Question
Which of the following compounds will exhibit $$cis-trans$$ (geometrical) isomerism?
A.
2-butene
B.
Butanol
C.
2-butyne
D.
2-butenol
Answer :
2-butene
Solution :
$$\eqalign{ & \mathop {C{H_3}C{H_2}C{H_2}C{H_2}OH}\limits_{{\text{Butanol}}} \Rightarrow {\text{no geometrical isomers}} \cr & C{H_3}C \equiv CC{H_3} \Rightarrow {\text{no geometrical isomers}} \cr & C{H_3}C{H_2}C\left( {OH} \right) = C{H_2} \Rightarrow {\text{no geometrical isomers}}{\text{.}} \cr} $$
$$\eqalign{ & \mathop {C{H_3}C{H_2}C{H_2}C{H_2}OH}\limits_{{\text{Butanol}}} \Rightarrow {\text{no geometrical isomers}} \cr & C{H_3}C \equiv CC{H_3} \Rightarrow {\text{no geometrical isomers}} \cr & C{H_3}C{H_2}C\left( {OH} \right) = C{H_2} \Rightarrow {\text{no geometrical isomers}}{\text{.}} \cr} $$