Which of the following cannot show geometrica | vedclass

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(C) For a molecule to show geometrical isomerism,each carbon atom of the double bond (or the substituted carbons in a ring) must be attached to two di

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(C) For a molecule to show geometrical isomerism,each carbon atom of the double bond (or the substituted carbons in a ring) must be attached to two different groups.$A$: The structure shown is a complex bicyclic ether. Due to the symmetry and the nature of the substituents,it does not possess the necessary conditions for geometrical isomerism.$B$: $CH_3-CH=CH-CH_3$ (but$-2-$ene) shows geometrical isomerism ($cis$ and $trans$ forms).$C$: In $1,1$-dimethylcyclopropane,the carbon atom at position $1$ is attached to two identical methyl groups,so it cannot show geometrical isomerism.$D$: The structure shown is a substituted bicyclic system where the terminal carbons have identical substituents ($Br, Br$ and $Cl, Cl$),thus failing the condition for geometrical isomerism.However,in the context of standard chemistry problems of this type,$1,1$-dimethylcyclopropane $(C)$ is the most classic example of a molecule that cannot show geometrical isomerism due to identical groups on the same carbon.

Which of the following cannot show geometrical isomerism?

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