Chapter 4, Problem 4.70AP

Interpretation Introduction

Interpretation:

The alcohol that undergoes dehydration faster is to be stated. The speed of dehydration of the faster alcohol is to be sated with explanation.

Concept introduction:

A thermodynamic potential that is utilized in the calculation of the highest reversible function taking place in the thermodynamic system at the constant value pressure and temperature is known as Gibbs-free energy. It is denoted by $\Delta G°$.

Answer to Problem 4.70AP

The rate of hydration of $2-\text{methyl}-2-\text{propene}$ is $411$ times faster as compared to the hydration rate of methylenecyclobutane.

Explanation of Solution

The given standard free energy of activation for hydration of $2-$ methylpropene to $2-\text{methyl}-2-\text{propanol}$ is $91.3\text{kJ}{\text{mol}}^{-1}$ or $21.8\text{ kcal}{\text{mol}}^{-1}$.

The given standard free energy $\Delta G°$ for hydration of $2-$ methylpropene is $-5.56\text{ kJ}{\text{mol}}^{-1}$$\left(-1.33\text{ kcal}{\text{mol}}^{-1}\right)$.

The given rate of hydration of methylenecyclobutane to give an alcohol is $0.6$ times the rate of hydration of $2-$ methylpropene.

The equilibrium constant for the hydration of methylenecyclobutane is about $250$ times higher than $2-$ methylpropene.

The hydration of methylenecyclobutane is supposed to be reaction 1.

The hydration of $2-$ methylpropene is supposed to be reaction 2.

The expression for the relationship between the rate for the reaction 1, reaction and their standard free activation energies is given below.

$\log \left(\frac{{\text{rate}}_{\text{1}}}{{\text{rate}}_{\text{2}}}\right)=\frac{\Delta G{°}_2^{\begin{array}{l}+\\ +\end{array}}-\Delta G{°}_1^{\begin{array}{l}+\\ +\end{array}}}{2.3RT}$ …(1)

Substitute the ratio of rate of reaction 1 and 2 as $0.6$, $\Delta G{°}_2^{\begin{array}{l}+\\ +\end{array}}$ as $91.3\text{kJ}{\text{mol}}^{-1}$ and $2.3RT$ as $-5.7\text{ kJ}{\text{mol}}^{-1}$ in the above expression.

$\begin{array}{rll}\log \left(\text{0}\text{.6}\right)& =& \frac{91.3\text{kJ}{\text{mol}}^{-1}-\Delta G{°}_1^{\begin{array}{l}+\\ +\end{array}}}{5.7\text{ kJ}{\text{mol}}^{-1}}\\ \Delta G{°}_1^{\begin{array}{l}+\\ +\end{array}}& =& 91.3\text{kJ}{\text{mol}}^{-1}-5.7\text{ kJ}{\text{mol}}^{-1}\times -0.222\\ & =& 92.6\text{kJ}{\text{mol}}^{-1}\end{array}$

The expression for the relationship between the rate for the reaction 1, reaction and their rate constants is given below.

$\log \left(\frac{{\text{rate}}_{\text{1}}}{{\text{rate}}_{\text{2}}}\right)=\log \left(\frac{K_1}{K_2}\right)$ …(2)

Substitute equation (1) in equation (2) as follows.

$\log \left(\frac{K_1}{K_2}\right)=\frac{\Delta G_2^{\text{o}}-\Delta G_1^{\text{o}}}{2.3RT}$

Substitute the ratio of rate constants of reaction 1 and 2 as $250$, $\Delta G{°}_2^{\begin{array}{l}+\\ +\end{array}}$ as $-5.56\text{kJ}{\text{mol}}^{-1}$ and $2.3RT$ as $-5.7\text{ kJ}{\text{mol}}^{-1}$ in the above expression.

$\begin{array}{rll}\log \left(\text{250}\right)& =& \frac{-5.56\text{kJ}{\text{mol}}^{-1}-\Delta G{°}_1^{\begin{array}{l}+\\ +\end{array}}}{5.7\text{ kJ}{\text{mol}}^{-1}}\\ \Delta G_1^{\text{o}}& =& -5.56\text{kJ}{\text{mol}}^{-1}-2.4\times 5.7\text{ kJ}{\text{mol}}^{-1}\\ & =& -19.2\text{kJ}{\text{mol}}^{-1}\end{array}$

The expression for $\Delta G{°}_1^{\begin{array}{l}+\\ +\end{array}}$ (reverse) is given below.

