The product (P) of the following reaction: | vedclass

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(C) The reaction involves two steps: $1$. The first step uses $Br_2/NaOH$,which is the condition for the Hofmann bromamide degradation reaction. This

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(C) The reaction involves two steps: $1$. The first step uses $Br_2/NaOH$,which is the condition for the Hofmann bromamide degradation reaction. This reaction converts an amide group $(-CONH_2)$ into a primary amine group $(-NH_2)$. $2$. The starting material is methyl $2-(carbamoylmethyl)benzoate$. The $-CONH_2$ group is converted to $-CH_2NH_2$. $3$. The intermediate formed is methyl $2-(aminomethyl)benzoate$. $4$. In the second step,heating $(\Delta)$ causes an intramolecular cyclization (nucleophilic acyl substitution) where the $-NH_2$ group attacks the ester carbonyl group $(-COOCH_3)$,resulting in the loss of a methanol molecule $(-CH_3OH)$ and the formation of a cyclic amide (lactam). $5$. The final product $(P)$ is $isoindolin-1-one$.

The product $(P)$ of the following reaction:

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