Question

How many $\mathbf{H}\mathbf{\leftrightarrow }\mathbf{H}$eclipsing interactions would be present if cyclopentanewere planar? Assuming an energy cost of 4.0 kJ/mol for each eclipsing interaction,how much torsional strain would planar cyclopentane have? Since themeasured total strain of cyclopentane is 26 kJ/mol, how much of the torsionalstrain is relieved by puckering?

Short answer

The torsional strain planar cyclopentane will have 40kJ/mol.

The torsional strain relieved by puckering is 35%.

Step-by-step solution

Definition of Torsional strain and Strain energy

Torsional strain or eclipsing strain is an increase in the potential energy of a molecule because of repulsion in electrons of bonds that do not share any atom.

Strain energy is the energy released when the molecule’s constituent atoms are allowed to rearrange themselves in a reaction.

Structures of cyclopentane

• Here, the ring has five-carbon.
• Each carbon forms 2 bonds with neighboring carbon atoms and two bonds with hydrogen.
• So, the total number of hydrogen in cyclopropane is 10.

Calculation of total torsional strain in cyclopentane

• As two faces have 5 H each, there are ten possible eclipsing interactions. Therefore, eclipsing interactions are present in cyclopentane.
• As each $\mathrm{H}\leftrightarrow \mathrm{H}$eclipsing interaction in ethane costs about 4.0 kJ/mol and ten hydrogen present, here is a total of ten eclipsing interactions. Torsional strain due to complete six eclipsing exchange is$10\times 4=40\mathrm{kJ}/\mathrm{mol}$
• Fraction of the overall strain energy reduced by puckering of cyclopentane is

$$\begin{gathered} \frac{40-26}{40}\times 100 \\ =\frac{14}{40}\times 100 \\ =35\% \end{gathered}$$