solvent and diethyl ether (C₂H₅OC₂H₅) is an aprotic solvent.

Question 18. How will you separate ether from a mixture of ether and alkane or alkyl halide?
Answer:

When a mixture of ether and alkane or alkyl halide is shaken with a cone. H₂SO₄, the alkane or alkyl halide does not dissolve in acid, but ether dissolves forming an oxonium salt. The acid layer is thus separated and ether is made free from acid by adding excess of water to it. The ether layer is then separated from the aqueous layer, dried and distilled.

Question 19. What is absolute ether? How can absolute ether be prepared from commercial ether? Give an example of a reaction In which such an ether is used.
Answer:

Pure ether which is free from water, is termed as absolute ether. Commercial ether is made free from peroxide by shaking it with a solution of ferrous sulphate. The moist ether is initially dried by keeping it in contact with anhydrous Calcium chloride for some days. It is then allowed to stand in the presence of sodium for some more days and distilled. Thus absolute ether is obtained.

Absolute ether is used in the preparation of Grignard reagents (RX + Mg→RMgX).

Question 20. Prove Alcohol, but not ether, contains an —OH group.
Answer:

HCl is produced in the reaction of PCl₅ with alcohol (white smoke is produced when a glass rod rinsed with cone. ammonia solution is held) but not in the reaction of PCl₅ with ether. This observation proves that alcohol contains —OH group but ether does not.

⇒ \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+\mathrm{PCl}_5 \rightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl}+\mathrm{POCl}_3+\mathrm{HCl}\)

⇒ \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OC}_2 \mathrm{H}_5+\mathrm{PCl}_5 \rightarrow 2 \mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl}+\mathrm{POCl}_3\)

Question 21. Predict the major products in each of the following reactions and give your reasoning.
Answer:

Answer:

Because of steric reason, Bre attacks the methyl carbon of the protonated ether to give these products.

Since the benzene ring does not take part in an S_N2 reaction, these two products are obtained.

The ortho-and para-positions of the ring are activated due to the +R effect of —OC₂H₅ group. For this reason, nitration occurs to give ortho- and para-nitro derivatives.

Since the 3° carbocation (CH₃)₃C⁺ is more stable than the 3° carbocation (CF₃)₃C⁺ (destabilized by the -I effect of the -CF₃ groups), the protonated ether dissociates (S_N1) to give (CH₃)₃C⁺ and (CF₃)₃COH. This carbocation then undergoes nucleophilic attack by I⁺ to give (CH₃)₃C—I.

Question 22. Give Williamson Synthesis of 2-ethoxy-3-methylpentane using ethanol and 3-methyipentan-2-ol.
Answer:

Since an S_N2 reaction is favourable for primary (1°) substrate, CH₃CH₂I is used instead of CH₃CH₂CH(CH₃)CH(I)CH₃.

Question 23. Give IUPAC names of the following compounds:

Answer:

1. 4-chloro-2, 3-dimethylpentan-l-ol
2. 2-ethoxypropane
3. 2, 6-dimethylphenol
4. 1-ethoxy-2-nitrocyclohexane

Question 24. Give the structures and IUPAC names of the products expected from the following reactions:

1. Catalytic reduction of butanal.
2. Hydration of propene in the presence of dilute sulphuric acid.
3. Reaction of propanone with methyimagnesium bromide followed by hydrolysis.

Answer:

Question 25. Arrange the following sets of compounds in order of their increasing boiling points:

1. Pentan-1-ol, butan-1-ol, butan-2-ol, ethanol, propan-1-ol, methanol.
2. Pentan-1-ol, n-butane, pentanal, ethoxyethane.

Answer:

1. Boiling point of normal alcohols increases as the length of the carbon chain increases due to the corresponding increase in their van der Waals forces of attraction. Again, among isomeric alcohols, 2° alcohols have lower boiling points than 1° alcohols due to corresponding decrease in the extent of H-bonding because of steric hindrance. Thus the increasing order of boiling point is: methanol < ethanol < propan-1-ol < butan-2-ol < butan-1-ol < pentan-1-ol.
2. Due to the existence of dipole-dipole interactions, the boiling points of lower ethers having molecular masses equal to or less than n-butane are higher than those of the corresponding n -alkanes. Thus boiling point of ethoxyethane is greater than that of n-butane. Again the boiling point of aldehydes are greater than those of the alkoxides having comparable molecular masses, because stronger dipole-dipole attractive forces operate in the former category of compounds.
3. So boiling point of pentanal is greater than that of ethoxyethane. Further, the boiling points of alcohols are greater than those of the aldehydes having comparable molecular masses because of the existence of intermolecular H-bonding in the former. Thus, the boiling point of the given compounds increase in the sequence—

n-butane < ethoxyethane < pentanal < pentan-1-ol

Question 26. Arrange the following compounds in increasing order of their acid strength: Propan-1-ol, 2,4,6-trlnitro-phenol, 3-nitrophenol, 3,5-dinitrophenol, phenol, 4-methylphenol.
Answer:

Phenolic compounds are stronger acids than alcoholic compounds, primarily because of the greater stabilisation of phenoxide ion relative to alkoxide ion. Thus, phenol itself is a stronger acid than propan-1-ol. However, the acid strength of phenolic compounds decreases by the presence of electrondonating alkyl group in the ring.

So 4-methylphenol is a weaker acid than phenol. On the other hand, the introduction of electron-withdrawing nitro group into the ring system increases the acid strength of the corresponding phenolic compound. Further, acid strength increases as the number of — NO2 groups increases in the ring system. The effect is more marked if such groups are present at ortho- or para-positions with respect to the —OH group. Thus acid strength of the given compounds follows the sequence—

propan-1-ol < 4-methylphenol < phenol < 3- nitrophenol < 3,5-dinitrophenol < 2, 4, 6-trinitrophenol.

Question 27. Write the structures of the major products expected from the following reactions:

1. Mononitration of 3-methylphenol
2. Dinitration of 3-methylphenol
3. Mononitration of phenyl methanoate.

Answer: The combined directing influence of —OH and —CH3 groups determine the position to be taken up by the incoming group.

Since both —OH and —CH₃ groups are ortho/paradirecting, therefore, positions 2, 4 and 6 are activated towards electrophilic substitution. However, due to steric hindrance — NO₂ group avoids its attack at C₂.

Question 28. The following is not an appropriate reaction for the preparation of f-butyl ethyl ether.

C₂H₅ONa + (CH₃)₃C—Cl—OC₂H₅

1. What would be the major product of this reaction?
2. Write a suitable reaction for the preparation of t-butylethyl ether.

Answer:

(CH₃)₃CClis a fert-alkyl halide. It does not undergo nucleophilic substitution with C₂H₅O^–N⁺a due to steric reasons. Instead, C₂H₂O^– acts as a base and under its influence the tert-alkyl halide undergoes bimolecular elimination (E2) to give isobutene as the major product.

Tert-butyl ethyl ether is prepared by the reaction between sodium f-butoxide and ethyl chloride involving S_N2 mechanism.

Question 29. Give the major products that are formed by heating each of the following ethers with HI.

1. CH₃—CH₂—CH(CH₃)—CH₂—O—CH₂—CH₃
2. CH₃—CH₂—CH₂—O—C(CH₃)₂—CH₂—CH₃
3. 

Answer:

Protonated ether reacts via S_N2 mechanism.

Protonated ether reacts via S_N1 mechanism because of its ability to form highly stable 3° carbocation.

Protonated ether reacts via S_N1 mechanism because of its ability to form highly stable benzyl cation.

Question 30. Classify the following as primary, secondary and tertiary alcohols:

1. CH₃—C(CH₃)₂—CH₂OH
2. H₂C=CH—CH₂OH
3. CH₃—CH₂—CH₂—OH
4. C₆H₅CH(OH)—CH₃

Answer:

Primary Alcohols: 1, 2 and 3. Secondary alcohols: 4 and 5. Tertiary alcohol: 6.

Question 31. Identify allylic alcohols in the previous examples.
Answer: Allylic alcohols: 2 and 6.

Question 32. Name the compounds according to IUPAC system.

Answer:

1. 3-chloromethyl-2-isopropylpentan-l-ol
2. 2, 5-dimethylhexane-l,3-diol
3. 3-bromocyclohexanol
4. Hex-l-en-3-ol
5. 2-bromo-3-methylbut-2-en-l-ol

Question 33. Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal?

Answer:

Question 34. Write structures of the products of the reactions:
Answer:

Answer:

Question 35. Give structures of the products you would expect when (1) butan-l-ol and (2) 2-methyIbutan-2-ol reacts with

1. HCl-ZnCl₂
2. HBr and
3. SOCl₂.

Answer: With HCl-ZnCl₂:

With HBr:

With SOCl₂:

Question 36. Predict the major product of acid-catalysed dehydration of

1. 1-methylcyclohexanol and
2. Butan-1-ol.

Answer:

The substrate contains two types of /? -hydrogens. It undergoes acid catalysed dehydration via the formation of a carbocation intermediate to give the more substituted alkene I (Saytzeff product) as the major product.

Butan-1-ol undergoes dehydration via the initial formation of a 1° carbocation, followed by a 2° carbocation and finally loss of proton to give 2-butene as the major product (more substituted Saytzeff product).

Question 37. ortho and para Nitrophenols are more acidic than phenol. Draw the resonance structures of the corresponding phenoxide ions.
Answer:

ortho- and para-nitrophenoxide ions are more stable than unsubstituted phenoxide ion because negative charge on the phenolic oxygen atom is involved in delocalisation not only with the benzene ring but also with the nitro group present at ortho- or para-positions of the ring.

Question 38. Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol.
Answer:

This synthesis involves three steps:

1. Conversion of ethanol to ethyl bromide:

⇒ \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}+\mathrm{HBr} \stackrel{\Delta}{\longrightarrow} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br}+\mathrm{H}_2 \mathrm{O}\)

2. Conversion of 3-methylpentan-2-ol to respective alkoxide:

3. Interaction between the alkyl bromide and the alkoxide:

In this synthesis, treatment of a primary alkyl halide with secondary alkoxide is preferred. Treatment of secondary alkyl halide [CH₃CH₂CH(CH₃)CH(CH₃)-Brl with primary alkoxide (CH₃CH₂ONa) is discouraged because in that case elimination reaction will preferably occur to give alkene as the predominant product.

Question 39. Which of the following is an appropriate set of reactants for the preparation of l-methoxy-4- nitrobenzene and why?

Answer:

Interaction between the 1st set of reactants involves an activated aromatic nucleophilic substitution to produce the desired product.

Interaction between the 2nd set of reactants involves nucleophilic attack of p-nitrophenoxide ion on methyl bromide via S_N2 pathway to produce the desired product.

But in this case the yield of the desired product is relatively low because of the low nucleophilicity of p -nitrophenoxide ion (as the -ve charge on the phenolic oxygen atom is involved in delocalisation with the aromatic ring).

So the 1st set of reactants is appropriate for the preparation of 1-methoxy-4-nitrobenzene.

Question 40. Predict the products of the following reactions:

Answer:

The ethoxy group (—OC₂H₅), which is capable of exhibiting +R effect, is an ortho- /para-directing group. So ethoxybenzene undergoes nitration to give a mixture of o-and p-nitro derivatives. Due to steric reason, paraproduct predominates.

Protonated ether reacts with the nucleophile (I^–) via S_N1 pathway because of its ability to form stable tertbutyl cation.

Question 41. Write IUPAC names of the following compounds:

Answer:

1. 2,2,4-trimethyIpentan-3-ol
2. 5-etliylheptane-2,4-dio
3. Butane-2, 3-diol
4. Propane-1, 2,3-triol
5. 2-methylphenol
6. 4-methylphenol
7. 2,5-dimethylpheno
8. 2,6-dimethylphenol
9. 1-methoxy-2-methylpropane
10. Ethoxybenzene
11. 1-phenoxyheptane
12. 2-ethoxybutane

Question 42. Write structures of the compounds whose IUPAC names are as follows:

1. 2-methylbutan-2-oI
2. 1-phenylpropan-2-ol
3. 3,5-dimethylhexane-1,3,5-triol
4. 2,3-dicthylphenol
5. 1-ethoxypropane
6. 2 ethoxy-3-methylpentane
7. Cyclohexylmethanol
8. 3-cyclohexylpentan-3-ol
9. Cyclopcnt-3-en-l-ol
10. 4-chloro-3-ethylbutan-l-ol.

Answer:

Question 43.

1. Draw the structures of all isomeric alcohols of molecular formula C₅H₁₀O and give their IUPAC names.
2. Classify the isomers of alcohols in question
3. As primary, secondary and tertiary alcohols.

Answer: Eight isomeric alcohols can be represented:

Question 44. Explain why propanol has higher boiling point than that of the hydrocarbon, butane?
Answer:

The molecules of propanol are held together by strong intermolecular forces involving H-bonds, while those of butane are held together only by weak van der Waals forces of attraction.

So propanol has a much higher boiling point than butane.

Question 45. Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Answer:

The solubility of alcohols in water is due to their ability to form H-bonds with water molecules. But hydrocarbons cannot form H-bonds with water molecules and hence, are insoluble in water.

Question 46. What Is meant by hydroboration-oxidation reaction? Illustrate It with an example.
Answer:

The addition of diborane (B₂H₆)to alkenes followed by their oxidation with alkaline H₂O₂ to form alcohols is called hydroboration-oxidation reaction. Alkenes undergo hydration in anti-Markownikoff’s manner by this reaction. For example,

Question 47. Give the structures and IUPAC names of monohydric phenols of molecular formula, C₇H₈O.
Answer:

Three isomeric phenols with formula C₇H₈O are—

Question 48. While separating a mixture of ortho and para-nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.
Answer:

P-nitrophenol exists in associated form as its molecules are intermolecularly H-bonded and its high boiling point. But o -nitrophenol exists as a single molecule due to intramolecular H-bonding involving —OH and — NO₂ groups. So it has a lower boiling point and hence, is steam volatile.

Question 49. Give the equations of reactions for the preparation of phenol from cumene.
Answer:

Question 50. Write a chemical reaction for the preparation of phenol from chlorobenzene.
Answer:

Question 51. Write the mechanism of hydration of ethene to yield ethanol.
Answer:

Question 52. You are given a benzene, cone. H₂SO₄ and NaOH. Write the equations for the preparation of phenol using these reagents.
Answer:

Question 53. Show how will you synthesise:acid

1. 1-phenylethanol from a suitable alkene.
2. Cyclohexylmethanol using an alkyl halide by an S_N2 reaction.
3. Pentan-1-ol using a suitable alkyl halide?

Answer:

Question 54. Give two reactions that show the acidic nature of phenol. Compare the acidity of phenol with that of ethanol.
Answer: Phenol reacts with metallic Na to liberate H₂ gas:

Phenol reacts with NaOH to form salt (sodium phenoxide) and water:

Acid strength: Phenol (which is a resonance hybrid involving separation of charge) loses a proton to form its conjugate base, phenoxide ion (which is stabilised by resonance involving no separation of charge).

So there is some increase in resonance stability on going from phenol to phenoxide ion. Thus phenol shows acidic behaviour. But there is no such increase in resonance stability when ethanol is converted to its conjugate base (C₂H₅O^–). So, it shows very poor acidic behaviour.

Question 55. Explain why is ortho-nitrophenol more acidic than ortho-methoxyphenol.
Answer:

In ortho-nitro phenoxide ion, the negative (-ve ) charge on oxygen is stabilised by the -R effect of the — NO₂ group, but in ortho-methoxy phenoxide ion, the -ve charge on oxygen is destabilised by the +R effect of the —OCH₃ group. Thus o-nitrophenol is a stronger acid than o-methoxyphenol.

Question 56. Explain how the -OH group attached to a carbon of benzene ring activate it towards electrophilic substitution?
Answer:

Attachment of the —OH group to benzene ring tends to decrease electron density in the ring by the -I effect. At the same time —OH group has a strong +R effect which tends to increase electron density (preferably at ortho and para-positions) in the ring. Since the +R effect of the —OH group is greater than that of its —I effect ( +R > -I ), therefore, there Is a net increase in electron density in the ring system of the phenol molecule and hence, the ring becomes activated towards electrophilic substitution.

Question 57. Give equations of the following reactions:

1. Oxidation of propan-1-ol with alkaline KMnO₄ solution.
2. Bromine In CS₂ with phenol.
3. Dilute HNO₃ with phenol.
4. Treating phenol with chloroform in the presence of aqueous NaOH.

