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Organic Chemistry 1 · Alkene and Alkyne Chemistry

Halogenation and Halohydrin Formation

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On this page 7 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Worked example
  6. Key takeaway
  7. Study tools

In 30 seconds

Alkenes react with Br₂ or Cl₂ by electrophilic addition to give (1,2-) dihalides. The reaction proceeds through a cyclic three-membered (a bromonium or chloronium ion), which forces the second halogen to add from the opposite face, giving . When water is present, it acts as the nucleophile and opens the halonium ion, producing a (a halogen and an OH on adjacent carbons) with Markovnikov-like regiochemistry. Because the halonium ion is a bridged cation rather than an open carbocation, no hydride or alkyl shifts (rearrangements) occur.

Why this matters

and halohydrin chemistry appear throughout drug synthesis: bromohydrins and chlorohydrins are versatile building blocks, and the stereospecific anti addition is exploited to install defined stereochemistry in drug candidates. (Br₂, Cl₂, and halogenated solvents are hazardous; handling, quantities, and disposal must follow approved institutional safety documentation — this topic is strictly conceptual.)

The college version

1. Halogenation and the Halonium Ion

Halogenation of an alkene is the addition of Br₂ or Cl₂ across the C=C bond to give a vicinal dihalide. The key intermediate is the halonium ion: a three-membered ring made of the two alkene carbons and one halogen (bromonium for Br₂, chloronium for Cl₂), with the positive charge on the halogen. This is a true intermediate (a species at an energy minimum), not a transition state.

2. Anti Addition and Stereospecificity

The halonium bridge blocks one face of the alkene, so the incoming halide (Br⁻ or Cl⁻) can only attack from the opposite face — anti addition. A reaction is stereospecific when different stereoisomers of the starting material give different stereoisomers of the product. Halogenation is stereospecific:

  • cis-2-butene → racemic mixture of (2R,3R)- and (2S,3S)-2,3-dibromobutane (enantiomers).
  • trans-2-butene → (2R,3S)-2,3-dibromobutane, a single meso (achiral) compound.

3. Halohydrin Formation

When the reaction is run in water (or another nucleophilic solvent), water acts as the nucleophile and attacks the halonium ion instead of the halide. Ring opening followed by deprotonation gives a halohydrin — a molecule bearing a halogen and an OH on adjacent carbons. The OH ends up on the more substituted carbon and the halogen on the less substituted carbon (Markovnikov-like regiochemistry), because the more substituted carbon of an unsymmetrical halonium ion carries more of the positive charge (better stabilized by hyperconjugation) and is therefore the site the nucleophile attacks.

How it works

  1. The π bond attacks the halogen, forming the halonium ion and releasing one halide.
  2. The bridge blocks one face, so the halide (or water) attacks from the opposite face (anti).
  3. In water, the nucleophile is water, which attacks the more substituted carbon.
  4. Deprotonation yields the neutral halohydrin; in pure halogen, the halide itself completes the dihalide.

Common confusions

Do not confuseWithDifference
Halonium ionOpen carbocationHalonium ion is bridged and does not rearrange; an open carbocation can shift H or alkyl groups
Anti additionSyn additionAnti = opposite faces; syn = same face
StereospecificStereoselectiveStereospecific means starting stereochemistry dictates product; stereoselective means one product is merely preferred
HalohydrinVicinal dihalideHalohydrin has an OH on one carbon; dihalide has a halogen on both
Bromonium ionBromide ionBromonium carries the positive charge in the ring; bromide is the free, negatively charged leaving group

Memory aids

"Bromonium Blocks Backside" — the bromonium bridge blocks one face, forcing the bromide to attack from the backside (anti). For halohydrins: "High side gets the Hydroxyl" — water (giving OH) attacks the more substituted (higher-energy-stabilized) carbon.

Quick review

Topic Recap

Alkenes add Br₂ or Cl₂ through a cyclic halonium-ion intermediate to give vicinal dihalides with anti, stereospecific addition. With water present, water opens the halonium ion to give a halohydrin (OH on the more substituted carbon). Because the intermediate is a bridged cation, no rearrangements occur.

Knowledge Check

  1. What is the key intermediate in alkene bromination, and why does it prevent rearrangements?
  2. Draw the product(s) of Br₂ addition to trans-2-butene and classify the product as meso or a racemic mixture.
  3. In halohydrin formation, which carbon receives the OH, and why?
  4. Why does anti addition give the trans dihalide when a cyclic alkene is brominated?
  5. What species acts as the nucleophile in (a) bromination and (b) halohydrin formation?

