In the given reaction sequence,what is Z?Anil | vedclass

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(A) $1$. The reaction of aniline with acetyl chloride $(CH_3COCl)$ in the presence of a base (like pyridine) leads to the formation of acetanilide $(X

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p-Bromoaniline

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(A) $1$. The reaction of aniline with acetyl chloride $(CH_3COCl)$ in the presence of a base (like pyridine) leads to the formation of acetanilide $(X)$,which is $C_6H_5NHCOCH_3$.$2$. The acetamido group $(-NHCOCH_3)$ is ortho/para directing,but due to steric hindrance,the para-isomer is the major product. Thus,$Y$ is p-bromoacetanilide $(p-Br-C_6H_4-NHCOCH_3)$.$3$. Acidic hydrolysis $(H_3O^+)$ of the acetamido group converts it back into the amino group $(-NH_2)$.$4$. Therefore,the final product $Z$ is p-bromoaniline $(p-Br-C_6H_4-NH_2)$.

In the given reaction sequence,what is $Z$?
Aniline $\xrightarrow{CH_3COCl} X$ $\xrightarrow{Br_2/AlBr_3} Y$ $\xrightarrow{H_3O^+} Z$

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In the given reaction sequence,what is $Z$?Aniline $\xrightarrow{CH_3COCl} X$ $\xrightarrow{Br_2/AlBr_3} Y$ $\xrightarrow{H_3O^+} Z$