Chapter 15, Problem 15.55P

Consider the following bromination: ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{CH}+{\text{Br}}_2\stackrel{\Delta }{\to }{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{CBr}+\text{HBr}$.

a. Calculate $\Delta H^o$ for this reaction by using the bond dissociation energies in Table 6.2.

b. Draw out a stepwise mechanism for the reaction, including the initiation, propagation, and termination steps.

c. Calculate $\Delta H^o$ for each propagation step.

d. Draw an energy diagram for the propagation steps.

e. Draw the structure of the transition state of each propagation step.

Interpretation Introduction

(a)

Interpretation: The value of $\Delta H^o$ for the given reaction by the use of bond dissociation energies is to be calculated.

Concept introduction: The value of $\Delta H^{\text{ο}}$ is calculated by the formula,

$\Delta H^{\text{ο}}=BDE\left(\text{B}\text{.B}\text{.}\right)-BDE\left(\text{B}\text{.F}\text{.}\right)$

Answer to Problem 15.55P

The value of $\Delta H^o$ for the given reaction by the use of bond dissociation energies is $-16\text{kcal}/\text{mol}$.

Explanation of Solution

The given bromination reaction is,

${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{CH}+{\text{Br}}_2\stackrel{\Delta }{\to }{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{CBr}+\text{HBr}$

In the given reaction, $\text{Br}-\text{Br}$ and $\text{C}-\text{H}$ bond is broken and $\text{C}-\text{Br}$ and $\text{H}-\text{Br}$ bond is formed.

The value of $\Delta H^{\text{ο}}$ is calculated by the formula,

$\Delta H^{\text{ο}}=BDE\left(\text{B}\text{.B}\text{.}\right)-BDE\left(\text{B}\text{.F}\text{.}\right)$

Where,

• $BDE\left(\text{B}\text{.B}\text{.}\right)$ is the sum of bond dissociation energy of bond broken.

• $BDE\left(\text{B}\text{.F}\text{.}\right)$ is the sum of bond dissociation energy of bond formed.

The bond dissociation energy of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{C}-\text{H}$ bond is $91\text{kcal}/\text{mol}$.

The bond dissociation energy of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{C}-\text{Br}$ bond is $65\text{kcal}/\text{mol}$.

The bond dissociation energy of $\text{Br}-\text{Br}$ bond is $46\text{kcal}/\text{mol}$.

The bond dissociation energy of $\text{H}-\text{Br}$ bond is $88\text{kcal}/\text{mol}$.

Substitute the bond dissociation energy of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{C}-\text{H}$, ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{C}-\text{Br}$, $\text{Br}-\text{Br}$ and $\text{H}-\text{Br}$ in the above formula.

$\begin{array}{rll}\Delta H^{\text{ο}}& =& \left(91\text{kcal}/\text{mol}+46\text{kcal}/\text{mol}\right)-\left(65\text{kcal}/\text{mol}+88\text{kcal}/\text{mol}\right)\\ & =& -16\text{kcal}/\text{mol}\end{array}$

Therefore, the value of $\Delta H^o$ for the given reaction by the use of bond dissociation energies is $-16\text{kcal}/\text{mol}$.

Conclusion

The value of $\Delta H^o$ for the given reaction by the use of bond dissociation energies is $-16\text{kcal}/\text{mol}$.

Interpretation Introduction

(b)

Interpretation: The stepwise mechanism for the given reaction which includes initiation, propagation and termination steps is to be drawn.

Concept introduction: The general steps followed by free-radical reaction are stated below:

1. First step is initiation that involves formation of radical.

2. Second step is propagation.

3. Third step is the termination that involves the formation of stable bond.

Answer to Problem 15.55P

The stepwise mechanism for the given reaction which includes initiation, propagation and termination steps is,

Figure 1

Explanation of Solution

The initiation step involves the generation of bromine radical by hemolytic cleavage of ${\text{Br}}_{\text{2}}$. In the propagation step, bromine radical on reaction with ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{C}-\text{H}$ generates tertiary radical which on reaction with ${\text{Br}}_{\text{2}}$ converts into ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{C}-\text{Br}$. In the termination step, all free radicals combine to form stable molecules. The stepwise mechanism for the given reaction which includes initiation, propagation and termination steps is shown below.

