Question

Which of the following compound(s) can exhibit geometrical isomerism?

• A. $C{H}_{3}CH=CHC{H}_3$
• B. $C_6{H}_{5}CH=C{H}_2$
• C. $HOOC-CH=CH-COOH$
• D. $Ph-N=N-Ph$

Answer 1

Answer: (A), (C), (D)

$Ph-N=N-Ph$

Conditions for Geometrical Isomerism:
1. Free rotation must be restricted
2. Both the terminals of the system must be attached with different atoms/groups.


Double bonds have restricted rotation so all given compound obeys condition (1).

In compound (a) and (c), we have different groups attached to the terminals of doubly bonded carbon. Hence, they show geometrical isomerism.

In compound (b), one of the terminal doubly bonded carbon has same hydrogen groups. Hence, it does not show geometrical isomerism.

In compound (d), each terminal of doubly bonded carbon attached to a phenyl group and a lone pair. Here, lone pair is considered as a group. Hence, this compound also shows geometrical isomerism.