Product (C) is: Ph-CH=CH_2 xrightarrow[CCl_4] | vedclass

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Question

(C) The reaction sequence is as follows:$1.$ Styrene $(Ph-CH=CH_2)$ reacts with $Br_2/CCl_4$ to give $1,2$-dibromo-$1$-phenylethane $(Ph-CHBr-CH_2Br)$

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$Ph-C \equiv C-CH_3$

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(C) The reaction sequence is as follows:$1.$ Styrene $(Ph-CH=CH_2)$ reacts with $Br_2/CCl_4$ to give $1,2$-dibromo-$1$-phenylethane $(Ph-CHBr-CH_2Br)$ as product $(A)$.$2.$ Dehydrohalogenation of $(A)$ with $alc. KOH$ followed by $NaNH_2$ yields phenylacetylene $(Ph-C \equiv CH)$ as product $(B)$.$3.$ Phenylacetylene reacts with $NaNH_2$ to form sodium phenylacetylide,which then reacts with $CH_3Cl$ via $S_N2$ mechanism to give $1$-phenylprop-$1$-yne $(Ph-C \equiv C-CH_3)$ as product $(C)$.

Product $(C)$ is: $Ph-CH=CH_2$ $\xrightarrow[CCl_4]{Br_2} (A)$ $\xrightarrow[(ii) NaNH_2]{(i) alc. KOH} (B)$ $\xrightarrow[(ii) CH_3-Cl]{(i) NaNH_2} (C)$

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