Radical addition of HBr (anti-Markovnikov): practice and mechanism
With peroxide the addition is radical: the bromine radical goes to the terminal carbon, so Br ends up on the less substituted carbon (anti-Markovnikov).
Example
Peroxides start a radical chain: the bromine radical adds first to the terminal carbon so that the more stable (more substituted) radical forms. Bromine therefore ends up on the less substituted carbon.
Mechanism step by step
- 1
The peroxide splits into radicals (RO·) that make a bromine radical Br· from HBr (initiation).
- 2
Br· adds to the terminal carbon so that the more stable, more substituted carbon radical forms.
- 3
The carbon radical takes an H from HBr, forming a new Br· (chain reaction).
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 HBr (Markovnikov) · Addition of bromine
Frequently asked questions
Why does this only work with HBr?
Only with HBr are both chain steps exothermic. In HCl the H–Cl bond is too strong; with HI the iodine radical addition is too unfavourable.
Isn't the result the same as hydroboration?
The regioselectivity yes (anti-Markovnikov), but here a bromoalkane forms, with hydroboration an alcohol.