Answer:

Oxidation of propan-1-ol with alkaline KMnO₄ produces a salt of propanoic acid, which on acidification gives propanoic acid.

In aprotic solvent like CS₂, phenol exists almost entirely in the undissociated form and hence, it is less reactive and undergoes only monobromination to give mainly the p-isomer.

Nitration occurs via nitrosation followed by oxidation to nitro derivative.

Question 58. Explain the following with an example. Unsymmetrical ether.
Answer:

If the alkyl or aryl groups attached to the oxygen atom are different, then such ethers are called unsymmetrical or mixed ethers.

Example: Ethylmethyl ether (C₂H₅ — O —CH₃).

Question 59. Write the mechanism of acid dehydration of ethanol to yield ethene.
Answer:

Question 60. How are the following conversions carried out?

1. Propene→propan-2-ol.
2. Benzyl chloride→Benzyl alcohol.
3. Ethyl magnesium chloride → propan-l-ol.
4. Methyl magnesium bromide→2-methylpropan-2-ol

Answer:

Question 61. Name the reagents used in the following reactions:

1. Oxidation of a primary alcohol to carboxylic acid.
2. Oxidation of a primary alcohol to aldehyde.
3. Bromination of phenol to 2,4,6-trlbromophenol.
4. Benzyl alcohol to benzoic acid.
5. Dehydration of propan-2-ol to propene.
6. Butan-2-one to butan-2-ol.

Answer:

1. K₂Cr₂O₇/H₂SO₄; or, acidic or alkaline KMnO₄ followed by acidification.
2. PCC (CrO₃-C₅H₅NHCl) in CH₂Cl₂ .
3. Bromine-water (Br₂/H₂O).
4. Alkaline KMnO₄, followed by acidification.
5. Conc.H,SO₄.Δ.
6. LiAlH₄ /ether, then H₂O ; or NaBH₄ /CH₃OH ; or H₂/Ni.

Question 62. Give reason for the higher boiling point of ethanol in comparison to methoxymethane.
Answer:

Ethanol exists in the associated form due to intermolecular H-bonding. So it has a high boiling point. On the other hand, weak dipole-dipole attraction and van der Waals attractive forces operate between the molecules of methoxy-methane (CH₃— O—CH₃). So it has a low boiling point.

Question 63. Give IUPAC names of the following ethers:

Answer:

1. 1-ethoxy-2-methylpropane.
2. 2-chloro-1 – methoxyethane.
3. 4-nitroanisole.
4. 1-methoxypropane.
5. 4-ethoxy-1, 1-dimethylcyclohexane.
6. Ethoxybenzene.

Question 64. Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis:

1. 1-propoxypropane
2. Ethoxybenzene
3. 2-methoxy-2-methylpropane
4. 1-methoxyethane

Answer:

Question 65. How is 1-propoxypropane synthesised from propanl-ol? Write mechanism of this reaction.
Answer:

1-propoxypropane is prepared by reaction between sodium propoxide and 1-bromopropane involving Williamson synthesis.

Question 66. Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.
Answer:

It is known that primary alcohols are converted to ethers by heating an excess of alcohol with a dehydrating agent like cone. H₂SO4. The reaction proceeds by nucleophilic attack (S_N2)of alcohol on protonated alcohol molecules:

To prepare ethers derived from 2° and 3° alcohols, it is found that alkenes are formed as predominant products instead of ethers. This is due to the fact that protonated alcohol loses water to form stable 2° or 3° carbocation which does not undergo nucleophilic attack by unprotonated alcohol in order to avoid reintroduction of steric hindrance. Instead, it loses a proton to give alkene with relief of crowding.

Thus, the preparation of ethers by acid-catalysed dehydration of 2° and 3° alcohols is not a suitable method.

Question 67. Write the equation of the reaction of hydrogen iodide with:

1. 1-propoxypropane
2. Methoxybenzene and
3. Benzyl ethyl ether.

Answer:

Question 68. Explain the fact that in aryl alkyl ethers

1. The alkoxy group activates the benzene ring towards electrophilic substitution and
2. It directs the incoming substituents to ortho- and porn-positions in benzene ring.

Answer:

In aryl alkyl ethers, the alkoxy group has a -I effect which tends to decrease electron density in the ring. At the same time, the alkoxy group has a strong +R effect which tends to increase electron density in the ring. Since +R effect of alkoxy group ( —OR) is greater than that of its -I effect (+R > -I), so there is a net increase in electron density in the aromatic ring. So the benzene ring in aryl alkyl ethers becomes activated towards electrophilic substitution.

The following resonance structures of aryl alkyl ethers indicate that ortho- and para-positions of the ring system are rich in electron density, and hence, incoming electrophile attacks preferably at ortho- and para-positions.

Question 69. Write the mechanism of the reaction of HI with methoxymethane.
Answer:

Methoxymethane and HI in 1:1 mole ratio react together to form a mixture of methyl alcohol and methyl iodide:

If methoxymethane and HI are used in 1: 2 mole ratio, then only methyl iodide is formed:

Question 70. Write equations of the following reactions:

1. FriedelCrafts reaction-alkylation of anisole.
2. Nitration of anisole.
3. Bromination of anisole in ethanoic acid medium.
4. Friedel-Crafts acetylation of anisole.

Answer:

Question 71. Show how would you synthesise the following alcohols from appropriate alkenes?

Answer:

It can be prepared by hydration of 2 different alkenes:

It is prepared by acid-catalysed hydration of 4-methylhept-3-ene:

The desired alcohol is obtained by acid-catalysed hydration of pent-1-ene.

The desired alcohol is obtained by acid-catalysed hydration of either of the following two alkenes.

Question 72. When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place:

Give a mechanism for this reaction.

(Hint: The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom.)

Answer:

The alcohol, on protonation, followed by loss of H₂O, generates a

2° carbocation. This undergoes rearrangement involving a 1, 2-shift of hydride to form more stable 3° carbocation. This is subsequendy attacked by the nucleophile Br- to give the desired product.

Question 73. What is the structure and IUPAC name of glycerol?
Answer: Propane-1,2,3-triol [HOCH₂CH(OH)CH₂OH],

Question 74. Write the IUPAC name of the following compounds.

Answer: A⇒ 3-ethyl-5-methyihexane-2, 4-diol, B⇒ 1-methoxy-3- nitrocyclohexane.

Question 75. Write the IUPAC name of the compound given below.

Answer: 3-methylpent -2-ene-l, 2-diol.

Question 76. Name the factors responsible for the solubility of alcohols in water.
Answer:

Solubility of alcohols in water depends on two factors:

1. Hydrogen bonding: The polar hydroxyl groups of alcohols form H-bonding with water and this is responsible for their solubility in water. Solubility increases as the number of -OH groups increases. Thus, CH₃CH₂CH₂CH₂CH₂OH<CH₃CH₂CH(OH)CH₂OH < HOH,C —CH(OH) — CH₂OH
2. Size of alkyl/aryl group: As the size of hydrocarbon part of alcohol increases, the extent of H-bonding decreases. Hence, solubility decreases. Thus, solubility decreases in the sequence: CH₃CH₂OH > CH₃CH₂CH₂OH > CH₃CH₂CH₂CH₂OH

Question 77. What is denatured alcohol?
Answer:

Alcohol made unfit for drinking purposes by adding poisonous substances like pyridine, copper sulphate, etc., is called denatured alcohol.

Question 78. Suggest a reagent for the following conversion.

Answer: PCC (Pyridinium chlorochromate. C₅H₅NHCrO₃Cle). It consists of CrO₃, pyridine and HCl.

Question 79. Out of 2-chloroethanol and ethanol which is more acidic and why?
Answer:

2-chloroethane is the stronger acid due to the electron-withdrawing -I effect of the Cl-atom attached to the carbon chain.

Question 80. Suggest a reagent for the conversion of ethanol to ethanal.
Answer: PCC (Pyridinium chlorochromate, C₅H₅NHCrO₃ClO)

Question 81. Suggest a reagent for conversion of ethanol to ethanoic acid.
Answer: Any strong oxidising agent, such as acidified K₂Cr₂Oy or acidified KMnO₄.

Question 82. Out of o-nitrophenol and p-nitrophenol, which is more volatile? Explain.
Answer:

O-nitrophenol exists as discrete molecules due to intramolecular H-bonding and so it is more volatile than p-nitrophenol which exists in the associated form due to intermolecular H-bonding.

Question 83. Out of o-nitrophenol and o-cresol which is more acidic?
Answer:

Due to -I and -R effects of — NO₂ group, electron density in the O—H bond decreases, while due to +1 and hyper conjugation effects of —CH₃ group, the electron density in the O—H bond increases. Thus the O—H bond in a nitrophenol is weaker than the O—H bond in O-cresol and hence, o-nitrophenol is a stronger acid than O-cresol.

Question 84. When phenol is treated with bromine water, a white precipitate is obtained. Give the structure and the name of the compound formed.
Answer:

Question 85. Arrange the following compounds in increasing order of acidity and give a suitable explanation. Phenol, o-nitrophenol, o-cresol
Answer:

Due to electron-withdrawing -I and -R effects of — NO₂ group, O -nitrophenol is a stronger acid than phenol, but due to electron-donating +1 and hyperconjugative effects of the —CH₃ group, o-cresol is a weaker acid than phenol. So acid strength follows the sequence: o -cresol < phenol < o-nitrophenol.

Question 86. Alcohols react with active metals e.g., Na, K, etc., to give corresponding alkoxides. Write down the decreasing order of reactivity of sodium metal towards primary, secondary and tertiary alcohols.
Answer:

Ongoing from 1° to 2° to 3° alcohols, the electron donating +1 effect of increasing the number of alkyl groups increases and hence, electron density in the O— H bond increases, thereby making this bond much stronger. Thus the acidic character of alcohols decreases in the sequence: 1°>2°>3°. Therefore, the reactivity of alcohols with Na-metal decreases in the same sequence: 1° > 2° > 3°.

Question 87. What happens when benzenediazonium chloride is heated with water?
Answer:

It is converted to phenol with the liberation of N₂ gas.

⇒ \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{~N}_2 \mathrm{Cl}+\mathrm{H}_2 \mathrm{O} \stackrel{\Delta}{\rightarrow} \mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}+\mathrm{N}_2 \uparrow+\mathrm{HCl}\)

Question 88. Arrange the following compounds in decreasing order of acidity. H₂O, ROH, HC = CH
Answer:

Due to electron-donating +1 effect of alkyl group, alcohols (R-s-OH) are weaker acids than H₂O. Further, the electronegativity of oxygen is greater than that of sphybridised carbon, hence alcohols are stronger acids than acetylene. So acid strength follows the sequence:

Question 89. Name the enzymes and write the reactions involved in the preparation of ethanol from sucrose by fermentation.
Answer:

Ethanol is prepared commercially by fermentation of sucrose by yeast, which contains the enzymes invertase and zymase. Sucrose is converted to a mixture of glucose and fructose by the enzyme invertase, while both glucose and fructose are converted to ethanol by the enzyme zymase.

Question 90. How can propan-2-one be converted into tert-butyl alcohol?
Answer:

Question 91. Write the structures of the isomers of alcohols with molecular formula C₄H₁₀O. Which of these exhibits optical activity?
Answer:

Question 92. Explain why is OH group in phenols more strongly held as compared to OH group in alcohols.
Answer:

In a phenol molecule, the lone pair of electrons on the O-atom of hydroxyl group is involved in delocalisation with the n-electrons of the benzene ring. This can be seen from the following resonance structures of phenol.

Thus, the carbon-oxygen bond of phenol has a partial double bond character. But in alcohols (RCH₂ —OH), the carbon-oxygen bond is a pure single bond. Therefore, —OH group in phenols is more strongly held as compared to —OH group in alcohols.

Question 93. Explain why nucleophilic substitution reactions are not very common in phenols.
Answer:

In phenolic compounds, the —OH group has a strong +R effect and this causes an increase in electron density in the aromatic ring system. As a result, any approaching nucleophile is strongly repelled by such aromatic rings, and hence, nucleophilic substitution reaction does not take place in phenolic compounds.

Question 94. Preparation of alcohols from alkenes involves the electrophilic attack on alkene carbon atom. Explain its mechanism.
Answer:

Alkenes undergo hydration in presence of an acid catalyst. In the first step of the reaction, H+ undergoes electrophilic addition to one of the olefinic carbons, thereby fonning a carbocation intermediate. This further undergoes nucleophilic attack by water to form alcohol.

Question 95. Explain why is O=C=0 nonpolar while R—O—R is polar.
Answer:

CO₂ is a linear molecule in which the moments of two C=O bonds cancel out each other because they are equal but opposite in direction. Thus CO₂ is a non-polar molecule, although each carbon-oxygen bond is polar.

In ether (R—O—R) molecule, the moments of two R — O bonds act at an angle of ≈110° with each other. So the resultant dipole moment (μ) will have a finite value. In other words, R—O—R is a polar molecule.

Question 96. Why is the reactivity of all the three classes of alcohols with cone. HCl and ZnCl₂ (Lucas reagent) different?
Answer:

The reaction of an alcohol with Lucas reagent (cone. HCl + ZnCI₂ ) proceeds via carbonium ion intermediate. Obviously, the more stable the carbocation, the faster the reaction. Since the stability of carbocations decreases in the sequence: 3°>2°>1°, therefore reactivity of three classes of alcohols towards Lucas reagent decreases in the same order, i.e., 3°>2°>1°.

Question 97. Write steps to carry out the conversion of phenol to aspirin.
Answer:

Question 98. Nitration is an example of aromatic electrophilic substitution and its rate depends upon the group already present in the benzene ring. Out of benzene and phenol, which one is more easily nitrated and why?
Answer:

Nitration of benzene or phenol involves an electrophilic substitution reaction. Obviously, higher the electron density in the aromatic ring, the higher the rate of nitration.

Thus phenol is more easily nitrated than benzene, as the presence of the -OH group in phenol increases the electron density at the ortho and para positions of the ring system by the +R effect.

Question 99. In Kolbe’s reaction, instead of phenol, phenoxide ion is treated with carbon dioxide. Why?
Answer:

Phenoxide ion (C₆H₅Oe)is more reactive than phenol (C₆H₅ —OH) towards electrophilic aromatic substitution because of the greater ability of -Oe ion (than OH group) to donate electrons to the benzene ring. Since in Kolbe’s reaction, the participating electrophile (CO₂)is very weak, therefore, it is treated preferably with phenoxide ion (having a highly activated ring system) rather than phenol, thereby enabling the reaction to proceed smoothly.

Question 100. The dipole moment of phenol is smaller than that of methanol. Why?
Answer:

In phenol, C—O bond is less polar due to the electron-withdrawing effect of the phenyl group whereas in methanol, C—O bond is more polar due to the electron-releasing +1 effect of -CH₃ group. So dipole moment of phenol is smaller than that of methanol.

Question 101. Ethers can be prepared by Williamson synthesis In which an alkyl halide is reacted with sodium alkoxide. Di-tert-butyl ether can’t be prepared by this method. Explain.
Answer:

In an attempt to prepare di-tert-butyl ether by Williamson synthesis using tert-butyl bromide as the substrate and sodium tert-butoxide as the nucleophile, we obtain isobutene as the ultimate product instead of the desired product.

This is due to the fact that the bulky nucleophile [(CH₃)₃CO^–] foils attack at the a -carbon of the bulky substrate [(CH₃)₃C^αBr] due to steric hindrance. In fact, this 3° alkyl halide prefers to undergo elimination rather than substitution. (CH₃)₃CO^– acts as a base rather than as a nucleophile.

Question 102. Why is the C —O —H bond angle in alcohols slightly less than the tetrahedral angle whereas the C—O—C bond angle in ether is slightly greater?
Answer:

The bond angle in alcohols is slightly less than the tetrahedral angle (109°28′). It is due to the repulsion between the unshared electron pairs of oxygen. In ethers, the four electron pairs i.e., the two bond pairs and two lone pairs of electrons on oxygen are arranged approximately in a tetrahedral arrangement. The bond angle is slightly greater than the tetrahedral angle due to the repulsive interaction between the two bulky (—R) groups.

Question 103. Explain why low molecular mass alcohols are soluble in water.
Answer:

The solubility of alcohols in water is due to their ability to form H-bonds with water molecules. The hydrocarbon part (i.e., R group) in alcohol tends to prevent the formation of H-bonds and thus tends to decrease the solubility of alcohols in water. In low molecular mass alcohols, the hydrocarbon part is small in size and hence, such alcohols are soluble in water.

