Chapter 5, Problem 5.41AP

Interpretation Introduction

(a)

Interpretation:

The $\text{C}-\text{S}$ bond dissociation energy for ethanethiol $\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}-\text{SH}\right)$ is to be calculated.

Concept introduction:

A process of polymerization in which the growth of chain involves free radicals is called free radical polymerization. The process includes three steps; the chain initiation step, the chain propagation step and the chain termination step.

The minimum amount of energy needed to dissociate a bond is called the bond dissociation energy. The term bond dissociation energy is only used for diatomic species. The covalent bonds have low value of bond dissociation energy.

Answer to Problem 5.41AP

The $\text{C}-\text{S}$ bond dissociation energy for ethanethiol is $310.6\text{kJ}{\text{mol}}^{-\text{1}}$ or $74.23\text{kcal}{\text{mol}}^{-1}$.

Explanation of Solution

The reaction required in the given problem is written below.

${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{-SH}\stackrel{\text{homolysis}}{\to }\cdot \text{SH}+\cdot {\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$

The expression for the term ${\text{BDE}}_{\left(\text{C}-\text{S}\right)}$ (enthalpy change) is given below.

${\text{BDE}}_{\left(\text{C}-\text{S}\right)}=\left(\Delta H_{f\left(\text{SH}\right)}+\Delta H_{f\left({\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}\right)}\right)-\left(\Delta H_{f\left({\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}-\text{SH}\right)}\right)$ ...(1)

Where,

• ${\text{BDE}}_{\left(\text{C}-\text{S}\right)}$ is the bond dissociation energy of carbon sulphur bond
• $\Delta H_{f\left(\text{SH}\right)}$ is the heat of formation of $\cdot \text{SH}$ radical.
• $\Delta H_{f\left({\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}\right)}$ is the heat of formation of $\cdot {\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$ radical.
• $\Delta H_{f\left({\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}-\text{SH}\right)}$ is the dissociation energy of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{-SH}$ molecule.

The value of $\Delta H_{f\left(\text{SH}\right)}$ is $143.1\text{kJ}{\text{mol}}^{-\text{1}}$.

The value of $\Delta H_{f\left({\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}\right)}$ is $121.3\text{kJ}{\text{mol}}^{-\text{1}}$.

The value of $\left(\Delta H_{f\left({\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}-\text{SH}\right)}\right)$ is $-46.15\text{kJ}{\text{mol}}^{-\text{1}}$.

Substitute the value of $\Delta H_{f\left(\text{SH}\right)}$, $\Delta H_{f\left({\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}\right)}$ and $\left(\Delta H_{f\left({\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}-\text{SH}\right)}\right)$ in the equation (1).

$\begin{array}{rll}{\text{BDE}}_{\left(\text{C}-\text{S}\right)}& =& \left(\Delta H_{f\left(\text{SH}\right)}+\Delta H_{f\left({\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}\right)}\right)-\left(\Delta H_{f\left({\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}-\text{SH}\right)}\right)\\ {\text{BDE}}_{\left(\text{C}-\text{S}\right)}& =& \left(143.1+121.3\right)-\left(-46.15\right)\text{kJ}{\text{mol}}^{-\text{1}}\\ & =& 310.6\text{kJ}{\text{mol}}^{-\text{1}}\end{array}$

The relation between $\text{kJ}{\text{mol}}^{-\text{1}}$ and $\text{kcal}{\text{mol}}^{-1}$ is given below.

$\text{1}\text{kcal}{\text{mol}}^{-\text{1}}=\text{4}\text{.184kJ}{\text{mol}}^{-\text{1}}$

The conversion of $310.6\text{kJ}{\text{mol}}^{-\text{1}}$ into $\text{kcal}{\text{mol}}^{-1}$ is done as shown below.

$\begin{array}{rll}{\text{BDE}}_{\left(\text{C}-\text{S}\right)}& =& 310.6\text{kJ}{\text{mol}}^{-\text{1}}\times \frac{1\text{kcal}{\text{mol}}^{-1}}{\text{4}\text{.184}\text{kJ}{\text{mol}}^{-\text{1}}}\\ & =& 74.23\text{kcal}{\text{mol}}^{-1}\end{array}$

The $\text{C}-\text{S}$ bond dissociation energy for ethanethiol $\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{-SH}\right)$ is $310.6\text{kJ}{\text{mol}}^{-\text{1}}$ and in $\text{kcal}{\text{mol}}^{-1}$, it is $74.23\text{kcal}{\text{mol}}^{-1}$.

Conclusion

The bond dissociation energy for the $\text{C}-\text{S}$ bond in $\text{kJ}{\text{mol}}^{-\text{1}}$ is $310.6\text{kJ}{\text{mol}}^{-\text{1}}$ and in $\text{kcal}{\text{mol}}^{-1}$, it is $74.23\text{kcal}{\text{mol}}^{-1}$.

Interpretation Introduction

(b)

Interpretation:

The reaction of a radical such as ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{CO}\cdot$ (from homolysis of a peroxide initiator) with ethanethiol is a good source of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{S}\cdot$ radicals is to be shown with the use of bond dissociation energies.

