`CH_4 + Cl_2 overset(hv)to CH_3 Cl + HCl` to obtain high yields of `CH_3Cl` , the ratio of `CH_4` to `Cl_2` must be

Options

A. high
B. low
C. equal
D. Can't be predicted

Answer

A

Explanation

In order to maximize the amount of monohalogenated product obtained, a radical substitution reaction should be carried out in the presence of excess of alkane. The presence of excess alkane in the reaction mixture ensures that there is a greater probability of the halogen radical colliding with a molecule of alkane than with a molecule of alkyl halide. This is true even toward the end of the reaction, by which time a considerable amount of alkyl halide will have been formed. If the halogen radical abstracts a hydrogen from a molecule of alkyl halide rather than from a molecule of alkane, a dihalogenated product will be obtained.
`{:(Cl^(*) + CH_(3)Cl rarr CH_(2)Cl + HCl),(CH_(2)Cl + Cl_(2) rarr CH_(2)Cl_(2) + Cl^(*)):}`
Bromination of alkanes follows the same mechanism as chlorination. The only difference is that chlorination produces alkyl chlorides, whereas bromination forms alkyl bromides.