Number of geometrical isomers possible for the given structure is/are _________.

Solution

Chemistry ma’am · 3 steps

Both end bonds count, but the middle bond counts only when its two arms differ in configuration.

1. End double bonds. Each terminal C=C has H and D on one carbon, and CH₃ and the CH₂ chain on the other. H and D are different groups, so each end bond can be E or Z.
2. Middle double bond. One carbon of the middle bond carries H and CH₃. The other carries two CH₂-C(CH₃)=CHD arms with the same atoms. So the middle bond shows geometrical isomerism only when those two arms have different configurations.

   End bonds	Two arms on middle carbon	Middle bond	Isomers
   E and E	identical	not stereogenic	1
   Z and Z	identical	not stereogenic	1
   one E, one Z	different	E or Z	2

3. So. Left E with right Z is the same molecule as left Z with right E, read from the other end. The total is 1 + 1 + 2 = 4 geometrical isomers.

The common slip: 8. Treating all three double bonds as independent gives 2³ = 8. That misses that the middle bond is stereogenic only when its two arms differ.

The idea to keep. When two groups on a double-bond carbon have the same atoms but can differ in configuration, that bond is stereogenic only in the isomers where they differ. Count case by case, not with 2ⁿ.

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