Chapter 10, Problem 16P

Carbon–carbon bond dissociation enthalpies have been measured for many alkanes.

Without referring to Table $\text{10}\text{.1}$, identify the alkane in each of the following pairs that

has the lower carbon–carbon bond-dissociation enthalpy, and explain the reason for your

choice.

Ethane or propane

Propane or $\text{2-methylpropane}$

$\text{2-Methylpropane}$ or $\text{2,2-dimethylpropane}$

Cyclobutane or cyclopentane

Interpretation Introduction

Interpretation:

In each of the given pairs, the alkane having lower carbon-carbon bond dissociation enthalpy is to be identified and the reason for this is to be explained.

Concept introduction:

Species that contain unpaired electrons are called free radicals.

Alkyl radicals are described by the presence of carbon with three bonds. The alkyl radicals are classified as primary, secondary or tertiary on the basis of the number of carbon atoms directly attached to the carbon atom bearing unpaired electron.

Similar to carbocation’s, free radicals are stabilized by alkyl substituents. The stability order of alkyl radicals is $\text{tertiary > secondary > primary > methyl}$.

In a hemolytic cleavage, each atom of the bond keeps one of the electrons in the bond.

The bond dissociation energy represents the stability of radical formed.

Answer to Problem 16P

Solution:

a) Propane has the lower carbon-carbon bond dissociation enthalpy because it produces more stable free radicals.

b) $\text{2-methylpropane}$ has the lower carbon-carbon bond dissociation enthalpy because it produces more stable free radicals.

c) $\text{2, 2 - dimethylpropane}$ has the lower carbon-carbon bond dissociation enthalpy because it produces more stable free radicals.

d) Cyclopentane has the lower carbon-carbon bond dissociation enthalpy because it produces more stable free radicals.

Explanation of Solution

a) The given alkanes are ethane and propane.

The cleavage of the carbon-carbon bond in propane produces one methyl radical and one ethyl radical.

The ethyl radical is a primary radical and is more stable than the methyl radical. Hence propane produces more stable radicals than ethane upon homolytic cleavage. Lower energy is required to generate free radicals in propane. Thus, propane has a lower bond dissociation enthalpy than ethane.

b) The given alkanes are propane and $\text{2-methylpropane}$.

The homolytic cleavage of the carbon-carbon bond in propane produces one methyl radical and one ethyl radical. The homolytic cleavage of the carbon-carbon bond in $\text{2-methylpropane}$ produces one methyl radical and one isopropyl radical.

The isopropyl radical is a secondary radical and is more stable than the ethyl radical, which is a primary radical. Hence $\text{2-methylpropane}$ produces more stable radicals than propane upon homolytic cleavage. Lower energy is required to generate free radicals in $\text{2-methylpropane}$. Thus, $\text{2-methylpropane}$ has a lower bond dissociation enthalpy than propane.

c) The given alkanes are $\text{2-methylpropane}$ and $\text{2, 2 - dimethylpropane}$.

The homolytic cleavage of the carbon-carbon bond in $\text{2-methylpropane}$ produces one methyl radical and one isopropyl radical. The homolytic cleavage of the carbon-carbon bond in $\text{2, 2 - dimethylpropane}$ produces one methyl radical and one tertiary butyl radical.

The tertiary butyl radical is a tertiary radical and is more stable than the isopropyl radical, which is a secondary radical. Hence $\text{2, 2 - dimethylpropane}$ produces more stable radicals than $\text{2-methylpropane}$ upon homolytic cleavage. Lower energy is required to generate free radicals in $\text{2, 2 - dimethylpropane}$ than $\text{2-methylpropane}$. Thus, $\text{2, 2 - dimethylpropane}$ has a lower bond dissociation enthalpy than $\text{2-methylpropane}$.

d) The given alkanes are cyclobutane and cyclopentane.

The homolytic cleavage of cyclobutane produces two radicals, which are attached to each other. Both the unpaired electrons are present on primary carbon atoms.

The homolytic cleavage of cyclopentane also produces two radicals, which are attached to each other. Both the unpaired electrons are present on primary carbon atoms.

In the radicals produced by cyclopentane, there is one alkyl substituent more as compared to the radicals in cyclobutane. The more the alkyl substituents, the more stable is the radical. Hence, in cyclopentane, the radicals are slightly more stable as compared to cyclobutane. Hence, cyclopentane has a lower bond dissociation enthalpy than cyclobutane.