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Organic Chemistry · Alkenes: Reactions and Synthesis

Halohydrins from Alkenes: Addition of HO-X

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On this page 9 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Worked example
  6. Key takeaway
  7. Check yourself
  8. Study tools
  9. Sources & references

In 30 seconds

When an alkene is treated with a halogen in the presence of water, the product is not a vicinal dihalide but a : a molecule with a halogen on one carbon and a hydroxyl group (–OH) on the adjacent carbon. The overall reaction adds HO–X across the double bond:

alkene + X2 + H2O ⟶ halohydrin + HX

For propene with bromine water, the equation is:

C3H6 + Br2 + H2O ⟶ C3H6BrOH + HBr

The reaction works because water is present in huge excess and outcompetes the bromide ion for the halonium ion intermediate. The result is a new functional group combination — a — with predictable : the –OH group ends up on the more substituted carbon, and the halogen on the less substituted carbon.

Why this matters

Halohydrins are valuable synthetic intermediates. Treating a halohydrin with base closes a three-membered ring to give an , one of the most versatile functional groups in organic synthesis (a reaction you will meet later in this chapter). Historically, propylene chlorohydrin was the key intermediate in the industrial manufacture of propylene oxide, a monomer for polyurethane foams. For exams, halohydrin formation is the first example of a reaction whose regiochemistry follows a "Markovnikov-like" pattern — the oxygen ends up where the carbocation would be most stable — but through a cyclic halonium ion rather than a free carbocation, so rearrangements never occur.

The college version

Core Concepts

The reaction and its regiochemistry

Water acts as both solvent and . The mechanism is the same two-part story as halogenation, with one swap: after the halonium ion forms, a water molecule (not Br-) attacks the ring, and a final deprotonation gives the neutral halohydrin:

alkene Br2, H2O⟶ bromohydrin

The regiochemistry is set by the halonium ion's asymmetry. In an unsymmetrical alkene, the more substituted carbon of the halonium ion carries more of the positive charge (it is more carbocation-like), so the water nucleophile attacks there. The result: the –OH adds to the more substituted carbon and the halogen to the less substituted carbon — the same carbon that would get the halogen in a Markovnikov HX addition, but with the roles reversed.

Mechanism walkthrough

  1. The alkene's π electrons attack one bromine atom of Br2, expelling Br- and forming a bromonium ion bridged between the two alkene carbons.
  2. A water molecule attacks the more substituted carbon of the bromonium ion from the back side, opening the ring and placing the water-derived oxygen on the opposite face from the bromine.
  3. A base (water or bromide ion) removes a proton from the bound water, giving the neutral halohydrin.

Because the water attacks from the back side, the addition is anti: the –OH and the halogen land on opposite faces of the former double bond.

Scope and variations

Bromine water gives bromohydrins; chlorine water gives chlorohydrins. The same chemistry works with alcohols as the solvent, in which case the alcohol intercepts the halonium ion and a β-halo ether (an alkoxy halide) forms instead. Iodine and fluorine follow the same reactivity limits as in halogenation: fluorine is uncontrollable and iodine is too unreactive. If the alkene is cyclic, anti addition again gives trans products — for example, cyclohexene gives trans-2-bromocyclohexan-1-ol.

Common Confusions

Do Not ConfuseWithDifference
Halogenation (X2, inert solvent)Halohydrin formation (X2 in water)In inert solvent you get the vicinal dihalide; in water the solvent traps the halonium ion and you get the halohydrin
"Markovnikov" for the halogen"Markovnikov" for the –OHIn halohydrins the –OH takes the more substituted carbon; the halogen takes the less substituted one — the opposite of HX addition
Anti addition in halohydrinsSyn addition in other hydration routesHalohydrins add –OH and X from opposite faces; hydroboration (later in this chapter) adds H and –OH from the same face
HalohydrinHydrateA halohydrin has C–X and C–OH on adjacent carbons; a hydrate is a water adduct of a carbonyl (geminal diol)
Bromine water colorProof of product identityDecolorization just shows the alkene reacted; structure must be established separately
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

Picture the double bond grabbing one acrobat (the bromine) just like before. This time, instead of the pushed-away acrobat coming back, a water "water balloon" is the first to arrive and attaches to the ring from underneath. The bromine stays on one side and the water's oxygen ends up on the other side, on the carbon that already has more friends (substituents). Then the water lets go of one of its hydrogen atoms, and you have a molecule with a bromine on one carbon and an –OH on the next — a halohydrin.