$\Delta G{°}_1^{\begin{array}{l}+\\ +\end{array}} \left(\text{reverse}\right)=\Delta G{°}_1^{\begin{array}{l}+\\ +\end{array}} \left(\text{forward}\right)+\Delta G_1^{\text{o}}$

Substitute the values for $\Delta G{°}_1^{\begin{array}{l}+\\ +\end{array}} \left(\text{forward}\right)$ and $\Delta G_1^{\text{o}}$ in the above equation.

$\begin{array}{rll}\Delta G{°}_1^{\begin{array}{l}+\\ +\end{array}} \left(\text{reverse}\right)& =& \Delta G{°}_1^{\begin{array}{l}+\\ +\end{array}} \left(\text{forward}\right)+\Delta G_1^{\text{o}}\\ & =& -19.2\text{kJ}{\text{mol}}^{-1}+92.6\text{kJ}{\text{mol}}^{-1}\\ & =& 111.8\text{kJ}{\text{mol}}^{-1}\end{array}$

The expression for $\Delta G{°}_2^{\begin{array}{l}+\\ +\end{array}}$ (reverse) is given below.

$\Delta G{°}_2^{\begin{array}{l}+\\ +\end{array}} \left(\text{reverse}\right)=\Delta G{°}_2^{\begin{array}{l}+\\ +\end{array}} \left(\text{forward}\right)+\Delta G_2^{\text{o}}$

Substitute the values for $\Delta G{°}_2^{\begin{array}{l}+\\ +\end{array}} \left(\text{forward}\right)$ and $\Delta G_2^{\text{o}}$ in the above equation.

$\begin{array}{rll}\Delta G{°}_2^{\begin{array}{l}+\\ +\end{array}} \left(\text{reverse}\right)& =& \Delta G{°}_2^{\begin{array}{l}+\\ +\end{array}} \left(\text{forward}\right)+\Delta G_1^{\text{o}}\\ & =& 91.3{\text{mol}}^{-1}+5.56\text{kJ}{\text{mol}}^{-1}\\ & =& 96.9\text{kJ}{\text{mol}}^{-1}\end{array}$

Substitute the values of $\Delta G{°}_1^{\begin{array}{l}+\\ +\end{array}} \left(\text{reverse}\right)$ and $\Delta G{°}_2^{\begin{array}{l}+\\ +\end{array}} \left(\text{reverse}\right)$ in equation (1).

$\begin{array}{rll}\log \left(\left(\frac{{\text{rate}}_{\text{1}}\left(\text{reverse}\right)}{{\text{rate}}_{\text{2}}\left(\text{reverse}\right)}\right)\right)& =& \frac{\Delta G{°}_2^{\begin{array}{l}+\\ +\end{array}} \left(\text{reverse}\right)-\Delta G{°}_1^{\begin{array}{l}+\\ +\end{array}} \left(\text{reverse}\right)}{2.3RT}\\ & =& \frac{111.8-96.9}{5.7}\\ & =& \frac{14.9}{5.7}\\ & =& 2.614\end{array}$

The further simplification of the above equation is given below.

$\begin{array}{rll}\log \left(\frac{{\text{rate}}_{\text{1}}\left(\text{reverse}\right)}{{\text{rate}}_{\text{2}}\left(\text{reverse}\right)}\right)& =& 2.614\\ \left(\frac{{\text{rate}}_{\text{1}}\left(\text{reverse}\right)}{{\text{rate}}_{\text{2}}\left(\text{reverse}\right)}\right)& =& \text{antilog}\left(2.614\right)\\ & =& 411\end{array}$

Therefore, the above equation suggest that the hydration rate of $2-\text{methyl}-2-\text{propene}$ is about $411$ times faster as compared to the hydration rate of methylenecyclobutane.

Conclusion

The hydration rate of $2-\text{methyl}-2-\text{propene}$ is about $411$ times faster as compared to the hydration rate of methylenecyclobutane.