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(i) Write the reaction… (Class 12 Chemistry 2023, Q34a)

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ChemistryCBSE Class 12 · Board exam 2023, set 56/5/1 · 5 marks · Long answer

CBSE Class 12 Chemistry 2023: question 34(a)

Set in the CBSE Class 12 board exam of 2023, paper 56/5/1, section E, question 34(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q34(a) · 5 marks · Long answerAldehydes, Ketones and Carboxylic Acids

(i) Write the reaction involved in Cannizaro’s reaction.
(ii) Why are the boiling point of aldehydes and ketones lower than that of corresponding carboxylic acids ?
(iii) An organic compound ‘A’ with molecular formula is reduced to n-pentane with hydrazine followed by heating with NaOH and Glycol. ‘A’ forms a dioxime with hydroxylamine and gives a positive Iodoform and Tollen’s test. Identify ‘A’ and give its reaction for Iodoform and Tollen’s test.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 56/5/1.

Hints

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

Hint 1

For (i), recall which aldehydes undergo Cannizzaro: those without α-hydrogen, in concentrated alkali.

Hint 2

For (iii), count the degrees of unsaturation. A dioxime means two carbonyl groups.

Hint 3

The Tollens' test needs a –CHO group and the iodoform test needs a CH3CO– group. Fit both on a straight 5-carbon chain.

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. (i) Cannizzaro reaction: an aldehyde with no α-hydrogen, heated with concentrated alkali, disproportionates. One molecule is oxidised to the carboxylate salt and the other is reduced to the alcohol:

    Other examples are also accepted, e.g. .

    1 mark

  2. (ii) Carboxylic acid molecules are held together by strong intermolecular hydrogen bonds and exist as dimers. Aldehydes and ketones cannot form H-bonds among themselves; they have only weaker dipole–dipole attractions. Less energy is therefore needed to separate their molecules, so their boiling points are lower.

    1 mark

  3. **(iii) Identifying A ():**

    • Degree of unsaturation , and A forms a dioxime, so A has two C=O groups.
    • Wolff–Kishner reduction gives n-pentane, so the carbon chain is straight (unbranched).
    • A positive Tollens' test shows a –CHO group.
    • A positive iodoform test shows a group.

    Hence **A = (4-oxopentanal)**.

    Check by Wolff–Kishner:

    1 mark

  4. Iodoform test (the methyl ketone end reacts):

    1 mark

  5. Tollens' test (the aldehyde end is oxidised):

    1 mark

Answer: (i) 2HCHO + conc. KOH → CH3OH + HCOOK; (ii) acids have strong intermolecular H-bonding, aldehydes/ketones only dipole–dipole attraction; (iii) A = CH3COCH2CH2CHO (4-oxopentanal), with iodoform and Tollens' equations as shown

Where the marks are

  • Correct Cannizzaro reaction equation with conc. alkali (1 mark)
  • Strong H-bonding in carboxylic acids vs weak dipole–dipole forces in aldehydes/ketones (1 mark)
  • Identifies A as 4-oxopentanal, CH3COCH2CH2CHO (1 mark)
  • Correct iodoform reaction of A (1 mark)
  • Correct Tollens' reaction of A (1 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 paper2023 board paper, set 56/5/1: 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.