Question

Which of the following compounds will show cis - trans isomerism?

• A. $(C{H}_3{)}_{2}C=CH{C}_2{H}_5$
• B. $H_{2}C=CC{l}_2$
• C. $C{H}_{3}HC=CClC{H}_3$
• D. $ClCH=C{H}_2$

Answer 1

The correct option is C $C{H}_{3}HC=CClC{H}_3$

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.

a. $(C{H}_3{)}_{2}C=CH{C}_2{H}_5$ has two methyl groups attached to the same carbon of the double bond so geometrical isomerism is not possible.

b. $H_{2}C=CC{l}_2$ has two chlorine atoms attached to the same carbon of the double bond so geometrical isomerism is not possible.

c. $C{H}_{3}HC=CClC{H}_3$ has different groups attached to both carbons of the double bond so geometrical isomerism is possible.

d. $ClCH=C{H}_2$ has two hydrogen atoms attached to the same carbon of the double bond so geometrical isomerism is not possible.

Since all compounds have the same groups attached to double bonded carbon atom except in 2-chlorobut-2-ene, thus only 2-chlorobut-2-ene exhibit cis-trans isomerism.