Question

Alkanes are halogenated in presence of sunlight. The reaction follows free radical mechanism which completes in three steps.
$\mathrm{R}-\mathrm{H}+{\mathrm{X}}_2\stackrel{\mathrm{h\nu }}{\to }\mathrm{R}-\mathrm{X}+\mathrm{H}-\mathrm{X}$
(i) Chain initiation (fast)
$\mathrm{X}-\mathrm{X}\stackrel{\mathrm{h\nu }}{\to }\mathrm{X}\mathrm{°}+\mathrm{X}\mathrm{°}$
(ii) Chain propogation (slow)
$\mathrm{R}-\mathrm{H}+\mathrm{X}\mathrm{°}\stackrel{\mathrm{RDS}}{\to }\mathrm{H}-\mathrm{X}+\mathrm{R}\mathrm{°}$
R°+X–X $\underset{}{\overset{}{\to }}$ R–X + X°
(iii) Chain termination (fast)
R° + X° $\underset{}{\overset{}{\to }}$ R – X
R°+ R° $\underset{}{\overset{}{\to }}$ R – R
X° + X° $\underset{}{\overset{}{\to }}$ X – X
Rate of this reaction increasing with stability of free radical formed and ease of its formation.
Rate 1° < 2°<3° (for same halogen)
Rate F₂ > Cl₂ > Br₂ > I₂ (for same alkane)
If excess of halogen is used one by one all H are replaced by X.