ChemistryCBSE Class 12 · Sample paper, session 2024-25 · 3 marks · Short answer

CBSE Class 12 Chemistry sample paper 2024-25: question 24(b)

From the official CBSE sample question paper for 2024-25, section C, question 24(b), for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q24(b) · 3 marks · Short answerAldehydes, Ketones and Carboxylic Acids

An organic compound ‘X’, does not undergo aldol condensation. However ‘X’ with compound ‘Y’ in the presence of a strong base react to give the compound 1,3-diphenylprop-2-en-1-one.
a. Identify ‘X’ and ‘Y’ (1)
b. Write the chemical reaction involved. (1)
c. Give one chemical test to distinguish between X and Y. (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

Aldol condensation needs an α-hydrogen. Which aromatic carbonyl compound has none?

Hint 2

Split the product at the C=C bond to find the two carbonyl partners.

Hint 3

One compound is an aldehyde and the other a methyl ketone: choose a test that only one of them passes.

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.

(a) X does not undergo aldol condensation, so it has no α-hydrogen: **X = benzaldehyde ( $C_{6} H_{5} C H O$ )**.
½ mark
The product $C_{6} H_{5} C H = C H - C O - C_{6} H_{5}$ needs a $C_{6} H_{5} C O - C H_{2}$ part, so **Y = acetophenone ( $C_{6} H_{5} C O C H_{3}$ )**.
½ mark
(b) Cross aldol (Claisen–Schmidt) condensation:
$C_{6} H_{5} C H O + C H_{3} C O C_{6} H_{5} d i l . N a O H, Δ C_{6} H_{5} C H = C H - C O - C_{6} H_{5} + H_{2} O$ (1,3-diphenylprop-2-en-1-one)
1 mark
(c) Tollens' test: warm each compound with ammoniacal silver nitrate. Benzaldehyde (X) gives a silver mirror, $C_{6} H_{5} C H O + 2 [A g (N H_{3})_{2}]^{+} + 3 O H^{-} \to C_{6} H_{5} C O O^{-} + 2 A g ↓ + 2 H_{2} O + 4 N H_{3}$ . Acetophenone (Y) does not react. (Alternatively, Y gives a yellow iodoform precipitate with $I_{2}$ /NaOH and X does not.)
1 mark

Answer: X = benzaldehyde, Y = acetophenone; Claisen–Schmidt condensation; Tollens' test (silver mirror with X only)

Where the marks are

X = benzaldehyde (½ mark)
Y = acetophenone (½ mark)
Correct cross-aldol (Claisen–Schmidt) equation (1 mark)
Suitable distinguishing test with observations (Tollens'/iodoform) (1 mark)

Revise the chapter

Chapter: Aldehydes, Ketones and Carboxylic Acids (notes, formulas, where marks are lost and every past question)

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 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.

CBSE Class 12 Chemistry sample paper 2024-25: question 24(b)

Hints

Step-by-step solution

Revise the chapter

More questions from Aldehydes, Ketones and Carboxylic Acids