At both SL and HL, alkanes undergo radical substitution because ultraviolet radiation causes halogen molecules to form highly reactive radicals, which replace hydrogen atoms in the alkane. The reaction is substitution rather than addition because alkanes contain only strong single bonds and no carbontocarbon double bond.
A radical contains an unpaired electron. Alkanes are normally relatively unreactive because their CtoC and CtoH bonds are strong and mostly non-polar, but radicals can initiate a chain reaction.
The key structural reason is that alkanes are saturated hydrocarbons: every carbon valence is already used in single covalent bonds. They therefore cannot undergo electrophilic addition, which requires an electron-rich C=C bond. A halogen radical instead abstracts a hydrogen atom, generating an alkyl radical; this reactive intermediate then forms a CtoX bond while regenerating another halogen radical. This repeated radical regeneration makes the process a chain reaction. The radical is consumed in one propagation step but produced in the next.
Using methane and chlorine as an example, the overall reaction is:
| Stage | What happens |
|---|---|
| Initiation | UV radiation causes homolytic fission of the CltoCl bond: . Each chlorine atom receives one bonding electron. |
| Propagation 1 | A chlorine radical removes hydrogen from methane: . |
A mixture of products often forms because chlorinated products can undergo further substitution. Larger alkanes may also form different positional products.
A common misconception is that UV radiation acts as a catalyst. It does not: UV supplies the energy needed for homolytic fission during initiation.
Exam technique: For an “explain” question, connect alkane structure to reactivity, state that UV causes homolytic fission, and give initiation, propagation and termination equations. Show radical dots clearly and ensure propagation regenerates a radical.
