In the reaction with HCl,an alkene reacts in | vedclass

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(C) Markovnikov's rule states that in the addition of a protic acid to an asymmetric alkene,the acid hydrogen attaches to the carbon with the smaller

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Both $(A)$ and $(B)$

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(C) Markovnikov's rule states that in the addition of a protic acid to an asymmetric alkene,the acid hydrogen attaches to the carbon with the smaller number of hydrogen atoms and the halide group attaches to the carbon with the larger number of hydrogen atoms. However,the product $1-$chloro$-1-$methylcyclohexane can be formed from both methylenecyclohexane and $1-$methylcyclohexene.For methylenecyclohexane: The protonation of the double bond leads to a $3^{\circ}$ carbocation at the $1-$position,which then reacts with $Cl^-$ to give $1-$chloro$-1-$methylcyclohexane.For $1-$methylcyclohexene: The protonation leads to a $2^{\circ}$ carbocation,which undergoes rearrangement to form a more stable $3^{\circ}$ carbocation,which then reacts with $Cl^-$ to give $1-$chloro$-1-$methylcyclohexane.Since both alkenes yield the same product,the correct answer is $(C)$.

In the reaction with $HCl$,an alkene reacts in accordance with the Markovnikov's rule to give a product $1-$chloro$-1-$methylcyclohexane. The possible alkene is

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