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(B) The reaction proceeds in two steps: $1$. Reduction of cyclobutyl methyl ketone with $NaBH_4/MeOH$ yields $1-cyclobutyl-1-ethanol$. $2$. Treatment

Vedclass · Accepted Answer

$1-$bromo$-1-$methylcyclopentane

Vedclass · Answer

(B) The reaction proceeds in two steps: $1$. Reduction of cyclobutyl methyl ketone with $NaBH_4/MeOH$ yields $1-cyclobutyl-1-ethanol$. $2$. Treatment with $HBr$ leads to the protonation of the alcohol,followed by the loss of water to form a secondary carbocation. $3$. This secondary carbocation undergoes ring expansion to form a more stable cyclopentyl carbocation. $4$. $A$ $1,2-hydride$ shift occurs to form a more stable tertiary carbocation. $5$. Finally,the nucleophilic attack by $Br^-$ yields $1-bromo-1-methylcyclopentane$ as the major product.

The major product formed in the following transformation is most likely to be

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