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(A) The reaction is an $E2$ elimination reaction. In $E2$ elimination,the leaving group $(-Br)$ and the $\beta$-hydrogen (or $\beta$-deuterium) must b

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(A) The reaction is an $E2$ elimination reaction. In $E2$ elimination,the leaving group $(-Br)$ and the $\beta$-hydrogen (or $\beta$-deuterium) must be in an anti-periplanar conformation. Looking at the Fischer projection: The molecule is $3$-bromo-$2$-deuteriobutane. To achieve the anti-periplanar transition state,the molecule must rotate such that $-Br$ and $-D$ are anti to each other. When $-Br$ and $-D$ are eliminated,the remaining groups on the two carbons are $CH_3$ and $H$ on one side,and $CH_3$ and $H$ on the other. Specifically,the $CH_3$ groups end up on opposite sides of the double bond,resulting in the trans-alkene product: $CH_3-CH=C(CH_3)-H$ (where $H$ and $CH_3$ are trans). Therefore,the major product is the trans-isomer.

The major product of the following reaction is:

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