Allylic bromination: practice and mechanism
NBS with light brominates radically at the allylic position (the carbon next to the double bond). The double bond survives because NBS keeps the Br₂ concentration very low.
Example
NBS keeps the Br₂ concentration very low, so instead of addition a radical substitution takes place: at the allylic position, because the allyl radical is resonance-stabilised. The double bond is kept.
Mechanism step by step
- 1
Light generates Br· radicals from a little Br₂ (supplied by NBS).
- 2
Br· abstracts an allylic H. The allyl radical is resonance-stabilised (spread over two carbons).
- 3
The radical reacts with Br₂ to the allyl bromide and a new Br·. The HBr formed regenerates Br₂ from NBS.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
Often confused with: Addition of bromine · Benzylic bromination
Frequently asked questions
Why no addition to the double bond?
The Br₂ concentration stays very low. Addition needs more Br₂ and cannot compete, while the radical chain can.
Can product mixtures form?
Yes, with unsymmetrical allyl radicals Br can bond at either end and the double bond shifts. Exercises usually use symmetric cases such as cyclohexene.