Chapter 11: Alcohols, Phenols and Ethers Class 12 Chemistry NCERT Solutions

NCERT Solutions for Class 12 Chemistry Chapter 11

Alcohols, Phenols and Ethers – Intext Questions with Answers

Question 11.1: Classify the following as primary, secondary, and tertiary alcohols.

Answer:
Primary alcohols: (i), (ii), (iii)
Secondary alcohols: (iv), (v)
Tertiary alcohol: (vi)

Question 11.2: Identify allylic alcohols among the above examples.

Answer:
Allylic alcohols: (ii) and (iv)
Structures:
H₂C=CH–CH₂OH and CH₃–CH=CH–CH₂OH

Question 11.3: Name the following compounds according to the IUPAC system.

Answer:
(i) 2-Methylpropan-2-ol
(ii) Butan-2-ol

Question 11.4: Show how the following alcohols are prepared by the reaction of a suitable Grignard reagent with methanal.

Answer:
(i) Ethanol from methyl magnesium bromide and methanal.
(ii) Propan-2-ol from ethyl magnesium bromide and methanal.

Question 11.5: Write the structures of the products of the following reactions.

Answer:
Example reaction:
CH₃CH₂CH₂OH + PCC → CH₃CH₂CHO (Propanal)

Question 11.6: Give structures of the products when each of the following alcohols reacts with:

• (a) HCl-ZnCl₂
• (b) HBr
• (c) SOCl₂

Answer:
(i) Butan-1-ol:
– (a) 1-Chlorobutane
– (b) 1-Bromobutane
– (c) 1-Chlorobutane

(ii) 2-Methylbutan-2-ol:
– (a) 2-Chloro-2-methylbutane
– (b) 2-Bromo-2-methylbutane
– (c) 2-Chloro-2-methylbutane

Question 11.7: Predict the major product of acid-catalysed dehydration of:

• (i) 1-Methylcyclohexanol
• (ii) Butan-1-ol

Answer:
(i) 1-Methylcyclohexene
(ii) But-1-ene (major) and But-2-ene (minor)

Question 11.8: Ortho and para-nitrophenols are more acidic than phenol. Draw the resonance structures of the corresponding phenoxide ions.

Answer:
The resonance structures show delocalization of negative charge into the aromatic ring.
Electron-withdrawing nitro groups stabilize the phenoxide ion more effectively at ortho and para positions.

Question 11.9: Write the equations involved in the following reactions:

• (i) Reimer-Tiemann reaction
• (ii) Kolbe’s reaction

Answer:
(i) Phenol + CHCl₃ + NaOH → Salicylaldehyde
(ii) Phenol + NaOH + CO₂ → Salicylic acid

Question 11.10: Write the reactions for Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol.

Answer:
Ethanol is converted to sodium ethoxide, which reacts with 3-methylpentan-2-yl chloride:
CH₃CH₂O⁻ + (CH₃)₂CHCH₂CHClCH₃ → CH₃CH₂OCH(CH₃)CH₂CH(CH₃)₂

Question 11.11: Which of the following is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene and why?

Answer:
Appropriate set: Sodium methoxide and 4-nitrochlorobenzene.
Reason: The electron-withdrawing nitro group activates the aryl halide for nucleophilic substitution.

Question 11.12: Predict the products of the following reactions.

Answer:
Reactions involve oxidation, dehydration, and substitution of alcohols, phenols, and ethers.
Example:
CH₃CH₂OH + H₂SO₄ (heat) → CH₂=CH₂ (Ethene)

NCERT Solutions for Class 12 Chemistry Chapter 11

Alcohols, Phenols and Ethers – Exercise Questions with Answers

Question 11.1: Write IUPAC names of the following compounds:

Answer:
(i) 2,2,4-Trimethylpentan-3-ol
(ii) 5-Ethylheptane-2,4-diol
(iii) Butane-2,3-diol
(iv) Propane-1,2,3-triol
(v) 2-Methylphenol
(vi) 4-Methylphenol
(vii) 2,5-Dimethylphenol
(viii) 2,6-Dimethylphenol
(ix) 1-Methoxy-2-methylpropane
(x) Ethoxybenzene
(xi) 1-Phenoxyheptane
(xii) 2-Ethoxybutane

Question 11.2: Write structures of the compounds whose IUPAC names are as follows:

Answer:
(i) 2-Methylbutan-2-ol
(ii) 1-Phenylpropan-2-ol
(iii) 3,5-Dimethylhexane-1,3,5-triol
(iv) 2,3-Diethylphenol
(v) 1-Ethoxypropane
(vi) 2-Ethoxy-3-methylpentane
(vii) Cyclohexylmethanol
(viii) 3-Cyclohexylpentan-3-ol
(ix) Cyclopent-3-en-1-ol
(x) 4-Chloro-3-ethylbutan-1-ol

Question 11.3: (a) Draw structural formulas and write IUPAC names of all the isomeric alkanols with the molecular formula C₅H₁₂O. (b) Classify the isomers as primary, secondary, or tertiary alcohols.

