Alkene Halogenation and Halohydrins

Introduction

An alkene can add a halogen molecule across its C=C bond, making a vicinal dihalide. When water is present during suitable halogen addition, it can capture the intermediate and produce a halohydrin instead. These reactions share an initial electrophilic step but differ in the nucleophile that opens the bridged halonium intermediate.

Core explanation

For ethene plus Br₂, the net equation is CH₂=CH₂ + Br₂ → BrCH₂–CH₂Br. Each former double-bond carbon gains one bromine, and the π component becomes part of new C–Br sigma bonding while the C–C sigma bond remains. The product is 1,2-dibromoethane, a vicinal dibromide. A similar pattern can apply with Cl₂ under suitable conditions. A visual bromine-color change is sometimes used as a qualitative test for unsaturation, but other substances can react with bromine, so color loss is not a unique proof of an alkene.

The mechanism is not usually a simple free-carbocation sequence. The alkene π electrons interact with Br₂, polarizing it and forming a three-membered bromonium ion spanning the two carbons. Br⁻ attacks from the side opposite the bridge and opens it. For a cyclic alkene, this often yields anti addition: the two bromines end up on opposite faces of the original double-bond plane. Product stereochemistry depends on substrate symmetry and conditions; do not infer one unique enantiomer from a flat formula alone.

If water is abundant relative to halide, water can attack the halonium intermediate and deprotonation gives a halohydrin with adjacent OH and halogen. For an unsymmetrical alkene, the OH often appears on the more substituted carbon because halonium opening there has greater positive-charge character under common conditions. The halogen stays on the other carbon, and anti stereochemistry is typical for the opening pathway. This product differs from a vicinal dihalide because water, rather than halide ion, supplies the second nucleophile.

Reagents and solvent must therefore be read together. Br₂ in a relatively non-nucleophilic solvent can favor dibromide formation; Br₂ in water can favor bromohydrin. The shorthand “alkene + Br₂” is incomplete for detailed product predictions if competing nucleophiles are present. Halogen addition is also distinct from alkane radical halogenation, which substitutes a hydrogen and commonly needs light or heat. In an alkene halogenation problem, mark the two original C=C carbons and check that new substituents land on adjacent positions.

Step-by-step reasoning

1. Identify the original alkene carbons and the halogen reagent. 2. Draw a bridged halonium intermediate for the standard polar pathway. 3. Choose attacking nucleophile from the reaction medium. 4. Open the bridge and check adjacent positions and stereochemistry.

Visual explanation

Draw C=C below Br₂, then a triangular C–Br–C bridge. Two arrows from this intermediate show Br⁻ attack to a dibromide and water attack to a bromohydrin.

Real-world analogy

A temporary bridge can block one approach direction while a second participant enters from the opposite side. Which participant enters depends on who is available in the surrounding medium.

Real-world example

An alkene in a controlled bromine-containing medium can form a vicinal dibromide; with water present, a bromohydrin may be obtained instead. The solvent changes the likely capture product.

Why?

Why is anti addition common in bromination? The bromonium bridge shields one face, so bromide tends to attack from the opposite side as it opens the intermediate.

Common misconception

“Alkene bromination is the same as methane chlorination.” Alkene addition consumes a π bond, whereas alkane radical halogenation replaces a C–H hydrogen.

Worked example

React propene, CH₃–CH=CH₂, with Br₂ in a non-nucleophilic solvent under ordinary addition conditions. The two original double-bond carbons each gain Br, giving CH₃–CHBr–CH₂Br, 1,2-dibromopropane. With Br₂ in abundant water, water can instead open the bromonium bridge and produce a bromohydrin such as CH₃–CH(OH)–CH₂Br as the expected major constitutional arrangement. Conditions explain why the two products differ.

Quick check

1. What does ethene plus Br₂ form in a simple halogen-addition model? Answer: 1,2-Dibromoethane, BrCH₂–CH₂Br.

Exam focus

Check whether water is present. Use the halonium intermediate to reason about anti stereochemistry and distinguish dibromide from halohydrin products.

Advanced insight

Halonium ions are unsymmetrical when the alkene carbons differ. Their charge distribution and steric accessibility both influence which carbon an external nucleophile attacks during opening.

Summary

Halogen addition makes vicinal dihalides through a bridged halonium intermediate. Water can capture the same intermediate to form a halohydrin instead, often with anti stereochemistry.

Practice questions

1. What is a vicinal dihalide? Answer: A molecule with halogens on adjacent carbons. 2. Which nucleophile opens a bromonium ion to make a bromohydrin? Answer: Water, followed by deprotonation. 3. Why does Br₂ color loss alone not prove the presence of an alkene? Answer: Other reactive compounds can also consume bromine under suitable conditions.