C_6H_5(CH_2)_5COCl xrightarrow[CS_2]{AlCl_3} | vedclass

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(C) $1$. The first step is an intramolecular Friedel-Crafts acylation of $C_6H_5(CH_2)_5COCl$ in the presence of $AlCl_3$ and $CS_2$. This reaction fo

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(C) $1$. The first step is an intramolecular Friedel-Crafts acylation of $C_6H_5(CH_2)_5COCl$ in the presence of $AlCl_3$ and $CS_2$. This reaction forms a cyclic ketone,specifically $\alpha$-tetralone derivative,which is $1$-oxo-$1,2,3,4$-tetrahydronaphthalene (or a substituted version depending on chain length). Given the formula $C_{12}H_{14}O$,the product $(A)$ is $1$-oxo-$1,2,3,4$-tetrahydronaphthalene.$2$. The second step involves the oxidation of the cyclic ketone $(A)$ using $KMnO_4$ in the presence of $H_3O^{\oplus}$ and heat. Strong oxidation of the alkyl side chain attached to the benzene ring results in the cleavage of the ring and the formation of a dicarboxylic acid.$3$. The oxidation of $1$-oxo-$1,2,3,4$-tetrahydronaphthalene with $KMnO_4$ leads to the formation of phthalic acid (benzene-$1,2$-dicarboxylic acid).$4$. Therefore,compound $(B)$ is phthalic acid,which corresponds to the structure shown in option $(C)$.

$C_6H_5(CH_2)_5COCl$ $\xrightarrow[CS_2]{AlCl_3}$ $(A)$ $(C_{12}H_{14}O)$ $\xrightarrow[H_3O^{\oplus}]{KMnO_4, \Delta}$ $(B)$. Compound $(B)$ is

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