Question 104. Explain why p-nitrophenol is more acidic than phenol.
Answer:

The electron density in the O —H bond of p -nitrophenol decreases relative to the O— H bond of phenol due to -I and -R effects of the -NO₂ group. As a result, O—H bond in p -nitrophenol is much weaker than the O—H bond in phenol and hence, p -nitrophenol is much more acidic than phenol.

Question 105. Explain why alcohols and ethers of comparable molecular mass have different boiling points.
Answer:

The O—H group in alcohols is involved in intermolecular H-bonding and hence, alcohol molecules exist in the associated form. Ethers, on the other hand, do not have an H-atom attached to the O-atom and hence, are unable to form H-bonds.

So they exist as discrete molecules which are held only by weak dipole-dipole attractions. As a result, less energy is required to break weak dipole-dipole attractions than to break H-bonds in associated molecules of alcohols. Obviously, alcohols have higher boiling points than ethers of comparable molecular mass.

Question 106. The carbon-oxygen bond in phenol is slightly stronger than that in methanol. Why?
Answer:

Phenol is a resonance hybrid of the following canonical forms.

Due to resonance, the carbon-oxygen bond in phenol has a partial double bond character. On the other hand, the C— O bond in methanol(CH₃—OH) is a pure single bond. Hence, the carbon-oxygen bond in phenol is stronger than that in methanol.

Question 107. Arrange water, ethanol and phenol in increasing order of acidity and give reason for your answer.
Answer:

Due to the electron-donating +1 effect of the ethyl group (CH₃CH₂—), the electron density in the O—H bond of CH₃CH₂OH is greater than that of the O—H bond in H₂O. Thus O—H bond of ethanol is stronger than that of H₂O, and hence, H₂O is a stronger acid than ethanol. On the other hand, due to the electron-withdrawing -I and -R effects of the phenyl group, the electron density in the O— H bond of phenol decreases appreciably, thereby making this bond relatively weak. So phenol is a stronger acid than H₂O. Thus acidity of the given compounds follows the sequence: CH₃CH₂OH < H₂O < C₆H₅OH.

Question 108. How can the presence of traces of water in a sample of alcohol be tested?
Answer:

If white anhydrous CuSO₄ does not turn blue when in contact with a sample of alcohol, then the alcohol must be anhydrous

Question 109. Propene is obtained on dehydration of two isomeric compounds A and B. Identify these two compounds.
Answer: 1-propanol (CH₃CH₂CH₂OH) and 2-propanol (CH₃CHOHCH₃)

Question 110. Ethanol cannot be used as a solvent in the test for unsaturation involving KMnO₄ or bromine-water—why?
Answer:

Alkaline KMnO₄ oxidises ethanol to acetic acid and bromine-water oxidises ethanol to acetaldehyde; as a result, both of these reagents are colourless

Question 111. Alcohol cannot be used as a solvent in the preparation of Grignard reagents—why?
Answer:

Grignard reagents readily react with alcohol-containing active hydrogen (an acid-base reaction) and as a result, they are destroyed (RMgX + ROH→RH + MgORX)

Question 112. Which is the compound obtained when acetylene is allowed to react with alcohol in the presence of Hg²⁺?
Answer:

Alcohols react with acetylene in the presence of Hg²⁺ ion (catalyst) to form acetals

Question 113. Ethanol is also called grain alcohol—why?
Answer:

As ethanol is produced by the fermentation of grains like maize, it is also called grain alcohol

Question 114. Both the alcohols obtained on reduction of an ester C5HIO°2 by LiAlH₄ respond to iodoform test. Identify the ester.
Answer:

Isopropyl ethanoate (MeCOOCHMe₂) ; LiAlH₄ reduction of this ester produces CH₃CH₂OH (ethanol) and CH₃CHOHCH₃ (2 -propanol); both the alcohols contain CH₃CHOH-group

Question 115. Severe conditions are required for oxidation of 3° alcohols —why?
Answer:

There is no H-atom at the carbinol carbon of 3° alcohol (R₃COH), therefore, its oxidation involving loss of two H-atoms is not possible. Oxidation actually occurs by the cleavage of stronger—C bond and for this reason, their oxidation requires severe conditions

Question 116. Give examples of two reactions of alcohols involving O—H and C—O bond cleavage.
Answer:

Esterification involves O— H bond cleavage and the reaction with PCl₅ involves C-O bond cleavage

Question 117. What happens when ethanol is allowed to react with CH₃C =CNa, CH₃(CH₂)₃Li and NaH. Explain.
Answer:

Propyne (CH₃C=CH), butane (CH₃CH₂CH₂CH₃) and H₂ are less acidic than ethanol; therefore, the given compounds are stronger bases than ethoxide. Since an acid-base reaction favours the formation of weaker acid and weaker base, these compounds react with ethanol to form weaker acids and weaker bases in each case. The products are CH₃C=CH + C₂H₅ONa, CH₃CH₂CH₂CH₃ + C₂H₅OLi and H₂ + C₂H₅ONa, respectively

Question 118. Dipole moment of methanol is higher than that of phenol—why?
Answer:

The bond moment is partially cancelled by the moment due to +R effect of the —OH group and hence, the dipole moment of phenol is less than that of methanol

Question 119. Give examples of two reactions of phenol involving O —H and C—O bond cleavage.
Answer:

The acetylation reaction involves O — H bond cleavage and zinc dust distillation (reduction) involves C—O bond cleavage

Question 120. Which compound is obtained when salicylic acid is treated with sodium bicarbonate solution and why?
Answer:

Salicylic acid contains a — COOH and a —OH group. Since a carboxylic acid is more acidic than H₂CO₃ but phenol is less acidic than H₂CO₃, the —COOH group, reacts with NaHCO₃ to form the corresponding sodium salt but the phenolic —OH group does not react. Therefore, the resulting compound is o-HOC₆H₄COONa.

Question 121. Explain why the C—O bond in phenol is stronger than the C—O bond in methanol.
Answer:

Due to resonance, the C—O bond in phenol possesses some double bond character. However, no such resonance occurs in methanol and for this reason, the C— O bond in phenol is somewhat stronger than that in methanol

Question 122. Explain why the same compound is obtained when phenol and salicylic acid are treated with Br₂/H₂O. What is that compound?
Answer:

Both phenol and salicylic acid react with Br₂/H₂O to form 2,4,6-tribromophenol as a white precipitate. The —COOH group of salicylic acid is replaced by Br (decarboxylation occurs)

Question 123. Explain why (S) -CH₃CHDOCH₃ is obtained when (R)-CH₃CHDBr is made to react with CH₃ONa.
Answer:

Williamson synthesis involves an S_N2 reaction since the priority of — Br and —OCH₃ is the same in the substrate and in the product. Thus, (R)-CH₃CHDBr reacts with CH₃ONa to yield (s)-CH₃CHDOMe

Question 124. Unlike diethyl ether, ferf-butyl methyl ether does not undergo oxidation by O₂ of air—why?
Answer:

There is no α -H in tert-butyl group and although there is a-H in the methyl group, the corresponding free radical (Me₃C—O—CH₂) is not stable; so, it is not formed easily. Hence, this ether does not undergo oxidation by O₂ of air

Question 125. Non-polar compounds are relatively more soluble in ether than in alcohol —why?
Answer:

These are insoluble in alcohol as it is not possible to break stronger H-bond that exist among alcohol molecules; weak dipole-dipole interactions and van der Waals forces operating among ether molecules are replaced by similar forces and therefore, non-polar compounds are more soluble in ether

Question 126. What are the products obtained when phenol is treated with HI and why?
Answer: Phenol and ethyl iodide

Question 127. Predict the product expected to be obtained when 2,2- dimethyloxirane is treated with dilute acid.
Answer: 2-methylpropane-1, 2-diol will be obtained

Question 128. Explain why tetrahydrofuran is more soluble in water than diethyl ether.
Answer:

Because of the cyclic structure, unshared electron pair on oxygen in tetrahydrofuran molecule is more available for making strong H-bonds with water molecules. However, due to the presence of two adjacent —C₂H₅ groups, unshared pair of electrons on oxygen is less exposed to form a strong H-bond with water molecules. Thus, tetrahydrofuran is more soluble in water than diethyl ether.

Question 129. Which C-O bond is stronger and why?
Answer: The bond ‘a’ is stronger than bond b as it gets some double bond character due to the -R effect of —NO₂ group.

Question 130. (A) and (B) are two isomeric organic compounds having the molecular formula C₃HgO. (A) and (B) when separately treated with cone. H₂SO₄ give the same organic product (C). (A) gives a yellow precipitate when treated with I₂/KOH, but (B) does not. The compound (A) on reaction with sodium provides (D). The compound (B) on reaction with PBr₃ gives (E). The compound (D) on reaction with the compound (E) gives (F). Identify (A) to (E) and write the chemical equation for the reaction of (D) with (E).
Answer:

Question 131. Indicate the reagent for the following transformation:

Answer:

Question 132.

1.  Arrange the following compounds in order of decreasing acidic strength:

Answer:

Question 133. Which of the following is produced when benzenediazonium chloride is coupled with phenol in an alkaline medium-

Answer:

Question 134. Identify A, B, C, D, E and F in the following reactions—

Answer:

Question 135. An organic compound A (C₂H₆O) reacts with sodium to form compound B and hydrogen gas. When heated with cone. H₂SO₄ at 413K, A produces C (C₄H₁₀O). C on reaction with cone. HI at 373K forms D. C Is aslo obtained when B is heated with D. Identify A, B, C and D and write chemical equations for the formation of B from A and the formation of C from B and D.
Answer:

1⇒ Ethanol (CH₃CH₂OH)

2 ⇒Sodium ethoxide (CH₃CH₂ONa)

3 ⇒ Diethyl ether (CH₃CH₂OCH₂CH₃)

4 ⇒ Ethyl iodide (CH₃CH₂I)

Class 12 Chemistry Chapter 11 Alcohols Phenols And Ethers Long Answer Question And Answer

Question 1. Explain why the alkoxide usually prepared from e ethanol is sodium ethoxide C₂H₅O^–Na⁺ while the alkoxide usually prepared from tort-butyl alcohol Is potassium tert-butoxide Me₃CO^–K⁺.
Answer:

In ethanol (CH₃CH₂OH), there is only one electron releasing (+1) methyl (CH₃—) group attached to the carbinol (a) carbon atom, whereas, in tort-butyl alcohol, there are three methyl groups attached to the a- carbon atom. Therefore, ethanol is relatively more acidic than tort-butyl alcohol. Hence, less electropositive metal like Na is sufficient to react with ethanol where as a more reactive metal like K is required to react with tort-butanol. As a result, the alkoxide prepared from ethanol is sodium ethoxide, while that from tert-butylalcohol is potassium tort-butoxide, (Me₃CO^–K⁺).

Question 2. An ester having molecular formula C₅H₁₀O₂ on Bouveault-Blanc reduction produces two different alcohols. Both of these alcohols give positive iodoform test. Identify the ester.
Answer:

Both the resulting alcohols give positive iodoform test So, both of them must contain CH₃CHOH -group. However, one of them must be a 1° alcohol, as these are obtained by reducing an ester. Therefore, one of them must be ethanol (CH₃CH₂OH), since it is a 1° alcohol containing CH₃CHOH -group.

The other alcohol must contain three carbon atoms as the ester contains five carbon atoms. Hence, the alcohol is 2-propanol (CH₃CHOHCH₃) which contains a CH₃CHOH -group. Since the 1° alcohol ethanol is obtained from R — CO—part and 2° alcohol, 2-propanol, from -OR part, the ester is isopropyl ethanoate [CH₃COOCH(CH₃)₂].

Question 3. Phenol cannot be converted into chlorobenzene but picric acid can be easily converted into picryl chloride by PCl₅. Explain these observations.
Answer:

First part: Phenol reacts with PCl₅ to form chlorophosphate ester.

The C—O bond in the chlorophosphate ester possesses some double bond character due to resonance. Also due to repulsive interaction caused by the ring n-electrons, the back side attack on the ring carbon by Cl^– does not take place. For these reasons, its dissociation leading to the formation of chlorobenzene does not take place.

Second part: In picric acid, the three electron-withdrawing — NO₂ groups at ortho-and para-positions reduce the electron density of the ring and particularly of the carbon attached to oxygen of the initially formed chlorophosphate ester.

As a result, nucleophilic attack on chlorophosphate ester occurs easily to form a stable (resonance-stabilized) carbanion. Expulsion of the chloro-phosphate ion as POCl₃ and Cl^– leads to the formation of picryl chloride. The reaction thus occurs by SNAr mechanism.

Question 4. At room temperature, tertiary alcohols (R₃COH) respond to Lucas test readily, but primary alcohols do not. Explain with reason.
Answer:

In the Lucas test, an alcohol (ROH) is converted into an alkyl chloride (RCl) through the formation of a carbocation (R^–). So the reactivity of an alcohol with lucas reagent is determined by the stability of the carbocation formed by it.

A 3° carbocation (R₃C) produced from a 3° alcohol (R₃COH) is very stable while a 1° carbocation (RCH₂) generated from a 1° alcohol (RCH₂OH) is very unstable. Hence, at room temperature, a 3° alcohol reacts with Lucas reagent readily but a 1° alcohol does not react at all.

Question 5. Simple ethers R— O — R cannot be cleaved by alkali. However, epoxides (three-membered cyclic ethers) can be easily cleaved by alkali. Explain.
Answer:

Since ROe (a strong base) is a very poor leaving group, S_N2 displacement by OH- is not possible. Epoxide molecules suffer from severe angle strain and so, they possess a tendency to open up the ring to release the angle strain. For this reason, epoxides undergo cleavage by alkali, despite having a similar leaving group (an alkoxy oxygen).

Question 6. Which ester produces a mixture of propan-l-ol and propan-2-ol when reduced with LiAlH₄?
Answer:

Isopropyl propanoate [CH₃CH₂COOCH(CH₃)₂]

Question 7. Which one of the following unsymmetrical ethers can be prepared conveniently by acid-catalysed dehydration of alcohols?

1. CH₃OCH₂CH₃
2. CH₃CH₂OCH(CH₃)₂
3. C₂H₅OC(CH₃)₃
4. C₂H₅OCH₂CH₂CH₂CH₃

Answer: C₂H₅OC(CH₃)₃

Question 8. (R)-CH₃CHDBr + CH₃ONa→A; Identify A.
Answer:

(S)-CH₃CHDOCH₃ because the inversion of configuration occurs and the priority of the leaving group is the same as that of the entering group.

Question 9. The C—O bond in phenol is stronger than the C—O bond In methanol—Why?
Answer:

Due to resonance, the C — O bond in phenol (C₆H₅OH) acquires some double bond character while no such resonance occurs in methanol (CH₃OH). Hence, the C—O bond in phenol is relatively stronger than the C—O bond in methanol.

Again, in phenol, the oxygen atom is attached to an sp²– hybridised carbon atom but in methanol, the oxygen atom is attached to an sp³-hybridised carbon atom. Since the bond strength increases with an increase in the s-character of the orbital involved, the C—O bond in phenol is stronger than that in methanol.

Question 10. Predict the product obtained when cleavage of 2, 2-dlmethyloxiranc is carried out with dil. acid.
Answer:

2- methylpropane-1, 2- diol will be obtained:

Question 11. Mention possible methods for the synthesis of 3-methylhexan-3-ol by Grignard reaction.
Answer:

The three possible methods are :

1. CH₃CH₂COCH₂CH₂CH₃ + CH₃MgI
2. CH₃COCH₂CH₂CH₃ + CH3CH₂MgBr
3. CH₃COCH₂CH₃ + CH₃CH₂CH₂MgBr

Question 12. THF is more soluble in water than diethyl ether-why?
Answer:

Due to ring structure, the unshared pair of electrons on oxygen in tetrahydrofuran is more available and hence, able to form a stronger H-bond with water molecules. On the other hand, in diethyl ether, the unshared electron pairs on oxygen, due to steric crowding caused by the two ethyl groups are unable to form stronger H-bond.

Question 13. Explain why the following ethers cannot be prepared by Williamson synthesis:

1. R₃C— O—CR₃
2. Ar—O—Ar
3. R₃CCH₂—O—CH₂CR₃

Answer:

1. Tertiary alkyl halides (R₃CX),
2. Aryl halides (ArX) and
3. Neopentyl type of halides (R₃CCH₂X) are unreactive towards S_N2 reaction.

Question 14. Which pair of the following reactants is to be used for preparing tert-butyl ethyl ether by Williamson synthesis and why?

