Among the following compounds, geometrical isomerism is exhibited by :
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The Correct Option is D
Solution and Explanation
Option (1), cyclohexylidene-CHCl: the two ring arms are identical, so no geometrical isomerism.
Option (2), 4-chloro-methylenecyclohexane, has a terminal \(\mathrm{=CH_2}\) (two identical H atoms), so no geometrical isomerism.
Option (3), 3,5-dimethylcyclohexylidene-CHCl: the two ring arms are constitutionally identical (each has a CH(CH3) at the second position), so no E/Z as drawn.
Option (4), 3-methylcyclohexylidene-CHCl: the arm through C-2 reaches the CH(CH3) carbon at once, while the arm through C-6 reaches CH2 first. The two arms are different, so the compound exists as E and Z isomers.
Answer: (4) 3-methylcyclohexylidene-chloromethane (option 4 structure)
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