JEE Advanced 2025 — Paper 2 — CHEM-I solution

JEE Advanced2025Paper 2CHEM-I

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

Solution

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.

$∴$ Option (C) is correct.

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