JEE Main 2026 Apr 8 Shift 2, Chemistry Q17: Carboxylic Acid Derivatives

Track each step on the bicyclic-acyl lactam substrate.

Step (i) NaBH${}_4$/MeOH: NaBH${}_4$ selectively reduces the more electrophilic carbonyl. The side-chain aldehyde –CHO is reduced to –CH${}_2$OH, while the two amide C=O groups (one in the ring, one connecting the ring N to the side chain) are untouched. So the side-chain becomes –CH(CH${}_3$)–CH${}_2$OH attached to the ring nitrogen through an amide linkage.

Step (ii) NaOH(aq.), heat: Aqueous base hydrolyses the amide bonds. Both the lactam ring C(=O)–N bond and the exocyclic N-acyl bond are cleaved, opening the ring to a carboxylate and a free secondary amine. The product is the sodium pentanoate, with a free –N(H)–CH(CH${}_3$)–CH${}_2$OH chain at the other terminus.

Step (iii) H${}_3$O${}^{+}$: Acid workup protonates the carboxylate to the free carboxylic acid –COOH.

Final structure: HOOC–(CH${}_2$)${}_3$–N(H)–CH(CH${}_3$)–CH${}_2$OH, matching option (3).