Bromohydrin Formation Then Epoxide Formation

Alkene Reactions: Bromohydrin Formation using Br2 and H2O, followed by Epoxide formation using NaOH

Bromohydrin Formation using Br2 and H2O, followed by Epoxide formation using NaOH

Bromine (Br2) or N-bromosuccinimide (NBS) in the presence of an alkene and water react to form a bromohydrin. A subsequent treatment with strong base (usually NaOH or KOH) deprotonates the alcohol and promotes intramolecular SN2 closure to the epoxide.

In the first step Br2 (or NBS) adds across the alkene via a bromonium ion; water opens the bridge anti at the more substituted carbon to give an anti halohydrin (OH on the more substituted carbon, Br on the less substituted carbon). In the second step base removes the alcohol proton to give an alkoxide which attacks back-side at the adjacent C-Br, ejecting Br- and forming the epoxide with inversion at that carbon.

Epoxide formation requires an anti-periplanar alignment between the alkoxide lone pair and the C-Br sigma* orbital (back-side attack). In acyclic halohydrins this geometry is usually accessible by rotation; in constrained rings the OH and Br must adopt an arrangement that allows anti-periplanar attack (for example, a trans-diaxial orientation in a cyclohexane chair).

Introduction

Regiochemistry: The bromohydrin step places OH on the more substituted (or resonance-stabilized) carbon and Br on the less substituted carbon because water opens the bromonium anti at the cationic site.
Stereochemistry: Bromohydrin formation is anti and the intramolecular SN2 requires a back-side approach, inverting the carbon bearing Br as the epoxide forms.
Reagents: Step 1 uses Br2 in water (or NBS/H2O) to generate the halohydrin. Step 2 uses a strong base such as NaOH or KOH to form the alkoxide and ring-close to the epoxide.

Quick Summary

Step 1: Alkene + Br2/H2O (or NBS/H2O) -> anti bromohydrin (OH on more substituted C, Br on less).
Step 2: NaOH or KOH deprotonates the alcohol -> intramolecular SN2 forms the epoxide (inversion at the C-Br carbon).
Key requirement: O- and the C-Br bond must align anti-periplanar for back-side attack; acyclic halohydrins can rotate to achieve this, but rings may need a trans-diaxial arrangement.
Leaving group placement: Because Br ends up on the less substituted carbon, the SN2 step is kinetically favorable.

Mechanism

Step 1: Alkene engages Br2 to form a bromonium — Step 1 - Alkene polarises Br2 to form a bromonium ion (Br-Br bond cleaves to Br-).

Step 2: Water opens bromonium anti — Step 2 - H2O attacks anti at the more substituted carbon, opening the bromonium to an oxonium.

Step 3: Water deprotonates the oxonium — Step 3 - A second water molecule deprotonates the oxonium, giving the anti bromohydrin.

Step 4: Base deprotonates alcohol — Step 4 - NaOH (or KOH) deprotonates the alcohol to make the alkoxide.

Step 5: Intramolecular SN2 forms epoxide — Step 5 - The alkoxide attacks back-side at C-Br (anti-periplanar), ejecting Br- and forming the epoxide.

Step 6: Epoxide product — Step 6 - Epoxide product (anti bromohydrin -> intramolecular SN2 with inversion at C-Br).

Mechanistic Checklist

Bromonium ion forms as Br2 approaches the alkene; Br-Br heterolysis delivers Br-.
Water opens the bromonium anti at the more substituted carbon -> oxonium intermediate.
Deprotonation by water furnishes the bromohydrin (OH on more substituted carbon, Br on less).
Strong base (NaOH or KOH) deprotonates the alcohol to produce an alkoxide.
Intramolecular SN2 closes the epoxide: the alkoxide attacks anti-periplanar to C-Br, expelling Br- with inversion at that center.

Worked Examples

Cyclohex-1-ene — Substrate - cyclohex-1-ene

Reagents Br2/H2O then NaOH — Reagents - Br2/H2O (or NBS/H2O), then NaOH

Cyclohexene oxide — Product - (1S,6R)-7-oxabicyclo[4.1.0]heptane (trans-diaxial halohydrin closes readily)

Propylene oxide — Product - (2S,3S)-2,3-dimethyloxirane (anti bromohydrin -> intramolecular SN2; racemic pair)

Scope & Limitations

Triggers: Br2/H2O or NBS/H2O delivers bromohydrin formation; follow with NaOH or KOH to close the epoxide.
Regioselectivity: Water opens the bromonium at the more substituted (or resonance-stabilized) carbon -> OH installs there, Br on the less substituted carbon.
Geometry: Epoxidation needs anti-periplanar O-/C-Br alignment. Acyclic halohydrins access this by rotation; in rings, trans-diaxial OH/Br is often required.
Alternative pathways: NBS with hv/ROOR and no water leads to allylic bromination (Wohl-Ziegler) rather than bromohydrin formation.
Competing reactions: Alcohol/water solvents or other nucleophiles can intercept the bromonium to give haloethers instead of bromohydrins.

Practical Tips

Run Step 1 (Br2/H2O or NBS/H2O) separately before adding base; NaOH destroys Br2/NBS.
Maintain low temperature during bromonium formation to minimise elimination or polymerisation.
Use aqueous NaOH or KOH for Step 2; phase-transfer conditions can accelerate intramolecular SN2 closure.
In cyclic systems, inspect conformations (e.g., chair flips) to verify an anti-periplanar OH/Br alignment before expecting rapid epoxidation.

Exam-Style Summary

Alkene + Br2/H2O (or NBS/H2O) -> bromonium -> anti water opening -> bromohydrin (OH on more substituted carbon).
NaOH/KOH deprotonates the alcohol -> alkoxide.
Alkoxide performs intramolecular SN2 on C-Br (anti-periplanar, inversion) -> epoxide + Br-.
Anti addition in Step 1 and backside attack in Step 2 control stereochemistry.

Interactive Toolbox

Explore halohydrin formation and epoxide closure in the Reaction Solver.
Use the Mechanism Solver to visualise bromonium opening and intramolecular SN2 transitions.

Study Tools