Chapter 22, Problem 28A

Interpretation Introduction

Interpretation:

The conversion of methane to bromomethane needs to be explained.

Concept introduction:

In a bromination reaction, Br₂ molecule is a reactant which reacts with other molecule form monobrominated product. Here, in Br₂ molecule, one bromine atom has partial positive and other has partial negative charged.

Answer to Problem 28A

Correct Answer: Option (c) is correct.

Explanation of Solution

In the presence of UV light, the reaction of methane takes place with bromine molecule which results in the formation of bromomethane and hydrogen bromide.

The reaction is represented as follows:

${\text{CH}}_{\text{4}}+{\text{Br}}_{\text{2}}^{}\to {\text{CH}}_{\text{3}}\text{Br}+\text{HBr}$

Here, tetrabromo product is possible if all the H atoms in methane are replaced by Br atom. The reaction can stop after the formation of bromomethane if large concentration of methane is taken with respect to Br₂.

Thus, the reactant which can be used to convert methane to btomonethane is Br₂ molecule.