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Solved: When a mixture of methane and bromine is exposed

Chapter 13, Problem 97P

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QUESTION:

When a mixture of methane and bromine is exposed to visible light, the following reaction occurs slowly:

\(\mathrm{CH}_{4}(g)+\mathrm{Br}_{2}(g) \rightarrow \mathrm{CH}_{3} \mathrm{Br}(g)+\mathrm{HBr}(g)\)

Suggest a reasonable mechanism for this reaction. (Hint: Bromine vapor is deep red; methane is colorless.)

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QUESTION:

When a mixture of methane and bromine is exposed to visible light, the following reaction occurs slowly:

\(\mathrm{CH}_{4}(g)+\mathrm{Br}_{2}(g) \rightarrow \mathrm{CH}_{3} \mathrm{Br}(g)+\mathrm{HBr}(g)\)

Suggest a reasonable mechanism for this reaction. (Hint: Bromine vapor is deep red; methane is colorless.)

ANSWER:

Step 1 of 3

Reactions that occur in the presence of light are called photochemical reactions.

Generally, photochemical reactions follow free radical mechanisms.

Free radicals are atoms, molecules or ions that have one unpaired valence electron.

Here

A mixture of methane and bromine under visible light yields bromomethane and hydrogen bromide.

We need to propose a suitable mechanism

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