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(A) The reaction of an alkyl halide with $alc. \ KOH$ proceeds via an $E2$ elimination mechanism.$E2$ elimination requires an anti-periplanar orientat

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(A) The reaction of an alkyl halide with $alc. \ KOH$ proceeds via an $E2$ elimination mechanism.$E2$ elimination requires an anti-periplanar orientation between the leaving group (in this case,$Cl$) and the $\beta$-hydrogen (or $\beta$-deuterium) being removed.In a cyclohexane ring,this means the leaving group and the $\beta$-substituent must both be in axial positions to be anti-periplanar to each other.In the given structure,the $Cl$ is equatorial. The molecule must undergo a ring flip to place the $Cl$ in an axial position.Once in the axial position,the $\beta$-substituents that are anti to the $Cl$ are the ones that will be eliminated.Since the $D$ atom is anti to the $Cl$ in the axial conformation,the $D$ atom is removed,leading to the formation of the alkene product where the $D$ is eliminated.

Predict the product for the following reaction:
$\xrightarrow[\Delta]{alc. \ KOH}$ product

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Predict the product for the following reaction:$\xrightarrow[\Delta]{alc. \ KOH}$ product