Product (B) of the reaction is | vedclass

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(C) The reaction proceeds through the following steps:$1$. Protonation of the hydroxyl group occurs, followed by the loss of water to form a carbocati

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(C) The reaction proceeds through the following steps:$1$. Protonation of the hydroxyl group occurs, followed by the loss of water to form a carbocation at the tertiary carbon.$2$. The double bond of the vinyl chloride group attacks the carbocation, leading to ring closure and the formation of a new carbocation at the carbon bearing the chlorine atom.$3$. Water attacks this carbocation to form a gem-chlorohydrin intermediate, which is unstable.$4$. The unstable intermediate undergoes elimination of $HCl$ to form the final ketone product, which is $8a$-methylhexahydro-$1H$-inden-$2(3H)$-one (or a related bicyclic ketone structure as shown in the options).

Product $(B)$ of the reaction is

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