JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Chapter_pending

The substrate carries a primary amide side chain, $-\mathrm{C}{\mathrm{H}}_2\mathrm{CON}{\mathrm{H}}_2$, and an ortho methyl ester, $-\mathrm{COOC}{\mathrm{H}}_3$, on the benzene ring. Treatment with bromine and sodium hydroxide is the Hofmann bromamide degradation, which converts the primary amide into a primary amine with loss of the carbonyl carbon, giving an ortho benzylic amine $-\mathrm{C}{\mathrm{H}}_2\mathrm{N}{\mathrm{H}}_2$. This amine, sitting next to the ester, attacks the ester carbonyl intramolecularly and closes a five-membered lactam ring fused to the benzene ring. The major product is isoindolin-1(3H)-one.