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Mix methane and chlorine gas together in a dark flask, and nothing happens for days. Shine a brief pulse of ultraviolet (UV) light into the flask, and it reacts immediatelyβ€”sometimes violentlyβ€”producing chloromethane and acidic hydrogen chloride gas.
The UV light provides the energy needed for homolytic fission, where a covalent bond breaks evenly so each atom keeps one electron, creating uncharged species with unpaired electrons called free radicals.
Why does the incoming UV light specifically break the chlorine bond instead of the methane bonds?

Step 1: Initiation

In the Initiation Step, ultraviolet (UV) light (hΞ½h\nu) provides the activation energy (EaE_a) to break the relatively weak halogen-halogen bond homolytically: Cl2β†’hΞ½2Clβˆ™\text{Cl}_2 \xrightarrow{h\nu} 2\text{Cl}^\bullet. The Cβˆ’H\text{C}-\text{H} bond (413Β kJ/mol413\text{ kJ/mol}) remains intact because the Clβˆ’Cl\text{Cl}-\text{Cl} bond (242Β kJ/mol242\text{ kJ/mol}) is much weaker.
Now that we have highly reactive chlorine radicals with an incomplete octet, how do they attack stable methane molecules?

Step 2: Propagation

Propagation is a self-sustaining two-step cycle where a radical reacts with a stable molecule to generate a product and a new radical.
In the first propagation step, a chlorine radical abstracts a hydrogen atom from methane to form HCl\text{HCl} and a methyl radical (CH3βˆ™\text{CH}_3^\bullet).
In the second propagation step, the newly formed methyl radical attacks a Cl2\text{Cl}_2 molecule, yielding the target haloalkane (CH3Cl\text{CH}_3\text{Cl}) and regenerating the chlorine radical to continue the chain.