Question

Which of the following biphenyls hinders $C-C$ single bond rotation?

• A.
• B.
• C.
• D.

Answer 1

Answer: (C), (D)

A biphenyl system consists of two phenyl rings attached by a sigma bond.
If there is steric hindrance due to bulky groups at ortho positions, then to minimise repulsions, the rings become mutually perpendicular to each other.

Perpendicular rings hinder the free rotation between them in the biphenyl system.

In compound (c) and (d), we have bulky groups at the ortho position of both the phenyl rings. Hence, to minimise the repulsion the ring becomes perpendicular to each other and restricts $C-C$ bond rotation.

In compound (a), $F$ group is attached to the phenyl ring which is not a bulky group so it does not create any steric hindrance so the molecule have $C-C$ free rotation.

Same goes for compound (b), it has only hydrogen atoms at ortho position of one phenyl ring.

Hence, (c) and (d) are the answers.

Theory:
Functional groups which contain more number of atoms and create steric effect on the molecule, they are attached to are known as bulky groups.
Bulkiness of an attached group is defined by its van der Waals' radii.

Types of groups based on steric effect:

1. Small $\to$ Less or no steric effect
2. Large $\to$ Moderate steric effect
3. Bulky $\to$ High steric effect

Examples:
Smaller groups: $-OH,-F,-H$ etc.
Large groups: $-N{O}_2,-COOH$ etc.
Bulky groups: $-S{O}_{3}H,-CM{e}_3$ etc.