Dibromide Formation

Alkene Reactions: Dibromide Formation using Br2 and Alkenes

Alkene Bromination with Br₂: Bromonium-Ion Anti Addition to Vicinal Dibromides

Bromine (Br₂) reacts with alkenes to give vicinal dibromides by a stereospecific anti addition. The transformation proceeds through a three-membered bromonium-ion intermediate rather than a free carbocation, so rearrangements are not observed. This guide focuses on classic bromination carried out under subdued light.

Introduction

Bromine reacts with alkenes to give vicinal dibromides. The transformation is anti stereospecific: the two bromine atoms add to opposite faces of the former double bond. The regiochemistry is not the focus (both carbons receive bromine), but the stereochemistry and intermediate are key: the alkene donates π electrons to Br₂ to generate a cyclic bromonium ion, and the bromide leaving group performs backside attack to open that three-membered ring. No carbocations, no hydride shifts, and no syn addition.


Quick Summary

  • Reagents: Br₂
  • Outcome: Anti addition — each vinylic carbon receives Br, giving a vicinal dibromide.
  • Mechanism: Bromonium-ion pathway. Alkene → bromonium; Br⁻ performs backside (anti) attack to open the ring.
  • Rearrangements: None. Bridged bromonium prevents hydride/alkyl shifts.
  • Stereochemistry: cis alkene → racemic enantiomeric pair; trans alkene (with symmetry) → meso product (both via anti addition).
  • Common pitfalls: Using ionic (two-electron) arrows for radical chemistry (not applicable here), proposing carbocations or syn addition, using nucleophilic solvents that give halohydrins.

Mechanism (Bromonium Anti Addition)

Step 1: bromonium ion formation from an alkene and molecular bromine.
Step 1 — Alkene polarizes Br₂ and forms a three-membered bromonium ion.

Step 1 — Bromonium formation. The alkene π bond polarizes Br₂ and donates into one bromine while the Br–Br bond heterolyzes. A bridged three-membered bromonium ion plus Br⁻ results; the positive charge is delocalized over both carbons and the bridging bromine.


Step 2: bromide attacks the more substituted carbon of the bromonium from the backside.
Step 2 — Br⁻ performs backside attack at the more substituted carbon, opening the bromonium anti.

Step 2 — Backside opening. Bromide approaches from the face opposite the bromonium bridge and attacks the carbon bearing greater positive character (typically the more substituted carbon). The SN2-like backside attack opens the ring with inversion at the attacked carbon and locks in anti stereochemistry.


Step 3: vicinal dibromide product showing anti addition across the double bond.
Step 3 — Anti vicinal dibromide product.

Step 3 — Product formation. Collapse of the bromonium delivers the neutral vicinal dibromide. The bromine atoms remain on opposite faces of the former double bond (anti relationship).


Mechanistic Checklist (Exam Focus)

  • Draw the bromonium ion explicitly; do not depict a free carbocation.
  • Use standard two-electron curved arrows for these polar steps (no fishhooks).
  • Show backside attack by Br⁻ opening the bromonium with inversion at the attacked carbon.
  • Omit 1,2-shifts or rearrangements — the bridge prevents them.
  • Mention that nucleophilic solvents (H2O, ROH) open the bromonium to halohydrins instead of dibromides.

Worked Example — Exocyclic Alkene

  • Substrate: Isopropenylcyclohexane (exocyclic C=C).
  • Reagents: Br₂
  • Pathway: The exocyclic double bond forms a bromonium; Br⁻ opens from the opposite face at the more substituted ring carbon.
  • Outcome: Vicinal dibromide with the two bromine atoms anti across the former double bond.
Substrate: isopropenylcyclohexane
Substrate — isopropenylcyclohexane
+
Reagent: liquid bromine
Reagent — Br₂
→
Product: anti vicinal dibromide
Product — anti vicinal dibromide

When Multiple Alkenes Are Present

Bromination occurs preferentially at the alkene that forms the more stabilized bromonium and/or is more electron-rich and accessible. Allylic or benzylic substitution can bias the outcome. Compare candidate bromonium ions and consider how easily Br⁻ can perform backside attack when justifying product choices.


Practical Tips & Pitfalls

  • Stereocontrol: Check stereochemical drawings (Newman, chair) to confirm the anti relationship in products.
  • Contrast reactions: In the presence of H2O or ROH, bromonium opening gives bromohydrins (anti Br/OH or Br/OR). NBS/H2O is a complementary halohydrin protocol rather than a dibromination.

Exam-Style Summary

  • Regiochemistry: Both vinylic carbons receive Br; regiochemistry is secondary to stereochemical control.
  • Mechanism: Bromonium formation followed by backside attack by Br⁻ (anti opening).
  • Intermediates: Bridged bromonium ion; no discrete carbocation.
  • Stereochemistry: Anti; cis → racemic pair, trans (sym.) → meso product.
  • Competing pathway: Nucleophilic solvents (H2O, ROH) yield halohydrins instead of dibromides.

Interactive Toolbox

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FAQ / Exam Notes

Why is the addition anti? The bromonium ion blocks one face of the alkene; Br⁻ must attack from the opposite face, enforcing anti addition.

What if water is present? Water would open the bromonium ion to give a bromohydrin (anti addition of Br and OH) rather than a dibromide.

Can Br₂ be replaced with Cl₂? Yes, the mechanism is analogous, but Cl₂ is more reactive and less selective; Br₂ is often preferred for teaching laboratories.


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