| Column 1 | Column 2 | Column 3 | | --- | --- | --- | | (I) Toluene | (i) $\mathrm{NaOH}/\mathrm{Br}_2$ | (P) Condensation | | (II) Acetophenone | (ii) $\mathrm{Br}_2/h\nu$ | (Q) Carboxylation | | (III) Benzaldehyde | (iii) $(\mathrm{CH}_3\mathrm{CO})_2\mathrm{O}/\mathrm{CH}_3\mathrm{COOK}$ | (R) Substitution | | (IV) Phenol | (iv) $\mathrm{NaOH}/\mathrm{CO}_2$ | (S) Haloform | Columns 1, 2 and 3 contain starting materials, reaction conditions, and type of reactions, respectively. The only CORRECT combination that gives two different carboxylic acids is

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Aldehydes Ketones and Carboxylic Acids

Column 1	Column 2	Column 3
(I) Toluene	(i) $NaOH / Br_{2}$	(P) Condensation
(II) Acetophenone	(ii) $Br_{2} / h ν$	(Q) Carboxylation
(III) Benzaldehyde	(iii) $(CH_{3} CO)_{2} O / CH_{3} COOK$	(R) Substitution
(IV) Phenol	(iv) $NaOH / CO_{2}$	(S) Haloform

Columns 1, 2 and 3 contain starting materials, reaction conditions, and type of reactions, respectively.

The only CORRECT combination that gives two different carboxylic acids is

(A)(II) (iv) (R)
(B)(IV) (iii) (Q)
(C)(III) (iii) (P)✓ Correct
(D)(I) (i) (S)

Answer: (C)

Solution

(III) Benzaldehyde with **(iii)

(CH_{3} CO)_{2} O / CH_{3} COOK

** undergoes (P) Perkin condensation:

C_{6} H_{5} CHO + (CH_{3} CO)_{2} O CH_{3} COOK C_{6} H_{5} CH = CHCOOH + CH_{3} COOH

This produces cinnamic acid and acetic acid — two different carboxylic acids.

Answer: (C)

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Single correct option

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JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Aldehydes Ketones and Carboxylic Acids

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