Chapter 7, Problem 7.39AP

Interpretation Introduction

(a)

Interpretation:

The standard free energy difference between the axial and equatorial forms of the chlorocyclohexane is to be calculated.

Out of axial and equatorial forms of the cyclohexane the most stable form is to be justified.

Concept introduction:

The most stable conformation for six membered ring compounds is chair conformation. There are two forms of chair conformation one is axial form in which the substituent is at axial position and other is equatorial form in which the substituent is at the equatorial position.

Answer to Problem 7.39AP

The standard free energy difference between the equatorial form and axial form is $-1.803\text{kJ}{\text{mol}}^{-1}$.

The most stable form is equatorial form as the standard free energy change of the reaction is negative.

Explanation of Solution

The equilibrium condition given for the axial and equatorial forms is shown below.

$\text{Axial}\left(\text{A}\right)⥂\text{Equitorial}\left(\text{E}\right)$

Equilibrium between the axial and equatorial form of the chlorocyclohexane is existing. It is given that the concentration of the equatorial form is $2.07$ times more than the axial position.

$\left[\text{E}\right]=2.07\left[\text{A}\right]$

Where,

• $\left[\text{E}\right]$ is the concentration of the equatorial form.

• $\left[\text{A}\right]$ is the concentration of the axial form.

Rearrange above expression as shown below.

$\frac{\left[\text{E}\right]}{\left[\text{A}\right]}=2.07$

The expression for the equilibrium constant for the above equilibrium is shown below.

${\text{K}}_{\text{eq}}=\frac{\left[\text{E}\right]}{\left[\text{A}\right]}$…(1)

Where,

• ${\text{K}}_{\text{eq}}$ is the equilibrium constant.

Substitute the value of $\frac{\left[\text{E}\right]}{\left[\text{A}\right]}$ in the equation (1) as shown below.

$\begin{array}{rll}{\text{K}}_{\text{eq}}& =& \frac{\left[\text{E}\right]}{\left[\text{A}\right]}\\ {\text{K}}_{\text{eq}}& =& 2.07\end{array}$

The value of equilibrium constant is $2.07$.

The standard free energy of the reaction is related to its equilibrium constant by the following expression.

$\text{Δ}{\text{G}}^{\text{ο}}=-\text{RT}\ln {\text{K}}_{\text{eq}}$…(2)

Where,

• $\text{Δ}{\text{G}}^{\text{ο}}$ is the standard free energy.

• $\text{R}$ is the universal gas constant.

• $\text{T}$ is the standard temperature of the reaction.

The value of $\text{R}$ is $8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$ and $\text{T}$ is $\text{25}{}^{\text{ο}}\text{C}$. Substitute the value of $\text{R}$, $\text{T}$ and ${\text{K}}_{\text{eq}}$ in the equation 2.

$\begin{array}{rll}\text{Δ}{\text{G}}^{\text{ο}}& =& -\text{RT}\ln {\text{K}}_{\text{eq}}\\ & =& -\left(8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\left(\left(25+273\right)\text{K}\right)\ln \left(2.07\right)\\ & =& -2477.57\text{J}{\text{mol}}^{-1}\left(0.7275\right)\\ & =& -1802.4\text{J}{\text{mol}}^{-1}\end{array}$

The standard free energy of the reaction is shown below.

$\text{Δ}{\text{G}}^{\text{ο}}=-1.802\text{kJ}{\text{mol}}^{-1}\left(\because \text{1 kJ}={\text{10}}^{\text{3}}\text{ J}\right)$

Therefore the standard free energy difference between the equatorial form and axial form is $-1.803\text{kJ}{\text{mol}}^{-1}$.

The most stable form is equatorial form as the standard free energy change of the reaction is negative which means reaction is spontaneous. Therefore, favours the equatorial form than the axial form.

Conclusion

The standard free energy difference between the equatorial form and axial form is $-1.802\text{kJ}{\text{mol}}^{-1}$.

The most stable form is equatorial form as the standard free energy change in going from axial to equatorial form is negative.

Interpretation Introduction

(b)

Interpretation:

The ratio of the two forms equatorial form and axial form of the isopropyl cyclohexane is to be calculated.

