Free Radical Halogenation Mechanism

Despite being described as relatively unreactive, alkanes do undergo one characteristic reaction

with the halogens (chlorine and bromine most readily; fluorine reacts explosively and iodine

barely reacts at all): a hydrogen is replaced by a halogen atom under the influence of ultraviolet

light or heat. This is a free-radical substitution, and it proceeds through a self-propagating

chain mechanism with three distinct stages, illustrated here for the monochlorination of methane.

Initiation. UV light or heat supplies enough energy to homolyse the weak $\text{Cl--Cl}$ bond,

splitting it symmetrically so that each chlorine atom keeps one electron of the shared pair and

becomes a highly reactive chlorine free radical:

Propagation (two repeating steps that keep the chain going without net radical consumption).

First, a chlorine radical abstracts a hydrogen atom from methane, forming HCl and a methyl

radical:

Second, the methyl radical attacks another chlorine molecule, forming the product chloromethane

and regenerating a chlorine radical that can start the cycle again:

These two steps alternate thousands of times per initiation event, which is why only a trace of

light/heat is needed to convert a large amount of methane; excess chlorine further substitutes

$\text{CH}_3\text{Cl}$ to $\text{CH}_2\text{Cl}_2$, $\text{CHCl}_3$ and eventually

$\text{CCl}_4$.

Termination. The chain eventually stops when two radicals collide and combine, consuming

radicals without producing new ones, e.g.

$\text{Cl}^{\bullet} + \text{Cl}^{\bullet} \rightarrow \text{Cl}_2$,

$\text{CH}_3^{\bullet} + \text{Cl}^{\bullet} \rightarrow \text{CH}_3\text{Cl}$, or

$\text{CH}_3^{\bullet} + \text{CH}_3^{\bullet} \rightarrow \text{C}_2\text{H}_6$ (the source

of trace ethane found in chlorinated methane).

Selectivity. When a higher alkane has more than one type of hydrogen (primary/1°,

secondary/2°, tertiary/3°), halogenation does not remove them statistically at random: a

tertiary $\text{C--H}$ bond is weaker and gives a more stable radical (more alkyl groups donate

electron density and hyperconjugate into the half-filled orbital) than a secondary, which in turn

is more stable than a primary radical, so the order of ease of radical formation -- and hence of

H-abstraction -- is $3° > 2° > 1°$. Chlorination is only mildly selective (roughly $5:3.8:1$ for …