Answers and Rationales

  1. A cyclic bromonium ion. It is a bridged cation, not an open carbocation, so there is no flat, electron-deficient carbon available for hydride or alkyl shifts — hence no rearrangement.
  2. (2R,3S)-2,3-dibromobutane, a meso compound. Anti addition to a trans alkene places the two bromines symmetrically, producing an achiral molecule with an internal plane of symmetry.
  3. The more substituted carbon. That carbon bears more of the halonium ion's positive charge and is better stabilized, so the nucleophilic water attacks there, placing OH on the more substituted carbon.
  4. The cyclic alkene's two carbons are already cis (held by the ring). Anti addition delivers the second bromine to the opposite face, yielding the trans relationship.
  5. (a) Bromide ion (Br⁻); (b) water.
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

Imagine a doorway with a big, wide bouncer standing in the middle. The first halogen atom pushes in and stands in the doorway (making the "halonium ion"), blocking that entire side of the room. The second halogen atom can only come in through the other door on the opposite side. That is anti addition: the two atoms enter from opposite sides.

A useful comparison: in HBr addition (Topic 28), the proton goes in first and leaves an open, flat carbocation, so bromine can attack from either side, and the flat cation can reshuffle (rearrange). In halogenation, the halogen bridge blocks one face, forcing anti attack, and there is no open cation to rearrange.

Where it stops being exact: the "bouncer" is not a solid wall — the bridged halogen is a three-membered ring whose positive charge is shared across the ring. It blocks the face only in the sense that the second nucleophile attacks from the backside to relieve the strained ring.

Simple Example

Cyclopentene + Br₂ → trans-1,2-dibromocyclopentane. The ring already holds the two carbons in a cis relationship, so anti addition places the two bromines on opposite faces of the ring, giving the trans product only.

Worked example

  1. The electron-rich π bond (nucleophile) attacks Br₂ (electrophile). Br₂ is polarizable, so the nearer Br develops a partial positive charge as the π electrons are donated into the Br–Br σ* orbital.
  2. The Br–Br bond breaks as one bromide (Br⁻) departs as the leaving group, and the remaining Br forms a three-membered bromonium ion bridging the two carbons (charge on Br).
  3. Bromide (Br⁻) now acts as a nucleophile and attacks one carbon of the bromonium ion from the face opposite the bridge. Electron movement: Br⁻ lone pair → carbon, while the C–Br bridge bond breaks (electrons return to the bridging Br), opening the ring.
  4. The product is a vicinal dibromide with anti stereochemistry. Charge balance: the bromonium ion (+1) plus bromide (−1) sum to neutral product. Atom balance: two Br and the alkene carbons are all accounted for; every carbon retains an octet.

For a halohydrin, step 3 is replaced by water attacking the more substituted carbon, followed by loss of a proton to give the neutral halohydrin.

Key takeaways

  • High yield: Halogenation goes through a cyclic halonium ion, NOT an open carbocation.
  • High yield: Anti addition is the stereochemical hallmark — cis-alkene → racemic mixture, trans-alkene → meso compound.
  • High yield: No rearrangements occur because there is no open carbocation.
  • High yield: In halohydrins, OH goes to the more substituted carbon (Markovnikov-like), halogen to the less substituted carbon.
  • Halohydrin formation uses water as the nucleophile.
  • Chlorination proceeds through a chloronium ion; bromination is the most common and best-behaved.

Keep learning

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Practice Organic Chemistry 1

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Study tools & related lessonsYou’ll learn to · Key vocabulary · Related

You’ll learn to

  • Draw the mechanism of alkene halogenation (bromination and chlorination) through a cyclic halonium ion, and state why anti addition results.
  • Explain the difference between a halonium-ion intermediate and an open carbocation, and connect that difference to the absence of rearrangements.
  • Predict the products of halohydrin formation, including the regiochemistry (where the OH and halogen end up) and stereochemistry.
  • Use stereospecificity to predict whether a given alkene stereoisomer gives a meso compound or a racemic mixture of enantiomers.

Key vocabulary

Halogenation
Addition of Br₂ or Cl₂ across an alkene to give a vicinal dihalide
Halonium ion
Cyclic three-membered cation with a bridging halogen (Br⁺ or Cl⁺)
Bromonium / chloronium ion
The specific halonium ions from Br₂ and Cl₂
Anti addition
Two new groups add to opposite faces of the alkene
Stereospecificity
Product stereochemistry is set by the starting alkene's stereochemistry
Halohydrin
Product with a halogen and OH on adjacent carbons
Vicinal
Two groups on adjacent carbons (1,2-relationship)

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