In the dehydrohalogenation of 2-bromobutane,w | vedclass

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(A) The dehydrohalogenation of $2$-bromobutane via the $E2$ mechanism requires an anti-periplanar arrangement of the leaving group $(-Br)$ and the $\b

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(A) The dehydrohalogenation of $2$-bromobutane via the $E2$ mechanism requires an anti-periplanar arrangement of the leaving group $(-Br)$ and the $\beta$-hydrogen atom. To form $cis$-but-$2$-ene,the methyl groups on the $C_2$ and $C_3$ carbons must be on the same side of the developing double bond in the transition state. Looking at the Newman projections,the conformation that places the $-Br$ atom anti to the $\beta$-hydrogen while keeping the two methyl groups on the same side is represented by the structure in option $A$. In this conformation,the $C_2$ methyl group and the $C_3$ methyl group are eclipsed or syn-like relative to each other when viewed along the $C_2-C_3$ bond,which leads to the $cis$ isomer upon elimination.

In the dehydrohalogenation of $2$-bromobutane,which conformation leads to the formation of $cis$-but-$2$-ene?

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