Bromine adds across propene to give 1,2-dibromopropane — the two bromines end up on adjacent carbons and, on a ring, trans to one another.
Alkenes are electron-rich: the exposed π bond makes an excellent nucleophile. Molecular halogens such as Br2 and Cl2 are polarizable enough to act as electrophiles toward that π cloud. The result is a two-step addition that runs through an unusual bridged intermediate — the halonium ion — which controls both the stereochemistry (always anti) and, when a solvent nucleophile competes, the regiochemistry. This page connects to the broader picture of electrophilic addition to alkenes.
1. The alkene attacks X2 to form a bridged halonium ion
As the π electrons approach one bromine of Br2, the Br–Br bond breaks heterolytically, expelling Br−. Rather than leaving a planar open carbocation, the remaining bromine uses one of its lone pairs to bond to both alkene carbons at once. This three-membered ring bearing a positive charge is the bromonium ion (chloronium for Cl2). Bridging locks the geometry and shields one entire face of the former double bond.
Because the bridging halogen blocks one face, any incoming nucleophile is forced to approach from the opposite side.
2. Backside opening of the halonium forces anti addition
The halonium ion is strained and electrophilic. A nucleophile attacks one of the bridged carbons from the back side, the face away from the halogen, in an SN2-like step that breaks that C–X bond. Because the first halogen sits on one face and the nucleophile arrives on the other, the two new groups add to opposite faces of the original π system. This is anti addition, and it is the signature stereochemical outcome of halogenation.
Ethylene plus Br2 gives 1,2-dibromoethane. Anti addition is invisible here because the product has no stereocenters — but the same backside mechanism operates.
3. X2 in an inert solvent gives a vicinal dihalide
When the reaction is run in a non-nucleophilic solvent such as dichloromethane, the only nucleophile available is the Br− (or Cl−) released in step 1. It opens the halonium ion to install the second halogen, delivering a vicinal dihalide — two halogens on adjacent carbons. Chlorine behaves the same way through a chloronium ion.
Propene plus Cl2 gives 1,2-dichloropropane — the chlorine analog of the opening bromination example.
4. On a ring the anti requirement gives a trans-1,2-dihalide
Ring substrates make the anti stereochemistry visible. Cyclohexene reacts with Br2 to give trans-1,2-dibromocyclohexane: the two bromines cannot end up cis because backside opening places them on opposite faces of the ring. A cis product is simply not formed, which is strong experimental evidence for the bridged halonium intermediate rather than a free carbocation.
Cyclohexene plus Br2 gives trans-1,2-dibromocyclohexane — the anti addition is locked in by the ring.
5. X2 in water gives a halohydrin with Markovnikov OH placement
Change the solvent to water and a second nucleophile enters the competition. Water is present in vast excess, so it out-competes the small amount of Br− and opens the halonium ion itself. After loss of a proton the product is a halohydrin — an –OH and a halogen on adjacent carbons. The stereochemistry is still anti (backside attack), but now regiochemistry matters.
Propene with Br2 in water gives the bromohydrin: OH on the more substituted carbon, Br on the less substituted one.
Why does OH land on the more substituted carbon? Although the halonium ion is bridged, the C–X bonds are unequal. The more substituted carbon carries more of the positive charge — it looks partway toward the more stable carbocation — so the C–X bond to that carbon is weaker and longer. Water therefore attacks the more substituted carbon, giving a Markovnikov-like outcome. This mirrors Markovnikov's rule: the nucleophile ends up where a cation would have been most stable.
2-Methylpropene gives a tertiary bromohydrin — OH on the fully substituted carbon, Br on the CH2. Running the reaction in methanol instead of water gives the corresponding ether, CC(OC)CBr.
6. Summary
Halogenation and halohydrin formation are two faces of the same mechanism. The alkene π bond attacks X2 to build a bridged halonium ion; a nucleophile then opens that ring by backside attack, guaranteeing anti addition. The identity of the nucleophile is set by the solvent:
- X2 in an inert solvent → nucleophile is X− → vicinal dihalide; on a ring this is a trans-1,2-dihalide.
- X2 in water → nucleophile is H2O → halohydrin, still anti, with OH on the more substituted carbon (Markovnikov). Using an alcohol solvent gives the analogous ether.
Halohydrins are more than a curiosity: treat a halohydrin with base and the alkoxide displaces the adjacent halide intramolecularly to close a three-membered ring. That is a standard route to an epoxide, and it opens the door to epoxide ring-opening chemistry downstream.
Quiz yourself
Tap a question to reveal the answer — free, no login.
The bromonium ion bridges one face of the ring, so bromide must attack from the opposite (back) side. The two bromines therefore land on opposite faces, giving trans-1,2-dibromocyclohexane and no cis product.
OH goes on the more substituted (tertiary) carbon and Br on the CH2. That carbon carries more positive charge in the halonium ion, so its C–Br bond is weaker and water attacks there — a Markovnikov-like result.
The solvent. Run the halogenation in water so H2O (in large excess) opens the halonium ion instead of X⁻. The mechanism and anti stereochemistry are unchanged; only the incoming nucleophile differs.
Add base. It deprotonates the OH to an alkoxide, which does an intramolecular backside attack on the adjacent C–X carbon, displacing halide and closing a three-membered epoxide ring.
Draw this on the whiteboard
Open the OChem Board whiteboard — benzene rings, wedge/dash bonds, and a clickable periodic table built in. No account needed.