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(B) The given reaction is a Hofmann elimination reaction of a quaternary ammonium hydroxide.In this reaction,the base $(HO^-)$ abstracts a proton from

Vedclass · Accepted Answer

methylenecyclohexane

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(B) The given reaction is a Hofmann elimination reaction of a quaternary ammonium hydroxide.In this reaction,the base $(HO^-)$ abstracts a proton from the $\beta$-carbon atom.According to the Hofmann rule,the less substituted alkene is formed as the major product because the base attacks the most accessible (least sterically hindered) $\beta$-hydrogen.In the given molecule,there are two types of $\beta$-hydrogens: those on the methyl group and those on the ring carbons.Removing a proton from the methyl group leads to the formation of the exocyclic double bond,resulting in methylenecyclohexane.Removing a proton from the ring carbon leads to the formation of $1-$methylcyclohexene.Since the exocyclic double bond (methylenecyclohexane) is less substituted than the endocyclic double bond ($1$-methylcyclohexene),methylenecyclohexane is the major product.

The major product $A$ is:

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