🧪
Chapter 7 · Class 12 Chemistry
Alcohols, Phenols and Ethers
1 exercises8 questions solved
Exercise 7.1Preparation, Properties and Reactions
Q1
Classify the following as primary, secondary and tertiary alcohols: (i) (CH₃)₂CHOH (ii) (CH₃)₂C(OH)CH₂CH₃ (iii) CH₃CH(OH)CH₂CH₃ (iv) (C₆H₅)₃COH
Solution
Primary alcohol: –OH on carbon attached to only one other carbon (or none)
Secondary alcohol: –OH on carbon attached to two other carbons
Tertiary alcohol: –OH on carbon attached to three other carbons
(i) (CH₃)₂CHOH — propan-2-ol:
The C–OH is attached to 2 CH₃ groups → Secondary alcohol
(ii) (CH₃)₂C(OH)CH₂CH₃ — 2-methylbutan-2-ol:
The C–OH is attached to 2 CH₃ and 1 CH₂CH₃ groups → Tertiary alcohol
(iii) CH₃CH(OH)CH₂CH₃ — butan-2-ol:
The C–OH is attached to CH₃ and CH₂CH₃ → Secondary alcohol
(iv) (C₆H₅)₃COH — triphenylmethanol:
The C–OH is attached to 3 phenyl groups → Tertiary alcohol
Q2
Identify the compound that does not undergo oxidation with alkaline KMnO₄: (i) propan-2-ol (ii) 2-methylpropan-2-ol (iii) propan-1-ol
Solution
Oxidation with alkaline KMnO₄:
• Primary alcohols → oxidised to carboxylic acids
• Secondary alcohols → oxidised to ketones
• Tertiary alcohols → NOT oxidised (no α-hydrogen on the carbon bearing –OH from the other side)
(i) Propan-2-ol: secondary alcohol → oxidised to propan-2-one (acetone)
(ii) 2-methylpropan-2-ol: TERTIARY alcohol → does NOT undergo oxidation
(iii) Propan-1-ol: primary alcohol → oxidised to propanoic acid
Answer: 2-methylpropan-2-ol does not undergo oxidation with alkaline KMnO₄.
Q3
Write the structures of the products of the following reactions: (i) Propenol + HBr (ii) Propan-2-ol + HBr (iii) Cyclohexanol + Lucas reagent
Solution
(i) Propenol (prop-2-en-1-ol, allyl alcohol) + HBr:
CH₂=CHCH₂OH + HBr → In two possible ways:
• Addition to double bond: CH₂BrCH₂CH₂OH or CH₂(OH)CHBrCH₃
• Substitution of OH: CH₂=CHCH₂Br (allyl bromide)
Under SN2 conditions: CH₂=CH–CH₂OH + HBr → CH₂=CH–CH₂Br + H₂O (allyl bromide)
(ii) Propan-2-ol + HBr:
(CH₃)₂CHOH + HBr → (CH₃)₂CHBr + H₂O
(2-bromopropane)
(iii) Cyclohexanol + Lucas reagent (ZnCl₂/conc. HCl):
Cyclohexanol is a secondary alcohol. With Lucas reagent:
Cyclohexanol + HCl → Chlorocyclohexane + H₂O
The secondary alcohol gives a milky turbidity after some time (not immediately).
Q4
Write the mechanism of acid-catalysed dehydration of ethanol to yield ethene.
Solution
Acid-catalysed dehydration of ethanol: CH₃CH₂OH → H₂SO₄, 170°C → CH₂=CH₂ + H₂O
Step 1 — Protonation of alcohol:
CH₃CH₂OH + H⁺ → CH₃CH₂OH₂⁺ (oxonium ion, protonated alcohol)
Step 2 — Loss of water to form carbocation:
CH₃CH₂OH₂⁺ → CH₃CH₂⁺ (ethyl carbocation) + H₂O
Step 3 — Loss of proton (E1 elimination):
CH₃CH₂⁺ + :B⁻ → CH₂=CH₂ + BH (B⁻ is base, e.g., HSO₄⁻)
The overall mechanism is E1 (unimolecular elimination).
