Question

Which of the following may exhibit conformational isomerism?
A.\[{\text{C}}{{\text{H}}_2} = {\text{C}}{{\text{H}}_2}\]
B.\[{\text{CH}} \equiv {\text{CH}}\]
C.\[{\text{C}}{{\text{H}}_3} - {\text{C}}{{\text{H}}_3}\]
D.\[{\text{C}}{{\text{H}}_2} = {\text{CHOH}}\]

Answer 1

Hint:All alkanes except Methane form conformers. As alkanes have sigma bonds about which free rotation can occur.

Complete step by step answer:
Conformers are also called conformational isomers or rotational isomers or rotamers. They are formed due to the free rotation about a sigma bond. They can't be isolated from each other. They have the same position of atoms, bond length, bond angle etc. Condition for conformer formation is that it must have at least 3 continuous sigma bonds. Compounds like methane and methyl chloride do not show conformational isomerism. All alkanes except methane form conformers. Alkenes and alkynes have double and triple bonds respectively and there is no free rotation which can occur or there is restricted rotation about double or triple bonds.
Option A is ethene which has a double bond and hence the rotation will be restricted.
Option B is an alkyne named as akyne which also has unrestricted rotation.
Option D also has a double bond in it and hence will have no free bond rotation.
Only option C that is ethane shows conformational isomerism.

Thus, option C is correct, as it is alkane.

Note:
There are two forms in which conformers exist, that is, staggered form and eclipsed form. Staggered form is the most stable form of conformer as there is maximum distance between groups and has least angle strain and torsional strain. Eclipsed form is the least stable form of conformer as there is minimal distance between groups causing more repulsion. Skew or Gauche forms are infinite forms that exist between staggered and eclipsed forms. Stability order: Staggered than Gauche and then eclipsed (generally).