meso-1,2-Dibromo-1,2-diphenylethane, $PhCHBr - CHBrPh$ (1R,2S), undergoes E2 with $KOH$ in ethanol to remove one HBr. The product is 1-bromo-1,2-diphenylethene in which:

Athe two phenyl groups are trans to each other
Bthe product is a mixture of cis and trans in equal amounts because of free rotation
Cno alkene is formed because both Br atoms are needed
Dthe two phenyl groups are cis to each other

Answer: D

E2 needs H on one carbon anti to Br on the other. In the meso (1R,2S) isomer, placing H(C1) anti to Br(C2) brings the two phenyl groups gauche to each other on the same side. After anti elimination the phenyls stay on the same side of the new double bond, so they are cis. The (1R,2R)/(1S,2S) racemate gives the isomer with trans phenyls.

Practice Organic Reaction Mechanisms IV: Elimination More on E2 mechanism and anti-periplanar requirement

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meso-1,2-Dibromo-1,2-diphenylethane, PhCHBr−CHBrPh (1R,2S), undergoes E2 with KOH in ethanol to remove one HBr. The product is 1-bromo-1,2-diphenylethene in which:

Answer: D

More questions on E2 mechanism and anti-periplanar requirement

meso-1,2-Dibromo-1,2-diphenylethane, $PhCHBr - CHBrPh$ (1R,2S), undergoes E2 with $KOH$ in ethanol to remove one HBr. The product is 1-bromo-1,2-diphenylethene in which: