JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Aldehydes Ketones and Carboxylic Acids

Sodium borohydride in methanol reduces only the aldehyde group and leaves the lactam (cyclic amide) untouched, so the side-chain $-\mathrm{CHO}$ becomes $-\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$ while the six-membered 2-piperidinone ring survives. Hot aqueous sodium hydroxide then hydrolyses the lactam, opening the ring at the amide linkage to give a carboxylate and a secondary amine; acidification converts the carboxylate to the free carboxylic acid. The major product is the open-chain amino acid $\mathrm{HOOC}-(\mathrm{C}{\mathrm{H}}_2{)}_4-\mathrm{NH}-\mathrm{CH}(\mathrm{C}{\mathrm{H}}_3)-\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$, bearing a terminal carboxylic acid, a secondary amine and a primary alcohol.