Among the following compounds, geometrical isomerism is exhibited by :

Solution

Chemistry ma’am · 3 steps

Walk around the ring from the double-bond carbon in both directions and see if the two arms match.

1. The condition. A C=C shows geometrical (cis-trans) isomerism only if each carbon carries two different groups. The exocyclic carbon here is =CHCl, and H and Cl differ. So everything depends on the ring carbon of the double bond, whose two groups are the two halves of the ring.
2. Compare ring arms. Number the ring carbon of the double bond as C1. Go round the ring one way through C2, and the other way through C6. If both paths meet the same atoms at the same distances, the two groups on C1 are identical and there is no geometrical isomerism.

   Option	Groups on exocyclic C	Two ring arms on C1	Geometrical isomerism?
   (a)	H, Cl	plain CH₂ chains both ways, identical	No
   (b)	H, H (=CH₂)	identical; Cl at C4 lies on the symmetry axis	No
   (c)	H, Cl	CH₃ at C3 and C5: each arm meets a methyl two bonds away, identical	No
   (d)	H, Cl	CH₃ at C3 only: one arm meets it two bonds away, the other four bonds away, different	Yes

3. So. Only in (d) does the single methyl make the two ring arms different, so Cl can be on the same side as the methyl-bearing arm or on the opposite side. The answer is (d).

The idea to keep. On a ring carbon of a C=C, the two groups are the two ring arms. They differ only if the ring is unsymmetrical about that carbon; a substituent at C4, or a matching C3 and C5 pair, keeps them equal.

Answer from the NTA final answer key. Solution by PrepWiser.