1. C₂H₅OK + (CH₃)₃CI
2. (CH₃)₃COK + CH₃CH₂I

Answer:

The second pair of reactants is to be used. This is because in Williamson synthesis, S_N2 reaction between an alkyl halide and an alkoxide occurs and since the reaction is very susceptible to steric hindrance, the alkyl halide should necessarily be a 1° alkyl halide. CH₃CH₂I is a 1° alkyl halide.

If the first pair of reactants is used, the S_N2 reaction involving (CH₃)₃CI (3° alkyl halide) will not take place due to severe steric hindrance. Instead, an elimination (E2) reaction will occur (EtO^– acts as a base) to form an alkene (2-methylpropene).

Question 15. PCl5 reacts with ethanol to form chloroethane but phenol does not produce chlorobenzene by a similar reaction—Explain.
Answer:

Both ethanol and phenol react with PCl5 to form the corresponding chlorophosphate esters.

Since the C—O bond in ethanol does not possess any double bond character, the chlorophosphate ester (I) undergoes easy nucleophilic attack by Cl- to form chloroethane.

On the other hand, the C—O bond in chlorophosphate ester (II) possesses some double bond character due to resonance. Also due to repulsive interaction caused by the ring n -electrons, the back side attack on the ring carbon by Cl^– does not take place. For these reasons, its dissociation leading to the formation of chlorobenzene does not take place.

Question 16. For the preparation of the following ethers by Williamson synthesis only one combination of alkyl halide and alkoxide is appropriate. Mention that combination in each case.

1. CH₃CH₂CH₂—O
2. P—CH₃C₆H₄—O—CH2CH₃
3. CH₂=CHCH₂ — O—CH(CH₃)₂
4. (CH₃)₃CCH₂ — O—CH₂CH₃
5. (CH₃)₂CH—O —CH₂CH₂CH₂—O—CH(CH₃)₂

Answer:

1. CH₃CH₂CH₂Br+ -ONa
2. P-CH₃C₆H₄ONa + CH₃CH₂Br
3. CH₂=CHCH₂Br + (CH₃)₂CHONa
4. (CH₃)₃CCH₂ONa + CH₃CH₂Br
5. 2(CH₃)₂CHONa + BrCH₂CH₂CH₂Br

Question 17. Ethers (R — O— R) cannot be cleaved by alkali but can easily be cleaved by acid (e.g., HI ). Explain.
Answer:

Alkoxide ion (RO^–) being a very strong base Is a very poor leaving group. So, displacement of the alkoxide ion (S_N2) by a nucleophile such as OH^– is not possible. For this reason, cleavage of ethers is not possible by alkali.

On the other hand, in presence of HI , ether undergoes protonation i.e., —OR group is converted into — HO⁺R group. Since the neutral R— OH molecule is a very good leaving group, therefore, it is easily displaced by the strong nucleophile Ie. That is why ethers can easily be cleaved by acid (HI).

Question 18. Arrange p -nitrophenol, p -cresol, m -nitrophenol and 2, 4-dinltrophenol in order of Increasing acidity.
Answer:

P- cresol < phenol < m- nitrophenol < p nitrophenol < 2, 4-dinitrophenol.

Question 19. How will you distinguish between the following compounds:

1. Phenol and ethanol
2. Phenol and 2,4-dinitrophenol
3. Terf-butyl alcohol and 1- butanol
4. 2-pentanol and 1-pentanol
5. Phenol and salicylic acid
6. 1-phenylethanol and 2-phenylethanol.

Answer:

These two compounds can be distinguished from each other by ferric chloride test. When a few drops of neutral FeCl₃ solution is added to phenol, the mixture becomes violet in colour. However, when a few drops of FeCl₃ solution is added to ethanol, no change in colour is observed.

⇒ \(6 \mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}+\mathrm{FeCl}_3 \rightarrow\left[\mathrm{Fe}\left(-\mathrm{OC}_6 \mathrm{H}_5\right)_6\right]^{3-}+3 \mathrm{H}^{+}+3 \mathrm{HCl}\) (Complex ion (violet))

These two compounds can be distinguished by the solubility test in aqueous solution of sodium bicarbonate. Phenol does not dissolve in NaHCO₃ solution and CO₂ is not liberated but 2,4 -dinitrophenol being more acidic reacts with and dissolves in sodium bicarbonate solution liberating CO₂.

Lucas test can be done to distinguish between these two compounds. When Lucas reagent (a mixture of anhy. ZnCI₂ and conc.HCl) is added to tert-butyl alcohol (a 3° alcohol), turbidity appears immediately. However, when Lucas reagent is added to 1 -butanol ( 1° ). turbidity does not appear.

Iodoform test can be used to distinguish between these two compounds. If 2-pentanol (CH₃CH₂CH₂CHOHCH₃) containing CH₃CH(OM)-group is heated with NaOH solution in the presence of iodine (I₂ ), yellow crystals of iodofornHCHlj) having characteristic smell are precipitated. However, when 1 pentanol (CH₃CH₂CH₂CH₂CH₂OH) containing no CH₃CHOH group is treated similarly no yellow precipitate is obtained.

These two compounds can be distinguished by the oil of the wintergreen test. When salicylic acid is heated with methyl alcohol in the presence of a few drops of cone. H₂SO₄, methyl salicylate having the smell of wintergreen oil (smell of iodex) is produced. However, when phenol is treated similarly, no such smell is produced.

Iodoform test can be used to distinguish between these two compounds. When 1-phenylethonol (C₆H₅CHOHCH₃) containing a CH₃CHOH -group is heated with NaOH solution in the presence of iodine, yellow crystals of iodoform having characteristic smell are produced. However, when 2-phenylethanol (C₆H₅CH₂CH₂OH) is treated similarly, no yellow crystals are produced.

Question 20. Give two reactions that show the acidic nature of phenol. Compare its acidity and nucleophilicity with that of ethanol.
Answer:

The following reactions exhibit the acidic nature of phenol:

Phenol reacts with active metals (e.g., sodium) to produce sodium phenoxide along with H₂ gas.

Phenol dissolves in NaOH solution to form sodium phenoxide and water.

Phenol is more acidic than ethanol. This can be explained in terms of resonance. The unshared electron pair on oxygen in phenol becomes involved in resonance interaction with the ring π-electrons. As a result, the oxygen atom acquires a partial positive charge. As a consequence, the oxygen atom attracts the O—H bonding electrons more towards itself with greater force, thereby favouring the dissociation of the O— H bond to liberate proton (H⁺).

On the other hand, similar resonance interaction is not possible in ethanol and so, the O-atom does not acquire a partial positive charge. Thus, the O—H bond in alcohol is relatively much stronger than the O—H bond in phenol and in fact, alcohol does not exhibit acidic property involving dissociation of a relatively stronger O— H bond.

Both ethanol and phenol behave as nucleophiles because they contain unshared pair of electrons on their hydroxyl O-atom. Now, higher is the availability of lone pair for donation, greater is nucleophilic power. Since unshared pair of electrons on oxygen in ethanol is not delocalised, it is readily available. On the other hand, unshared pair of electrons on the O-atom in phenol is less available due to resonance interaction with the ring n-electrons. Hence, ethanol is a stronger nucleophile.

Question 21. Give the possible methods by which 4-nitroanisole can be prepared. Give mechanisms of the reactions involved.
Answer:

4-nitroanisole can be prepared by the reaction between 1-bromo-4-nitrobenzene and sodium methoxide. The reaction occurs by SNAr mechanism involving two steps.

1st Step: Nucleophilic attack by CH₃O^– ion on the ring carbon attached to bromine forms a resonance-stabilised carbanion intermediate.

2nd Step: Expulsion of Br^– from the intermediate carbanion to give the substitution product (desired ether).

It may also be prepared by the reaction(S_N2) of sodium 4- nitrophenoxide with methyl bromide.

Question 22. Propose a mechanism for the following reaction:

Answer: The mechanism of the reaction involves the following steps:

In the first step, instead of displacement (S_N2) of bromine, ring opening occurs as the ring is under severe strain. In the second step, an internal S_N2 takes place to give a five membered cyclic ether.

Question 23. Predict the product expected to be formed in each of the following reactions:

Answer:

Question 24. Identify the products:

Answer:

1. A: (CH₃)₂CHBr;

B : (CH₃)₂CHMgBr ;

C: (CH₃)₂CHCH₂CH₂OMgBr;

D:(CH₃)₂CHCH₂CH₂OH

2. E:C₆H₅CH(CH₃)₂;

F:C₆H₅C(CH₃)₂OOH; G: C₆H₅OH

3. H: C₆H₅N₂Cl; L: C₆H₅OH;

4. K: C₆H₅SO₃Na ; LiC₆H₅ONa ; M:C₆H₅OH:

5. O:CH₃CH=CH₂;P:(CH₃CH₂CH₂)₃B;

Q: CH₃CH₂CH₂OH;

6. R: C₆H₅ONa ; S: C₆H₅OH ;

T+U:O- + P-HO₃SC₆H₄OH;

(by intramolecular claisen rearrangement)

Question 25. Which one out of the four isomeric compounds having molecular formula C₇H₇OH do not produce violet colouration with neutral FeCl₃ solution?
Answer:

(Benzyl alcohol). The compound contains no enol group.

Question 26. An organic compound A(C₂H₆O) reacts with sodium ; to evolve H₂ gas and compound B. A forms a yellow coloured compund C by reacting with I₂/NaOH. A in heated at 140CC with cone. H₂SO₄ to yield compound D(C₄H₁₀O) which is further heated with HI at 100°C to produce E. Upon heating B and E together form D. Identity A to E and write down the given reactions.
Answer:

Since A reacts with Na to yield H₂ gas, it is not the ether CH₃OCH₃ but the alcohol CH₃CH₂OH. The reaction is:

Question 27. Write names and structures:

1. A Lewis acid that cleave ethers;
2. An ether which undergoes cleavage by HCI at ordinary temperature;
3. An ether which does not form peroxide easily in the presence of air;
4. An ether which undergoes cleavage by HI to yield phenol and benzyl iodide;
5. An ether which cannot be prepared by Williamson synthesis;
6. An ether which reacts with excess HI to yield O-HOC₆H₄CH₂CH₂CH₂I;
7. A nitrogenous compound that reacts with ROH in presence of HBF₄ to give ROCH₃;
8. An ether which is a metamer of diethyl ether;
9. A cyclic diether (formula: C₄H₈O₂ ) which possesses no dipole moment;
10. An ether which is obtained when 2-methyl propene reacts with methanol in presence of H₂SO₄.

Answer:

1. Boron tribromide (BBr₃);
2. Di-tert-butyl ether (Me₃C—O —CMe₃);
3. Tert-butyl methyl ether (Me₃COCH₃);
4. Benzyl phenyl ether (PhCH₂OPh);
5. Di-terf-butyl ether (Me₃COCMe₃) ;
6. Benzooxane
7. Diazomethane (CH₂N₂);
8. 1-methoxypropane (CH₃CH₂CH₂OCH₃) ;
9. 1, 4- dioxane
10. Tert-butyl methyl ether (Me₃COCH₃).

Question 28. Identify the major organic products (A, B, C…. etc.)

Answer:

1. A⇒CH₃CH₂OH (ethanol),

B⇒CH₃CH₂ONa (Sodium ethoxide),

C⇒CH₃CH₂OCH₃ (Ethyl methyl ether);

2. D⇒CH₃OH (Methanol),

E⇒(CH₃)₃CI (terf-butyl iodide);

3. F⇒C₆H₅OH (Phenol),

G⇒BrCH₂CH₂CH(CH₃)CH₂Br

(1,4-dibromo-2-methyIbutane),

H⇒CH₃CH₂Br (Ethyl bromide);

4. (1,4-dioxane);

5. J⇒(CH₃)₃COCH₃ (tert-butyl methyl ether),

K ⇒(CH₃)₃CBr (ferf-butyl bromide),

L⇒CH₃OH (Methanol);

6.

N⇒Br(CH₂)4Br (1, 4-dibromobutane);

7.  (1-chloro-2,4-dinitrobenzene);

(Phenyl 2, 4-dinitrophenyl ether);

8. Q⇒CH₃CH₂OCH₂CH₃ (Diethyl ether),

R⇒CCl₃CCl₂OCCI₂CCl₃ (Perchlorodiethyl ether);

9. S⇒CH₃CH₂CH₂OCH₃ (1- methoxypropane),

T⇒CH₃I (methyl iodide);

U⇒CH₃CH₂CH₂OH (1-propanol);

10. V⇒C₆H₅OH (Phenol); W=>C₆H₅OCH₃ (Anisole);

X⇒C₆H₅OH (Phenol); Y=> CH₃I (Methyl iodide);

11. (2 -bromo-2- methylcyclohexanol).

Question 30. Predict the product(s) expected to be obtained when trans-2-butene is allowed to react with m chloroperoxybenzoic acid and comment on the optical activity of the mixture.
Answer:

Trans-2-butene reacts with m-chloroperoxybenzoic acid to yield a mixture of two enantiomeric 1,2-dimethyloxirane since epoxidation (addition of O-atom to the double bond) takes place from both the faces of the alkene.

Question 31. Write the mechanism of the reaction of HI with methoxybenzene.

1. Name the starting material used In the Industrial preparation of phenol.
2. Write complete reaction for the bromination of phenol in aqueous and non-aqueous medium.
3. Explain why Lewis acid is not required in the bromination of phenol.

Answer:

Question 32. How can phenol be converted to aspirin?

Answer: (1) Phenol is prepared industrially starting from cumene (isopropyl benzene).

(2) In bromination reaction, the purpose of Lewis acid is to polarise the bromine molecule (i.e., brominating agent) to generate the reactive electrophile Br⁺. But in the case of phenol, the +R effect of —OH group is so strong that o- and p-positions of the ring become highly electron-rich, and these electron-rich centres are responsible for polarising bromine molecules. Thus no Lewis acid catalyst is required in the bromination of phenol.

Question 79. How CH₃MgI could be used to prepare methoxyethane? (Write arrowhead equation only)
Answer:

2. CH₃MgI will react with water. Methane is formed as an organic compound.

3. CH₃CH₂OCH₂CH₃ is immiscible and unreactive towards water.

4. CH₃CH₂OH on dehydration produces an alkene. Here, cone. H₂SO₄ /170°C can be used as a dehydrating agent.

Class 12 Chemistry Chapter 11 Alcohols Phenols And Ethers Multiple Choice Questions And Answers

Question 3. Consider the following and answer the questions: CH₃MgI, CH₃OH, CH₃CH₂OCH₂CH₃, CH₃CH₂OH

1. Which will form oil of wintergreen? Give the arrowhead equation of the reaction for the formation of oil of wintergreen.
2. Which will react with water? What is the organic compound formed in the reaction?
3. Which is immiscible and unreactive with water?
4. Which on dehydration produces an alkene? Name the reagent for dehydration.

Answer:

1. CH₃OH will form oil of wintergreen.

Question 11. Which of the following compounds will respond to the iodoform test—

1. CH₃CH₂CH₂OH
2. 
3. CH₃OCH₂CH
4. CH₃OH

Answer: 2

Question 12. Which of the following compounds is formed when phenol Is heated with CCl₄ and NaOH —

1. Salicylic add
2. Sodium salicylate
3. Salicylaldehyde
4. Pom-hydnixyben/ahlelyde

Answer: 2. Sodium salicylate

Question 13. Identify A, B, C, D, E and F In the following reactions:

Answer:

Question 14. How would you distinguish between the following pair of compounds by a chemical reaction?

or,

How would you convert?

Write down the products of the following reaction:

⇒ \(\mathrm{CH}_3 \mathrm{OCH}_2 \mathrm{CH}_3 \underset{\text { Heat }}{\stackrel{\text { conc } \cdot \mathrm{HI}}{\longrightarrow}}\)

Answer:

Question 15. Draw the structure and name the product formed if the following alcohols are oxidized. Assume that an excess of oxidizing agent is used. (1) CH₃CH₂CH₂CH₂OH (2) 2-butenol (3) 2-methyl-1-propanol
Answer:

Question 16. Draw the structure of major monohalo product in the following reaction:

Answer:

Question 17. Complete the following reaction equation:

C₆H₅N₂CI + H₃PO₂ + H2O→

Answer:

Question 18. Explain the mechanism of dehydration steps of ethanol:

Answer: Dehydration of ethanol by cone. H₂SO₄/443K follows E2 pathway.

