Draw the structures of major product(s) in each of the following reactions : (a) 4-oxocyclohexane-carbaldehyde (cyclohexanone ring with CHO at the 4-position) [Ag(NH_3)_2]^+OH^- (b) (CH_3)(H)C=C(CH_3)-C(=O)-CH_3 NaOI (c) 2-formylbenzoic acid (benzene ring with ortho CHO and COOH) NaCN/HCl

ChemistryCBSE Class 12 · Board exam 2024, set 56/5/1 · 3 marks · Short answer

CBSE Class 12 Chemistry 2024: question 24

Set in the CBSE Class 12 board exam of 2024, paper 56/5/1, section C, question 24, for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q24 · 3 marks · Short answerAldehydes, Ketones and Carboxylic Acids

Draw the structures of major product(s) in each of the following reactions :

(a)
4-oxocyclohexane-carbaldehyde (cyclohexanone ring with CHO at the 4-position) $[Ag (N H_{3})_{2}]^{+} O H^{-}$
1 mark
(b)
$(C H_{3}) (H) C = C (C H_{3}) - C (= O) - C H_{3}$ $NaOI$
1 mark
(c)
2-formylbenzoic acid (benzene ring with ortho CHO and COOH) $NaCN/HCl$
1 mark

Figure in the paper: Three reaction schemes: (a) cyclohexanone ring with CHO substituent at 4-position, reagent [Ag(NH3)2]+ OH-; (b) CH3 and H on one alkene carbon, CH3 and C(=O)CH3 on the other, reagent NaOI; (c) benzene ring with ortho CHO and COOH, reagent NaCN/HCl. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 56/5/1.

Hints

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

Hint 1

Tollens' reagent is a mild oxidant: which carbonyl can it oxidise?

Hint 2

NaOI reacts with $- C O C H_{3}$ groups, and HCN adds to C=O groups.

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) Tollens' reagent oxidises only the –CHO group; the ring ketone is unaffected. The product is 4-oxocyclohexane-1-carboxylic acid (formed as its carboxylate, acid after acidification): cyclohexanone ring with –COOH at C-4.
1 mark
(b) A methyl ketone undergoes the haloform (iodoform) reaction with NaOI. The $- C O C H_{3}$ group becomes $- C O O^{-} N a^{+}$ , the C=C is untouched, and yellow iodoform separates:
$C H_{3} C H = C (C H_{3}) C O C H_{3} N a O I C H_{3} C H = C (C H_{3}) C O O N a + C H I_{3}$
(the acid is 2-methylbut-2-enoic acid)
1 mark
(c) HCN (from NaCN/HCl) adds across the aldehyde C=O to form a cyanohydrin: benzene ring with –COOH and an ortho –CH(OH)CN group, i.e. 2-[cyano(hydroxy)methyl]benzoic acid.
1 mark

Answer: (a) 4-oxocyclohexane-1-carboxylic acid; (b) $C H_{3} C H = C (C H_{3}) C O O N a$ + $C H I_{3}$ ; (c) 2-[cyano(hydroxy)methyl]benzoic acid (cyanohydrin)

Where the marks are

Only –CHO oxidised to –COOH/–COO⁻ (1 mark)
Iodoform reaction product: sodium 2-methylbut-2-enoate + $C H I_{3}$ (1 mark)
Cyanohydrin at the aldehyde carbon (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 paper2024 board paper, set 56/5/1: every question, by section.

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CBSE Class 12 Chemistry 2024: question 24

Hints

Step-by-step solution

Revise the chapter

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