Answer:
(a) Eight isomeric alkanols are possible.

(b) Classification:
Primary alcohols: (i), (ii), (iii), (iv)
Secondary alcohols: (v), (vi), (viii)
Tertiary alcohol: (vii)

Question 11.4: Explain why propanol has a higher boiling point than butane.

Answer:
Propanol molecules form strong intermolecular hydrogen bonds, requiring more energy to break. Butane, with only van der Waals forces, has a lower boiling point.

Question 11.5: Why are alcohols more soluble in water than hydrocarbons of comparable molecular mass?

Answer:
Alcohols form hydrogen bonds with water molecules, increasing solubility. Hydrocarbons lack hydrogen bonding and thus are less soluble.

Question 11.6: What is meant by hydroboration–oxidation reaction? Illustrate with an example.

Answer:
Hydroboration–oxidation involves the addition of diborane (B₂H₆) to an alkene to form trialkylborane, followed by oxidation with hydrogen peroxide to give alcohol.
Example:
CH₃CH=CH₂ + B₂H₆ → CH₃CH₂CH₂BH₂
CH₃CH₂CH₂BH₂ + H₂O₂/OH⁻ → CH₃CH₂CH₂OH

Question 11.7: Give structures and IUPAC names of monohydric phenols with molecular formula C₇H₈O.

Answer:
Isomers:
(i) o-Methylphenol (2-Methylphenol)
(ii) m-Methylphenol (3-Methylphenol)
(iii) p-Methylphenol (4-Methylphenol)

Question 11.8: Which isomer of nitrophenol is steam volatile and why?

Answer:
o-Nitrophenol is steam volatile because of intramolecular hydrogen bonding. In contrast, p-nitrophenol has intermolecular bonding, making it non-volatile.

Question 11.9: Give the reaction for the preparation of phenol from cumene.

Answer:
Step 1: Cumene is oxidized with oxygen to form cumene hydroperoxide.
Step 2: Cumene hydroperoxide on treatment with acid yields phenol and acetone.

Question 11.10: Write the chemical reaction for the preparation of phenol from chlorobenzene.

Answer:
C₆H₅Cl + NaOH (fused) → C₆H₅ONa + NaCl (at 623 K, 300 atm)
C₆H₅ONa + HCl → C₆H₅OH + NaCl

Question 11.11: Write the mechanism of hydration of ethene to yield ethanol.

Answer:
Step 1: Protonation of ethene to form a carbocation.
Step 2: Water attacks the carbocation to form ethanol after deprotonation.

Question 11.12: Write equations to prepare phenol from benzene, conc. H₂SO₄ and NaOH.

Answer:
Step 1: Benzene + H₂SO₄ → Benzene sulfonic acid
Step 2: Benzene sulfonic acid + NaOH → Phenol + Na₂SO₃

Question 11.13: Show how to synthesise:

Answer:
(i) 1-Phenylethanol: Styrene + BH₃·THF → Organoborane → 1-Phenylethanol (via oxidation)
(ii) Cyclohexylmethanol: Cyclohexylmethyl bromide + NaOH → Cyclohexylmethanol
(iii) Pentan-1-ol: 1-Bromopentane + NaOH → Pentan-1-ol

Question 11.14: Give two reactions showing acidic nature of phenol. Compare its acidity with ethanol.

Answer:
(i) Phenol + Na → Sodium phenoxide + H₂
(ii) Phenol + NaOH → Sodium phenoxide + H₂O
Phenol is more acidic than ethanol due to resonance stabilization of the phenoxide ion.

Question 11.15: Why is o-nitrophenol more acidic than o-methoxyphenol?

Answer:
The electron-withdrawing -NO₂ group stabilizes the phenoxide ion in o-nitrophenol, increasing acidity. The -OCH₃ group in o-methoxyphenol is electron-releasing, decreasing acidity.

Question 11.16: Explain how the –OH group attached to a carbon of the benzene ring activates it towards electrophilic substitution.

Answer:
The –OH group shows +R effect by donating electrons into the benzene ring, increasing electron density at ortho and para positions, making it more reactive to electrophilic substitution.

Question 11.17: Give equations of the following reactions:

Answer:
(i) CH₃CH₂CH₂OH + 2[O] → CH₃CH₂COOH + H₂O
(ii) Phenol + 3Br₂ (in CS₂) → 2,4,6-Tribromophenol + 3HBr
(iii) Phenol + HNO₃ (dilute) → o-Nitrophenol + p-Nitrophenol + H₂O
(iv) Phenol + CHCl₃ + 3NaOH → Salicylaldehyde + 3NaCl + 2H₂O

Question 11.18: Explain the following with examples:

Answer:
(i) Kolbe’s Reaction: Sodium phenoxide + CO₂ → Salicylic acid
(ii) Reimer–Tiemann Reaction: Phenol + CHCl₃ + NaOH → Salicylaldehyde
(iii) Williamson Synthesis: R–X + R’O⁻Na⁺ → R–O–R’ + NaX
(iv) Unsymmetrical Ether Example: CH₃–O–C₂H₅ (Ethyl methyl ether)

Question 11.19: Write the mechanism of acid dehydration of ethanol to yield ethene.

