Which one of the following will show geometrical isomerism?

Solution

Chemistry ma’am · 3 steps

Check both ends of the C=C. Each carbon needs two different groups, and that includes the ring carbon.

1. The condition. A double bond shows geometrical (cis-trans) isomerism only if each of its two carbons carries two different groups. If either carbon has two identical groups, swapping them gives back the same molecule.
2. Two ends to check. Here every C=C is exocyclic. One end is the outside carbon, which is =CH₂ or =CHBr. The other end is the ring carbon, whose two groups are the two arms of the ring. Those arms differ only if the ring Br is closer to one side than the other.

   Option	Outside carbon	Ring carbon (two ring arms)	Geometrical isomerism?
   (a)	H and Br, different	Br at C3, nearer one arm, so different	Yes
   (b)	H and H, same	different	No
   (c)	H and H, same	Br at C4, so the same	No
   (d)	H and Br, different	Br at C4, directly opposite, so the same	No

3. So. Only in (a) are both ends of the double bond unsymmetrical. The CHBr Br can sit on the same side as the C3 ring arm or on the opposite side, giving two geometrical isomers. The answer is (a).

The idea to keep. For an exocyclic C=C, treat the two ring arms as the two groups on the ring carbon. A substituent exactly opposite the double bond makes the arms identical, so there is no cis-trans isomerism.

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