JEE Main 2026 — Aldehydes and Ketones Question with Solution

The reaction sequence proceeds as follows:

The reagent ${\text{NaBH}}_4$ is a mild reducing agent that selectively reduces the aldehyde group ($-\text{CHO}$) to a primary alcohol ($-{\text{CH}}_2\text{OH}$). The lactam (cyclic amide) ring remains unaffected.

Heating with aqueous $\text{NaOH}$ causes the base-catalyzed hydrolysis of the lactam ring. The amide bond breaks, forming a carboxylate salt ($-{\text{COO}}^{-}$) and a secondary amine ($-\text{NH}-$).

Acidification with ${\text{H}}_3{\text{O}}^{+}$ protonates the carboxylate ion to form a carboxylic acid ($-\text{COOH}$).

The final product is an open-chain amino acid derivative with the structure $\text{HOOC}-{\text{CH}}_2-{\text{CH}}_2-{\text{CH}}_2-{\text{CH}}_2-\text{NH}-{\text{CH(CH}}_3)-{\text{CH}}_2\text{OH}$. This matches the structure given in option (C).

Answer: