The reaction of the compound shown in the ima | vedclass

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Question

(A) The given compound is $2-$methylbromocyclohexane (specifically the trans isomer based on the wedge/dash representation). Reaction with $alc. KOH$

Vedclass · Accepted Answer

$1-$methylcyclohexene (major product)

Vedclass · Answer

(A) The given compound is $2-$methylbromocyclohexane (specifically the trans isomer based on the wedge/dash representation). Reaction with $alc. KOH$ is a dehydrohalogenation reaction ($E2$ mechanism). In an $E2$ mechanism,the leaving group $(Br)$ and the $\beta$-hydrogen must be anti-periplanar. There are two types of $\beta$-hydrogens available: one at $C-1$ (bearing the $CH_3$ group) and one at $C-3$. Removal of the $\beta$-hydrogen from $C-1$ leads to the more substituted alkene,$1-$methylcyclohexene,which is the Saytzeff product and is more stable. Removal of the $\beta$-hydrogen from $C-3$ leads to the less substituted alkene,$3-$methylcyclohexene,which is the Hofmann product. According to Saytzeff's rule,the more substituted alkene is the major product. Therefore,$1-$methylcyclohexene is the major product.

The reaction of the compound shown in the image with $alc. KOH$ yields:

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