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Chapter 11: Problem 72

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

Short Answer

Expert verified
The mechanism consists of: initiation stage where bromine molecules are broken down into radicals by sunlight; propagation stage where the radicals react with methane and bromine molecules forming the products; termination stage where the reaction is terminated by the recombination of radicals to form the final products.

Step by step solution

01

Analyse the reactants and products

The reactants are methane (\( \mathrm{CH}_{4} \)) and bromine (\( \mathrm{Br}_{2} \)). The products are methyl bromide (\( \mathrm{CH}_{3}Br \)) and hydrogen bromide (\( \mathrm{HBr} \)). The loss of one hydrogen from methane and the addition of one bromine to form the products is indicative of a substitution reaction.
02

Postulate the reaction mechanism

In the first stage of the reaction, free radicals are created. This is called the initiation stage. The bromine molecule is split by sunlight into two bromine atoms. This is represented by the equation: \( \mathrm{Br}_{2} \rightarrow 2 \mathrm{Br}. \) Next, in the propagation stage, The bromine atom reacts with the methane molecule to form a methyl radical and hydrogen bromide. This is depicted by the equation: \( \mathrm{Br} + \mathrm{CH}_{4} \rightarrow \mathrm{CH}_{3} + \mathrm{HBr}. \) The methyl radical then reacts with another molecule of bromine to form the product methyl bromide and another bromine radical, continuing the chain: \( \mathrm{CH}_{3} + \mathrm{Br}_{2} \rightarrow \mathrm{CH}_{3}Br + \mathrm{Br}. \) Lastly, in the termination stage, the radicals (either two bromine radicals or a bromine and methyl radical) combine: \( 2 \mathrm{Br} \rightarrow \mathrm{Br}_{2} \) or \( \mathrm{CH}_{3} + \mathrm{Br} \rightarrow \mathrm{CH}_{3}Br \)
03

Verify the mechanism with the given reaction

The proposed mechanism accounts for the observed reaction. The bromine radicals formed in the initiation stage react with methane and bromine molecules in the propagation stage to create the observed products. The termination stage concludes the reaction with the recombination of radicals.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Substitution Reaction
Understanding the basics of a substitution reaction is pivotal for those diving into the intricacies of organic chemistry. In such a reaction, one atom or group in a molecule is replaced by another atom or group. An illustrative example of this process involves methane reacting with bromine vapor to create methyl bromide and hydrogen bromide.

Think of substitution reactions as an exchange between two 'partners', where one leaves and another comes in to take its place. This is particularly common in hydrocarbons, such as methane, where bromine, a more reactive element, substitutes a hydrogen atom. The versatility of this reaction allows for a range of organic compounds to be synthetically altered, enabling the creation of numerous useful products.
Free Radicals
Free radicals are atoms or groups of atoms that possess an unpaired electron, making them highly reactive and unstable. They seek stability by pairing their unpaired electron, often leading them to engage in chemical reactions that can dramatically alter the structure of molecules.

In this context, the reaction starts with the generation of bromine free radicals when bromine molecules absorb light and split apart. These reactive particles then go on to initiate a chain of events that defines the mechanism of the reaction. Comprehending free radicals is essential, as they play a central role not only in organic chemistry reactions but also in biological processes and health.
Initiation Stage
The initiation stage marks the launch of the substitution reaction. This phase involves the formation of free radicals, which act as the reaction's workhorses. As bromine absorbs light, it breaks into two solitary bromine atoms, each with an unpaired electron.

Represented by the equation: \( \text{Br}_2 \rightarrow 2 \text{Br} \cdot \), this stage is the spark that ignites the reaction. Without this initial split caused by light, the remaining stages wouldn't occur, highlighting the importance of this 'kickstarting' phenomenon in reactions involving free radicals.
Propagation Stage
Following initiation, the propagation stage is where the action truly unfolds. This sequence sees a series of successive reactions where the generated free radicals react with stable molecules to form new radicals. This creates a domino effect; each new radical then goes on to participate in further reactions.

During this stage, a bromine free radical interacts with methane, producing a methyl radical and hydrogen bromide: \( \text{Br} \cdot + \text{CH}_4 \rightarrow \text{CH}_3 \cdot + \text{HBr} \). Subsequently, the methyl radical reacts with another bromine molecule, sustaining the reaction cycle by creating methyl bromide and another bromine radical: \( \text{CH}_3 \cdot + \text{Br}_2 \rightarrow \text{CH}_3\text{Br} + \text{Br} \cdot \). This intricate interplay is fundamental in understanding how the molecules transform and why the products of the reaction emerge.
Termination Stage
Eventually, all good things must come to an end, and in the realm of chemical reactions, the termination stage represents this conclusion. Here, free radicals finally find partners, merging to form stable, non-radical compounds. This effectively brings the substitution reaction to a halt.

For instance, bromine radicals can combine to recreate a bromine molecule: \( 2 \text{Br} \cdot \rightarrow \text{Br}_2 \). Alternatively, a methyl radical could join forces with a bromine radical to form another molecule of methyl bromide: \( \text{CH}_3 \cdot + \text{Br} \cdot \rightarrow \text{CH}_3\text{Br} \). This stage is as crucial as the previous ones, as it prevents the reaction from running indefinitely, which would result in the consumption of all reactants and possibly the formation of unwanted by-products.

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