The rate-determining step is the formation of carbocation (Step 2).
Q5
Phenol is more acidic than ethanol. Why?
Solution
Phenol (C₆H₅OH) is more acidic than ethanol (C₂H₅OH).
Reason:
1. Resonance stabilisation of phenoxide ion:
When phenol loses a proton, the phenoxide ion (C₆H₅O⁻) formed is stabilised by resonance.
The negative charge on oxygen is delocalised into the benzene ring through 5 contributing resonance structures.
This stabilises the conjugate base significantly.
2. In contrast, the ethoxide ion (C₂H₅O⁻) has no resonance stabilisation — the negative charge remains localised on oxygen.
3. Greater stability of conjugate base → easier ionisation → higher acidity
Ka(phenol) ≈ 10⁻¹⁰ vs Ka(ethanol) ≈ 10⁻¹⁶
The resonance stabilisation of phenoxide ion makes phenol considerably more acidic than alcohols.
Q6
Explain why the C–O–H bond angle in alcohols is slightly less than tetrahedral angle.
Solution
In water and alcohols, the oxygen is sp³ hybridised.
In a perfect tetrahedral arrangement, all bond angles would be 109.5°.
In alcohols (C–O–H):
The oxygen has 2 bond pairs and 2 lone pairs.
Lone pair–lone pair repulsion > lone pair–bond pair repulsion > bond pair–bond pair repulsion
The two lone pairs on oxygen repel the bonding pairs more strongly than the bonding pairs repel each other. This compresses the C–O–H bond angle to slightly less than the ideal tetrahedral value.
Actual C–O–H angle in methanol ≈ 108.9° (< 109.5°)
This is the same reason the H–O–H angle in water is 104.5° (even smaller, due to two lone pairs on O).
Q7
Write IUPAC names for the following ethers: (i) ClCH₂OCH₂Cl (ii) (CH₃)₃C–O–C₂H₅ (iii) C₆H₅OCH₃ (iv) CH₃OCH₂CH₂OH
Solution
(i) ClCH₂OCH₂Cl:
(Chloromethoxy)chloromethane or bis(chloromethyl) ether
IUPAC: 1-chloro-2-(chloromethoxy)methane → dichloro(dimethyl) ether?
Simpler: ClCH₂–O–CH₂Cl = bis(chloromethyl) ether or 1-chloro-1-(chloromethoxy)methane
(ii) (CH₃)₃C–O–C₂H₅:
The larger group: ethyl, smaller: tert-butyl
As alkoxy compound: 2-ethoxy-2-methylpropane or 1,1-dimethylethyl ethyl ether
IUPAC: 2-ethoxy-2-methylpropane
(iii) C₆H₅OCH₃:
Methoxybenzene (anisole)
IUPAC: methoxybenzene
(iv) CH₃OCH₂CH₂OH:
2-methoxyethanol
IUPAC: 2-methoxyethanol
Q8
Write the names and structures of the products obtained by the action of HI on: (i) methoxybenzene (ii) 1-methoxypropane (iii) 4-methoxytoluene
Solution
(i) Methoxybenzene (anisole) + HI:
C₆H₅OCH₃ + HI → C₆H₅OH + CH₃I
(Phenol + iodomethane)
Aromatic C–O bond is stronger (resonance), so alkyl–O bond breaks preferentially.
(ii) 1-methoxypropane + HI:
CH₃OCH₂CH₂CH₃ + HI → CH₃I + CH₃CH₂CH₂OH
(Iodomethane + propan-1-ol)
With excess HI: CH₃CH₂CH₂OH + HI → CH₃CH₂CH₂I + H₂O
(iii) 4-methoxytoluene (p-methoxytoluene) + HI:
CH₃–C₆H₄–OCH₃ + HI → CH₃–C₆H₄–OH + CH₃I
(4-methylphenol/p-cresol + iodomethane)
More chapters
← All chapters: Class 12 Chemistry