The following sequence of reactions on A give | vedclass

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(C) The reaction sequence is as follows:$1$. The starting material is methyl $2$-(carbamoylmethyl)benzoate.$2$. Treatment with $Br_2/NaOH$ (Hofmann br

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$A$ cyclic amide (lactam)

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(C) The reaction sequence is as follows:$1$. The starting material is methyl $2$-(carbamoylmethyl)benzoate.$2$. Treatment with $Br_2/NaOH$ (Hofmann bromamide degradation) converts the amide group $(-CONH_2)$ into a primary amine group $(-NH_2)$.$3$. The resulting intermediate is methyl $2$-(aminomethyl)benzoate.$4$. Upon heating,this intermediate undergoes intramolecular cyclization (nucleophilic acyl substitution) where the amino group attacks the ester carbonyl,eliminating methanol $(CH_3OH)$ to form a cyclic amide,specifically a lactam (isoindolin$-1-$one).

The following sequence of reactions on $A$ gives:

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