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(A) $1$. The reaction of benzamide $(C_6H_5CONH_2)$ with benzenesulfonyl chloride $(C_6H_5SO_2Cl)$ in the presence of pyridine leads to the dehydratio

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$3-$bromobenzoic acid

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(A) $1$. The reaction of benzamide $(C_6H_5CONH_2)$ with benzenesulfonyl chloride $(C_6H_5SO_2Cl)$ in the presence of pyridine leads to the dehydration of the amide to form benzonitrile $(C_6H_5CN)$ as intermediate $X$.$2$. The subsequent hydrolysis of benzonitrile $(C_6H_5CN)$ with $H_3O^+$ yields benzoic acid $(C_6H_5COOH)$.$3$. The $-COOH$ group is a deactivating and meta-directing group. Therefore,electrophilic aromatic substitution with $Br_2/FeBr_3$ directs the bromine atom to the meta position.$4$. The final product $Y$ is $3-$bromobenzoic acid.

What is the major product $Y$ in the following reaction sequence?
$C_6H_5CONH_2$ $\xrightarrow[\text{Pyridine } 70^{\circ}C]{C_6H_5SO_2Cl} X$ $\xrightarrow[(ii) Br_2, FeBr_3]{(i) H_3O^+} Y$

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What is the major product $Y$ in the following reaction sequence?$C_6H_5CONH_2$ $\xrightarrow[\text{Pyridine } 70^{\circ}C]{C_6H_5SO_2Cl} X$ $\xrightarrow[(ii) Br_2, FeBr_3]{(i) H_3O^+} Y$