Answer:
Step 1: Protonation of ethanol to form ethyl oxonium ion.
Step 2: Loss of water to form a carbocation.
Step 3: Elimination of a proton to form ethene.

Question 11.20: How are the following conversions carried out?

Answer:
(i) Propane → Propene → Propan-2-ol (hydration)
(ii) Benzyl chloride + NaOH → Benzyl alcohol
(iii) Ethyl magnesium chloride + HCHO → Propan-1-ol
(iv) Methyl magnesium bromide + Acetone → 2-Methylpropan-2-ol

Question 11.21: Name the reagents used in the following reactions:

Answer:
(i) Acidified KMnO₄ or K₂Cr₂O₇
(ii) PCC (Pyridinium chlorochromate)
(iii) Br₂ in water
(iv) Acidified KMnO₄
(v) 85% H₂SO₄ at 440 K
(vi) NaBH₄ or LiAlH₄

Question 11.22: Give reason for the higher boiling point of ethanol compared to methoxymethane.

Answer:
Ethanol forms strong hydrogen bonds due to its –OH group, leading to a higher boiling point compared to methoxymethane, which lacks hydrogen bonding.

Question 11.23: Give IUPAC names of the following ethers:

Answer:
(i) 1-Ethoxy-2-methylpropane
(ii) 2-Chloro-1-methoxyethane
(iii) 1-Methoxy-4-nitrobenzene
(iv) 1-Methoxypropane
(v) 1-Ethoxy-4,4-dimethylcyclohexane
(vi) Ethoxybenzene

Question 11.24: Write reagents and equations for preparing the following ethers using Williamson synthesis:

Answer:
(i) 1-Propoxypropane: Propyl bromide + Sodium propoxide
(ii) 2-Methoxy-2-methylpropane: tert-Butyl bromide + Sodium methoxide
(iii) Ethoxybenzene: Bromobenzene + Sodium ethoxide
(iv) Methoxyethane: Methyl bromide + Sodium ethoxide

Question 11.25: Illustrate with examples the limitations of Williamson synthesis.

Answer:
Tertiary alkyl halides prefer elimination over substitution in Williamson synthesis. Example: tert-Butyl bromide + Sodium ethoxide → Isobutylene instead of ether.

Question 11.26: How is 1-propoxypropane synthesised from propan-1-ol?

Answer:
Propan-1-ol is converted to sodium propoxide using Na metal. It then reacts with propyl bromide to form 1-propoxypropane.

Question 11.27: Why is acid dehydration of secondary or tertiary alcohols not suitable for ether preparation?

Answer:
Secondary and tertiary alcohols form stable carbocations that prefer elimination reactions to form alkenes rather than ethers.

Question 11.28: Write reactions of HI with:

Answer:
(i) 1-Propoxypropane → 2C₃H₇I + H₂O
(ii) Methoxybenzene → C₆H₅OH + CH₃I
(iii) Benzyl ethyl ether → C₆H₅CH₂I + C₂H₅OH

Question 11.29: Explain the behavior of alkoxy group in aryl ethers.

Answer:
The –OR group shows +R effect, increasing electron density at ortho and para positions, thus activating the benzene ring towards electrophilic substitution.

Question 11.30: Write the mechanism for the reaction of HI with methoxymethane.

Answer:
Protonation of methoxymethane forms oxonium ion → C–O bond cleavage → Formation of CH₃I and CH₃OH via SN2 mechanism.

Question 11.31: Write equations for the following reactions:

Answer:
(i) Friedel-Crafts Alkylation: Anisole + R–Cl + AlCl₃ → ortho- and para-alkylated anisole
(ii) Nitration: Anisole + HNO₃/H₂SO₄ → o- and p-nitroanisole
(iii) Bromination: Anisole + Br₂ → o- and p-bromoanisole
(iv) Friedel-Crafts Acylation: Anisole + RCOCl + AlCl₃ → o- and p-acyl anisole

Question 11.32: Show how you would synthesise the following alcohols from alkenes:

Answer:
(i) 1-Methylcyclohexene → Alcohol via acid-catalyzed hydration
(ii) 4-Methylpent-3-ene → Alcohol via Markovnikov addition
(iii) Pent-2-ene → Pentan-2-ol via acid-catalyzed hydration
(iv) 2-Cyclohexylbut-2-ene → Corresponding alcohol via hydration

Question 11.33: Give a mechanism for the reaction of 3-methylbutan-2-ol with HBr.

Answer:
Step 1: Protonation of the alcohol → Formation of water leaving group.
Step 2: Carbocation formation and rearrangement.
Step 3: Attack of Br⁻ nucleophile to give final alkyl bromide.