🥼Organic Chemistry Unit 8 Review

Halohydrin formation adds both a halogen (X) and a hydroxyl group (OH) across an alkene double bond in a single reaction. The result is a 1,2-halohydrin, a compound that serves as a versatile intermediate for further transformations like epoxide synthesis. This reaction relies on a halonium ion intermediate, anti addition stereochemistry, and Markovnikov regiochemistry.

Formation of Halohydrins from Alkenes

Halohydrins form through electrophilic addition of $X_2$ ( $Cl_2$ , $Br_2$ , or $I_2$ ) to an alkene in aqueous solution. The water is critical here: it acts as the nucleophile that delivers the hydroxyl group. Without water, you'd just get a vicinal dihalide.

The mechanism proceeds in distinct steps:

The $\pi$ electrons of the alkene attack $X_2$ , displacing one halide ion and forming a cyclic halonium ion (a three-membered ring with the halogen bridging both carbons).
Water performs a backside attack on the more substituted carbon of the halonium ring. It attacks the more substituted carbon because that carbon bears more positive charge (it can better stabilize partial carbocation character).
Loss of a proton from the water gives the halohydrin product.

Because water attacks from the opposite face of the halonium ion, the halogen and hydroxyl group end up anti to each other (on opposite sides of the plane of the former double bond).

Regiochemistry follows Markovnikov's rule:

The halogen ends up on the less substituted carbon (it's delivered first when the halonium ion forms, bridging both carbons).
The hydroxyl group ends up on the more substituted carbon, because water attacks the carbon that bears more positive character in the halonium ion.

A common point of confusion: the halogen bonds to the less substituted carbon, and OH bonds to the more substituted carbon. This is Markovnikov in the sense that the nucleophile (water/OH) goes to the more substituted position, but be careful not to mix up which group goes where.

Example: Addition of $Br_2$ in water to 2-methylbut-2-ene places Br on C-3 (less substituted) and OH on C-2 (more substituted, tertiary position), giving 3-bromo-2-methylbutan-2-ol as the major product.

Formation of halohydrins from alkenes, Organic chemistry 18: Electrophilic addition to alkenes

NBS as an Alternative to Bromine

N-Bromosuccinimide (NBS) is often used instead of $Br_2$ for halohydrin formation. NBS is a solid, making it easier and safer to handle than liquid bromine.

In aqueous solution, NBS reacts with water to generate a low, steady concentration of HOBr (hypobromous acid). This HOBr is the actual electrophile that reacts with the alkene. The low concentration of HOBr minimizes side reactions like dihalide formation.

The mechanism with NBS is analogous to the $X_2$ /water pathway:

HOBr acts as the electrophilic bromine source, forming a bromonium ion intermediate on the alkene.
Water attacks the bromonium ion from the back side at the more substituted carbon.
Deprotonation gives the anti halohydrin product.

Example: NBS in aqueous DMSO with cyclohexene produces trans-2-bromocyclohexanol. The "trans" designation reflects the anti addition: Br and OH are on opposite faces of the ring.

Formation of halohydrins from alkenes, 10.7. Additions involving cyclic intermediates | Organic Chemistry 1: An open textbook

Predicting Products of Halohydrin Reactions

When working through a halohydrin problem, follow these steps:

Identify the alkene and reagent. Is it $X_2/H_2O$ or NBS/ $H_2O$ ? Which halogen is involved?
Apply Markovnikov regiochemistry. Place the halogen on the less substituted carbon and OH on the more substituted carbon of the double bond.
Assign anti stereochemistry. The halogen and hydroxyl group must be on opposite faces. For cyclic substrates, this means a trans product. For acyclic substrates, draw the anti relationship using wedge-dash notation.
Check for competing reactions. In dilute aqueous conditions, halohydrin formation dominates. In non-aqueous conditions or with concentrated $HX$ , simple hydrohalogenation or dihalide formation may compete.

Example: Addition of $Cl_2/H_2O$ to cyclopentene gives trans-2-chlorocyclopentanol, with Cl and OH on opposite faces of the ring.

Watch for carbocation rearrangements in substrates prone to 1,2-hydride or 1,2-methyl shifts. For instance, addition of $Br_2/H_2O$ to 3,3-dimethylbut-1-ene can produce a rearranged bromohydrin via a 1,2-methyl shift to form a more stable carbocation.

Reaction Mechanism and Intermediates

The halonium ion is the key intermediate in this reaction. It's a bridged, three-membered ring where the halogen sits on top of both carbons. This bridging is what enforces anti stereochemistry: the nucleophile must attack from the opposite face.

An important distinction: the halonium ion is not a free carbocation. The halogen's lone pairs partially stabilize both carbons through bridging. However, the more substituted carbon still carries more positive character because alkyl groups stabilize that partial charge through hyperconjugation. That's why the nucleophile (water) preferentially attacks the more substituted carbon.

The full sequence:

$\pi$ bond attacks $X_2$ → halonium ion + $X^-$
$H_2O$ attacks the more substituted carbon of the halonium ion (backside, anti to X)
Proton transfer to solvent → neutral halohydrin product

This mechanism explains both the regiochemistry (Markovnikov) and the stereochemistry (anti addition) in a single, consistent picture.

🥼Organic Chemistry Unit 8 Review