Chapter 7, Problem 7.73AP

Interpretation Introduction

(a)

Interpretation:

The one which is larger among methyl or phenyl is to be stated.

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 7.73AP

The value of ${\text{ΔG}}^{°}{}_{\text{Ph-H}}$ shows that compound $\text{B}$ is less stable than compound $\text{A}$. Therefore, phenyl is larger.

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 two and in conformation $\text{B}$, the number of phenyl-hydrogen $1,3$-diaxial interactions are two.

The phenyl group is more bulky than the methyl group. So, it will have a destabilizing effect, and therefore, the equilibrium is favored towards conformation $\text{A}$.

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

$\Delta {\text{G}}^{°}{}_{\text{B}}-\Delta {\text{G}}^{°}{}_{\text{A}}=\left[\left(2\times \Delta {\text{G}}^{°}{}_{\text{Ph-H}}\right)-\left(\text{2}\times \Delta {\text{G}}^{°}{}_{{\text{CH}}_{\text{3}}\text{-H}}\right)\right]\text{kJ}{\text{mol}}^{-1}$

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 $4.73\text{kJ}{\text{mol}}^{-1}$.

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

$\begin{array}{rll}4.73\text{kJ}{\text{mol}}^{-1}& =& \left[\left(2\times {\text{ΔG}}^{°}{}_{\text{Ph-H}}\right)-\left(2\times 3.7\right)\right]\text{kJ}{\text{mol}}^{-1}\\ 4.73\text{kJ}{\text{mol}}^{-1}& =& \left[\left(2\times {\text{ΔG}}^{°}{}_{\text{Ph-H}}\right)-\left(7.4\right)\right]\text{kJ}{\text{mol}}^{-1}\\ {\text{ΔG}}^{°}{}_{\text{Ph-H}}& =& \frac{\left(4.73+7.4\right)\text{kJ}{\text{mol}}^{-1}}{2}\\ {\text{ΔG}}^{°}{}_{\text{Ph-H}}& =& 6.065\text{kJ}{\text{mol}}^{-1}\end{array}$

The value of ${\text{ΔG}}^{°}{}_{\text{Ph-H}}$ shows that compound $\text{B}$ is less stable than compound $\text{A}$, so, phenyl is larger.

Conclusion

The equilibrium is favored towards compound $\text{A}$, therefore, phenyl is larger.

Interpretation Introduction

(b)

Interpretation:

The value of $\Delta {\text{G}}^{°}$ for two equilibriums 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 7.73AP

(1) The value of $\Delta {\text{G}}^{°}$ is $12.13\text{kJ}{\text{mol}}^{-1}$.

(2) The value of $\Delta {\text{G}}^{°}$ is $2.67\text{kJ}{\text{mol}}^{-1}$.

Explanation of Solution

(1) The two conformations of cyclohexane are given below.

Figure 2

In the above shown figure, in conformation $\text{B}$, the number of phenyl-hydrogen $1,3$-diaxial interactions are two. In conformation $\text{A}$, there is no $1,3$-diaxial interaction.

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

$\Delta {\text{G}}^{°}{}_{\text{B}}-\Delta {\text{G}}^{°}{}_{\text{A}}=\left[\left(2\times \Delta {\text{G}}^{°}{}_{\text{Ph-H}}\right)-0\right]\text{kJ}{\text{mol}}^{-1}$

The value of ${\text{ΔG}}^{°}{}_{\text{Ph-H}}$ is $6.065\text{kJ}{\text{mol}}^{-1}$.

Substitute the value of energy into the above equation.

$\begin{array}{rll}\Delta {\text{G}}^{°}{}_{\text{B}}-\Delta {\text{G}}^{°}{}_{\text{A}}& =& \left[\left(2\times \Delta {\text{G}}^{°}{}_{\text{Ph-H}}\right)-0\right]\text{kJ}{\text{mol}}^{-1}\\ \Delta {\text{G}}^{°}& =& \left[\left(2\times 6.065\right)-0\right]\text{kJ}{\text{mol}}^{-1}\\ \Delta {\text{G}}^{°}& =& 12.13\text{kJ}{\text{mol}}^{-1}\end{array}$

(2) The two conformations of cyclohexane are given below.

Figure 3

In the above shown figure, in conformation $\text{D}$, the number of phenyl-hydrogen $1,3$-diaxial interactions are four. In conformation $\text{C}$, there are two phenyl-hydrogen $1,3$-diaxial interaction.

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

$\Delta {\text{G}}^{°}{}_{\text{D}}-\Delta {\text{G}}^{°}{}_{\text{C}}=\left[\left(4\times \Delta {\text{G}}^{°}{}_{{\text{CH}}_3\text{-H}}\right)-\left(2\times \Delta {\text{G}}^{°}{}_{\text{Ph-H}}\right)\right]\text{kJ}{\text{mol}}^{-1}$

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

The value of ${\text{ΔG}}^{°}{}_{\text{Ph-H}}$ is $6.065\text{kJ}{\text{mol}}^{-1}$.

Substitute the values of energy into the above equation.

$\begin{array}{rll}\Delta {\text{G}}^{°}{}_{\text{D}}-\Delta {\text{G}}^{°}{}_{\text{C}}& =& \left[\left(4\times 3.7\right)-\left(2\times 6.065\right)\right]\text{kJ}{\text{mol}}^{-1}\\ \Delta {\text{G}}^{°}& =& \left(14.8-12.13\right)\text{kJ}{\text{mol}}^{-1}\\ & =& 2.67\text{kJ}{\text{mol}}^{-1}\end{array}$

Conclusion

The value of $\Delta {\text{G}}^{°}$ is $2.67\text{kJ}{\text{mol}}^{-1}$.