Question 19. Name the following according to IUPAC system:

Answer: Butan-2-ol

Question 20. How are the following conversions carried out? Benzyl chloride to benzyl alcohol,
Answer:

Question 21. How are the following conversions carried out? Ethyl magnesium chloride to propan-1-ol.
Answer:

Question 22. Write the major product in the following equations:

Answer:

Question 23. Write the main product(s) in each of the following reactions:

1. 
2.  \(\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2 \frac{\text { (1) } \mathrm{B}_2 \mathrm{H}_6}{\text { (2) } 3 \mathrm{H}_2 \mathrm{O}_2 / \mathrm{OH}^{-}}\)
3. \(\mathrm{C}_6 \mathrm{H}_5-\mathrm{OH} \underset{\text {(1) aq. } \mathrm{NaOH}}{\stackrel{\text { (2) } \mathrm{CO}_2 / \mathrm{H}^{+}}{\longrightarrow}}\)

Answer:

Question 24. (1) Write the product(s) in each of the following reactions: 1.C₆H₅—O—CH₃ + HI

⇒ \(\mathrm{C}_6 \mathrm{H}_5-\mathrm{OH} \stackrel{\text { Zn dust }}{\longrightarrow}\)

Answer: \(\mathrm{C}_6 \mathrm{H}_5-\mathrm{O}-\mathrm{CH}_3+\mathrm{HI} \rightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}+\mathrm{CH}_3 \mathrm{I}\)

Question 25. Write the chemical equations involved in the following reactions:

1. Reimer-Tiemann reaction
2. Friedel-Crafts alkylation of anisole

or, What happens when:

1. Phenol reacts with cone.HNO₃
2. Salicylic acid reacts with (CH₃CO)₂O/H+
3. Ethyl chloride reacts with NaOCH₃ write the chemical equations involved in the above reactions.

Answer:

Question 26. Write the product(s) in the following reactions:

Answer:

Question 27. Write the formula of reagents used in the following reactions:

1. Bromination of phenol to 2, 4, 6-tribromophenol
2. Hydroboration of propene and then oxidation to propanol

Answer:

1. Br₂/H₂O
2. B₂H₆ followed by H₂O₂/NaOH

Question 28. Arrange the following compounds in the increasing order of their acid strength: p-cresol, p-nitrophenol, phenol
Answer:

The increasing order of acidity of the given compounds is p-cresol < phenol < p-nitrophenol

Question 29. Write the mechanism (using curved arrow notation) of the following reaction:

⇒ \(\mathrm{CH}_2=\mathrm{CH}_2 \longrightarrow \mathrm{CH}_3-\mathrm{CH}_2^{+}+\mathrm{H}_2 \mathrm{O}\)

Answer:

Question 30. Write the IUPAC name of the following:

Answer: 3-phenylprop-2-en-1-ol

Question 31. Write the IUPAC name of the following:

Answer: 3,3-dimethylbutan-2-ol

Question 32. Write the structures of the main products in the following reactions:

Answer:

Question 33. Among the alkenes, which one produces tertiary butyl alcohol on acid hydration—

1. CH₃CH₂—CH=CH₂
2. CH₃—CH=CH—CH₃
3. (CH₃)₂C=CH₂
4. CH₃—CH=CH₂

Answer:

Question 34. Which one of the following properties is exhibited by phenol—

1. It is soluble in aqueous NaOH and produces bubbles of CO₂ in the presence of aqueous NaHCO₃
2. It is soluble is aqueous NaOH and does not produce bubbles of CO₂ in the presence of aqueous NaHCO₃
3. It is insoluble in aqueous NaOH but produces bubbles in the presence of NaHCO₃
4. It is insoluble in aqueous NaOH and does not produce bubbles in the presence of NaHCO₃

Answer: It is soluble is aqueous NaOH and does not produce bubbles of CO₂ in the presence of aqueous NaHCO₃

Question 35. Which one of the following methods is used to prepare Me₃COEt with a good yield—

1. Mixing EtONa with Me₃CCl
2. Mixing Me3CONa with EtCl
3. Heating a mixture of (1: 1) EtOH and Me3COH in presence of cone. H₂SO₄
4. Treatment of Me₃COH with EtMgl

Answer:

Question 36. The correct order of acid strength of the following substituted phenols in water at 28°C is—

1. P-nitrophenol < p-fluorophenol < p-chlorophenol
2. P-chlorophenol < p-fluorophenol < p-nitrophenol
3. P-fluorophenol < p-chlorophenol < p-nitrophenol
4. P-fluorophenol < p-nitrophenol < p-chlorophenol

Answer: 3. P-fluorophenol < p-chlorophenol < p-nitrophenol

Electron withdrawing effect of —NO₂,—Cl and —F groups is as follows: —NO₂ > —Cl > —F. Thus, the conjugate base of p-nitrophenol will be stabilised to the maximum amount followed by the conjugate base of p-chlorophenol and p-fluorophenol. Thus the order of acidity of the given substituted phenols is—

Question 37. When phenol is treated with D₂SO₄/D₂O, some of the hydrogens get exchanged. The final product in this exchange reaction is—

Answer: 1.

As —OH group in phenol is o-/p-directing, only H-atom in o-/p-positive will be replaced by deuterium.

Question 38. Which of the following compounds would not react with Lucas reagent at room temperature—

1. H₂C=CHCH₂OH
2. C₆H₅CH,OH
3. CH₃CH₂CH₂OH
4. (CH₃)₃COH

Answer: 3 CH₃CH₂CH₂OH

Primary alcohols do not respond to the Lucas test. Though benzyl alcohol is a primary alcohol, it responds to Lucas test readily.

Question 39. Which of the following will be dehydrated most readily in alkaline medium—

Answer: 2

The presence of C=O group stabilises the -ve charge of the carbanion.

Question 40. The structure of the product P of the following reaction is—

Answer: 3

—OH group is stronger o/p-directing compared to —OMe group.

Question 41. Reduction of lactol with sodium borohydride gives—

Answer: 3

Question 42.

here N is-

Answer: 2

Question 43. Methoxybenzene on treatment with HI produces—

1. Iodobenzene and methanol
2. Phenol and methyl iodide
3. Iodobenzene and methyl iodide
4. Phenol and methanol

Answer: 2. Phenol and methyl iodide

Question 44. The number of alkene(s) which can produce 2-butanol by the successive treatment of (1) B₂H₆ in tetrahydrofuran solvent and (2) alkaline H₂O₂ solution is—

1. 1
2. 2
3. 3
4. 4

Answer: 1.1

Question 45. When phenol is heated with KBr and KBrO₃, the compound obtained as major product is—

1. 3-bromophenol
2. 4-bromophenol
3. 2,4,6-tribromophenol
4. 2-bromophenol

Answer: 3. 2,4,6-tribromophenol

Question 46. Which one of the following reagents can be used to distinguish between phenol and benzonic acid.

1. Tollen’s reagent
2. Molish reagent
3. Neutral FeCl₃ solution
4. Aqueous NaOH

Answer: 3. Neutral FeCl₃ solution

Phenol in presence of neutral FeCl₃ solution exhibits blue colour while benzoic acid does not.

Question 47. An unknown alcohol is treated with the “Lucas reagent” to determine whether the alcohol is primary, secondary or tertiary. Which alcohol reacts fastest and by what mechanism—

1. 2° alcohol by S_N1
2. 3° alcohol by S_N1
3. 2° alcohol by S_N2
4. 3° alcohol by S_N2

Answer: 2. 3° alcohol by S_N1

3° alcohol readily reacts with Lucas reagent via S_N1 mechanism.

Question 48. Arrange the following compounds in order of decreasing acidity—

1. 2 > 4 > 1 > 3
2. 1 > 2 > 3 > 4
3. 3 > 1 > 2 > 4
4. 4 > 3 > 1 > 2

Answer: 3. 3 > 1 > 2 > 4

Stability of the conjugate bases of the given phenols are in the order:

∴ Thus the order of acidity is 3 > 1 > 2 > 4

Question 49. The most suitable reagent for the conversion of R—CH₂—OH—CHO is-

1. PCC
2. KMnO4
3. K₂Cr₂O₇
4. CrO₃

Answer: 1. PCC

PCC or pyridinium chlorochromate is a mild oxidising agent which selectively oxidises alcohols to aldehydes.

Question 50. The product of the reaction given below is—

Answer: 2

Question 51. 2-chloro-2-methylpentane on reaction with sodium methoxlde in methanol yields—

1. All of these
2. 1 and 3
3. 3 only
4. 1 and 2

Answer: 1. All of these

Question 52. Which of the following, upon treatment with tert-BuONa followed by addition of bromine water, fails to decolourise the colour of bromine—

Answer: 3

The above product does not have any C=C or C = C bond, so it will not give Br2 -water test.

Question 53. The correct sequence of reagents of the following conversion will be—

⇒ \(\mathrm{CH}_3 \mathrm{MgBr},\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+} \mathrm{OH}^{-}, \mathrm{H}^{+} / \mathrm{CH}_3 \mathrm{OH}\)

⇒ \(\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+} \mathrm{OH}^{-}, \mathrm{CH}_3 \mathrm{MgBr}, \mathrm{H}^{+} / \mathrm{CH}_3 \mathrm{OH}\)

⇒ \(\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+} \mathrm{OH}^{-}, \mathrm{H}^{+} / \mathrm{CH}_3 \mathrm{OH}, \mathrm{CH}_3 \mathrm{MgBr}\)

⇒ \(\mathrm{CH}_3 \mathrm{MgBr}, \mathrm{H}^{+} / \mathrm{CH}_3 \mathrm{OH},\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+} \mathrm{OH}^{-}\)

Answer: 3

Question 54. Phenol on treatment with CO₂ in the presence of NaOH followed by acidification produces compound X as the major product. X on treatment with (CH₃CO)₂O in the presence of catalytic amount of H₂SO₄ produces—

Answer: 3

Question 55. The major product formed in the following reaction is—

Answer: 2

Question 56. Phenol reacts with methyl chloroformate in the presence of NaOH to form product A. A reacts with Br₂ to form product B. A and B are respectively—

Answer: 1

Question 57. In the following reactions:

Answer: 3

Question 58. In the following reaction:

Answer: 3

Question 59. Which of the following compounds will give a yellow precipitate with iodine and alkali—

1. Methyl acetate
2. Acetamide
3. 2-hydroxypropane
4. Acetophenone

Answer: 3. 2-hydroxypropane

Question 60. Identify Z in the sequence of reactions—

1. CH₃—(CH₂)₃—O—CH₂CH₃
2. (CH₃)₂CH— O—CH₂CH₃
3. CH₃(CH₂)₄—O—CH₃
4. CH₃CH₂—CH(CH₃)—O—CH₂CH₃

Answer: 1. CH₃—(CH₂)₃—O—CH₂CH₃

Question 61. Among the following sets of reactants which one produces anisole—

1. CH₃CHO ; RMgX
2. C₆H₅OH; NaOH; CH₃I
3. C₆H₅OH; neutral FeCl₃
4. C₆H₅—CH₃; CH₃COCl; AlCl₃

Answer: 2

Question 62. Which of the following is not the product of dehydration of  

Answer: 2

The intermediate carbocation formed, stable. That is why rearrangement does not occur.

Question 63. Which of the following reaction(s) can be used for the preparation of alkyl halides—

⇒ \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}+\mathrm{HCl} \stackrel{\text { anhy. } \mathrm{ZnCl}_3}{\longrightarrow}\)

⇒ \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}+\mathrm{HCl} \longrightarrow\)

⇒ \(\left(\mathrm{CH}_3\right)_3 \mathrm{COH}+\mathrm{HCl} \longrightarrow\)

⇒ \(\left(\mathrm{CH}_3\right)_2 \mathrm{CHOH}+\mathrm{HCl} \stackrel{\text { anhy. } \mathrm{ZnCl}_2}{\longrightarrow}\)

1. 1, 3 and 4
2. 1 and 2
3. Only 4
4. 3 and 4

Answer: 1. 1, 3 and 4

1 and 4 react with Lucas reagent and III reacts with HCl to produce alkyl halides.

Question 64. Reaction of phenol with chloroform in presence of dilute sodium hydroxide finally introduces which one of the following functional group—

1. —CH₂CI
2. —COOH
3. —CHCl₂
4. —CHO

Answer: 4. —CHO

Question 65. Which of the following reagents would distinguish cis-cyclopenta-1,2-diol from the trans-isomer—

1. Aluminium isopropoxide
2. Acetone
3. Ozone
4. MnO₂

Answer: 2. Acetone

Trans-cyclopenta-1, 2-diol does not form cyclic ketal with acetone whereas cis-cyclopenta-1,2-diol does.

Question 37. The reaction:

can be classified as—

1. Williamson alcohol synthesis reaction
2. Williamson ether synthesis reaction
3. Alcohol formation reaction
4. Dehydration reaction

Answer: 2. General concept

Question 66. The heating of phenyl methyl ether with HI produces—

1. Iodobenzene
2. Phenol
3. Benzene
4. Ethyl chloride

Answer: 2. Ethyl chloride

Question 67. Which one is the most acidic compound—

Answer: 3

The presence of electrons with the drawing group increases the acidity of phenol. Thus 2, 4, 6-trinitrophenol is the most acidic among the given substituted phenols.

Question 68. Compound A, C₂H₁₀O is found to react with NaOI (produced by reacting Y with NaOH ) and yields a yellow precipitate with characteristic smell. A and T are respectively—

Answer: 3

Question 69. The compound A on treatment with Na gives B and with PCI₅ gives C. B and C react together to give diethyl ether. A, B and C are in the order —

1. C₂H₅OH, C₂H₅ONa, C₂H₅Cl
2. C₂H₅OH, C₂H₆, C₂H₅CI
3. C₂H₅CI, C₂H₆, C₂H₅OH
4. C₂H₅OH, C₂H₅Cl, C₂H₅ONa

Answer: 1. C₂H₅OH, C₂H₅ONa, C₂H₅Cl

Question 70. Find the product for \(\mathrm{CH}_3 \mathrm{CH}_2-\mathrm{O}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{O}-\mathrm{CH}_2-\mathrm{C}_6 \mathrm{H}_5+\mathrm{HI}\)

1. HO—CH₂CH₂—OH, C₆H₅CH₂—I, CH₃CH₂—I
2. C₆H₅CH₂—OH, CH₃CH₂—I, I—CH₂CH₂—OH
3. I— CH₂CH₂—I, C₆H₅CH₂—I, CH₃CH₂—OH
4. HO—CH₂CH₂—OH ,C₆H₅CH₂—I ,CH₃CH₂—OH

Answer: 3. I— CH₂CH₂—I, C₆H₅CH₂—I, CH₃CH₂—OH

Presence of excess of HI favours S_N1 mechanism. So, formation of products is controlled by the stability of the carbocation formed due to the cleavage of C—O bond in protonated ether. Thus the products for given reaction are C₆H₅CH₂I , CH₃CH₂I , HOCH₂CH₂OH .

Question 71. But-l-ene \(\underset{\mathrm{H}_3 \mathrm{O}^{+}}{\stackrel{\left(\mathrm{CH}_3 \mathrm{COO}\right)_2 \mathrm{Hg}}{\longrightarrow}}\) The product in the above reaction is—

Answer: 2

Question 72. A compound containing two —OH groups attached with one carbon atom is unstable but which one of the following is stable—

Answer: 3.

Chloral hydrate is stable due to the presence of intramolecular hydrogen bonding.

Question 73. Which of the following reaction will not produce ethylene glycol—

Answer: 3

In all other reactions, ethylene glycol is formed.

Question 74.

Answer: 2

Question 45. A (major product). A is—

Answer: 1

Question 76. During the dehydration of alcohols to alkenes by heating with cone. H₂SO₄, the initiating step is—

1. Elimination of water
2. Protonation of an alcohol molecule
3. Formation of an ester
4. Formation of carbocation

Answer: 2. Protonation of an alcohol molecule

The mechanism of dehydration of alcohol involves the following steps—

1. Formation of protonated alcohol
2. Formation of carbocation
3. Formation of alkene by elimination of a proton

Question 77. Isopropyl benzene on air oxidation in the presence of dilute acid gives—

1. C₆H₅COOH
2. C₆H₅COCH₃
3. C₆H₅CHO
4. C₆H₅OH

Answer: 4. C₆H₅OH

Question 78. The reaction,

Suggest that-

1. Phenol is a stronger acid than carbonic acid
2. Carbonic acid is a stronger acid than phenol
3. Water is a stronger acid than phenol
4. None of these

Answer: 2. Carbonic acid is a stronger acid than phenol

Since a stronger acid displaces a weaker acid from its salts, therefore, carbonic acid (CO₂ + H₂O→H₂CO₃) is a stronger acid than phenol.