Concept introduction:

The most stable conformation for six membered ring compounds is chair conformation. There are forms of chair conformation one is axial form if the substituent is at axial position and one is equatorial position.

Answer to Problem 7.39AP

The ratio between the two forms equatorial and axial of the isopropylcyclohexane is $1.31$.

Explanation of Solution

The equilibrium condition given for the axial and equatorial forms is shown below.

$\text{Equitorial}\left(\text{E}\right)⥄\text{Axial}\left(\text{A}\right)$

Equilibrium between the axial and equatorial form of the isopropylcyclohexane is existing. It is given that the standard free energy difference between the two conformation of isopropylcyclohexane is $9.2\text{kJ}{\text{mol}}^{-1}$.

$\text{Δ}{\text{G}}^{\text{ο}}=9.2\text{kJ}{\text{mol}}^{-1}$.

The standard free energy of the reaction is related to its equilibrium constant by the following expression.

$\text{Δ}{\text{G}}^{\text{ο}}=-\text{RT}\ln {\text{K}}_{\text{eq}}$…(1)

Where,

• $\text{Δ}{\text{G}}^{\text{ο}}$ is the standard free energy.

• $\text{R}$ is the universal gas constant.

• $\text{T}$ is the standard temperature of the reaction.

• ${\text{K}}_{\text{eq}}$ is the equilibrium constant.

The value of $\text{R}$ is $8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$ and $\text{T}$ is $\text{25}{}^{\text{ο}}\text{C}$.

Substitute the value of $\text{R}$, $\text{T}$ and $\text{Δ}{\text{G}}^{\text{ο}}$ in the equation (1).

$\begin{array}{rll}\text{Δ}{\text{G}}^{\text{ο}}& =& -\text{RT}\ln {\text{K}}_{\text{eq}}\\ 9.2\text{kJ}{\text{mol}}^{-1}& =& -\left(8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\left(\left(25+273\right)\text{K}\right)\ln {\text{K}}_{\text{eq}}\\ 9.2\text{kJ}{\text{mol}}^{-1}& =& -2477.57\text{J}{\text{mol}}^{-1}\ln {\text{K}}_{\text{eq}}\end{array}$

Rearrange above expression to find out ${\text{K}}_{\text{eq}}$

$\begin{array}{rll}\ln {\text{K}}_{\text{eq}}& =& \frac{-2477.57\text{J}{\text{mol}}^{-1}}{9.2\text{kJ}{\text{mol}}^{-1}}\\ & =& \frac{-2477.57\text{J}{\text{mol}}^{-1}}{9200\text{J}{\text{mol}}^{-1}}(\because 1\text{kJ}& =& {10}^3\text{J})\\ & =& -0.2693\end{array}$

Take antilog on both sides.

$\begin{array}{rll}{\text{K}}_{\text{eq}}& =& e^{-0.2693}\\ & =& 0.764\end{array}$

The expression for the equilibrium constant for the above equilibrium is shown below.

${\text{K}}_{\text{eq}}=\frac{\left[\text{A}\right]}{\left[\text{E}\right]}$…(2)

Where,

• $\left[\text{E}\right]$ is the concentration of the equatorial form.

• $\left[\text{A}\right]$ is the concentration of the axial form.

Substitute the value of ${\text{K}}_{\text{eq}}$ in the equation 2.

$\begin{array}{rll}{\text{K}}_{\text{eq}}& =& \frac{\left[\text{A}\right]}{\left[\text{E}\right]}\\ 0.764& =& \frac{\left[\text{A}\right]}{\left[\text{E}\right]}\end{array}$

Rearrange above expression as shown below.

$\begin{array}{rll}\left[\text{E}\right]& =& \frac{\left[\text{A}\right]}{0.764}\\ & =& 1.31\left[\text{A}\right]\end{array}$

Rearrange again as shown below.

$\frac{\left[\text{E}\right]}{\left[\text{A}\right]}=1.31$

Therefore, the ratio between the equatorial and axial forms is $1.31$.

Conclusion

The ratio between the two forms equatorial and axial of the isopropylcyclohexane is calculated as $1.31$.