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(A) The reaction of a haloalkane with alcoholic $KOH$ proceeds via an $E2$ elimination mechanism. For $E2$ elimination to occur,the leaving group $(-B

Vedclass · Accepted Answer

$3-$methylcyclohexene

Vedclass · Answer

(A) The reaction of a haloalkane with alcoholic $KOH$ proceeds via an $E2$ elimination mechanism. For $E2$ elimination to occur,the leaving group $(-Br)$ and the $\beta$-hydrogen (or deuterium) must be in an anti-periplanar conformation. In the given molecule,the $-Br$ atom is in an axial position. The adjacent carbon has a deuterium atom in an axial position (anti to $-Br$) and a hydrogen atom in an equatorial position. Since the deuterium is anti to the $-Br$ group,the $E2$ elimination will involve the removal of $D$ and $Br$ ($-DBr$ elimination). Therefore,the deuterium atom is removed,and the double bond forms between the carbon that held the $-Br$ and the carbon that held the $-D$. The resulting product is $3$-methylcyclohexene.

The major product of the following reaction is:

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