Radical bromination of alkanes: practice and mechanism
Bromine is very selective in the radical step: it almost only abstracts the H from the most substituted carbon (tertiary > secondary > primary), because the most stable radical forms there.
Example
Bromine radicals are very selective: they abstract the hydrogen that gives the most stable radical. Tertiary radicals are the most stable, so the bromine ends up on the tertiary carbon.
Mechanism step by step
- 1
Initiation: light splits Br₂ homolytically into two Br· radicals.
- 2
Propagation 1: Br· abstracts an H from the alkane, preferably from the tertiary carbon. HBr and the most stable alkyl radical form.
- 3
Propagation 2: the alkyl radical reacts with Br₂ to the bromoalkane and a new Br·, which continues the chain. Termination when two radicals meet.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- A secondary C–H is only brominated if there is no tertiary one. Bromine picks the carbon that gives the most stable radical, here the tertiary one.
- Primary radicals are the least stable. Unlike chlorine, bromine practically never brominates the CH₃ group while a more substituted C–H is available.
Often confused with: Benzylic bromination
Frequently asked questions
Why is bromine more selective than chlorine?
H abstraction is endothermic for Br·, so the transition state is late and radical-like. Differences in radical stability therefore count fully. For Cl· the step is exothermic and hardly selective.
Which radicals are most stable?
Tertiary > secondary > primary > methyl, because alkyl groups offset the electron deficiency by hyperconjugation. Allyl and benzyl radicals are even more stable through resonance.