In the following reaction sequence: Ph-CH=CH- | vedclass

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Question

(A) The reaction sequence is as follows:$1.$ Addition of $Br_2$ in $CCl_4$ to stilbene $(Ph-CH=CH-Ph)$ gives $1,2$-dibromo-$1,2$-diphenylethane $(A)$.

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cis-$Ph-CH=CH-Ph$

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(A) The reaction sequence is as follows:$1.$ Addition of $Br_2$ in $CCl_4$ to stilbene $(Ph-CH=CH-Ph)$ gives $1,2$-dibromo-$1,2$-diphenylethane $(A)$.$2.$ Dehydrohalogenation of $(A)$ with $2NaNH_2$ removes two molecules of $HBr$ to form diphenylacetylene $(B)$,which is $Ph-C \equiv C-Ph$.$3.$ Partial hydrogenation of diphenylacetylene $(B)$ using Lindlar's catalyst $(H_2/Pd-CaCO_3)$ gives the $cis$-alkene,which is $cis$-$Ph-CH=CH-Ph$ $(C)$.

In the following reaction sequence: $Ph-CH=CH-Ph$ $\xrightarrow{Br_2/CCl_4} (A)$ $\xrightarrow{2NaNH_2} (B)$ $\xrightarrow{H_2/Pd-CaCO_3} (C)$,the product $(C)$ is:

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