Product (A) is:(cyclobutylidene)methylcyclopr | vedclass

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(A) The reaction proceeds via electrophilic addition of $HBr$ to the alkene.$1$. Protonation of the double bond occurs to form the most stable carboca

Vedclass · Accepted Answer

$1-$bromo$-1-$methylcyclopentane

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(A) The reaction proceeds via electrophilic addition of $HBr$ to the alkene.$1$. Protonation of the double bond occurs to form the most stable carbocation. The initial carbocation formed is a tertiary carbocation on the cyclobutane ring.$2$. The four-membered ring undergoes ring expansion to form a more stable five-membered ring carbocation.$3$. $A$ $1,2$-methyl shift occurs to generate an even more stable tertiary carbocation.$4$. Finally,the bromide ion $(Br^-)$ attacks this tertiary carbocation to form the major product,$1$-bromo-$1$-methylcyclopentane.

Product $(A)$ is:
(cyclobutylidene)methylcyclopropane + $HBr \rightarrow A$ (Major)

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Product $(A)$ is:(cyclobutylidene)methylcyclopropane + $HBr \rightarrow A$ (Major)