JEE Advanced 2025 Paper 2 Q04 Chemistry Organic Reactions & Mechanisms Carboxylic Acids & Derivatives Hard

JEE Advanced 2025 Paper 2 · Q04 · Carboxylic Acids & Derivatives

The correct reaction/reaction sequence that would produce a dicarboxylic acid as the major product is:

(A) A chloroethanol/chlorohydrin ($HO-CH_2-CH_2-Cl$ substrate): (i) $NaCN$, (ii) $HO^-$, $H_2O$, (iii) $H_3O^+$ (gives a hydroxy-acid). (B) Glucose $CHO-(CHOH)_4-CH_2OH$ treated with $Br_2/H_2O$ (mild oxidation of the aldehyde -> gluconic acid). (C) A bromocycloalkene/alkyl bromide: (i) $KOH/EtOH$ (E2 elimination -> cycloalkene), then (ii) hot acidic $KMnO_4$ (oxidative cleavage of the $C=C$). (D) A cycloalkanol with a carbonyl elsewhere treated with $H_2CrO_4$ (Jones / Cr(VI) oxidation -> keto-acid).

  1. A. Sequence (A)
  2. B. Sequence (B)
  3. C. Sequence (C)
  4. D. Sequence (D)
Reveal answer + step-by-step solution

Correct answer:C

Solution

(A) $NaCN$/$S_N2$ then alkaline hydrolysis converts $-CH_2Cl$ to $-CH_2COOH$; the $-OH$ stays. Product = alpha-hydroxy carboxylic acid (one $-COOH$ only). Not dicarboxylic. (B) $Br_2/H_2O$ is a mild oxidant that oxidises only the $-CHO$ of an aldose to $-COOH$ (gluconic acid). One $-COOH$ only. (C) (i) $KOH/EtOH$ eliminates $HBr$ to give a cycloalkene; (ii) hot acidic $KMnO_4$ oxidatively cleaves $C=C$ to two carboxyl groups. Starting from cyclohexene this affords adipic acid (a 1,6-dicarboxylic acid). DICARBOXYLIC ACID — major product. CORRECT. (D) Cr(VI) oxidation of a hydroxy-cyclic-ketone gives a keto-acid (one $-COOH$, one $C=O$), not dicarboxylic. Answer: (C).

Want to drill deeper? Open this question inside SolveKar AI and tap "Why?" on any step — the AI Mentor reads your full solution context and explains until it clicks. Download SolveKar AI →