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(A) The reaction is an $E2$ elimination reaction using alcoholic $KOH$.In an $E2$ mechanism,the base abstracts a $\beta$-hydrogen that is anti-peripla

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(A) The reaction is an $E2$ elimination reaction using alcoholic $KOH$.In an $E2$ mechanism,the base abstracts a $\beta$-hydrogen that is anti-periplanar to the leaving group $(Br^-)$.The starting material is $1$-bromo-$1,2$-diphenylpropane. In the Newman projection or the eclipsed conformation,the $H$ on the $\beta$-carbon and the $Br$ on the $\alpha$-carbon must be anti to each other.Upon elimination,the groups that are on the same side in the transition state will end up on the same side in the final alkene product.Given the stereochemistry of the reactant,the $CH_3$ and the $H$ (from the $\beta$-carbon) will be on the same side of the double bond,while the two $C_6H_5$ groups will be on opposite sides.This corresponds to the structure where $CH_3$ and $H$ are cis to each other.

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