Question

The restricted rotation about carbon-carbon double bond in $2$- butene is due to.
A. overlap of two p orbitals.
B. overall of one p and one $s{p^2}$ hybridised orbitals.
C. overlap of two $s{p^2}$ hybridised orbitals.
D. overlap of one s and one $s{p^2}$ hybridised orbitals.

Answer 1

Hint: Carbon atoms having single bonds can rotate freely due to the end to end approach of overlap of the two orbitals. $2$- butene has carbon-carbon double bond and rotation around double bond is restricted due to lateral overlap of the unhybridized orbitals. The pi bond prevents rotation due to the electron overlap both below and above the plane of the atoms.

Complete step by step answer:
We know that $2$-butene has the structure $C{H_3} - CH = CH - C{H_3}$ having a carbon-carbon double bond and the rotation around it is restricted. The two carbon atoms of the $C = C$ double bond are $s{p^2}$ hybridized. The other three hybrid $s{p^2}$ orbitals lie in one plane. These three hybrid orbitals form three coplanar bonds.
The unhybridized p orbital of each p orbital forms a pie bond. There is maximum overlapping in the two unhybridized p orbitals in the case when they are parallel to each other. This is possible when all the atoms along with groups are in the same plane, i.e. they are coplanar.
This configuration is extremely stable and any change in this configuration by rotation around the carbon-carbon double bond would make the molecule unstable. Thus this restriction is due to the formation of a pi bond, which is formed by the overlap of two p orbitals.
Therefore the rotation around the carbon-carbon double bond in $2$-butene is restricted due to the overlap of two p orbitals.

So, the correct answer is Option A.

Note: Single bonds can rotate around the single bond due to the molecules having rotational energy at a certain temperature. The double bonds are rigid so it can’t be rotated freely. The electron density is above the axis and below the axis and it can’t rotate as it would break the pie bond interaction.