Halohydrin formation is an anti addition. When a halogen reacts with an alkene in the presence of water, the halogen and the hydroxyl group end up on opposite faces of the original double bond. This stereochemical outcome is a direct consequence of the cyclic halonium ion intermediate that forms during the reaction.
Why the Addition Is Anti
The key to understanding halohydrin stereochemistry is the three-membered ring intermediate that forms in the first step. When bromine or chlorine approaches the alkene, it doesn’t simply form a carbocation on one of the carbon atoms. Instead, the halogen bridges across both carbons, creating a cyclic bromonium or chloronium ion. This ring sits on one face of the molecule and physically blocks anything from attacking that same side.
In the second step, water acts as the nucleophile. Because the halonium ion is sitting on one face, water can only approach from the opposite face, in what chemists call a backside attack. This forces the hydroxyl group onto the side opposite the halogen, producing anti addition with a dihedral angle of 180° between the two new bonds. On a cyclohexene ring, for example, this means the bromine and hydroxyl group end up in a trans relationship.
The bridged intermediate also prevents carbocation rearrangements. Because the positive charge is delocalized across the three-membered ring rather than sitting on a single carbon, the skeleton of the molecule stays intact. You won’t see the methyl or hydride shifts that plague other reactions involving carbocation intermediates.
Enantiomers but Not Syn Products
One subtlety worth understanding: while the addition is always anti, the halogen can initially approach the alkene from either the top or the bottom face. This means you get a mixture of enantiomers (mirror-image products) when the substrate allows it. What you do not get is any syn addition product. The reaction is stereoselective for anti but not enantioselective, so expect a racemic mixture of trans products rather than a single enantiomer.
Where the Halogen and OH End Up
On an unsymmetrical alkene, the halogen and hydroxyl group don’t attach randomly. The positive charge in the halonium ion is concentrated at the more substituted carbon, because that position better stabilizes a positive charge. Water, being the nucleophile, attacks at that more substituted carbon. The halogen therefore ends up on the less substituted carbon, and the hydroxyl group lands on the more substituted one. This gives a Markovnikov-like orientation for the OH group.
Reagents and Conditions
Halohydrins form when you treat an alkene with bromine or chlorine in the presence of water. The water competes with the halide ion as a nucleophile. Because water is present at a much higher concentration than the halide released in step one, it is more likely to be in the right position to attack the halonium ion. This is what tips the reaction toward a halohydrin rather than a vicinal dihalide.
In practice, most alkenes aren’t very soluble in water, so the reaction is typically run in a mixed solvent like aqueous DMSO. A commonly used reagent is N-bromosuccinimide (NBS), which slowly decomposes in water to release bromine at a controlled rate. NBS is safer and easier to handle than molecular bromine, and it produces the same bromonium ion intermediate. Cyclohexene treated with NBS in aqueous DMSO, for instance, gives a racemic mixture of trans-bromohydrin.
When Anti Selectivity Breaks Down
The strong preference for anti addition can erode under certain conditions. Highly polar solvents or alkenes bearing electron-donating groups can stabilize an open carbocation intermediate rather than the bridged halonium ion. When the ring opens to a free carbocation, nucleophilic attack is no longer restricted to one face, and some syn product can form. For most standard substrates and reaction conditions in an introductory organic chemistry course, though, you can treat halohydrin formation as exclusively anti.
Connection to Epoxide Synthesis
The anti relationship between the halogen and the hydroxyl group is not just a stereochemical curiosity. It is exactly what makes halohydrins useful as precursors to epoxides. When you treat a halohydrin with a base, the oxygen is deprotonated to form an alkoxide. That alkoxide then performs an intramolecular backside attack on the adjacent carbon, kicking out the halide and closing a three-membered epoxide ring. This works precisely because the oxygen and the halogen are anti to each other, which is the geometric requirement for a backside displacement. If the addition had been syn, the oxygen would be on the same face as the leaving group and the ring closure couldn’t happen.