Halohydrin formation: practice and mechanism
Br₂ forms a bromonium ion. Water is in large excess and attacks the more substituted carbon (it carries more positive charge). OH ends up on the more substituted carbon, Br on the less substituted one, both anti.
Example
Bromine forms a bromonium ion with the alkene. Because water is present in large excess, it opens the ring instead of bromide: at the more substituted carbon (Markovnikov) and from the back (anti). A bromo alcohol forms.
Mechanism step by step
- 1
Br₂ approaches the double bond and the π electron pair attacks: a cyclic bromonium ion and Br⁻ form.
- 2
Water is in large excess and attacks the bromonium ion from the back at the more substituted carbon.
- 3
Deprotonation of the oxonium ion gives the bromohydrin: OH and Br are anti.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- Regiochemistry swapped: water attacks the bromonium ion where there is more positive charge, i.e. at the more substituted carbon. The Br stays on the CH₂.
Often confused with: Addition of bromine · Acid-catalysed hydration
Frequently asked questions
Why does water attack rather than bromide?
Water is the solvent and so in huge excess. Although bromide is the better nucleophile, quantity wins.
What are halohydrins good for?
With base the alkoxide closes the ring to the epoxide (intramolecular SN2). That gives epoxides without a peracid.