When a mixture of methane and bromine is exposed to visible 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.)

Radical halogenation is a chemical reaction where a halogen atom is introduced into an organic molecule in the presence of sufficient energy, typically from heat or light. This process involves the creation of highly reactive species known as radicals, which then trigger a chain reaction that results in the halogenation of the organic compound.

In the context of the given exercise, where methane reacts with bromine under the influence of light, the color change mentioned (from deep red to colorless) is indicative of the bromine molecules reacting and being consumed. The bromine molecule dissociates into two bromine radicals, which are the key reactive intermediates that drive the reaction forward. As this process unfolds, various steps ensure the reaction's progression from reactants to products.

The initiation step in radical reactions is where the first reactive intermediates are formed. It is essentially the 'starting gun' for the reaction. This step is crucial because it generates the radicals that will participate in further steps of the reaction.

For the exercise involving methane and bromine, the initiation occurs when the bromine molecule (\r\( \r\text{Br}_2 \r\)) absorbs visible light and splits into two bromine radicals (\r\( \r\text{Br} \r\text{•} \r\)). This is a homolytic cleavage, meaning that the electrons in the bromine-bromine bond are evenly divided between the two bromine atoms, each taking one electron and thereby forming two radicals.

After initiation, the reaction enters the propagation step. This is where the bulk of the reaction occurs and the radicals generated in the initiation step react with stable molecules to form new radicals, perpetuating the chain reaction.

Within the methanol and bromine example, there are two propagation steps. The first is when a bromine radical reacts with methane (\r\( \r\text{CH}_4 \r\)), resulting in the formation of a methyl radical (\r\( \r\text{CH}_3 \r\text{•} \r\)) and hydrogen bromide (\r\( \r\text{HBr} \r\)). The newly formed methyl radical is highly reactive and doesn’t remain idle. It swiftly reacts with another bromine molecule (\r\( \r\text{Br}_2 \r\)), forming bromomethane (\r\( \r\text{CH}_3\text{Br} \r\)) and another bromine radical. This newly regenerated bromine radical can then react with another methane molecule, creating a cycle that continues to feed the reaction.

Finally, the termination step caps off the radical chain reaction. At this point, radical species combine to form stable, non-radical compounds, thus effectively halting the chain propagation. The combination of radicals happens as their concentration decreases and the likelihood of their encounter increases.

Several terminations are possible in the methane bromination example. Two bromine radicals can recombine to form a bromine molecule (\r\( \r\text{Br} \r\text{•} + \text{Br} \r\text{•} \rightarrow \text{Br}_2 \r\)). Alternatively, two methyl radicals can join to create ethane (\r\( \r\text{CH}_3 \r\text{•} + \text{CH}_3 \r\text{•} \rightarrow \text{CH}_3\text{CH}_3 \r\)). Another possibility is a methyl radical combining with a bromine radical to give bromomethane (\r\( \r\text{CH}_3 \r\text{•} + \text{Br} \r\text{•} \rightarrow \text{CH}_3\text{Br} \r\)). These termination reactions ensure that the radical species are depleted, thereby concluding the overall chemical process.