Question

Which of the following will show geometrical?
1.
2.

3.
$ClBrC=C=C=C\left(C{H}_3\right)\left(C_2{H}_5\right)$
4.
$C{H}_{3}HC=C=CBrH$

• A. 1, 2, 3
• B. 2,3 only
• C. 3,4 only
• D. 2, 3, 4

Answer 1

The correct option is A 1, 2, 3

Conditions for geometrical isomers:
1) Free rotation must be restricted.
2) Both the terminals must be attached with different groups.
3) Terminals must not be in perpendicular planes.
Even numbered cumulenes donot show Geometrical isomerism thus 4th will not show geometrical isomerism.
In even cumulenes, groups on the terminal carbons exist in a perpendicular plane. now for compounds to exhibit geometrical isomerism, groups on the terminal carbons must be in the same plane. Hence, even cumulenes do not show geometrical isomerism.

Odd numbered cumulenes show Geometrical isomerism. When there is odd number of consecutive double bonds, terminals lie in the same plane giving geometrical isomerism. thus structure 1,2, 3 will show geometrical isomerism. Hence correct option is a.