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

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benzoic acid

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(C) $1$. The reaction of benzene with $n$-propyl chloride 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 $X$ is isopropylbenzene (cumene). $3$. Oxidation of alkylbenzenes with $KMnO_4$ oxidizes the alkyl side chain to a carboxylic acid group,provided there is at least one benzylic hydrogen atom. $4$. In isopropylbenzene,the benzylic carbon has one hydrogen atom,so it gets oxidized to a benzoic acid group. $5$. Therefore,the final product $Y$ is benzoic acid $(C_6H_5COOH)$.

Identify the product $Y$ in the following reaction sequence:
$C_6H_6 + CH_3CH_2CH_2Cl$ $\xrightarrow{AlCl_3} X$ $\xrightarrow{KMnO_4} Y$

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Identify the product $Y$ in the following reaction sequence:$C_6H_6 + CH_3CH_2CH_2Cl$ $\xrightarrow{AlCl_3} X$ $\xrightarrow{KMnO_4} Y$