(1R,2S)-1,2-Dibromo-1,2-diphenylethane (the meso isomer) is treated with one equivalent of $NaOEt$ in ethanol, and the E2 anti elimination of HBr takes place. The product is (CIP priority: Br > C6H5 > H on the carbon bearing Br; C6H5 > H on the other carbon)

Adiphenylethyne (tolane)
B(E)-1-bromo-1,2-diphenylethene
Ca 1:1 mixture of the (E) and (Z) alkenes
D(Z)-1-bromo-1,2-diphenylethene

Answer: B

In the conformation with H and Br anti, the two phenyl groups of the meso isomer are gauche to each other, so after anti elimination they end up cis on the double bond (checked with a 3D model). On one carbon Ph > H, on the other Br > Ph; the Ph on the first carbon is then trans to Br, which makes the alkene E. The (R,R)/(S,S) isomer would give the Z alkene.

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

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(1R,2S)-1,2-Dibromo-1,2-diphenylethane (the meso isomer) is treated with one equivalent of NaOEt in ethanol, and the E2 anti elimination of HBr takes place. The product is (CIP priority: Br > C6H5 > H on the carbon bearing Br; C6H5 > H on the other carbon)

Answer: B

More questions on E2 mechanism and anti-periplanar requirement

(1R,2S)-1,2-Dibromo-1,2-diphenylethane (the meso isomer) is treated with one equivalent of $NaOEt$ in ethanol, and the E2 anti elimination of HBr takes place. The product is (CIP priority: Br > C6H5 > H on the carbon bearing Br; C6H5 > H on the other carbon)