Question: Calculate the barrier to rotation for each bond highlighted in pink.

a.

b.

The energy of the conformation is calculated by its substituents arranged in an eclipsed manner or staggered manner.

If the conformation is an eclipsed one, the energy of the substituents is:

$$\begin{gathered} H/H=4kJmo{l}^{-1} \\ C{H}_3/H=6kJmo{l}^{-1} \end{gathered}$$

If the conformation is a staggered one, it will be more stable.

The potential barrier rotation is calculated from the difference in energies between the most stable staggered conformation and unstable eclipsed conformation.

a. The molecule has eclipsed conformation and staggered conformation. The energy in the eclipsed conformation is the sum of the individual energies.

$$\begin{gathered} 1H/H=4kJmo{l}^{-1} \\ 2C{H}_3/H=2\times 6kJmo{l}^{-1}=12kJmo{l}^{-1} \end{gathered}$$

The total energy of the eclipsed conformation is 16 kJ per mole. Therefore, the potential energy barrier for the rotation is 16 kJ per mole.

Conformation of compound a

b. The molecule has eclipsed conformation and staggered conformation. The potential barrier energy is calculated as:

role="math" localid="1648619616035" $3C{H}_3/H=3\times 6kJmo{l}^{-1}$$=$$18kJmo{l}^{-1}$

The total energy of the eclipsed conformation is 18 kJ per mole. Therefore, the potential energy barrier for the rotation is 18 kJ per mole.

Conformation of compound b