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(A) The reaction of the given substituted cyclohexane with alcoholic $KOH$ is an $E2$ elimination reaction. For an $E2$ reaction to occur,the leaving

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(A) The reaction of the given substituted cyclohexane with alcoholic $KOH$ is an $E2$ elimination reaction. For an $E2$ reaction to occur,the leaving group $(-Br)$ and the $\beta$-hydrogen (or $\beta$-deuterium) must be in an anti-periplanar conformation. In the given chair conformation,the $-Br$ group is in an axial position. The $\beta$-hydrogens and $\beta$-deuterium are also present. Specifically,the $\beta$-deuterium atom is in an anti-periplanar position relative to the axial $-Br$ group. Therefore,the base removes the deuterium atom,leading to the formation of the alkene where the double bond is formed between the carbon bearing the $-Br$ and the carbon bearing the $-D$. The product is $3$-methylcyclohexene.

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