Among the following compounds,geometrical iso | vedclass

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(B) For a compound to exhibit geometrical isomerism,the groups attached to each carbon of the double bond must be different. In the given options,we a

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$3-$methyl$-1-$chloromethylenecyclohexane

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(B) For a compound to exhibit geometrical isomerism,the groups attached to each carbon of the double bond must be different. In the given options,we analyze the exocyclic double bond: $(A)$ $4$-chloromethylenecyclohexane: The exocyclic carbon is attached to two $H$ atoms,so it does not show geometrical isomerism. $(B)$ $3$-methyl-$1$-chloromethylenecyclohexane: The exocyclic carbon is attached to one $H$ and one $Cl$ atom. The ring carbon is attached to the rest of the ring and a $CH_3$ group. Since the groups on both sides of the double bond are different,it exhibits geometrical isomerism. $(C)$ $1$-chloromethylenecyclohexane: The exocyclic carbon is attached to $H$ and $Cl$,but the ring carbon is attached to two identical sides of the cyclohexane ring,so it does not show geometrical isomerism. $(D)$ $3,5$-dimethyl-$1$-chloromethylenecyclohexane: Due to the symmetry of the $3,5$-dimethyl substitution,the ring carbon is attached to two identical paths,thus it does not show geometrical isomerism.

Among the following compounds,geometrical isomerism is exhibited by

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