JEE Advanced 2025 Paper 2, Chemistry Section 1 Q4: Mechanism of Some Important Reactions of Alkenes

Check each route for the number of -COOH groups in the major product.

(A) HO-CH${}_2$-CH${}_2$-Cl with NaCN gives HO-CH${}_2$-CH${}_2$-CN; hydrolysis yields HO-CH${}_2$-CH${}_2$-COOH (3-hydroxypropanoic acid), a monocarboxylic acid.

(B) Br${}_2$/H${}_2$O on an aldose oxidises only the -CHO terminus to -COOH (aldonic acid), leaving -CH${}_2$OH intact. Monocarboxylic acid (not the dicarboxylic aldaric acid, which would require HNO${}_3$).

(C) KOH/EtOH on bromocyclohexane eliminates HBr to cyclohexene; hot acidic KMnO${}_4$ then oxidatively cleaves the C=C, giving the open-chain dicarboxylic acid HOOC-(CH${}_2$)${}_4$-COOH (adipic acid). This is a dicarboxylic acid.

(D) H${}_2$CrO${}_4$ oxidises the primary alcohol (-CH${}_2$OH) to -COOH but leaves the existing ketone unchanged, giving a keto acid, not a dicarboxylic acid.

$\therefore$ Option (C) is correct.