Chapter 9.3, Problem 10E

(a)

Interpretation Introduction

Interpretation: The reason for the bromination to be much more regioselective than chlorination needs to be explained.

Concept Introduction : Hess's law states that when the reactants are converted to products the change of enthalpy (i.e. the heat of reaction at constant pressure) does not dependent on the pathway between the initial and final states.

Hess’s law states that enthalpy changes are additive. Thus, for a single reaction,

  $\Delta H°\left(\text{reaction}\right)={\displaystyle \sum \Delta H°f\left(\text{Products}\right)-{\displaystyle \sum \Delta H°f\left(\text{Reactants}\right)}}$

Answer to Problem 10E

The radical construction depends on the statistical factor and bromination reaction the reaction is endothermic and hence the transition state is nearer to the radical generated than the alkane.

The formation of radical depends on the stability of the radical and more selectivity is attained .

Explanation of Solution

Considering the bromination and chlorination of an alkane to check the thermodynamic nature of these reactions.

In the propagation process, the free radical of the reagent in step by step reacts with the alkene with regeneration of the halogen free radical hence the reaction can propagate.

  

The initial stage controls the generation of a specific kind of radical.

The reaction enthalpy of this stage of bromination is measured as

  $\begin{array}{l}\Delta H\text{ = }\left(HH-Br-HR-H\right)\\ =\left(-366+410\right)kJmo{l}^{-1}\\ =44kJmo{l}^{-1}\end{array}$

The reaction enthalpy of this stage of chlorination is measured as follows:

  $\begin{array}{l}\Delta H=\left(HH-Cl-HR-H\right)\\ =\left(-431+410\right){\text{ kJ mol}}^{-1}\\ =-21{\text{ kJ mol}}^{-1}\end{array}$

Hence, the chlorination reaction is exothermic whereas the stage for bromination reaction is endothermic.

Hammond’s postulate the transition state of a reaction and would be nearer to the reactant in case of an exothermic reaction.

In chlorination reaction, the reaction is exothermic and thus the transition state is closer to the alkane than the radical generated.

The radical construction depends on the statistical factor.

In bromination reaction the reaction is endothermic and hence the transition state is nearer to the radical generated than the alkane.

The radical formation depends on the stability of the radical and more selectivity is attained.

(b)

Interpretation Introduction

Interpretation: The reason behind the dangerousness of fluorination needs to be explained.

Concept Introduction : Hess's law states that when the reactants are converted to products the change of enthalpy (i.e. the heat of reaction at constant pressure) does not dependent on the pathway between the initial and final states.

Hess’s law states that enthalpy changes are additive. Thus, for a single reaction,

  $\Delta H°\left(\text{reaction}\right)={\displaystyle \sum \Delta H°f\left(\text{Products}\right)-{\displaystyle \sum \Delta H°f\left(\text{Reactants}\right)}}$

Answer to Problem 10E

High exothermic nature of the bond formation, a huge amount of heat is liberated during fluorination hence the fluorination is dangerous.

Explanation of Solution

The bond formation enthalpy of carbon fluorine bond formation is as follows:

  $\left(C-F\right)\Delta H=-451kJmo{l}^{-1}$

Due to this high exothermic nature of the bond formation, huge amount of heat is liberated during fluorination. Thus, the process is dangerous.

(C)

Interpretation Introduction

Interpretation: The reason behind the difficulty in the generation of an alkyl iodide by free radical chain halogenations needs to be explained.

Concept Introduction : Hess's law states that when the reactants are converted to products the change of enthalpy (i.e. the heat of reaction at constant pressure) does not dependent on the pathway between the initial and final states.

Hess’s law states that enthalpy changes are additive. Thus, for a single reaction,

  $\Delta H°\left(\text{reaction}\right)={\displaystyle \sum \Delta H°f\left(\text{Products}\right)-{\displaystyle \sum \Delta H°f\left(\text{Reactants}\right)}}$

Explanation of Solution

The iodination reaction of an alkane can be expressed as follows

  $CH3-H+I-I\to CH3-I+H-I$

The enthalpy change in this reaction is calculated as

  $\begin{array}{l}\Delta H=\left(HCH3-I+HH-I\right)-\left(HCH3-H+HI-I\right)\\ =\left[\left(-234–297\right)-\left(-435-150\right)\right]kJmo{l}^{-1}\\ =54kJmo{l}^{-1} \end{array}$

As the reaction is endothermic in nature the reaction is difficult to carry out at room temperature.

The reverse reaction can also occur which further decreases the yield of the reaction.