Halohydrin formation template

Propene and Br₂ in water form a bromonium ion; water attacks the more substituted carbon (anti) and loses H⁺ to give 1-bromopropan-2-ol.

Mechanism step by step

Bromine water: water, not bromide, opens the bromonium ion at the more substituted carbon

1. Bromonium ion – The π bond attacks Br; Br closes the three-membered ring
2. Water opens the ring – Anti attack at the more substituted carbon
3. Deprotonation – Water removes H⁺ from the oxonium ion

Key points

• Water is in huge excess, so it beats Br⁻ to the bromonium ion (only a little 1,2-dibromide forms).
• Markovnikov-like: the more substituted carbon carries more δ+ in the bromonium ion, so OH ends up there.
• Anti addition: cyclohexene gives trans-2-bromocyclohexanol.
• Base closes a halohydrin to an epoxide (intramolecular Williamson).

Drawn with 44 atoms, 11 arrows, 7 shapes – every one editable in MolDraw.

Sources

• Clayden, Greeves & Warren, Organic Chemistry, 2nd ed., OUP 2012, ch. 19 Textbook
• Smith, March's Advanced Organic Chemistry, 8th ed., Wiley 2020, ch. 15 Textbook