Worked example

Example 1: 1-Methylcyclohexene plus bromine water

Predict the product when 1-methylcyclohexene reacts with Br2 in water.

Reasoning. The bromonium ion forms between C1 and C2 of the ring. C1 is the more substituted carbon (it bears the methyl group), so it carries more positive charge in the bromonium ion. Water therefore attacks C1, placing –OH there, while the bromine ends up on C2. The water attack is back-side, so the –OH and Br are trans.

Answer. The product is 2-bromo-1-methylcyclohexan-1-ol (SMILES: CC1(O)C(Br)CCCC1, trans relationship between –OH at C1 and Br at C2). The –OH lands on the more substituted carbon as expected.

Example 2: Propene plus chlorine water

Predict the product of propene with Cl2 in water.

Reasoning. Propene has the double bond between C1 (terminal, less substituted) and C2 (internal, more substituted, bearing the methyl group). The chloronium ion forms, and water attacks the more substituted carbon C2.

Answer. The product is 1-chloro-2-propanol (SMILES: CC(O)CCl), with –OH at C2 and Cl at C1. Note this is the opposite placement of the halogen from a simple chlorination, which would give 1,2-dichloropropane with Cl at both carbons.

Example 3: Stoichiometry with dimensional analysis

How many grams of Br2 are required to convert 0.200 mol of propene to its bromohydrin in water?

Formula first. The reaction consumes one mole of Br2 per mole of alkene, so n(Br2) = n(alkene). Mass is:

m = n × M

Substitution. M(Br2) = 2 × 79.90 g/mol = 159.80 g/mol, and n = 0.200 mol:

m = 0.200 mol × 159.80 gmol = 31.96 g

Answer. About 32.0 g of bromine is needed. (In practice, bromine water is used in excess to drive the reaction and compensate for losses.)

Key takeaways

  • Alkene + X2 in water gives a halohydrin (β-halo alcohol); the water is the nucleophile that traps the halonium ion.
  • Regiochemistry: the –OH goes to the more substituted carbon, the halogen to the less substituted carbon (a "Markovnikov-like" result for oxygen).
  • Addition is anti: –OH and X end up on opposite faces; cyclic alkenes give trans halohydrins.
  • The mechanism uses a cyclic halonium ion, so no carbocation rearrangements occur.
  • Halohydrins are precursors to epoxides: treatment with base closes the ring (intramolecular Williamson-type reaction).
  • Using an alcohol as solvent instead of water gives a β-halo ether (alkoxy halide).
  • General lab-safety principle: bromine and chlorine water are corrosive and irritating; work in a fume hood with gloves and eye protection and never mix bleach with other cleaners.

Check yourself

6 review questions from the chapter. Try each one, then open the answer.

  1. Why does water, rather than bromide ion, open the halonium ion in halohydrin formation?

    Show answer

    Water is the solvent, so its concentration is enormous compared with the small amount of bromide ion released — water wins the competition for the halonium ion.

  2. Where do the –OH and the halogen end up on an unsymmetrical alkene?

    Show answer

    The –OH adds to the more substituted carbon; the halogen adds to the less substituted carbon.

  3. Is halohydrin formation syn or anti? What product does cyclohexene give?

    Show answer

    Anti addition. Cyclohexene gives trans-2-bromocyclohexan-1-ol.

  4. Can halohydrin formation show carbocation rearrangements? Why or why not?

    Show answer

    No. The intermediate is a cyclic halonium ion, not a free carbocation, so hydride and alkyl shifts cannot occur.

  5. What happens when a halohydrin is treated with base, and why is that useful?

    Show answer

    Base removes the –OH proton and the alkoxide closes a three-membered ring, giving an epoxide — a versatile synthetic intermediate.

  6. What is the product of 2-methylpropene with Br2 in water?

    Show answer

    2-Methylpropene (CH2=C(CH3)2) gives 1-bromo-2-methylpropan-2-ol: the –OH goes to the central, more substituted carbon and the Br to the terminal carbon (SMILES: CC(C)(O)CBr).

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

halohydrin
A molecule with a halogen on one carbon and an –OH on the adjacent carbon
β-halo alcohol
Another name for a halohydrin; the halogen and alcohol are on neighboring carbons
regiochemistry
Which of two possible constitutional isomers forms
nucleophile
An electron-rich species that donates electron density
epoxide
A three-membered ring containing an oxygen atom

Sources & references

  1. openstax.org — Organic Chemistry

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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