R3.3.3 Substitution reactions with alkanes Notes

Radical Substitution Reactions: Mechanism and Stability of Alkanes

The Mechanism of Radical Substitution

1. Radical substitution reactions proceed through three key stages: initiation, propagation, and termination.
2. Let’s break these down with the example of methane ($CH_4$) reacting with chlorine ($Cl_2$) under UV light.

1. Initiation: Formation of Radicals

1. The initiation step generates radicals, which are essential to start the chain reaction.
2. For $Cl_2$, UV light provides the energy needed to break the chlorine-chlorine bond via homolytic fission: $$
   Cl_2 \xrightarrow{\text{UV light}} Cl• + Cl•
   $$
3. Each chlorine radical now has one unpaired electron, making it highly reactive.

2. Propagation: The Chain Reaction

1. In the propagation stage, radicals react with stable molecules to form new radicals, perpetuating the chain reaction.
2. For methane and chlorine, this involves two key steps:

Step 1:

A chlorine radical reacts with a methane molecule, abstracting a hydrogen atom and forming hydrogen chloride ($HCl$) and a methyl radical ($CH_3•$):

$$
Cl• + CH_4 \rightarrow CH_3• + HCl
$$

Step 2:

The methyl radical reacts with another chlorine molecule, forming chloromethane ($CH_3Cl$) and regenerating a chlorine radical:

$$
CH_3• + Cl_2 \rightarrow CH_3Cl + Cl•
$$

3. Termination: Stopping the Reaction

1. The termination step occurs when two radicals combine, forming a stable molecule and effectively removing radicals from the reaction mixture.
2. In the chlorination of methane, possible termination reactions include: $$
   Cl• + Cl• \rightarrow Cl_2
   $$
   $$
   CH_3• + Cl• \rightarrow CH_3Cl
   $$
   $$
   CH_3• + CH_3• \rightarrow C_2H_6
   $$
3. Termination reduces the concentration of radicals, eventually bringing the reaction to a halt.

Why Are Alkanes Stable but Reactive in Radical Reactions?

1. Alkanes, such as methane ($CH_4$), are known for their chemical stability. This stability arises from two key factors:
2. Strong Bonds:
   1. The carbon-hydrogen ($C-H$) bond has a bond enthalpy of 414 $\mathrm{kJ \, mol}^{-1}$, and the carbon-carbon ($C-C$) bond has a bond enthalpy of 346 $\mathrm{kJ \, mol}^{-1}$.
   2. Breaking these bonds requires significant energy.
3. Nonpolar Nature:
   1. Alkanes consist of nonpolar bonds, making them unreactive toward polar reagents.
   2. However, under the right conditions, such as exposure to UV light, alkanes can participate in radical reactions.

This is because:

1. Radical Reactions Are Thermodynamically Favorable:
   1. Although alkanes are kinetically stable, radical reactions are thermodynamically favorable due to the formation of strong new bonds (e.g., $C-Cl$ and $H-Cl$).
2. Radicals Are Highly Reactive:
   1. Once radicals are formed (e.g., $Cl•$), they can overcome the stability of alkanes by abstracting hydrogen atoms.