Question 79. CH₃OC₂H₅ and (CH₃)₃C-OCH₃ are treated with hydroiodic acid. The fragments after reactions obtained are—

1. CH₃I + HOC₂H₅; (CH₃)₃C—I + HOCH₃
2. CH₃OH + C₂H₅I; (CH₃) CI + HOCH₃
3. CH₃OH + C₂H₅I; (CH₃)₃C—OH + CH₃I
4. CH₃I + HOC₂H₅; CH₃I + (CH₃)₃C—OH

Answer: 1. CH₃I + HOC₂H₅; (CH₃)₃C—I + HOCH₃

When mixed ethers are used, the alkyl iodide produced depends on the nature of alkyl groups. If one group is Me— and the other primary or secondary alkyl group, then methyl iodide is produced. Here reaction occurs via SN2 mechanism and because of the steric effect of the larger group, I” attacks the smaller (Me) group.

⇒ \(\mathrm{CH}_3 \mathrm{OC}_2 \mathrm{H}_5+\mathrm{HI} \rightarrow \mathrm{CH}_3 \mathrm{I}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\)

When the substrate is a methyl t-alkyl ether, the products are t-R—I and MeOH. Here reaction occurs via S_N1 mechanism and formation of products is controlled by the stability of carbocation. Since carbocation stability order is 3° > 2° > 1° > CH₃, alkyl halide is always derived from the tert-alkyl group.

Question 80.

The product P is—

Answer: 1

Question 81. Decreasing order or reactivity in Williamson synthesis of the following is—

1. Me₃CCH₂Br

2. CH₃CH₂CH₂Br

3. CH₂=CHCH₂CI

4. CH₃CH₂CH₂Cl

1. 3 > 2 > 4 > 1
2. 1 > 2 > 4 > 3
3. 2 > 3 > 4 > 1
4. 1 > 3 > 2 > 4

Answer: 3. 2 > 3 > 4 > 1

C—Br bond is weaker than the C—Cl bond, therefore, alkyl bromide (2) reacts faster than alkyl chlorides, (3) and (4). Since CH₂=CH— is electron withdrawing group, therefore, ar-C-atom has more +ve charge on 3 than on 4.

In other words, nucleophilic attack occurs faster on 3 than on 4. Further, since Williamson synthesis occurs by S_N2 mechanism, therefore, due to steric hindrance alkyl bromide (1) is the least reactive. Thus, the decreasing order of reactivity is 2 > 3 > 4 > 1.

Question 82. Lucas test is given by an alcohol within 5 minutes, 0.22 g of which liberates 56 mL of CH₄ at STP on treating with CH₃MgI. The structure of alcohol is—

Answer: 2

Lucas test will be given by all the given alcohols within 5 minutes as all are 2° alcohols. Exact structure can be known by molecular mass or formula of alcohol which will be obtained as follows: Let molecular mass ofalcohol be M.

Thus, \(\frac{56}{22400}=\frac{0.22}{M} \quad \text { or, } M=\frac{22400 \times 0.22}{56}=88\)

The general formula of alcohols is C_nH_2n+1 OH. The molecular mass 88 corresponds to the value of n = 5. Thus, the 2° alcohol is (CH₃)₂CHCHOHCH₃.

Question 83. Identify the final product of the given reaction—

Answer: 3

Question 84.

Answer: 1

Question 85. Monochlorination of toluene in sunlight followed by hydrolysis with aq. NaOH yields—

1. O-cresol
2. M-cresol
3. 2,4-dihydroxytoIuene
4. Benzyl alcohol

Answer: 4. Benzyl alcohol

Question 86. How many alcohols with molecular formula C₄H₁₀O are chiral in nature—

1. 1
2. 2
3. 3
4. 4

Answer: 1

Hint: Only CH₃CH₂CH(OH)CH₃ is chiral.

Question 87. What is the correct order of reactivity of alcohols in the following reaction?

⇒ \(\mathrm{R}-\mathrm{OH}+\mathrm{HCl} \stackrel{\mathrm{ZnCl}_2}{\longrightarrow} \mathrm{RCl}+\mathrm{H}_2 \mathrm{O}\)

1. 1° > 2° > 3°
2. 1° < 2° > 3°
3. 3° > 2° > 1°
4. 3° > 1° > 2°

Answer: 3. 3° > 2° > 1°

Question 88. CH₃CH₂OH can be converted into CH₃CHO by—

1. Catalytic hydrogenation
2. Treatment with LiAlH₄
3. Treatment with pyridinium chlorochromate
4. Treatment with KMnO₄

Answer: 3. Treatment with pyridinium chlorochromate

Question 89. Converting alkyl halides into alcohols involves—

1. Addition reaction
2. Substitution reaction
3. Dehydrohalogenation reaction
4. Rearrangement reaction

Answer: 2. Substitution reaction

Question 90. Which of the following compounds is aromatic alcohol—

1. 1, 2, 3, 4
2. 1, 4
3. 2, 3
4. 1

Answer: 3. 2,3

Hint: In aromatic alcohols, the -OH group is attached to side chain of an aromatic ring system.

Question 91. Give IUPAC name of the compound given below—

1. 2-chIoro-5-hydroxyhexane
2. 2-hydroxy-5-chlorohexane
3. 5-chlorohexan-2-ol
4. 2-chlorohexan-5-ol

Answer: 3. 5-chlorohexan-2-ol

Question 92. IUPAC name of m -cresol is—

1. 3-methylphenol
2. 3-chlorophenoI
3. 3-methoxyphenol
4. Benzene-1, 3-diol

Answer: 1. 3-methylphenol

Question 93. IUPAC name of the compound

1. 1-methoxy-l-methylethane
2. 2-methoxy-2-methylethane
3. 2-methoxypropane
4. Isopropylmethyl ether

Answer: 3. 2-methoxypropane

Question 94. Which of the following can act as the strongest base—

Answer: 2. OR

Hint: The weakest acid ROH has the strongest conjugate base RO^–.

Question 95. Which of the following compounds will react with sodium hydroxide solution in water-

1. C₆H₅OH
2. C₆H₅CH₂OH
3. (CH₃)₃COH
4. C₂H₅OH

Answer: 1. C₆H₅OH

Hint: Phenol is an acidic compound and so it reacts with base (NaOH).

Question 96. Phenol is less acidic than—

1. Ethanol
2. O-nitrophenol
3. O-methylphenol
4. O-methoxyphenol

Answer: 2. O-nitrophenol

Hint: Electron-donating groups such as CH₃— or — OCH₃ decrease the acid strength while electron-withdrawing groups (e.g.,-NO2) increase the acid strength of phenolic compounds.

Question 97. Which of the following is most acidic—

1. Benzyl alcohol
2. Cyclohexanol
3. Phenol
4. M-chlorophenol

Answer: 4. M-chlorophenol

Hint: Cl -atom in the meta position has electron withdrawing -I effect and hence, increases the acid strength of phenolic compounds.

Question 98. Mark the correct order of decreasing acid strength of the following compounds —

1. 5>4>2>1>3
2. 2>4>3>1>5
3. 4>5>3>2>1
4. 5>4>3>2>1

Answer: 2. 2>4>3>1>5

Hint: Presence of electron-withdrawing -NO₂ group in o-, m- or p- position causes an increase in acid strength of phenol. However, the presence of the -OCH₃ group in the meta-position increases acidity, while its presence in o-, or p- positions decreases the acidity of phenol.

Question 99. Mark the correct increasing order of reactivity of the following compounds with HBr/HCl—

1. 1 < 2 < 3
2. 2 < 1 < 3
3. 2 < 3 < 1
4. 3 < 2 < 1

Answer: 3. 2 < 3 < 1

Hint: Reaction with HBr/HCl proceeds via the formation of benzylic cations. The stability of cations increases in the sequence:

Question 100. Arrange the following compounds in increasing order of boiling point. Propan-l-ol, butan-l-ol, butan-2-ol, pentan-l-ol—

1. Propan-1-ol, butan-2-oI. butan-1-ol, peman-l-ol
2. Propan-1-ol, butan-1-ol, butan-2-ol, pentan-1-ol
3. Pentan-1-ol, butan-2-ol, butan-1-ol, propan-1-ol
4. Pentan-1-ol, butan-1-ol, butan-2-ol, propan-1-ol

Answer: 1. Propan-1-ol, butan-2-oI. butan-1-ol, peman-1-ol

Hint: Boiling points increase as the molecular masses of alcohols increase. Among isomeric alcohols, 1° alcohols have higher boiling point than 2° alcohols.

Question 101. Which of the following are used to convert RCHO into RCH₂OH —

1. H₂/Pd
2. LiAlH₄
3. NaBH₄
4. Reaction with RMgX followed by hydrolysis

Answer: 1,2 and 3

Hint: RMgX reacts with aldehydes to form 2° alcohols.

Question 102. Which of the following reactions will yield phenol—

Answer: 1,2 and 3

Hint: Chlorobenzene does not undergo nucleophilic substitution with aq. NaOH at 298K and 1 atm pressure.

Question 103. Which of the following reagents can be used to oxidise primary alcohols to aldehydes—

1. CrO₃ in anhydrous medium
2. KMnO₄ in an acidic medium
3. Pyridinium chlorochromate
4. Heat in the presence of Cu at 573 K

Answer: 1,3,4

Hint: KMnO₄/H+ oxidises 1° alcohols to carboxylic acids.

Question 104. Phenol can be distinguished from ethanol by the reactions with—

1. Br₂ /water
2. Na
3. Neutral FeCl₃
4. All the above

Answer: 1 and 3

Hint: Na reacts with both phenol and alcohols to liberate H2gas.

Question 105. Which of the following are benzylic alcohols—

Answer: 2,3

Hint: In benzylic alcohols, —OH group is attached to the carbon next to a carbon atom of the aromatic ring.

Question 106. Which one of the following compounds reacts with NaOH to yield methanol most rapidly—

1. (CH₃)₄N +I^–
2. CH₃OCH₃
3. (CH₃)₃3S+I^–
4. (CH₃)₃Cl

Answer: 1. (CH₃)₄N +P

Question 107. The compound obtained on acid-catalysed hydration of 2 -phenylpropene is —

1. 2 -phenyl- 2 -propanol
2. 2 -phenyl- 1 -propanol
3. 3 -phenyl- 1 -propanol
4. 1 phenyl- 2 -propanol

Answer: 1. 2 -phenyl- 2 -propanol

Question 108. Which compound is to be introduced in the body of a patient poisoned by methanol—

1. 2 -propanol
2. 1 -propanol
3. Ethanol
4. Ethylene glycol

Answer: 3. Ethanol

Question 109. The compounds obtained on treatment of CH₂=CH —OCH₃ with HBr at room temp, are—

1. CH₃CHO and CH₃Br
2. BrCH₂CHO and CH₃OH
3. BrCH₂CH₂OCH₃
4. CH₃CHBrOCH₃

Answer: 4. CH₃CHBrOCH₃

Question 110. Benzene diazonium chloride reacts with a weakly alkaline solution of phenol to yield—

1. Diphenyl ether
2. P-hydroxyazobenzene.
3. Chlorobenzene
4. Benzene

Answer: 2. P-hydroxyazobenzene

Question 111. Phenol + chloroform Electrophile involved in this reaction Is—

1. C⁺HCl₂
2. : CCl₂
3. C^–Cl₃
4. C⁺HO

Answer: 2. : CCl₂

Question 112. Which one of the following compounds will react with HBr most rapidly—

1. C₆H₅CH₂OH
2. P-O₂NC₆H₄CH₂OH
3. P-CH₃OC₆H₄CH₂OH
4. P-ClC₆H₄CH₂OH

Answer: 3. P-CH₃OC₆H₄CH₂OH

Question 113. Tert-butyl alcohol is heated with Cu at 350°C to form—

1. 1 -butanol
2. Butanol
3. 2 -butene
4. 2 methylpropene

Answer: 4. 2 methylpropene

Question 114. Which compound is responsible for the appearance of turbidity in Lucas test of alcohol—

1. Aldehyde
2. Ketone
3. Acid chloride
4. Alkyl chloride

Answer: 4. Alkyl chloride

Question 115. The order of reactivity of alcohols towards active metal is—

1. 3° > 2° > 1°
2. 3° < 2° < 1°
3. 3° < 1° < 2°
4. 2° < 3° < 1°

Answer: 2. 3° < 2° < 1°

Question 116. Which reagent is to be used for the preparation of cyclohexane from cyclohexanol—

1. Al₂O3, 350°C
2. Cone. HCI/ZnCI₂
3. Conc.HCI
4. Cone. HBr

Answer: 1. Al₂O3, 350°C

Question 117. Ph₂CHCH₂OH reacts with HBr to form—

1. Ph₂CHCH,Br
2. PhCHBrCH₂Ph
3. Ph₂C=CH₂
4. PhCH —CHPh

Answer: 2. PhCHBrCH₂Ph

Question 118. Which one of the following compounds has lowest value of pKa —

CF₃CH₂OH

HC=CH

Answer: 3

Question 119. Which one of the following compounds reacts with glycerol at 250°C to form allyl alcohol—

1. Formic add
2. Oxalic acid
3. Both
4. None of these

Answer: 3. Both

Question 120. In Liebcrmann’s test for phenol, the change of colour follows the order—

1. Reddish-brown → green → deep blue
2. Green→reddish-brown→deep blue
3. Red→green →white
4. Red→blue green

Answer: 2. Green→reddish-brown→deep blue

Question 121. The violet coloured compound obtained when FeCl₃ reacts with phenol is-

1. [Fe(OC₆H₅)₆]³⁺
2. [Fe(OC₆H₅)₆]^3-
3. [Fe(OC₆H₅)₃]³⁺
4. [Fe(OC₆H₅)₃]^3-

Answer: 2. [Fe(OC₆H₅)₆]^3-

Question 122. Which one of the following compounds will be obtained by the reaction between acetylene and formaldehyde In the presence of copper acetylide—

1. Butyne- 1,4 diol
2. 2 -butyne
3. Ethylene- 1, 4 -diol
4. None of these

Answer: 1. Butyne- 1,4 diol

Question 123. The reducing agent used in the following reaction is—

1. LiAlH₄
2. NaBH₄
3. H₂/Pt
4. Both 1 and 3

Answer: 3. H₂/Pt

Question 124. Which of the following reaction produce 2-phenyl butane- 2-ol—

1. C₆H₅COCH₃ + C₂H₅MgBr
2. C₂H₅COCH₃ + C₆H₅MgBr
3. C₆H₅COC₂H₅ + CH₃MgBr
4. All of these reactions.

Answer: 4. All of these reactions

Question 125.

The compound Z is—

1. CH₂=CH₂
2. CH₃CH₂OH
3. CH₃CH₂OCH₂CH₃
4. CH₃CH₂SO₃H

Answer: 2. CH₃CH₂OH

Question 126. Glycerol reacts with PClg to form—

1. 1,2,3-trichloropropane
2. Glycerol monochlorohydrine
3. Glycerol dichlorohydrine
4. All of these compounds

Answer: 1. 1,2,3-trichloropropane

Question 127. The suitable reducing agent to be used in the following conversion is —

1. (1) LiAlH₄ , (2) H⁺
2. (1) NaBH₄, (2) H⁺
3. H₂/Pt, carbon
4. Both 1 and 3

Answer: 4. Both 1 and 3

Question 128. Which one of the following compounds produce only ether when treated with sodium tert-butoxide—

1. CH₃CH₂Br
2. CH₃X
3. 
4. (CH₃)₃C-X

Answer: 2. CH₃X

Question 129.

