Skip to content

Compound (A) (C_6H12O_2) on… (Class 12 Chemistry 2024, Q28)

questionbuddy.in/q/cbse-12-chemistry-2024-56-1-1-q28-compound-reduction-gives-compounds-compound

ChemistryCBSE Class 12 · Board exam 2024, set 56/1/1 · 3 marks · Short answer

CBSE Class 12 Chemistry 2024: question 28

Set in the CBSE Class 12 board exam of 2024, paper 56/1/1, section C, question 28, for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q28 · 3 marks · Short answerAldehydes, Ketones and Carboxylic Acids

Compound (A) () on reduction with gives two compounds (B) and (C). The compound (B) on oxidation with PCC gives compound (D) which upon treatment with dilute NaOH and subsequent heating gives compound (E). Compound (E) on catalytic hydrogenation gives compound (C). The compound (D) is oxidized further to give compound (F) which is found to be a monobasic acid (Molecular weight = 60). Identify the compounds (A), (B), (C), (D), (E) and (F).

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 56/1/1.

Hints

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

Hint 1

Start from the end: which monobasic acid has molar mass 60?

Hint 2

Work backwards — an aldehyde with dil. NaOH and heat gives an α,β-unsaturated aldehyde; LiAlH₄ reduces an ester into two alcohols.

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. (F) is a monobasic acid of molar mass 60 → (ethanoic acid). So (D), which oxidises to F, is (ethanal), and (B), which gives D with PCC, is (ethanol).

    1½ marks

  2. (D) with dil. NaOH undergoes aldol condensation; heating dehydrates the aldol: , so (E) is but-2-enal (crotonaldehyde).

    ½ mark

  3. Catalytic hydrogenation of E: , so (C) is butan-1-ol.

    ½ mark

  4. (A), C₆H₁₂O₂, gives ethanol and butan-1-ol with LiAlH₄, so it is the ester ethyl butanoate : . (Butyl ethanoate, , also fits the data.)

    ½ mark

Answer: A = ethyl butanoate (or butyl ethanoate); B = ethanol; C = butan-1-ol; D = ethanal; E = but-2-enal; F = ethanoic acid

Where the marks are

  • F = CH₃COOH (½ mark)
  • D = CH₃CHO (½ mark)
  • B = CH₃CH₂OH (½ mark)
  • E = CH₃CH=CHCHO (½ mark)
  • C = CH₃CH₂CH₂CH₂OH (½ mark)
  • A = CH₃CH₂CH₂COOC₂H₅ / CH₃COOC₄H₉ (½ mark)

Revise the chapter

Practise this chapterBoard-style questions from Aldehydes, Ketones and Carboxylic Acids, one at a time, marked against the answer key and marking points.Back to the full paper2024 board paper, set 56/1/1: 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.