The major product formed in the following reaction is

The reaction proceeds in two distinct steps.

In the first step, $$Br_2$$ and $$NaOH$$ (on heating) bring about the Hoffmann bromamide degradation.

This reaction converts a primary amide ($-CONH_2$) into a primary amine ($-NH_2$) with the loss of one carbon atom.

For the given substrate,

$$-CH_2CONH_2 \xrightarrow{Br_2/NaOH,\ \Delta} -CH_2NH_2.$$

After this transformation, the intermediate contains an ortho-substituted benzene ring with the following groups:

• Position 1: $$-CH_2NH_2$$
• Position 2: $$-COOCH_3$$

In the second step, the amino group acts as a nucleophile.

The lone pair on the nitrogen atom attacks the carbonyl carbon of the neighboring ester group, resulting in an intramolecular nucleophilic acyl substitution.

During this process, the methoxy group ($-OCH_3$) leaves, and a new bond is formed between the nitrogen atom and the carbonyl carbon, producing a cyclic amide (lactam).

To determine the ring size, count the atoms involved in ring formation:

1. Benzene carbon attached to the carbonyl group
2. Carbonyl carbon
3. Nitrogen atom
4. $$CH_2$$ carbon
5. Benzene carbon attached to the $$CH_2$$ group

Thus, a five-membered fused lactam ring is formed.

Examining the given options:

• Option (A): An anhydride derivative and therefore incorrect.
• Option (B): Represents the five-membered fused lactam formed through intramolecular cyclization and is correct.
• Option (C): Does not contain a nitrogen atom and is therefore incorrect.
• Option (D): Represents a six-membered ring and does not account for the loss of one carbon atom during Hoffmann degradation.

Hence, the major product formed in the reaction is option (B).