 The product X is—

1. CH₃CH₂OH
2. (CH₃)₂CHOH
3. CH₃CH₂CH₂OH
4. HOCH₂CH₂CH₂CH₂OH

Answer: 3. CH₃CH₂CH₂OH

Question 130. The compound obtained on heating a mixture of phenol, phthalic anhydride and cone. H₂SO₄ followed by the addition of the mixture to NaOH solution is—

1. Alizarin
2. Methyl orange
3. Fluorescein
4. Phenolphthalein

Answer: 4. Phenolphthalein

Question 131. Which one of the following compounds produces ethyl propanoate when heated with BF3 at 150°C under 500 atm pressure—

1. C₂H₅OH
2. CH₃OCH₃
3. C₂H₅OC₂H₅
4. CH₃OC₂H₅

Answer: 3. C₂H5OC₂H₅

Question 132. Which one of the following substances is called denatured alcohol—

1. Ethanol + methanol
2. Rectified spirit + methanol + naptha
3. Unrefined ethanol
4. Rectified spirit

Answer: 2. Rectified spirit + methanol + naptha

Question 133. The catalysts involved in the conversion of ethanol from starch are—

1. Maltase, diastase
2. Diastase, maltase, zymase
3. Invertase, zymase
4. Invertase, diastase, maltase

Answer: 2. Diastase, maltase, zymase

Question 134. 1,4 -hexadiene-3 -ol reacts with H₂SO₄ to form—

1. 3, 5-hexadiene-2-ol
2. 2 -4 -hexadiene-1-ol
3. A mixture of 1 and 2
4. No reaction occurs

Answer: 3. A mixture of 1 and 2

Question 135. CH₃—O—CH(CH₃)₂ + HI→Products. The major product (s) obtained is this reaction is (are) —

Answer: 1. CH₃I + (CH₃)₂CHOH

Question 136. P-CH₃OC₆H₄CH₂OH(A), p-ClC₆H₄CH₂OH(B) and p-O₂NCgH₄CH₂OH(C). The order of reactivity of these benzyl alcohols towards HBr is—

1. A> B> C
2. A> C> B
3. C> A> B
4. C> B> A

Answer:  1. A> B> C

Question 137. The reagent used to detect the presence of peroxide in a sample of old ether is—

1. Sodium
2. Dil.HCl
3. Aq.ferrous ammonium sulphate followed by ammonium thiocyanate
4. Dil. NaOH

Answer: 3. Aq.ferrous ammonium sulphate followed by ammonium thiocyanate

Question 138.

The final product is—

Answer: 3

Question 139. The boiling points of n -pentanol (1), n-pentane (2), 3 -pentanol (3) and 2, 2 -dimethyl-1-propanol (4) decrease in the order—

1. 1, 3, 4, 2
2. 2, 4, 3, 1
3. 3, 1, 4, 2
4. None of these

Answer: 1. 1, 3, 4, 2

Question 140. Which one of the following compounds does not react j with Br₂/H₂O to form 2, 4, 6-tribromophenol—

Answer: 4

Question 141. The two gases obtained when benzenesulphonic acid and p-nitrophenol are subjected to react with NaHCO₃ are respectively—

1. SO₂,NO₂
2. NO₂,NO₂
3. SO₂,CO₂
4. CO₂,CO₂

Answer: 4. CO₂, CO₂

Question 142. Which of the following compounds cannot be converted into methyl ether or methyl ester by treating with diazomethane—

1. C₂H₅OH
2. C₆H₅OH
3. CH₃COOH
4. C₆H₅COOH

Answer: 1. C₂H₅OH

Question 143. Which of the following reactions can be used to prepare acetophenone—

1. Reimer-Tiemann reaction
2. Wurtz-Fittig reaction
3. Friedel-Crafts reaction
4. Cannizzaro reaction

Answer: 3. Friedel-Crafts reaction

Question 144. The rate of reaction of HBr is highest with—

1. 3° alcohol
2. 2° alcohol
3. Phenol
4. Ethanol

Answer: 1. 3° alcohol

Question 145. Which one of the following is known as wood spirit—

1. Methanol
2. Ethanol
3. Acetone
4. Benzene

Answer: 1. Methanol

Question 146. Picric acid is—

1. 2,4,6 -tribromophenol
2. 2,4,6 -trinitrotoluene
3. 2,4,6 – trinitrophenol
4. None of these

Answer: 3. 2,4,6 – trinitrophenol

Question 147. Mark the correct order of acidity of the following compounds:

1. 3 > 4 > 2 > 1
2. 4 > 3 > 1 > 2
3. 3 > 2 > 1 > 4
4. 2 > 3 > 4 > 1

Answer: 1. 3 > 4 > 2 > 1

Question 148. Conversion of m-nitrophenol to resorcinol involves —

1. Hydrolysis, diazotisation and reduction
2. Diazotisation, reduction and hydrolysis
3. Hydrolysis, reduction and diazotisation
4. Reduction, diazotisation and hydrolysis

Answer: 4. Reduction, diazotisation and hydrolysis

Question 149. The acidic hydrolysis of ether (X) is fastest when—

1. One phenyl group is replaced by a methyl group
2. One phenyl group is replaced by a p-methoxyphenyl group
3. Two phenyl groups are replaced by two p-methoxyphenyl group
4. No structural change is made to X

Answer: 3. Two phenyl groups are replaced by two p-methoxyphenyl group

Question 150. Among the following compounds, the most acidic is—

1. P-nitrophenol
2. P-hydroxybenzotc acid
3. O-hydroxybenzoic acid
4. P-toluic acid

Answer: 3. O-hydroxybenzoic acid

Question 151. The IUPAC name of the compound 

1. 4-bromo-3-cyanophenol
2. 2-bromo-5-hydroxybenzonitrile
3. 2-bromo-4-hydroxybenzonitrile
4. 6-bromo-3-hydroxybenzonitrile

Answer: 2. 2-bromo-5-hydroxybenzonitrile

Question 152. Chlorobenzene\(\stackrel{\boldsymbol{X}}{\longrightarrow}\) Phenol \(\stackrel{\boldsymbol{Y}}{\longrightarrow}\) , Salicyiaidehyde X and Y reactions are respectively—

1. Fries rearrangement and Reimer-Tiemann
2. Cumene and Reimer-Tiemann
3. Dow and Reimer-Tiemann
4. Dow and Friedel-Crafts

Answer: 3. Dow and Reimer-Tiemann

Question 153. To which one of the following does phenol react to form bakelite—

1. CH₃COCH₃
2. HCHO
3. (CH₂OH)₂
4. CH₃CHO

Answer: 2. HCHO

Question 154. Which one of the following alcohols reacts fastest with cone. HCl and anhy. ZnCl₂ —

1. 2-butanol
2. 2-methylpropan-2 ol
3. 2-methylpropan-l-ol
4. 1-butanol

Answer: 2. 2-methylpropan-2 ol

Question 155. Among the following, which one will produce methyl alcohol on treatment with hot cone. HI—

1. (CH₃)₂CHCH₂OCH₃
2. CH₃(CH₂)₃OCH₃
3. 
4. (CH₃)₃COCH₃

Answer: 4. (CH₃)₃COCH₃

Question 156. Consider the following reaction:

The product Z is—

1. Toluene
2. Benzaldehyde
3. Benzoic acid
4. Benzene

Answer: 3. Benzoic acid

Question 157. Which one of the following reactions involves C —C bond formation—

1. Reimer-Tiemann reaction
2. Hydroboration-oxidation
3. Cannizzaro reaction
4. PCC-reaction of 1° alcohol

Answer: 1. Reimer-Tiemann reaction

Question 158. Alkyl aryl ethers, when heated with HI, give—

1. Alcohol and phenol
2. Alcohol and aryl halide
3. Phenol and alkyl halide
4. Alkyl halide and aryl halide

Answer: 3. Phenol and alkyl halide

Question 159.

Answer: 2

Question 160. Identify the correct product (Y) formed during the following reaction—

Answer: 4

Question 161. Which of the following compounds form a bubble when 5% NaHCO₃ solution is added to them—

Answer: 2,4

Question 162. Which of the following will give the iodoform test—

1. 2-hydroxypropane
2. 2-iodobutane
3. 1-phenylethanol
4. 1-phenylbromoethane

Answer: 1,2,3,4

Question 163. Find reactant P and RX for Williamson synthesis—

1. PhBr and NaOCH₂—O—CH₃
2. CH₃OH and PhCH₂Br
3. PhCH₂OH and CH₃Br
4. PhCH₂Br and CH₃Br

Answer: 2 and 3

Question 164. Which of the following reagents/conditions can convert 2-propanol to acetone—

1. LiAlH₄
2. CU/575K
3. K₂Cr₂O₇/H⁺
4. H₂/Pd

Answer: 2 and 3

Question 165. Which of the following compounds would give a ketone on oxidation—

1. Iso-butyl alcohol
2. Iso-propyl alcohol
3. Iso-pentyl alcohol
4. Sec-butyl alcohol

Answer: 2 and 4

Question 166. Which of the following are less acidic than phenol—

Answer: 3 and 4

Question 167. Choose the correct order of acidity—

Answer: 3 and 2

Question 168. Which of the following compounds respond to the Lucas test readily—

1. CH₃CH₂OH
2. (CH₃)₃COH
3. (CH₃)₂CHOH
4. PhCH₂OH

Answer: 2 and 4

Question 169. Which of the reagents required for preparing bakelite (a thermosetting plastic) —

1. Salicylic acid
2. Chloroform
3. Phenol
4. Formaldehyde

Answer: 3 and 4

Question 170. The acidity of phenol increases when the ring contains the groups—

1. -NO₂
2. -CH₃
3. -CN
4. -OR

Answer: 1 and 3

Question 171. Which of the following alcohols undergoes acid catalyzed dehydration involving rearrangement—

Answer: 2 and 4

Question 172. Which of the following alcohols produce alkene when passed through Cu heated at 300°C—

Answer: 1 and 4

Question 173. Which of the following contains ethanol —

1. Rectified spirit
2. Power alcohol
3. Methylated spirit
4. Wood spirit

Answer: 1 and 2

Question 174. Which of the following reactions addition of water occurs in an anti-Markownikoff fashion—

⇒ \(\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_3 \frac{\text { (1) } \mathrm{HI}}{\text { (2) dil. } \mathrm{KOH}}\)

⇒ \(\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{CH}_2 \frac{\text { (i) } \mathrm{HBr}}{\text { (ii) dil. } \mathrm{KOH}}\)

⇒ \(\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{CH}_2 \quad \frac{\text { (i) } \mathrm{B}_2 \mathrm{H}_6}{\text { (ii) } \mathrm{H}_2 \mathrm{O}_2 / \mathrm{OH}^{\ominus}}\)

Answer: 3 and 4

 

Class 12 Chemistry Chapter 11 Alcohols Phenols And Ethers Match The Following

Question 1.

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

Question 2.

Answer: A-4, B-5, C-2, D-1.

Question 3.

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

Question 4.

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

Class 12 Chemistry Chapter 11 Alcohols Phenols And Ethers Assertion-Reason Type

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 choices given below.

1. Both A and R are true, R is the correct explanation of A.
2. Both A and R are true but R is not the correct explanation of A.
3. A is true but R is false.
4. A is false but R is true.
5. Both A and R are false.

Question 1. Assertion(A): The additional reaction of water to but-1-ene in acidic medium yields butan-1-ol.

Reason (R): The addition of water in acidic medium proceeds through the formation of primary carbocation.

Answer: 5. Both A and R are false.

Hint: Correct assertion: But-l-ene undergoes hydration in acid medium to yield butan-2-ol. Correct reason: But-l-ene undergoes hydration in acid medium via the formation of more stable 2° carbocation.

Question 2. Assertion (A): p-nitrophenol is more acidic than phenol.

Reason (R): The Nitro group helps in the stabilisation of the phenoxide ion by dispersal of negative charge due to resonance.

Answer: 1. Both A and R are true, R is the correct explanation of A.

Question 3. Assertion (A): IUPAC name of the compound

Reason (R): In IUPAC nomenclature, ether is regarded as a hydrocarbon derivative in which a hydrogen atom is replaced by —OR or — OAr group [where R = alkyl group and Ar = aryl group].

Answer: 4. A is false but R is true.

Hint: IUPAC name of the given compound is l-(2-propoxy) propane.

Question 3. Assertion (A): The bond angle in ethers is slightly less than the tetrahedral angle.

Reason (R): There is a repulsion between the two bulky (-R) groups.

Answer: 4. A is false but R is true.

Hint: Bond angle in ethers is slightly more than the tetrahedral angle.

Question 4. Assertion (A): Boiling points of alcohols and ethers are high.

Reason (R): They can form intermolecular hydrogen¬ bonding.

Answer: 5. Both A and R are false.

Hint: Correct assertion: The boiling points of alcohols are higher than ethers of comparable molecular masses. Correct reason: Alcohols form intermolecular H-bonding but ethers do not.

Question 5. Assertion (A): Like bromination of benzene, bromination of phenol is also carried out in the presence of Lewis acid.

Reason (R): Lewis acid polarises bromine molecules.

Answer: 4. A is false but R is true.

Hint: Bromination of benzene is carried out in the presence of a Lewis acid catalyst, but bromination of phenol does not require any Lewis acid catalyst

Question 6. Assertion (A): o-nitrophenol is less soluble in water than the m -and p -isomers.

Reason (R): m-and p-nitrophenols exist as associated molecules.

Answer: 2. Both A and R are true but R is not the correct explanation of A.

Hint: —OH and —NO₂ groups of o-nitrophenol are involved in intramolecular H-bonding and hence, onitrophenol does not form H-bonds with H₂O but m -and p -nitrophenols do.

Question 7. Assertion (A): Ethanol is a weaker acid than phenol.

Reason (R): Sodium ethoxide may be prepared by the reaction of ethanol with aqueous NaOH.

Answer: 3. A is true but R is false.

Hint: Correct reason: In phenoxide ion, -ve charge is stabilised by resonance but in ethoxide ion -ve charge is not stabilised by resonance.

Question 8. Assertion (A): Phenol forms 2,4,6-tribromophenol on treatment with Br, in carbon disulphide at 273K.

Reason (R): Bromine polarises in carbon disulphide.

Answer: 5. Both A and R are false.

Hint: Correct reason: Phenol forms 2, 4, 6- tribromophenol on treatment with Br2 -water. Correct reason: In H₂O, phenol undergoes partial ionisation to form phenoxide ion, which has highly activated ring system for electrophilic substitution.

Question 9. Assertion (A): Phenols give o- and p-nitrophenol on nitration with cone. HNO₃ and H₂SO₄ mixture.

Reason (R): —OH group in phenol is o-, p-directing.

Answer: 4. A is false but R is true.

Hint: Correct reason: Phenol gives a mixture of o-and p-nitrophenols on nitration with dil.HNO₃.

Class 12 Chemistry Chapter 11 Alcohols Phenols And Ethers Fill In The Blanks

Question 1. A tertiary alcohol is a ____ acid as compared to its isomeric primary and secondary alcohol.
Answer: weak

Question 2. Due to ____ lower alcohols are highly soluble in water.
Answer: H-bond

Question 3. Absolute alcohol can be prepared by ____ distillation of rectified spirit.
Answer: Azeotropic

Question 4. Phenol forms coloured complex ion with neutral ____
Answer: FeCl₃ solution

Question 5. A 3° alcohol undergoes’ ____ when its vapours are passed through heated copper.
Answer: Dehydration

Question 6. ____ can be prepared from phenol by ReimerTiemann reaction.
Answer: Salicylaldehyde

Question 7. Alcohols are ____ acids than phenols but are ____ nucleophiles.
Answer: stronger, weaker

Question 8. The reaction between phenol and ____ presence of sodium hydroxide is called SchottenBaumann reaction.
Answer: Benzoyl chloride

Question 9. The suitable oxidising agent used to prepare aldehydes from l ° alcohols is ____
Answer: Pyridinium chlorochromate (PCC)

Question 10. 3,3-Dimethylbutan-2-ol undergoes dehydration in the presence of concentrated H₂SO₄ to form ____
Answer: 2, 3-dimethyl but-2-ene

Question 11. Alkenes react successively with ____ and ____ to undergo anti-Markownikoff hydration.
Answer: 1. B₂H₆ 2. H₂O₂/OH^–

Question 12. Grignard reagents react with ketones to form ____ alcohols.
Answer: Tertiary alcohol

Question 13. Phenol on bromination in CS₂ produces mainly ____
Answer: P-bromophenol

Question 14. The reaction of phenol with benzenediazonium chloride in weakly alkaline medium is called ____ reaction.
Answer: Coupling

Question 15. Among the nitrophenols,____ is least soluble in water.
Answer: O-nitrophenol

Question 16. The boiling points of ethers are much ____ of alcohols having comparable molecular masses.
Answer: lower

Question 17. Dehydration of alcohols is not a suitable method for the preparation of ____ ethers.
Answer: Unsymmetrical

Question 18. Peroxides can be removed from ether by shaking the latter with a solution of ____.
Answer: FeSO₄

Question 19. Ethers dissolve in cone. H₂SO₄ by forming ____ salts.
Answer: Oxonium

Question 20. The melting point of diethyl ether is very ____
Answer: low

Question 21. tert-Butyl methyl ether reacts with HI to form ____ and ____
Answer: Tert-butyl iodide, methyl alcohol

