The major product formed in the following reaction is

Solution

Chemistry ma’am · 4 steps

Br2 with NaOH is Hofmann: the amide becomes an amine, and that amine closes a ring onto the ester next door.

1. Hofmann bromamide. With Br₂ and NaOH, a primary amide loses its carbonyl carbon and becomes a primary amine. The side chain -CH₂CONH₂ therefore turns into -CH₂NH₂, one carbon shorter.

   R-CONH₂ + Br₂ + 4NaOH → R-NH₂ + Na₂CO₃ + 2NaBr + 2H₂O

2. Amine attacks ester. The new -NH₂ group sits right next to the -COOCH₃ group on the ring. On heating, the nitrogen attacks the ester carbonyl and pushes out CH₃OH. This intramolecular aminolysis forms a cyclic amide, a lactam.
3. Count the ring. The new ring contains two benzene carbons, the CH₂, the N-H and the C=O, so it has five atoms. This product is isoindolin-1-one: one C=O and an NH next to a CH₂.
4. So. The major product is the five-membered lactam shown in option (d).

The idea to keep. When you see Br₂/NaOH on an amide, do the Hofmann step first and drop one carbon. Then check whether the new amine sits near an ester or acid that it can cyclise onto.

Answer from the NTA final answer key. Solution by PrepWiser.