Concept introduction:

A process of polymerization in which the growth of chain involves free radicals is called free radical polymerization. The process includes three steps; the chain initiation step, the chain propagation step and the chain termination step.

The minimum amount of energy needed to dissociate a bond is called the bond dissociation energy. The term bond dissociation energy is only used for diatomic species. The covalent bonds have low value of bond dissociation energy.

Answer to Problem 5.41AP

The reaction of a radical such as ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{CO}\cdot$ (from homolysis of a peroxide initiator) with ethanethiol is a good source of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{S}\cdot$ radicals as the value of $\Delta H$ is $-72.0\text{kJ}{\text{mol}}^{-1}$ (negative) which makes it highly exothermic and favorable.

Explanation of Solution

The reaction required in the given problem is written below.

${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{CO}\cdot +{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{S-H}\stackrel{\text{homolysis}}{\to }{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{S}\cdot +{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{CO-H}$

In the above reaction, it is shown that one $\text{S}-\text{H}$ bond is breaks and new $\text{O}-\text{H}$ bond is formed

The expression for the term $\Delta H$ (enthalpy change) is given below.

$\Delta H=\Delta H_{\text{S}-\text{H}}-\Delta H_{\text{O}-\text{H}}$ …(2)

Where,

• $\Delta H_{\text{S}-\text{H}}$ is the bond dissociation energy of sulfur and hydrogen bond.
• $\Delta H_{\text{O}-\text{H}}$ is the bond formation energy of oxygen and hydrogen bond.

The value of $\Delta H_{\text{S}-\text{H}}$ is $366\text{kJ}{\text{mol}}^{-1}$.

The value of $\Delta H_{\text{O}-\text{H}}$ is $438\text{kJ}{\text{mol}}^{-1}$.

Substitute the value of $\Delta H_{\text{S}-\text{H}}$ and $\Delta H_{\text{O}-\text{H}}$ in the equation (2) and solve for $\Delta H$.

$\begin{array}{rll}\Delta H& =& \Delta H_{\text{S}-\text{H}}-\Delta H_{\text{O}-\text{H}}\\ & =& \left(366-438\right)\text{kJ}{\text{mol}}^{-1}\\ & =& -72.0\text{kJ}{\text{mol}}^{-1}\end{array}$

The relation between $\text{kJ}{\text{mol}}^{-\text{1}}$ and $\text{kcal}{\text{mol}}^{-1}$ is given below.

$\text{1}\text{kcal}{\text{mol}}^{-\text{1}}=\text{4}\text{.184kJ}{\text{mol}}^{-\text{1}}$

The conversion of $-72.0\text{kJ}{\text{mol}}^{-\text{1}}$ into $\text{kcal}{\text{mol}}^{-1}$ is done as shown below.

$\begin{array}{rll}\Delta H& =& -72.0\text{kJ}{\text{mol}}^{-\text{1}}\times \frac{1\text{kcal}{\text{mol}}^{-1}}{\text{4}\text{.184}\text{kJ}{\text{kcal}}^{-1}}\\ & =& -17.21\text{kcal}{\text{mol}}^{-1}\end{array}$

The value of $\Delta H$ is negative which means the reaction is highly exothermic. Therefore, the reaction is feasible for the generation of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{S}\cdot$ radicals.

Conclusion

The reaction of a radical such as ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{CO}\cdot$ (from homolysis of a peroxide initiator) with ethanethiol is highly exothermic as the $\Delta H$ is negative for the reaction. Therefore, the reaction is feasible for the generation of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{S}\cdot$ radicals.

Interpretation Introduction

(c)

Interpretation:

Each propagation step of thiol addition to an alkene such as ethylene is exothermic and therefore favorable is to be shown with the help of bond dissociation energies.

Concept introduction:

A process of polymerization in which the growth of chain involves free radicals is called free radical polymerization. The process includes three steps; the chain initiation step, the chain propagation step and the chain termination step.

The minimum amount of energy needed to dissociate a bond is called the bond dissociation energy. The term bond dissociation energy is only used for diatomic species. The covalent bonds have low value of bond dissociation energy.

Answer to Problem 5.41AP

The bond dissociation energy of first step is $-67.6\text{kJ}{\text{mol}}^{-1}$ and for the second step is $-57.0\text{kJ}{\text{mol}}^{-1}$. Therefore the first and second propagation steps of thiol addition to an alkene such as ethylene are exothermic as the values obtained are negative and therefore each propagation step is favorable.

Explanation of Solution

The reaction for the second propagation step of thiol addition to ethylene is shown below.

${\text{CH}}_{\text{2}}={\text{CH}}_{\text{2}}+{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{S}-\text{H}\stackrel{\text{homolysis}}{\to }{\text{CH}}_{\text{2}}-{\text{CH}}_{\text{2}}-{\text{SCH}}_{\text{2}}{\text{CH}}_{\text{3}}$

In the above reaction, it is shown that one $\text{C}=\text{C}$ bond is breaks and new $\text{C}-\text{S}$ bond is formed

The expression for the term $\Delta H$ (enthalpy change) is given below.

