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A. Give reasons for… (Class 12 Chemistry SQP 2024-25, Q32a)

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ChemistryCBSE Class 12 · Sample paper, session 2024-25 · 5 marks · Long answer

CBSE Class 12 Chemistry sample paper 2024-25: question 32(a)

From the official CBSE sample question paper for 2024-25, section E, question 32(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q32(a) · 5 marks · Long answerAlcohols, Phenols and Ethers

a. Give reasons for the following:
(i) The reaction of ethanol with acetyl chloride is carried out in the presence of pyridine. (1)
(ii) Cresols are less acidic than phenol. (1)
b. Williamson’s process is used for the preparation of ethers from alkyl halide. Identify the alkyl bromide and sodium alkoxide used for the preparation of 2- Ethoxy-3-methylpentane (1)
c. Convert:
(i) Toluene to 3-nitrobenzoic acid. (1)
(ii) Benzene to m-nitroacetophenone. (1)

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2024-25.

Hints

Open one hint at a time and try again before the next.

Hint 1

What by-product forms with acetyl chloride, and why would a base help?

Hint 2

Acidity depends on the stability of the anion. How does an electron-donating group affect phenoxide?

Hint 3

For conversions, decide which group directs meta and introduce it before nitration. For Williamson, keep the halide primary.

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. a(i) . Pyridine, an organic base, neutralises the HCl as it forms. This prevents the reverse reaction and shifts the equilibrium forward, increasing the yield of ester.

    1 mark

  2. a(ii) In cresols, the –CH₃ group is electron-releasing (+I effect and hyperconjugation). It increases electron density on the phenoxide oxygen and destabilises the cresoxide ion relative to phenoxide. Cresols therefore lose less readily and are less acidic than phenol.

    1 mark

  3. b. The target is . Williamson synthesis needs a primary alkyl halide to avoid elimination, so use **bromoethane, , and sodium 3-methylpentan-2-olate, **:

    1 mark

  4. c(i) Toluene benzoic acid 3-nitrobenzoic acid. Oxidation is done first because –COOH is meta-directing.

    1 mark

  5. c(ii) Benzene acetophenone m-nitroacetophenone. Acylation is done first because –COCH₃ is meta-directing; nitrobenzene would not undergo Friedel–Crafts acylation.

    1 mark

Answer: (i) Pyridine removes HCl and shifts equilibrium forward; (ii) +I/hyperconjugation of –CH₃ destabilises the phenoxide ion; b. C₂H₅Br + sodium 3-methylpentan-2-olate; c(i) KMnO₄ oxidation then nitration; c(ii) Friedel–Crafts acetylation then nitration

Where the marks are

  • Pyridine as base removing HCl, shifting equilibrium (1 mark)
  • Electron-releasing CH₃ destabilises the phenoxide ion, so lower acidity (1 mark)
  • Correct alkyl bromide (C₂H₅Br) and sodium alkoxide (sodium 3-methylpentan-2-olate) (1 mark)
  • Toluene → benzoic acid → 3-nitrobenzoic acid with reagents (1 mark)
  • Benzene → acetophenone → m-nitroacetophenone with reagents (1 mark)

Revise the chapter

Practise this chapterBoard-style questions from Alcohols, Phenols and Ethers, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2024-25: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.