8.6b Halohydrin Formation
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|Two Groups Added|| || |
|Reagents Added|| X2 (Cl2 or Br2), H2O|
|Other Characteristics||Intermediate is a 3-membered ring (halonium ion).|
Halohydrin formation (specifically chlorohydrin and bromohydrin formation) is the result of the addition of a halogen (Cl or Br--less substituted side) and a hydroxyl group (more substituted side) across an alkene. The reaction proceeds with Anti stereospecificity, and as the intermediate is NOT a carbocation rearrangements are NOT possible. The reagents are either Cl2 or Br2 in H2O.
This reaction is considered Markovnikov addition as the electrophile (Cl or Br) ends up on the less substituted side, and the nucleophile (OH group from water) ends up on the more substituted side.
Halohydrin Formation Mechanism
I show the mechanism for bromohydrin formation specifically but the mechanism for chlorohydrin formation is analogous in every respect with Cl2 being used in place of Br2.
Step 1: The pi electrons of the alkene attack one of the bromine atoms which attacks one the of the alkene carbons back forming a three-membered ring with Br referred to as a bromonium ion (generically referred to as a halonium ion and specifically a chloronium ion in the reaction with Cl2). The bond between the bromine atoms also breaks in this step forming a bromide ion.
While the bromonium ion is more stable than a carbocation it is still a rather high energy intermediate, and this first step is the rate determining step (slow step) of the reaction. Also, the bromonium intermediate does not undergo rearrangements such as are possible with carbocations.
Step 2: Rather than the bromide ion, H2O (which is present in a much higher concentration being the solvent) carries out back-side attack (as in SN2) on the more substituted carbon of the bromonium ion resulting in the bond to the Br breaking (analogous to loss of a leaving group in SN2 even though it is still attached to the adjacent carbon atom) and the opening of the 3-membered ring.
The back-side attack results in inversion of configuration at the carbon attacked explaining why the bromine and hydroxyl group add to opposite sides of the alkene (i.e. anti addition).
Step 3: The intermediate is deprotonated by another water molecule in the solution yielding the halohydrin product.
Halohydrin Formation Alternative; Alcohol Instead of H2O
Halogenation can also take place using an alcohol as the solvent instead of water. The reaction and mechanism is directly analogous to that of halohydrin formation with the alcohol replacing the role of water in the mechanism. The result is the addition of a halogen (Cl or Br--less substituted side) and an alkoxy group (more substituted side) across an alkene.