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Halogenation &
Halohydrin Formation

Master the electrophilic addition of halogens and water to alkenes: mechanism, stereochemistry, regiochemistry, and real-world applications.

Organic Chemistry 1

What is Electrophilic Addition?

Alkenes contain a \(\pi\) bond: a region of high electron density above and below the plane of the double bond. This makes alkenes nucleophilic and reactive toward electrophiles.

C C \(\pi\) electron cloud \(\text{E}^+\) electrophile

Halogenation: Adding \(\text{X}_2\) to Alkenes

When an alkene reacts with a molecular halogen such as \(\text{Br}_2\) or \(\text{Cl}_2\), both halogen atoms add across the double bond, forming a vicinal dihalide.

Reagent
\(\text{Br}_2\) or \(\text{Cl}_2\) in an inert solvent such as \(\text{CH}_2\text{Cl}_2\)
Product
Vicinal dihalide (1,2-dihalide): both halogens on adjacent carbons
Stereochemistry
Anti addition: the two halogen atoms add to opposite faces of the alkene

Halohydrin Formation: Adding \(\text{X}_2\) + \(\text{H}_2\text{O}\)

When halogenation is performed in water instead of an inert solvent, water acts as the nucleophile. The product is a halohydrin: one carbon gets the halogen, the adjacent carbon gets an \(\text{OH}\) group.

Reagent
\(\text{Br}_2 / \text{H}_2\text{O}\) or \(\text{Cl}_2 / \text{H}_2\text{O}\)
Product
Halohydrin: \(\text{X}\) and \(\text{OH}\) on adjacent carbons
Regiochemistry
Markovnikov-like: \(\text{OH}\) goes to the more substituted carbon

The Halonium Ion Intermediate

Both halogenation and halohydrin formation proceed through a cyclic halonium ion: a three-membered ring where the halogen bridges both carbons. This intermediate is the key to understanding anti stereochemistry.

\(\text{Br}^+\) C C \(\delta+\) \(\delta+\) Nu: (\(\text{Br}^-\) or \(\text{H}_2\text{O}\)) Nucleophile attacks from the opposite face

Halogenation Mechanism

1
Electrophilic attack: The \(\pi\) electrons of the alkene attack \(\text{Br}_2\), causing the \(\text{Br{-}Br}\) bond to break heterolytically. A cyclic bromonium ion forms.
2
Nucleophilic opening: The displaced \(\text{Br}^-\) attacks the back side of one of the ring carbons (\(\text{S}_\text{N}2\)-like), opening the three-membered ring.
3
Product: A vicinal dibromide with anti stereochemistry (the two Br atoms are on opposite faces).

Halohydrin Mechanism

1
Bromonium ion forms: Same Step 1 as halogenation: \(\pi\) electrons attack \(\text{Br}_2\), creating a cyclic bromonium ion and releasing \(\text{Br}^-\)
2
Water attacks: In aqueous solution, \(\text{H}_2\text{O}\) is present in vast excess over \(\text{Br}^-\) Water performs a back-side attack on the more substituted carbon of the bromonium ion.
3
Deprotonation: The oxonium ion loses a proton to solvent, yielding the halohydrin with anti stereochemistry.

Halogenation vs. Halohydrin

Halogenation Halohydrin Formation
Reagent \(\text{X}_2\) in \(\text{CH}_2\text{Cl}_2\) \(\text{X}_2\) in \(\text{H}_2\text{O}\)
Nucleophile \(\text{X}^-\) \(\text{H}_2\text{O}\)
Product Vicinal dihalide Halohydrin
Stereochemistry Anti Anti
Regiochemistry N/A (both atoms same) OH on more substituted C
Key intermediate Halonium ion Halonium ion

Why Anti Addition?

alkene plane Br top face blocked Nu: attacks bottom C1 C2 Br Nu anti (trans) product

Regiochemistry: Why OH Goes to the More Substituted Carbon

Br \(\text{CH}_2\) less substituted less \(\delta+\) \(\text{CR}_2\) more substituted more \(\delta+\) \(\text{H}_2\text{O}\)

How to Apply These Reactions

1
Identify the alkene and draw its structure. Note any asymmetry (substituent groups on each carbon).
2
Identify the reagent and solvent. \(\text{X}_2\) in inert solvent = halogenation. \(\text{X}_2\) in \(\text{H}_2\text{O}\) = halohydrin.
3
Draw the halonium ion intermediate. Mark the more substituted carbon.
4
Attack with the nucleophile from the back side (anti). For halohydrin, \(\text{OH}\) goes to the more substituted carbon.
5
Assign stereochemistry. Both reactions give anti products. If chiral centers form, determine R/S or draw the trans arrangement.

Example 1: Cyclohexene + \(\text{Br}_2\)

double bond + \(\text{Br}_2\) \(\text{CH}_2\text{Cl}_2\) Br Br trans-1,2-dibromocyclohexane anti addition (trans diaxial)

Example 2: Propene + \(\text{Br}_2 / \text{H}_2\text{O}\)

\(\text{CH}_3\) CH \(\text{CH}_2\) C2 (more sub.) C1 + \(\text{Br}_2/\text{H}_2\text{O}\) \(\text{CH}_3\) C OH \(\text{CH}_2\) Br 1-bromo-2-propanol

Example 3: Stereochemistry with 1-Methylcyclohexene

When 1-methylcyclohexene reacts with \(\text{Br}_2 / \text{H}_2\text{O}\), we must predict both regiochemistry and stereochemistry.

1
Bromonium ion forms on the double bond (C1-C2). C1 bears a methyl group, making it more substituted.
2
Water attacks C1 (the more substituted carbon) from the opposite face of the bromine bridge.
3
Product: trans-2-bromo-1-methylcyclohexanol. The OH and Br are trans (anti) to each other.

Common Mistakes

Drawing Syn Instead of Anti
Both halogenation and halohydrin formation give anti products via the halonium ion. Never draw both groups on the same face.
Putting OH on the Wrong Carbon
In halohydrin formation, \(\text{OH}\) goes to the more substituted carbon, not the less substituted one.
Confusing Solvent Effects
Inert solvent (\(\text{CH}_2\text{Cl}_2\)) = dihalide. Aqueous solvent (\(\text{H}_2\text{O}\)) = halohydrin. The solvent is the key variable.

Summary Table

Feature Halogenation Halohydrin
Solvent Inert (\(\text{CH}_2\text{Cl}_2\)) Aqueous (\(\text{H}_2\text{O}\))
Intermediate Halonium ion Halonium ion
Nucleophile \(\text{X}^-\) \(\text{H}_2\text{O}\)
Product type Vicinal dihalide Halohydrin
Stereo Anti (trans) Anti (trans)
Regio N/A (both atoms same) OH to more sub. C
Rearrangements? No No

Question 1 of 5

When cyclohexene reacts with \(\text{Br}_2\) in \(\text{CH}_2\text{Cl}_2\), what is the major product?

Question 2 of 5

In halohydrin formation from propene with \(\text{Br}_2 / \text{H}_2\text{O}\), where does the \(\text{OH}\) group end up?

Question 3 of 5

What is the key intermediate shared by both halogenation and halohydrin formation?

Question 4 of 5

What determines whether the product is a dihalide or a halohydrin?

Question 5 of 5

Why does anti addition occur in these reactions instead of syn addition?