Chlorohydrin Formation
Chlorohydrin formation adds Cl and OH across an alkene's C=C double bond. Step-by-step mechanism, reagents and practice.
What happens
Chlorohydrin formation adds Cl and OH across an alkene's C=C double bond. The reaction uses Cl₂ in water (H₂O): the double bond breaks, Cl is added to one carbon, and OH is added to the other. The OH usually goes to the more substituted carbon, while Cl goes to the less substituted carbon.
Reaction summary
Propene + Cl₂/H₂O → 1-Chloropropan-2-ol
Reagents and conditions
- Starting material: Propene (alkene)
- Reagents: Cl₂; H₂O
- Products: 1-Chloropropan-2-ol (chlorohydrin)
Mechanism
- Chloronium Ion Formation: The alkene's π bond attacks one chlorine of Cl₂; the Cl–Cl bond breaks toward the other chlorine, which leaves as chloride. This forms a bridged three-membered chloronium ion.
- Anti Attack by Water: Water's lone pair attacks the more substituted carbon of the chloronium ion from the face opposite the bridging chlorine; as the new C–O bond forms, the C–Cl bond to that same carbon breaks, opening the ring and leaving Cl on the other carbon.
- Deprotonation: A second water molecule's lone pair removes a proton from the oxonium intermediate; the O–H bond breaks back onto oxygen, giving the neutral chlorohydrin and a hydronium ion.
- Product: The neutral chlorohydrin, 1-chloropropan-2-ol.
Selectivity
- Regioselectivity: OH is added to the more substituted carbon, while Cl ends up on the less substituted carbon, matching the regioselectivity stated in the source.
- Stereoselectivity: Anti addition: Cl and OH end up on opposite faces of the former double bond, because water attacks the chloronium ion from the side opposite the bridging Cl.
Common mistakes
- There is no free carbocation here, a bridged chloronium ion forms instead, so hydride/alkyl-shift rearrangements (unlike HBr addition or acid-catalyzed hydration) do not occur.