`CH_(4) + Cl_(2) overset(hv)(to)CH_(3)(Cl) + HCl`
To obtain high yields of `CH_(3)Cl`, 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 reation mixture ensure 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 to .^(**)CH_(2)Cl + HCl`
`.^(**)CH_(2)Cl + Cl_(2) to 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.