Question 22. Tetrahydrofuran reacts with excess of hydrobromic acid to form _____
Answer: 1, 4-dibromo butane

Class 12 Chemistry Chapter 11 Alcohols Phenols And Ethers Warm Up Exercise

Question 1. Designate the following alcohols as 1°, 2° and 3°:

Answer:

1. 1°
2. 2°
3. 3°
4. 1 0
5. 2°
6. 2°
7. 3°

Question 2. Identify the allylic and benzylic alcohols of question no.1
Answer: Allylic alcohol: (4) and (7); Benzylic alcohol: (5)

Question 3. Write the structures of all the isomeric alcohols and ethers having molecular formula C₃H₈O and C₄H₁₀O.
Answer:

Alcohols with molecular formula C₃H₈O CH₃CH₂CH₂OH and CH₃CHOHCH₃. Ether with molecular formula C₃H₈O: CH₃CH₂OCH₃. Alcohols with molecular formula C₄H₁₀O: (CH₃)₂CHCH₂OH,CH₃CH₂CH₂CH₂OH (CH₃)₃COH and CH₃CH₂CHOHCH₃ Ethers with molecular formula C₄H₁₀O: CH₃CH₂OCH₂CH₃, CH₃CH₂CH₂OCH₃ and (CH₃)₂CHOCH₃

Question 4. Give an example of a phenolic ether having two phenyl groups.
Answer:

Question 5. Write the IUPAC names of the following compounds:

1. CH₃CHClCH₂CH₂OH
2. (CH₃)₂CHOCH₂CH₃

Answer:

1. 3-chlorobutanol-l-ol;
2. 2-ethoxy propane;
3. 2-bromo-6-methyl phenol;
4. l-ethoxy-3-nitro cyclohexane;
5. 2,4-dinitrophenol;
6. 2-methyl cyclo pentanol;
7. 2,4-dihydroxy phenol

Question 6. Write the structures of the following compounds:

1. 1 -phenoxy heptane,
2. 2-ethoxy-1, 1 dimethylcylohexane.
3. Ethoxy-benzene,
4. Isobutyl alcohol,
5. Glycerol
6. Hex 1 -en-3-ol
7. 2-bromo-3 methylbut-2-en-1-o

Answer:

Question 7. Explain why the C—O—C bond angle in di-tert-butyl ether (130°) is much greater than the C—O—C bond angle in dimethyl ether (111.7°).
Answer:

The tert-butyl group is much larger than the methyl group. Due to the larger size of the tert-butyl group, two tert-butyl groups in di-tert-butyl ether are sterically hindered for which the C— O—C bond angle is greater than that of dimethyl ether.

Question 8. How will you prepare the following alcohols by acid-catalyzed hydration of appropriate alkenes?

Answer:

Question 9. Arrange ethylene, propene and 2-methylpropene in the order of decreasing reactivity towards acid-catalyzed hydration and explain the reason for such an arrangement.
Answer:

In acid-catalyzed hydrolysis, the C =C double bond gets protonated first to form a carbocation. The higher the stability of carbocation, the greater is the rate of the reaction. The stability order of carbocation, generated from ethylene, propane, and 2-methylpropene is CH3CH2 (1° carbocation) < CH3CHCH3 (2° carbocation) < (CH3)3C (3° carbocation). Thus, the order of decreasing reactivity towards acid-catalyzed hydration is 2- methylpropene > propene > ethylene.

Question 10. Out of LiAlH₄ and NaBH₄ which would you use to carry out the following transformations and why?
Answer:

1. LiAlH₄ (NaBH₄ cannot reduce acid group)
2. NaBH₄ ( LiAlH₄ can reduce both the groups)
3. NaBH₄ (LiAlH₄ can reduce both the groups)

Question 11. A mixture of 1-propanol and 2-propanol is obtained when an ester is reduced by LiAlH₄. Identify the ester.
Answer:

Question 15.

Write the structure of A and B (two optical isomers).
Answer:

Question 12.

Identify the compound X.

Answer:

Question 13. Convert by using the Grignard reaction:

1. C6H5COOC2H5→(C6H5)3COH
2. CH3CH2Br→CH3CH2CH2CH2OH
3. HC02C2H5→CH3CH0HCH3
4. 

Answer:

Question 14. Prepare A-methylphenol from sodium toluenesulphanate.
Answer:

Question 15.

Identify the compound X

Answer:

Question 16. Which member of each of the following pairs is more soluble in water and why?

1. CH₂Cl₂ and CH₃OH
2. CH₃CH₂CH₂SH and CH₃CH₂CH₂OH
3. CH₃COCH₃ and (CH₃)₂C=CH₂
4. CH₃CH₂CH₂CH₂OH and (CH₃)₃COH

Answer:

1. CH₃OH, as it forms H-bonds with water molecules.
2. CH₃CH₂CH₂OH, as it forms H-bonds with water molecules.
3. CH₃COCH₃, as it forms H-bonds with water molecules.
4. (CH₃)₃COH, as with increasing the area of the side chain of an aliphatic group, solubility in water decreases.

Question 17. Arrange in the order of increasing boiling point and explain:

CH₃CH₂OH, CH₃OCH₃, CH₃CH₂CH₃, CH₃COOH

Answer:

In CH₃CH₂CH₃ molecules, there are weak van der Waals forces of attraction present whereas in CH₃OCH₃, along with weak van der Waals forces of attraction weak dipole-dipole attraction forces are also present. Thus CH₃OCH₃ has a higher boiling point than CH₃CH₂CH₃. Again, CH₂CH₂OH is a polar compound that forms intermolecular H-bonds with other CH₃CH₂OH molecules for which CH3CH₂OH has a higher boiling point than CH₃OCH₃.

On the other hand, CH₃COOH also forms intermolecular H-bonds with other CH₃COOH molecules. But, in the CH₃COOH molecule, the O—H bond is more polar in CH₃COOH than in CH₃CH₂OH. Thus H-bonds formed by CH₃COOH are much stronger than those formed by CH₃CH₂OH molecule. Hence boiling point of CH₃COOH is higher than CH₃CH₂OH. Thus, the order of increasing boiling point is— CH₃CH₂CH₃ < CH₃OCH₃ < CH₃CH₂OH < CH₃COOH

Question 18. Alcohols cannot be dried with any. CaCl₂ or MgCl₂ — why?
Answer:

Alcohols cannot be dried with anhydrous CaCl₂ or MgCl₂ because alcohols form additional compounds. For example, C₂H5OH forms CaCl₂.3C₂H5OH and MgCl₂.6C₂H5OH with CaCl₂ and MgCl₂, respectively.

Question 19. Which alcohol having molecular formula C₄H₁₀O cannot be obtained by reducing an aldehyde or a ketone?
Answer:

Since the given alcohol cannot be prepared by reducing an aldehyde or ketone, it is a tertiary alcohol. The tertiary alcohol having the molecular formula C4H₁₀O is tertbutyl alcohol (Me₃COH).

Question 20. Give a reaction, where alcohol acts as a nucleophile and protonated alcohol acts as an electrophile.
Answer:

Here, in the first step, alcohol acts as a nucleophile whereas in the second step, protonated alcohol acts as an electrophile.

Question 21. Out of propane- 2 -ol and 2 -methyldopa-2 -ol, which will undergo dehydration in the presence of one.H₂SO₄ at a faster rate and why?
Answer:

2-methylpropan-2-ol produces more stable 3° carbocation. So, the rate of dehydration of 2-methylpropan- 2-ol is faster than that ofpropan-2-ol.

Question 22. Arrange the following in the order of increasing acidic strength and explain: CH₃CH₂OH, CF₃CH₂OH, CCI₃CH₂OH.
Answer:

With increasing electronegativity of the halogen atom (-I effect) attached with β-carbon, the acidity of alcohol increases. Thus the order of increasing acidic strength of given alcohols will be— CH₃CH₂OH < CCI₃CH₂OH < CF₃CH₂OH

Question 23. How can you distinguish between phenol and 2,4,6- trinitrophenol by a solubility test?
Answer:

2,4,6-trinitrophenol is soluble in Na₂CO₃ solution whereas phenol is insoluble as the acidity of 2,4,6-trinitrophenol is much higher than phenol.

Question 24. A and B are two isomeric compounds having molecular formula C₇H₇OH. A produces a violet coloration with a neutral FeCl₃ solution but B does not. Identify A and B.
Answer:

Question 25. How can o-nitrophenol be separated from p-nitrophenol in their mixture? Give reason.
Answer:

In o-nitrophenol intramolecular H-bonding is present. But in the case of p-nitrophenol, there is intermolecular H-bonding. Hence p-nitrophenol molecules are strongly associated with each other. Thus o-nitrophenol is steam volatile and the mixture is separated by steam distillation.

Question 26. Which member of the following pairs is more acidic and why?

1. Phenol and cyclohexanol
2. 3,5-dimethyl-4 -nitrophenol and 4- nitrophenol

Answer:

1. The conjugate base of phenol is much more stable than the conjugate base of cyclohexanol as the phenoxide ion is resonantly stabilized. So, phenol is a much stronger acid than cyclohexanol.
2. 4-nitrophenol is more acidic than 3,5-dimethyl-4- nitrophenol as it has two electron-donating methyl groups in the ring which suppress the electron-withdrawing effect of the —NO2 group.

Question 27. Unlike alcohols, phenols cannot be used to prepare esters from carboxylic acids using cones. H₂SO₄ as a catalyst—why?
Answer:

Phenol has less nucleophilic character than alcohol because the unshared electron pair is delocalized into the benzene ring. So, in acidic medium phenol cannot react with carboxylic acid to form ester.

Question 28. What is picric acid? Explain why it is named as an acid. How can picric acid be prepared from phenol?
Answer:

1st Part: 2,4,6-trinitrophenol is known as picric acid. 2nd Part: In the presence of three electron withdrawing — N02 groups, the negative charge on the O-atom of the conjugate base of 2,4,6-trinitrophenol is highly stabilized. Hence, the removal of H+ ions is feasible. Thus it is named an acid

Question 29. Which one of the following reagents can be used to distinguish between phenol and cyclohexanol acid why? Na₂CO₃, CH₃COCI, NaOH
Answer:

Phenol is soluble in NaOH solution but cyclohexanol is insoluble.

Question 30. Mention the electrophile involved in the Reimer-Tiemann reaction. Explain why it acts as an electrophile.
Answer:

In the Reimer-Tiemann reaction, dichlorocarbene (SCCl₂), acts as an electrophile as the C-atom in: CCl₂ has an incomplete octet.

Question 31. Predict the structure of the compound C₆H₂Br₄O obtained when phenol is treated with excess of bromine water.
Answer:

Question 32. What happens when phenol is reduced with Na/liq? NH₃ and the resulting compound is treated with dilute acid?
Answer:

Question 33. How will you distinguish between the two members of each pair?

Ethanol and methanol

Phenol and 2,4-dinitrophenol.

Answer:

Ethanol forms yellow-coloured iodoform but methanol does not.

2,4-dinitrophenol reacts with NaHCO₃ with which phenol does not react.

Question 34. Predict whether each of the following compounds will respond to the iodoform test or not. Explain.

1. CH₃CH₂CHOHCH₂CH₃
2. C₆H₅CHOHCH₂I
3. CH₃CH₂CHOHCHOHCH₃
4. C6H₅CHOHCH₂CH₃

Answer:

1. Since the keto-methyl group is absent, it will not respond to the iodoform test.
2. Since the keto-methyl group is present, it will respond to the iodoform test.
3. Since the keto-methyl group is present, it will respond to the iodoform test.
4. Since the keto-methyl group is absent, it will not respond to the iodoform test.

Question 35. Alcohol reacts with I₂/NaOH to form iodoform and sodium propanoate. Identify the alcohol.
Answer: CH₃CH₂CHOHCH₃

Question 36. Cany out the following transformations: 2-propanol→ethanol

Answer:

Question 37. Give two possible combinations of reactants for preparing isopropyl methyl ether by Williamson synthesis. Which combination of reactants is preferable and why?
Answer:

Both the reactions follow the S_N2 mechanism. In process: 1, smaller nucleophile CH₃O^– attacks comparatively larger 2° alkyl halide, CH₃CHBrCH₃. Here, the attack of the nucleophile to the alkyl halide becomes somewhat difficult because of steric hindrance by the presence of two —CH₃ groups. But in process: 2, alkyl halide has no such steric hindrance. Thus, process: 2 is preferable.

Question 38. Is it possible to prepare anisole (C₆H₅OCH₃) using bromobenzene and sodium methoxide? Explain.
Answer:

Aryl halide can not undergo nucleophilic substitution reaction as a backside attack is not possible. Thus, from this given method anisole preparation is not possible.

Question 39. Which of the following ethers can/cannot be prepared by Williamson synthesis and why?

1. 
2. C₆H₅-O-CH₂CH₃
3. Me₃CCH₂-O-CH₂CMe₃
4. 

Answer:

1. Can be prepared as CH₃— X will be used as reactant and cyclopentoxide ion will act as nucleophile.
2. Can be prepared as C₆H₅O- will be used as a nucleophile and CH₃CH₂—X will be used as a reactant.
3. Cannot be prepared as neopentyl halide can not undergo nucleophilic substitution reaction.
4. Can be prepared as C₆H₆O- and can act as a nucleophile to substitute halogen from 2,4-dinitrohalobenzene because in the presence of two electrons withdrawing —NO₂ groups attached to a benzene ring.

Question 40. How is (S)-CH₃CHD—O—CH₂CH₃ prepared from CH₃CH₂OH?
Answer:

Question 41. Di-tert-butyl ether cannot be prepared from tert-butyl alcohol by acid-catalyzed dehydration method—why?
Answer:

Cannot be prepared as tertiary alkyl halides are completely unreactive towards S_N2 reaction.

Question 42. How can 1,2-dimethoxyethane (CH₃OCH₂CH₂OCH₃) be prepared by Williamson synthesis?
Answer:

Question 43. Suggest a mechanism that explains the formation of tetrahydrofuran (THF) from the reaction of 4-chloro-1-butanol and aqueous sodium hydroxide.
Answer:

Question 44. trans-2-chlorocyclohexanol reacts readily with NaOH to yield cyclohexene oxide while the cis-isomer does not undergo the reaction. Explain.
Answer:

Question 45. Diethyl ether is taken separately in two test tubes. In one test tube, concentrated H2S04 is added and in the other, concentrated aqueous solution of NaOH is added. Both of these mixtures are shaken well. What will be observed in both cases? Explain these observations with proper reasons.
Answer:

Diethyl ether (C₂H₅OC₂H₂) dissolves in cone. H₂SO₄ but remains insoluble in cone. NaOH solution forms two layers, ether being the upper layer. Due to the presence of lone pairs of electrons on oxygen atoms in diethyl ether molecules, it can act as an electron pair donor, i.e., it can behave as a Lewis base.

So, ether does not react with alkali and remains insoluble in it. On the other hand, ether dissolves in acid because it reacts with H2S04 to form an oxonium salt which is soluble in acid. The lone pair of electrons on the oxygen atom is donated to the proton (H*) of the acid to form a coordinate covalent bond and thus the salt is produced.

Question 46. Identify A and R and explain their formations.

Answer:

Question 47. Why the cleavage of ethers cannot be affected by using HCl?
Answer:

Since Cl^– is a weak nucleophile, HCl can not cleave the ether linkage.

Question 48. Cleavage of ethers is possible by HI but not by NaOH — why?
Answer:

Since alkoxide is a strong base, it is a poor leaving group. Thus it is not possible to substitute the alkoxide group with a nucleophile like HO^–. Thus, cleavage of ethers is not possible by NaOH.

On the other hand, HI protonates ether to convert the RO— group to a good leaving group — RO⁺H. Since ROH is neutral and a good leaving group, I^– can substitute to undergo the reaction. Thus, HI can cleave ether linkages.

Question 49. An ether reacts with an excess of HBr to form 1,4- dibromobutane. Identify the ether.
Answer:

Question 50. Identify the products C and D in the following reaction and designate the optically active compound as R or S:

Answer:

Question 51. Identify the product(s) in each of the following reactions:

Answer:

Question 52. C₆H₅MgCl dissolves in THF but not in benzene—why?
Answer:

Lone pairs of oxygen atoms of ether (THF) can coordinate with Mg of C₆H₅MgCl to form chelate compounds making C₆H₅MgCl soluble. Since benzene is a non-polar solvent, it can not form that kind of linkage with C₆H₅MgCl making C₆H₅MgCl insoluble in benzene.