CBSE Class 12 Chemistry 2023: question 23(a)
Set in the CBSE Class 12 board exam of 2023, paper 56/1/1, section B, question 23(a), for 2 marks. Try it first; the hints and our step-by-step solution are below.
Q23(a) · 2 marks · Very short answerAlcohols, Phenols and Ethers
Account for the following :
(i) Phenol is a stronger acid than an alcohol.
(ii) The boiling point of alcohols decreases with increase in branching of alkyl chain.
Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 56/1/1.
Hints
Open one hint at a time and try again before the next.
Hint 1
Compare the stability of the anions formed after each compound loses H⁺.
Hint 2
Can the negative charge on the oxygen be spread out in either anion?
Hint 3
For (ii), think about how molecular shape affects intermolecular forces.
Step-by-step solution
Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.
- 1 mark
(i) When phenol loses , it forms the phenoxide ion. The negative charge of phenoxide is delocalised over the benzene ring by resonance, so the ion is stabilised. In the alkoxide ion the charge stays on oxygen, and the +I effect of the alkyl group makes it even less stable. Hence phenol releases more easily and is the stronger acid.
- 1 mark
(ii) Branching makes the molecule more compact, so its surface area decreases. This weakens the van der Waals forces between molecules, so less energy is needed to boil the alcohol. Hence the boiling point falls as branching increases.
Answer: (i) Phenoxide ion is resonance-stabilised, alkoxide is not; (ii) branching lowers surface area and hence van der Waals forces.
Where the marks are
- (i) Resonance stabilisation of phenoxide ion compared with alkoxide ion (1 mark)
- (ii) Branching reduces surface area, so van der Waals forces decrease (1 mark)
Revise the chapter
- Chapter: Alcohols, Phenols and Ethers (notes, formulas, where marks are lost and every past question)
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