Figure 1

Conclusion

The stepwise mechanism for the given reaction which includes initiation, propagation and termination steps is shown in Figure 1.

Interpretation Introduction

(c)

Interpretation: The value of $\Delta H^o$ for each propagation step is to be calculated.

Concept introduction: The general steps followed by free-radical reaction are stated below:

1. First step is initiation that involves formation of radical.

2. Second step is propagation.

3. Third step is the termination that involves the formation of stable bond.

Answer to Problem 15.55P

The value of $\Delta H^o$ for each propagation step is $3\text{kcal}/\text{mol}$ and $-19\text{kcal}/\text{mol}$.

Explanation of Solution

The propagation step of given reaction involves two steps. In the first step ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{C}-\text{H}$ bond breaks and $\text{H}-\text{Br}$ bond is formed. In the second step, $\text{Br}-\text{Br}$ bond breaks and ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{C}-\text{Br}$ bond is formed.

The value of $\Delta H^{\text{ο}}$ is calculated by the formula,

$\Delta H^{\text{ο}}=BDE\left(\text{B}\text{.B}\text{.}\right)-BDE\left(\text{B}\text{.F}\text{.}\right)$

Where,

• $BDE\left(\text{B}\text{.B}\text{.}\right)$ is the sum of bond dissociation energy of bond broken.

• $BDE\left(\text{B}\text{.F}\text{.}\right)$ is the sum of bond dissociation energy of bond formed.

The bond dissociation energy of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{C}-\text{H}$ bond is $91\text{kcal}/\text{mol}$.

The bond dissociation energy of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{C}-\text{Br}$ bond is $65\text{kcal}/\text{mol}$.

The bond dissociation energy of $\text{Br}-\text{Br}$ bond is $46\text{kcal}/\text{mol}$.

The bond dissociation energy of $\text{H}-\text{Br}$ bond is $88\text{kcal}/\text{mol}$.

The value of $\Delta H^{\text{ο}}$ for first propagation step is,

$\begin{array}{rll}\Delta H^{\text{ο}}& =& 91\text{kcal}/\text{mol}-88\text{kcal}/\text{mol}\\ & =& 3\text{kcal}/\text{mol}\end{array}$

The value of $\Delta H^{\text{ο}}$ for second propagation step is,

$\begin{array}{rll}\Delta H^{\text{ο}}& =& 46\text{kcal}/\text{mol}-65\text{kcal}/\text{mol}\\ & =& -19\text{kcal}/\text{mol}\end{array}$

Conclusion

The value of $\Delta H^o$ for each propagation step is $3\text{kcal}/\text{mol}$ and $-19\text{kcal}/\text{mol}$.

Interpretation Introduction

(d)

Interpretation: The energy level diagram for the propagation steps is to be drawn.

Concept introduction: The graphical representation of chemical reaction in which x-axis represents energy of the reaction and y-axis represents the reaction coordinate is called energy profile diagram.

Answer to Problem 15.55P

The energy level diagram for the propagation steps is,

Figure 2

Explanation of Solution

The energy level diagram for the propagation steps is shown in Figure 2.

Figure 2

Conclusion

The energy level diagram for the propagation steps is shown in Figure 2.

Interpretation Introduction

(e)

Interpretation: The structure of transition state of each propagation step is to be drawn.

Concept introduction: The general steps followed by free-radical reaction are stated below:

1. First step is initiation that involves formation of radical.

2. Second step is propagation.

3. Third step is the termination that involves the formation of stable bond.

Answer to Problem 15.55P

The structure of transition state of first and second propagation step is,

Figure 3

Explanation of Solution

The propagation step of given reaction involves two steps. In the first step ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{C}-\text{H}$ bond breaks and $\text{H}-\text{Br}$ bond is formed. In the second step, $\text{Br}-\text{Br}$ bond breaks and ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{C}-\text{Br}$ bond is formed.

The transition state of the first propagation step is,

Figure 4

The transition state of the first propagation step is,

Figure 5

Conclusion

The structure of transition state of first and second propagation step is shown in Figure 4 and Figure 5, respectively.