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Electrophilic addition — anti-selective

Halohydrin formation from an alkene + Br₂/H₂O

How the Halohydrin formation from an alkene + Br₂/H₂O works, in 3 steps. Each step shows the structures and the curved arrows that move the electrons.

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In short. A 3-step mechanism. Typical reagents: Br₂ in H₂O (or NBS, H₂O/DMSO). Conditions: Water as solvent and nucleophile, 0–25 °C. Bonds formed: C–Br and C–O σ bonds; bonds broken: C=C π bond; Br–Br. Stereochemistry: Anti addition (trans Br/OH via bromonium). Rate and selectivity: Faster for electron-rich alkenes; OH at the more substituted carbon.

At a glance

ReagentsBr₂ in H₂O (or NBS, H₂O/DMSO)
Solvent / conditionsWater as solvent and nucleophile, 0–25 °C
Bonds formedC–Br and C–O σ bonds
Bonds brokenC=C π bond; Br–Br
StereochemistryAnti addition (trans Br/OH via bromonium)
Rate / selectivityFaster for electron-rich alkenes; OH at the more substituted carbon

Step 1: π bond attacks Br₂, bromonium forms

+BrBrδ⁺δ⁻1 · Π BOND ATTACKS BR₂, BROMONIUM FORMSpropeneC₃H₆NUCLEOPHILEbromineBr₂ELECTROPHILEπ attacks BrNo open carbocation: bromine bridges both carbons.

The alkene's π electrons attack Br₂, but instead of an open carbocation, they form a bridged, three-membered bromonium ion — bromine partially bonded to both carbons at once.

Step 2: Water opens the ring from the back

+Brδ⁺OH₂OH₂Br2 · WATER OPENS THE RING FROM THE BACKbromonium ionC₃H₆Br⁺ELECTROPHILEwaterH₂ONUCLEOPHILEprotonated halohydrinC₃H₈BrO⁺INTERMEDIATEanti to Brmore substituted C(more δ⁺)SN2-like backside attack at the carbon best able to carry positive charge.

Water attacks the bromonium ion from the face opposite bromine (anti), preferentially at the more substituted carbon where positive character is greatest, opening the ring.

Step 3: Deprotonation gives the halohydrin

++OHHBrOH₂OHBrOH₃3 · DEPROTONATION GIVES THE HALOHYDRINprotonated halohydrinC₃H₈BrO⁺INTERMEDIATEwaterH₂OBASE1-bromopropan-2-olC₃H₇BrOPRODUCThydroniumH₃O⁺BYPRODUCTBr and OH are anti, as set by backside attack.

Loss of a proton gives the halohydrin. Because the nucleophile is forced to attack from the opposite face, the Br and OH end up anti to each other on the final product.

Abbreviations and terms

DMSO
Dimethyl sulfoxide. Polar aprotic solvent; also the oxidant in Swern/Kornblum-type reactions.
H2
Hydrogen gas. Reductant for catalytic hydrogenation.
NBS
N-Bromosuccinimide. Source of a low, steady concentration of Br₂/Br• for allylic/benzylic bromination.
SN2
Bimolecular nucleophilic substitution. One concerted step with backside attack and inversion; rate = k[RX][Nu].

How it compares with related reactions

Halohydrin formation from an alkene + Br₂/H₂OBr₂ addition to an alkene (vicinal dihalide)Acid-catalyzed hydration of an alkene
ReagentsBr₂ in H₂O (or NBS, H₂O/DMSO)Br₂Dilute H₂SO₄ or H₃PO₄ (H₃O⁺), H₂O
ConditionsWater as solvent and nucleophile, 0–25 °CCH₂Cl₂ or CCl₄, 25 °C, dark25–100 °C; heat and excess water favour the alcohol
StereochemistryAnti addition (trans Br/OH via bromonium)Anti addition via bromonium; trans product (racemic)Planar carbocation: racemic; Markovnikov regiochemistry
Rate / selectivityFaster for electron-rich alkenes; OH at the more substituted carbonBromonium formation rate-limiting; fastProtonation of the alkene is rate-determining

Test yourself

Try to answer each question before opening it.

Which reagents and conditions are typical for the Halohydrin formation from an alkene + Br₂/H₂O?

Br₂ in H₂O (or NBS, H₂O/DMSO). Conditions: Water as solvent and nucleophile, 0–25 °C.

Which bonds are formed and which are broken?

Formed: C–Br and C–O σ bonds. Broken: C=C π bond; Br–Br.

What is the stereochemical outcome?

Anti addition (trans Br/OH via bromonium)

What controls the rate or the selectivity?

Faster for electron-rich alkenes; OH at the more substituted carbon

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