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(C) $1$. The reaction of benzene with $n$-propyl chloride $(CH_3CH_2CH_2Cl)$ in the presence of $AlCl_3$ is a Friedel-Crafts alkylation.$2$. Due to th

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Benzoic acid

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(C) $1$. The reaction of benzene with $n$-propyl chloride $(CH_3CH_2CH_2Cl)$ in the presence of $AlCl_3$ is a Friedel-Crafts alkylation.$2$. Due to the rearrangement of the primary carbocation $(CH_3CH_2CH_2^+)$ to a more stable secondary carbocation $(CH_3CH^+CH_3)$,the major product $X$ formed is isopropylbenzene (cumene).$3$. When isopropylbenzene is treated with $KMnO_4$ (a strong oxidizing agent),the alkyl side chain is oxidized to a carboxylic acid group,regardless of the length of the chain,provided there is at least one benzylic hydrogen atom.$4$. Therefore,the isopropyl group is oxidized to a carboxyl group $(-COOH)$,resulting in the formation of benzoic acid $(C_6H_5COOH)$ as the final product $Y$.

Identify the product $Y$ in the following reaction sequence:
$Benzene + CH_3CH_2CH_2Cl$ $\xrightarrow{AlCl_3} X$ $\xrightarrow{KMnO_4} Y$

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Identify the product $Y$ in the following reaction sequence:$Benzene + CH_3CH_2CH_2Cl$ $\xrightarrow{AlCl_3} X$ $\xrightarrow{KMnO_4} Y$