$\Delta H=\Delta H_{{\text{CH}}_{\text{2}}={\text{CH}}_{\text{2}}}-\Delta H_{\text{C}-\text{S}}$ …(3)

Where,

• $\Delta H_{{\text{CH}}_{\text{2}}={\text{CH}}_{\text{2}}}$ is the bond dissociation energy of $\text{C}=\text{C}$ bond.
• $\Delta H_{\text{C}-\text{S}}$ is the bond formation energy of carbon-sulfur bond.

The value of $\Delta H_{{\text{CH}}_{\text{2}}={\text{CH}}_{\text{2}}}$ is $243\text{kJ}{\text{mol}}^{-1}$.

The value of $\Delta H_{\text{C}-\text{S}}$ is $310.6\text{kJ}{\text{mol}}^{-1}$.

Substitute the value of $\Delta H_{\text{S}-\text{H}}$ and $\Delta H_{\text{O}-\text{H}}$ in the equation (3).

$\begin{array}{rll}\Delta H& =& \Delta H_{{\text{CH}}_{\text{2}}={\text{CH}}_{\text{2}}}-\Delta H_{\text{C}-\text{S}}\\ & =& \left(243-310.6\right)\text{kJ}{\text{mol}}^{-1}\\ & =& -67.6\text{kJ}{\text{mol}}^{-1}\end{array}$

The relation between $\text{kJ}{\text{mol}}^{-\text{1}}$ and $\text{kcal}{\text{mol}}^{-1}$ is given below.

$\text{1}\text{kcal}{\text{mol}}^{-\text{1}}=\text{4}\text{.184kJ}{\text{mol}}^{-\text{1}}$

The conversion of $-16.15\text{kJ}{\text{mol}}^{-\text{1}}$ into $\text{kcal}{\text{mol}}^{-1}$ is done as shown below.

$\begin{array}{rll}\Delta H& =& -16.15\text{kJ}{\text{mol}}^{-\text{1}}\times \frac{1\text{kcal}{\text{mol}}^{-1}}{\text{4}\text{.184}\text{kJ}{\text{kcal}}^{-1}}\\ & =& -3.861\text{kcal}{\text{mol}}^{-1}\end{array}$

The value of $\Delta H$ is negative. The reaction is highly exothermic and therefore, the propagation step is favorable.

The reaction for the second propagation step of thiol addition to sulfide is shown below.

${\text{CH}}_3{\text{CH}}_{\text{2}}{\text{SCH}}_{\text{2}}{\text{CH}}_3+{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{S}-\text{H}\stackrel{\text{homolysis}}{\to }{\text{H-CH}}_{\text{2}}{\text{-CH}}_{\text{2}}{\text{-SCH}}_{\text{2}}{\text{CH}}_{\text{3}}+{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{S}$

In the above reaction, it is shown that one $\text{S}-\text{H}$ bond is breaks and new $\text{C}-\text{H}$ bond is formed.

The expression for the term $\Delta H$ (enthalpy change) is given below.

$\Delta H=\Delta H_{\text{S}-\text{H}}-\Delta H_{\text{C}-\text{H}}$ …(4)

Where,

• $\Delta H_{\text{S}-\text{H}}$ is the bond dissociation energy of sulfur-hydrogen bond
• $\Delta H_{\text{C}-\text{H}}$ is the bond formation energy of carbon-hydrogen bond.

The value of $\Delta H_{\text{S}-\text{H}}$ is $366\text{kJ}{\text{mol}}^{-1}$.

The value of $\Delta H_{\text{C}-\text{H}}$ is $423\text{kJ}{\text{mol}}^{-1}$.

Substitute the value of $\Delta H_{\text{S}-\text{H}}$ and $\Delta H_{\text{O}-\text{H}}$ in the equation (4).

$\begin{array}{rll}\Delta H& =& \Delta H_{\text{S}-\text{H}}-\Delta H_{\text{C}-\text{H}}\\ & =& \left(366-423\right)\text{kJ}{\text{mol}}^{-1}\\ & =& -57.0\text{kJ}{\text{mol}}^{-1}\end{array}$

The relation between $\text{kJ}{\text{mol}}^{-\text{1}}$ and $\text{kcal}{\text{mol}}^{-1}$ is given below.

$\text{1}\text{kcal}{\text{mol}}^{-\text{1}}=\text{4}\text{.184kJ}{\text{mol}}^{-\text{1}}$

The conversion of $-57.0\text{kJ}{\text{mol}}^{-\text{1}}$ into $\text{kcal}{\text{mol}}^{-1}$ is done as shown below.

$\begin{array}{rll}\Delta H& =& -57.0\text{kJ}{\text{mol}}^{-\text{1}}\times \frac{1\text{kcal}{\text{mol}}^{-1}}{\text{4}\text{.184}\text{kJ}{\text{kcal}}^{-1}}\\ & =& -13.62\text{kcal}{\text{mol}}^{-1}\end{array}$

The value of $\Delta H$ is negative which means the reaction is highly exothermic and therefore, the second propagation step is favorable.

Conclusion

The first and second propagation steps of thiol addition to an alkene such as ethylene are exothermic (the value of $\Delta H$ is negative) and each the propagation steps are favorable.