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(A) The reaction sequence is as follows: $1$. The starting material $A$ reacts with $NH_3$ to form an amide $(C_8H_8ClNO)$. $2$. The amide then underg

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$4-$chloro$-2-$methylbenzoyl chloride

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(A) The reaction sequence is as follows: $1$. The starting material $A$ reacts with $NH_3$ to form an amide $(C_8H_8ClNO)$. $2$. The amide then undergoes the Hofmann bromamide degradation reaction with $Br_2$ and $NaOH$ to form an amine. $3$. Looking at the final product,which is $3-chloro-4-methylaniline$ (or a similar substituted aniline derivative),we can trace back the structure. $4$. The final product has a $Cl$ group at the ortho position and a $CH_3$ group at the meta position relative to the $NH_2$ group. $5$. The starting material $A$ must be $4-chloro-2-methylbenzoyl$ chloride,which upon reaction with $NH_3$ gives $4-chloro-2-methylbenzamide$,and subsequent Hofmann degradation yields $3-chloro-4-methylaniline$.

Consider the above reaction,the compound $A$ is

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