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(A) $1$. In Reaction $1$,methylenecyclohexane reacts with $dil. H_2SO_4$ (acid-catalyzed hydration). The electrophile $H^+$ attacks the double bond to

Vedclass · Accepted Answer

$(P)$ and $(Q)$ are same

Vedclass · Answer

(A) $1$. In Reaction $1$,methylenecyclohexane reacts with $dil. H_2SO_4$ (acid-catalyzed hydration). The electrophile $H^+$ attacks the double bond to form a stable tertiary carbocation at the ring carbon. Water then attacks this carbocation to form $1-$methylcyclohexanol $(P)$.$2$. In Reaction $2,1-$methylcyclohexanol reacts with $dil. H_2SO_4$ (acid-catalyzed dehydration). The hydroxyl group is protonated to form a good leaving group $(-OH_2^+)$,which leaves to form a tertiary carbocation. Since this is already a stable tertiary carbocation,it undergoes elimination to form the most stable alkene,which is $1-$methylcyclohexene $(Q)$.$3$. Comparing $(P)$ ($1$-methylcyclohexanol) and $(Q)$ ($1$-methylcyclohexene),they are different compounds (an alcohol and an alkene). Therefore,option $(A)$ is incorrect as they are not the same. Option $(B)$ is incorrect because they are not isomers of each other. However,in the context of identifying the 'incorrect' statement,$(A)$ is the most fundamentally incorrect statement.

Select the incorrect option based on the given reactions:
Reaction $1$: Methylenecyclohexane + $dil. H_2SO_4 \rightarrow (P)$ (major product)
Reaction $2$: $1-$Methylcyclohexanol + $dil. H_2SO_4 \rightarrow (Q)$ (major product)

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Select the incorrect option based on the given reactions:Reaction $1$: Methylenecyclohexane + $dil. H_2SO_4 \rightarrow (P)$ (major product)Reaction $2$: $1-$Methylcyclohexanol + $dil. H_2SO_4 \rightarrow (Q)$ (major product)