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

There are three steps in the reaction mechanism. First, the bromine molecule splits into two bromine atoms due to the effect of light. Secondly, the methane molecule reacts with a bromine atom to form HBr and a methyl radical. Lastly, the methyl radical reacts with another bromine molecule to form CH_3Br and a bromine atom.

Step-by-step solution

Breakdown of Reagents

To start, separate the molecules of the reagents. The bromine molecule \(Br_2\) breaks down into two bromine atoms in the presence of light. This step is known as photochemical homolysis. This process is indicated by the change of color from deep red to colorless as each bromine molecule is split into two bromine atoms. So, it can be represented as follows: \(Br_2 \xrightarrow{h\nu} 2Br\).

First Reactive Stage

In the second step, the methane molecule reacts with one of the bromine atoms resulting from the previous step. The bromine atom is very reactive, it will combine with a hydrogen atom from methane. This results in the formation of Hydrogen Bromide (HBr) and a methyl radical. It can be depicted as: \(CH_4 + Br \longrightarrow CH_3 + HBr\).

Second Reactive Stage

The methyl radical formed from the previous step, \(CH_3\), is highly reactive and will react rapidly with another \(Br_2\) molecule to form \(CH_3Br\) and another bromine atom \(Br\). This reaction can be summarized as follows: \(CH_3 + Br_2 \longrightarrow CH_3Br + Br\).

Key concepts

Photochemical Homolysis

When light interacts with certain chemical bonds, it can provide the energy needed to break these bonds in a process called photochemical homolysis. This is central to understanding reactions involving light, such as the reaction between methane and bromine. Photochemical homolysis refers specifically to the cleavage of a chemical bond where the bond's two electrons are split evenly between the two atoms.

For example, in the presence of light (represented as \( hu \)), a bromine molecule (\(Br_2\)) undergoes homolysis to form two highly reactive bromine atoms (\(Br\cdot\)). The equation \(Br_2 \xrightarrow{hu} 2Br\cdot\) illustrates this step, showing how each bromine atom retains one electron from the broken bond. This initial stage leads to color change—from deep red (as seen in bromine gas) to colorless—hinting at the molecular change.

Bromine and Methane Reaction

The reaction between bromine and methane, when exposed to light, is a classic example of a photochemical process in organic chemistry. When methane (\(CH_4\)) is mixed with bromine gas (\(Br_2\)), under the influence of light, they undergo a reaction to yield bromomethane (\(CH_3Br\)) and hydrogen bromide (\(HBr\)).

The reaction kicks off with the creation of reactive bromine atoms through photochemical homolysis. Following this, a bromine atom approaches methane and abstracts one hydrogen atom from it, resulting in the formation of \(HBr\) and a methyl radical (\(CH_3\cdot\)). The equation \(CH_4 + Br\cdot \rightarrow CH_3\cdot + HBr\) depicts this exchange. The reactive methyl radical is then ready to engage in further reactions, illustrating the complexity and chain-reaction nature of this chemical process.

Mechanism of Halogenation

Halogenation is a type of chemical reaction where a halogen atom, such as bromine, is introduced into an organic molecule. This process typically involves the generation of reactive intermediates, such as free radicals, cations, or anions, which can quickly react with other species to create a halogenated compound.

In the context of the reaction between methane and bromine, the halogenation mechanism is a chain reaction involving three main steps: initiation, propagation, and termination. The initiation step, as previously described, is the photochemical homolysis of bromine to form bromine atoms. During the propagation steps, a bromine atom reacts with methane to form \(HBr\) and a methyl radical, which then reacts with another bromine molecule to form \(CH_3Br\) and regenerate a bromine atom (\(CH_3 \cdot + Br_2 \rightarrow CH_3Br + Br\cdot\)). The propagation phase continues as long as there are reactants available. The termination steps, which are not detailed in the provided exercise, involve the combination of free radicals to form a stable product, effectively ending the chain reaction.