The product of the following reaction is:

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(A) The reaction is a dehalogenation of a vicinal dihalide using zinc dust $(Zn/\Delta)$.This reaction proceeds via an anti-elimination mechanism.In t

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(A) The reaction is a dehalogenation of a vicinal dihalide using zinc dust $(Zn/\Delta)$.This reaction proceeds via an anti-elimination mechanism.In the given reactant,the two chlorine atoms are on adjacent carbons.Looking at the Fischer projection or the given structure,the two $Ph$ groups are on the same side and the two $Me$ groups are on the same side (erythro-like configuration).Upon anti-elimination of $Cl_2$ by $Zn$,the groups that are anti to each other will end up on opposite sides of the resulting double bond.Specifically,the $Ph$ group on one carbon and the $Me$ group on the other carbon are anti to each other,and the $Ph$ group on the other carbon and the $Me$ group on the first carbon are anti to each other.Therefore,the resulting alkene will have the $Ph$ and $Me$ groups on opposite sides,forming the trans-isomer (or the $E$-isomer).This corresponds to the structure where $Ph$ and $Me$ are trans to each other across the $C=C$ bond.

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