Chapter 7, Problem 7.72AP

Interpretation Introduction

(a)

Interpretation:

The energy cost of a $1,3$-diaxial interaction between two methyl groups is to be calculated.

Concept introduction:

In a cyclohexane ring, the steric interactions that takes place between an axial substituent located on the carbon atom 1 and the hydrogen atoms located on carbon 3 and 5 is known as 1, 3 diaxial interaction. In the cyclohexane derivatives, each 1, 3-diaxial interaction between the methyl and hydrogen increases the enthalpy of the ring by $3.7\text{kJ}{\text{mol}}^{-1}$.

Answer to Problem 7.72AP

The energy cost of a $1,3$-diaxial interaction between two methyl groups is $15.8\text{kJ}{\text{mol}}^{-1}$.

Explanation of Solution

The two conformations of cyclohexane are given below.

Figure 1

In the above shown figure, in conformation $\text{A}$, the number of methyl-hydrogen $1,3$-diaxial interactions are four and in conformation $\text{B}$, the number of methyl-hydrogen $1,3$-diaxial interactions are two and methyl-methyl $1,3$-diaxial interaction is one.

The formula to calculate $\Delta {\text{G}}^{°}$ for the above reaction is shown below.

$\Delta {\text{G}}^{°}{}_{\text{B}}-\Delta {\text{G}}^{°}{}_{\text{A}}=\left[\left(2\times \text{Energy}\text{of}{\text{H-CH}}_{\text{3}}\right)+\left(\text{Energy}\text{of}{\text{CH}}_{\text{3}}{\text{-CH}}_{\text{3}}\right)\right]-\left(4\times \text{Energy}\text{of}{\text{H-CH}}_{\text{3}}\right)$ The energy for every methyl-hydrogen $1,3$-diaxial interaction is $3.7\text{kJ}{\text{mol}}^{-1}$.

The $\Delta {\text{G}}^{°}$ for the equilibrium between $\text{A}$ and $\text{B}$ is $8.4\text{kJ}{\text{mol}}^{-1}$.

Substitute the values of energy and $\Delta {\text{G}}^{°}$ into the above equation.

$\begin{array}{rll}8.4\text{kJ}{\text{mol}}^{-1}& =& \left[\left(2\times \text{3}\text{.7}\text{kJ}{\text{mol}}^{-1}\right)+\left(\text{Energy}\text{of}{\text{CH}}_{\text{3}}{\text{-CH}}_{\text{3}}\right)\right]-\left(4\times \text{3}\text{.7}\text{kJ}{\text{mol}}^{-1}\right)\\ 8.4\text{kJ}{\text{mol}}^{-1}& =& \left(7.4\text{kJ}{\text{mol}}^{-1}+\text{Energy}\text{of}{\text{CH}}_{\text{3}}{\text{-CH}}_{\text{3}}\right)-14.8\text{kJ}{\text{mol}}^{-1}\\ \text{Energy}\text{of}{\text{CH}}_{\text{3}}{\text{-CH}}_{\text{3}}& =& 23.2\text{kJ}{\text{mol}}^{-1}-7.4\text{kJ}{\text{mol}}^{-1}\\ & =& 15.8\text{kJ}{\text{mol}}^{-1}\end{array}$

Therefore, the energy between two methyl groups is $15.8\text{kJ}{\text{mol}}^{-1}$.

Conclusion

The energy between two methyl groups is $15.8\text{kJ}{\text{mol}}^{-1}$.

Interpretation Introduction

(b)

Interpretation:

The value of $\Delta {\text{G}}^{°}$ for the equilibrium between $\text{C}$ and $\text{D}$ is to be calculated.

Concept introduction:

In a cyclohexane ring, the steric interactions that takes place between an axial substituent located on the carbon atom 1 and the hydrogen atoms located on carbon 3 and 5 is known as 1, 3 diaxial interaction. In the cyclohexane derivatives, each 1, 3-diaxial interaction between the methyl and hydrogen increases the enthalpy of the ring by $3.7\text{kJ}{\text{mol}}^{-1}$.

Answer to Problem 7.72AP

The $\Delta {\text{G}}^{°}$ for the equilibrium between $\text{C}$ and $\text{D}$ is $23.2\text{kJ}{\text{mol}}^{-1}$.

Explanation of Solution

The two conformations of cyclohexane are given below.

Figure 2

In the above shown figure, in conformation $\text{D}$, the number of methyl-hydrogen $1,3$-diaxial interactions are two and methyl-methyl $1,3$-diaxial interaction is one. In conformation $\text{C}$, there is no $1,3$-diaxial interaction.

The formula to calculate $\Delta {\text{G}}^{°}$ for the above reaction is shown below.

$\Delta {\text{G}}^{°}{}_{\text{D}}-\Delta {\text{G}}^{°}{}_{\text{C}}=\left[\left(2\times \text{Energy}\text{of}{\text{H-CH}}_{\text{3}}\right)+\left(\text{Energy}\text{of}{\text{CH}}_{\text{3}}{\text{-CH}}_{\text{3}}\right)\right]-0$

The energy for every methyl-hydrogen $1,3$-diaxial interaction is $3.7\text{kJ}{\text{mol}}^{-1}$.

The energy for methyl-methyl $1,3$-diaxial interaction is $15.8\text{kJ}{\text{mol}}^{-1}$.

Substitute the values of energy into the above equation.

$\begin{array}{rll}\Delta {\text{G}}^{°}& =& \left[\left(2\times \text{3}\text{.7}\text{kJ}{\text{mol}}^{-1}\right)+\left(\text{15}\text{.8}\text{kJ}{\text{mol}}^{-1}\right)\right]-\left(0\right)\\ \Delta {\text{G}}^{°}& =& \left(7.4\text{kJ}{\text{mol}}^{-1}+\text{15}\text{.8}\text{kJ}{\text{mol}}^{-1}\right)\\ \Delta {\text{G}}^{°}& =& 23.2\text{kJ}{\text{mol}}^{-1}\end{array}$

Conclusion

The value of $\Delta {\text{G}}^{°}$ is $23.2\text{kJ}{\text{mol}}^{-1}$.