Question

When a mixture of methane and bromine is exposed to light, the following reaction occurs slowly: $$ \mathrm{CH}_{4}(g)+\mathrm{Br}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{Br}(g)+\mathrm{HBr}(g) $$ Suggest a reasonable mechanism for this reaction. (Hint: Bromine vapor is deep red; methane is colorless.)

Short answer

The reaction mechanism involves initiation, propagation, and termination steps involving radicals.

Step-by-step solution

Initiation

In the initiation step, light provides the energy needed to break the bromine molecule into two bromine radicals. This can be represented as: \[ \text{Br}_2 \xrightarrow{hu} 2\,\text{Br}\cdot \] Here, \(hu\) represents the energy from light, and \(\text{Br}\cdot\) signifies a bromine radical.

Propagation - First Reaction

In the first propagation step, a bromine radical reacts with a methane molecule (\( \text{CH}_4 \)) to form methyl radical and hydrogen bromide. \[ \text{Br}\cdot + \text{CH}_4 \rightarrow \text{CH}_3\cdot + \text{HBr} \] The methyl radical, \(\text{CH}_3\cdot\), is generated along with the formation of hydrogen bromide \(\text{HBr}\).

Propagation - Second Reaction

The methyl radical from the previous step reacts with another bromine molecule to form bromomethane and regenerate a bromine radical. \[ \text{CH}_3\cdot + \text{Br}_2 \rightarrow \text{CH}_3\text{Br} + \text{Br}\cdot \]

Termination

The reaction can end when two radicals combine to form a stable molecule. Possible termination steps include: 1. \( \text{Br}\cdot + \text{Br}\cdot \rightarrow \text{Br}_2 \)2. \( \text{CH}_3\cdot + \text{Br}\cdot \rightarrow \text{CH}_3\text{Br} \)3. \( \text{CH}_3\cdot + \text{CH}_3\cdot \rightarrow \text{C}_2\text{H}_6 \)These termination steps reduce the number of reactive radicals in the system.

Key concepts

Initiation Step

In a chemical reaction mechanism, the initiation step is where it all begins. For the reaction between methane and bromine, light acts as a trigger. It provides the energy needed to break the bonds in a bromine molecule. This process is called photolysis. When the bromine molecule ( \( \text{Br}_2 \)) absorbs light energy, it splits into two bromine radicals ( \( \text{Br}\cdot \)).
This step is crucial as it sets off a chain of subsequent reactions. Without the formation of any radicals, the reaction would not proceed. Bromine radicals are highly reactive species due to their unpaired electron. This makes them ready to engage in further reactions with the methane molecules.

Propagation Step

The propagation step of a reaction proceeds in a chain-like fashion, where the generated radicals react further to sustain the reaction. There are usually several propagation steps in a reaction mechanism.
In the first propagation step of the given reaction, a bromine radical reacts with a methane molecule, \( \text{CH}_4 \), to form a methyl radical ( \( \text{CH}_3\cdot \)) and hydrogen bromide (\( \text{HBr} \)). In simpler terms:

• The bromine radical extracts a hydrogen atom from methane, resulting in the formation of a methyl radical.
• At the same time, hydrogen bromide is formed.

The second propagation step involves this new methyl radical reacting with another \( \text{Br}_2 \) molecule:

• It leads to the production of bromomethane (\( \text{CH}_3\text{Br} \)).
• The cyclical nature of propagation reactions results in the regeneration of the bromine radical, which can continue to react with new methane molecules.

Termination Step

Termination steps occur when reactive radicals combine to form stable molecules, effectively halting further reaction progression. In the context of the methane-bromine reaction, there are several potential termination avenues.
Some important possibilities include:

• Two bromine radicals can combine to reform a \( \text{Br}_2 \) molecule.
• A bromine radical may combine with a methyl radical, forming bromomethane (\( \text{CH}_3\text{Br} \)).
• Two methyl radicals may unite to create ethane (\( \text{C}_2\text{H}_6 \)).

These termination reactions are stabilizing, as they reduce the number of radicals, thereby diminishing the reaction rate. Understanding these mechanisms is crucial to predict the outcomes and scalability of various chemical reactions.

Bromine Radical

A bromine radical is a bromine atom with an unpaired electron, making it extremely reactive. It plays a pivotal role in radical chain reactions, like the one between methane and bromine.
Here’s why bromine radicals are significant:

• They initiate propagation by abstracting hydrogen from methane, leading to the creation of methyl radicals.
• They are regenerated during the propagation steps, continuing the reaction cycle.
• Any combination with another radical results in termination, reducing their concentration.

These characteristics explain why bromine radicals are essential for the continuation of the reaction but can abruptly halt it when involved in termination steps.

Methyl Radical

The methyl radical (\( \text{CH}_3\cdot \)) is a carbon-centered radical resulting from the hydrogen abstraction from methane by a bromine radical. It's an essential intermediate in the mechanistic pathway of the methane-bromine reaction.
Key points about methyl radicals include:

• They are generated in the first propagation step of the reaction.
• They react with bromine molecules to form bromomethane and regenerate a bromine radical.
• They can also be involved in termination reactions, combining with themselves or bromine radicals.

Understanding methyl radicals is crucial, as they help to sustain the reaction by continuing the chain propagation and contributing to the formation of the final products.