Reaction of trans-2-phenyl-1-bromocyclopentan | vedclass

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Question

(B) The reaction follows the $E2$ mechanism. In an $E2$ elimination,the leaving group $(-Br)$ and the $\beta$-hydrogen atom must be in an anti-peripla

Vedclass · Accepted Answer

$3-$phenylcyclopentene

Vedclass · Answer

(B) The reaction follows the $E2$ mechanism. In an $E2$ elimination,the leaving group $(-Br)$ and the $\beta$-hydrogen atom must be in an anti-periplanar (trans) orientation to allow for the formation of the double bond. In trans-$2-$phenyl$-1-$bromocyclopentane,the $\beta$-hydrogen at the $C-3$ position is anti to the $-Br$ group at the $C-1$ position. Therefore,the elimination occurs between $C-1$ and $C-3$,resulting in the formation of $3-$phenylcyclopentene.

Reaction of trans-$2-$phenyl$-1-$bromocyclopentane with alcoholic $KOH$ produces:

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