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(C) The reaction of $2$-chlorocyclohexanol with alcoholic $KOH$ proceeds via an $E_2$ elimination mechanism.In this reaction,the base abstracts a prot

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Cyclohexanone

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(C) The reaction of $2$-chlorocyclohexanol with alcoholic $KOH$ proceeds via an $E_2$ elimination mechanism.In this reaction,the base abstracts a proton from the carbon adjacent to the chlorine atom,leading to the formation of cyclohex$-2-$en$-1-$ol.This product exists in tautomeric equilibrium with cyclohexanone.Since cyclohexanone is the more stable keto form,it is the final major product $A$.

Product $A$ is

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