Alkene Reactions: Bromohydrin Formation using Br2 and H2O
Bromohydrin Formation of Alkenes with Br₂ and H₂O
Bromohydrin formation converts alkenes into vicinal Br/OH products using Br₂ in water. The alkene polarizes Br₂, forming a three-membered bromonium ion while Br⁻ is generated (equivalently described as Br⁺/HOBr formation). Water then attacks the more substituted carbon from the backside (anti), giving Markovnikov placement of OH (and Br on the less substituted carbon). Deprotonation yields the neutral bromohydrin. Because the mechanism does not involve free carbocations, rearrangements do not occur, and the anti relationship of Br and OH gives predictable stereochemical outcomes.
Introduction
Br₂/H₂O reacts with alkenes to produce bromohydrins through anti addition. The alkene π bond polarizes Br₂, forming a bromonium ion while releasing Br⁻. Water, the solvent nucleophile, attacks the more substituted carbon from the backside, opening the bridge. Deprotonation yields the neutral bromohydrin with OH installed at the Markovnikov position and Br on the less substituted carbon. The absence of carbocation intermediates prevents rearrangements, and the anti relationship between Br and OH provides predictable stereochemical outcomes.
Quick Summary
- Reagents: Br₂ in H₂O (or aqueous solvent mixtures). NBS/H₂O or NaOBr (aq) are common alternatives.
- Outcome: Anti addition of Br and OH; OH occupies the more substituted carbon (Markovnikov), Br the less substituted carbon.
- Mechanism: Bromonium ion formation → water anti attack → deprotonation to the bromohydrin.
- Rearrangements: None (bridged bromonium prevents shifts).
- Stereochemistry: Anti addition ensures Br and OH are trans; when two stereocenters form, products appear as enantiomeric pairs (no meso because Br and OH differ).
- Common pitfalls: Forgetting the bromonium intermediate, having OH add syn with Br, or using an anhydrous solvent (leads to dibromide).
Mechanism (Bromohydrin Formation)
Step 1 — Bromonium formation. The alkene π bond polarizes and donates into Br₂ while the Br–Br bond heterolyzes, generating a three-membered bromonium ion and Br⁻.
Step 2 — Anti opening by water. Water approaches the bromonium from the opposite face of the bridge. Attack occurs at the more substituted carbon (greater positive character), opening the ring in an SN2-like fashion and giving an oxonium intermediate.
Step 3 — Deprotonation. A base (often Br⁻ or another water molecule) removes the proton from oxygen, yielding the neutral bromohydrin with Br and OH anti to one another.
Step 4 — Product formation. The anti bromohydrin is formed with OH at the more substituted carbon and Br at the less substituted carbon.
Mechanistic Checklist (Exam Focus)
- Draw the bromonium ion explicitly; no free carbocation.
- Show water attacking anti to the bridge at the more substituted carbon.
- Include the deprotonation step that yields the neutral bromohydrin.
- Highlight Markovnikov regiochemistry (OH to the more substituted carbon) and anti stereochemistry.
- Note that absence of water leads to dibromide rather than bromohydrin.
Worked Examples
Example A — Bromohydrin Formation from 1-Methylcyclohexene
- Substrate: 1-Methylcyclohexene.
- Reagents: Br₂ (aqueous).
- Pathway: Br₂ forms a bromonium ion; water attacks the tertiary carbon anti to Br; deprotonation delivers the bromohydrin.
- Outcome: Markovnikov bromohydrin with OH at the tertiary ring carbon and Br on the adjacent, less substituted ring carbon; Br/OH anti.
When Multiple Alkenes Are Present
Br₂/H₂O reacts fastest at the alkene that forms the most stabilized bromonium ion (greater substitution, resonance). Consider steric approach of water: less hindered attack on the more substituted carbon increases selectivity. If two alkenes generate bromonium ions of comparable stability, mixtures can form.
Practical Tips & Pitfalls
- Water required: Ensure aqueous conditions; otherwise dibromide forms instead of bromohydrin.
- Conditions: Run at 0–25 °C in subdued light to suppress radical pathways.
- Alternatives: NBS/H₂O or NaOBr (aq) also provide bromohydrins via the same anti mechanism.
- Workup: Quench excess Br₂ with aqueous Na₂S₂O₃.
- Avoid nucleophiles: Competing nucleophiles (ROH, N₃⁻, etc.) open the bromonium and give different products.
- Stereochemistry: Products are anti; draw Br and OH on opposite faces and expect enantiomeric pairs when two stereocenters form.
Exam-Style Summary
- Regiochemistry: Markovnikov — OH on the more substituted carbon, Br on the less substituted carbon.
- Mechanism: Bromonium formation → anti attack by H₂O → deprotonation.
- Intermediates: Bromonium ion; no free carbocation.
- Stereochemistry: Anti addition gives enantiomeric pairs when stereocenters are formed (Br/OH differ, so meso products do not form).
- Competing pathways: Absence of water (or presence of inert solvent) gives dibromides via Br⁻ attack instead of bromohydrins.
Interactive Toolbox
- Model bromohydrin vs. dibromide outcomes — compare solvent effects on the product.
- Generate mechanism diagrams in Mechanism Solver — choose the Br₂/H₂O template for stepwise visuals.
FAQ / Exam Notes
Why is the OH Markovnikov? The more substituted carbon in the bromonium carries greater positive character, so water attacks there preferentially.
Do rearrangements occur? No. The bromonium ion prevents carbocation shifts; the carbon attacked remains fixed.
What happens without water? Br⁻ opens the bromonium to give a vicinal dibromide instead of a bromohydrin.