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ChemistryCBSE Class 12 · Board exam 2023, set 56/1/1 · 3 marks · Short answer

CBSE Class 12 Chemistry 2023: question 29(b)

Set in the CBSE Class 12 board exam of 2023, paper 56/1/1, section C, question 29(b), for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q29(b) · 3 marks · Short answerAlcohols, Phenols and Ethers

An organic compound 'A', having the molecular formula on treatment with Cu at 573 K, gives 'B'. 'B' does not reduce Fehling's solution but gives a yellow precipitate of the compound 'C' with /NaOH. Deduce the structures of A, B and C.

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

Cu at 573 K removes H₂ from an alcohol; which type of alcohol gives a ketone?

Hint 2

Not reducing Fehling's solution rules out an aldehyde.

Hint 3

A yellow precipitate with I₂/NaOH means a CH₃CO– group is present.

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. Cu at 573 K dehydrogenates alcohols, so A () is an alcohol. B does not reduce Fehling's solution, so B is a ketone, not an aldehyde. A must therefore be a secondary alcohol: **A = (propan-2-ol).**

    1 mark
  2. . **B = (propanone).** It is a methyl ketone, which explains the positive iodoform test.

    1 mark
  3. . **C = (iodoform),** the yellow precipitate.

    1 mark

Answer: A = CH₃CH(OH)CH₃ (propan-2-ol); B = CH₃COCH₃ (propanone); C = CHI₃ (iodoform)

Where the marks are

  • A = propan-2-ol (1 mark)
  • B = propanone (1 mark)
  • C = iodoform CHI₃ (1 mark)

Revise the chapter

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.