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(C) $1$. The reaction of $\gamma$-butyrolactone with $HCl$ followed by $SOCl_2$ results in the ring opening to form $4-$chlorobutanoyl chloride,$(A) =

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Cyclopropyl phenyl ketone

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(C) $1$. The reaction of $\gamma$-butyrolactone with $HCl$ followed by $SOCl_2$ results in the ring opening to form $4-$chlorobutanoyl chloride,$(A) = Cl-CH_2-CH_2-CH_2-COCl$.$2$. Friedel-Crafts acylation of benzene $(Ph-H)$ with $(A)$ in the presence of $AlCl_3$ yields $4-$chlorobutyrophenone,$(B) = Ph-CO-CH_2-CH_2-CH_2-Cl$.$3$. Treatment of $(B)$ with $KOH/MeOH$ leads to an intramolecular nucleophilic substitution reaction. The base abstracts an $\alpha$-proton to form an enolate,which then attacks the carbon bearing the chlorine atom,resulting in the formation of a cyclopropyl ring via Neighboring Group Participation $(NGP)$.$4$. The final product $(C)$ is cyclopropyl phenyl ketone.

Identify the compound $(C)$ in the following reaction sequence:
$\gamma$-butyrolactone $\xrightarrow[(ii) SOCl_2]{(i) HCl} (A)$ $\xrightarrow[AlCl_3]{Ph-H} (B)$ $\xrightarrow{KOH/MeOH} (C)$

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Identify the compound $(C)$ in the following reaction sequence:$\gamma$-butyrolactone $\xrightarrow[(ii) SOCl_2]{(i) HCl} (A)$ $\xrightarrow[AlCl_3]{Ph-H} (B)$ $\xrightarrow{KOH/MeOH} (C)$