Bromohydrin Formation
Bromohydrin formation adds Br and OH across the C=C double bond of an alkene. Step-by-step mechanism, reagents and practice.
What happens
Bromohydrin formation adds Br and OH across the C=C double bond of an alkene. The reaction uses Br₂ in water (H₂O). The double bond becomes a single bond, with Br added to one carbon and OH added to the other. The OH goes to the more substituted carbon, while Br goes to the less substituted carbon.
Reaction summary
2-Methylpropene + Br₂/H₂O → 1-Bromo-2-methylpropan-2-ol
Reagents and conditions
- Starting material: 2-Methylpropene (alkene)
- Reagents: Br₂; H₂O
- Products: 1-Bromo-2-methylpropan-2-ol (bromohydrin)
Mechanism
- Bromonium Ion Formation: The alkene's π bond attacks one bromine of Br₂; the Br–Br bond breaks toward the other bromine, which leaves as bromide. This forms a bridged three-membered bromonium ion.
- Anti Attack by Water: Water's lone pair attacks the more substituted carbon of the bromonium ion from the side opposite the bridging bromine; as the new C–O bond forms, the C–Br bond to that same carbon breaks, opening the ring and leaving Br on the other carbon.
- Deprotonation: A proton is removed from the oxonium intermediate by another water molecule, giving the neutral bromohydrin.
- Product: The neutral bromohydrin, 1-bromo-2-methylpropan-2-ol.
Selectivity
- Regioselectivity: OH is added to the more substituted carbon, while Br ends up on the less substituted carbon.
- Stereoselectivity: Anti addition: Br and OH are added from opposite sides of the double bond, because H₂O attacks the bromonium ion from the side opposite the bridging Br.
Common mistakes
- There is no free carbocation here, a bridged bromonium ion forms instead, so carbocation rearrangements do not occur.