Reactions of Alkenes with Halogens
A concise revision guide to the addition reaction of alkenes with bromine or chlorine, including the functional group change, reaction conditions and the electrophilic addition mechanism.
- 6C.22ii
- 6C.24i
- 6C.24iii-a
What these spec points say
- 6C.22ii understand the addition reactions of alkenes with: halogens to produce dihalogenoalkanes
- 6C.24i understand the mechanism of the electrophilic addition reactions between alkenes and: halogens
- 6C.24iii-a use curly arrow notation in electrophilic addition mechanisms
Before You Start
This page builds on the bromine water test you met at GCSE and on the π bond in alkenes.
- Alkenes decolourise bromine water; alkanes do not.
- A C=C bond contains an electron-rich π bond.
- Bromine is a halogen with the formula Br2; chlorine is Cl2.
- An addition reaction joins two molecules into one product.
GCSE Recap: Halogens and the Bromine Test
Four quick GCSE questions on halogen molecules and testing for alkenes before you look at the reaction in detail.
Watch: how a non-polar bromine molecule becomes an electrophile
Bromine is non-polar, so how can it be an electrophile? Watch it become one as it approaches the double bond.
The Overall Reaction
Alkenes react with halogens such as bromine, Br2, or chlorine, Cl2, in an addition reaction. The C=C double bond opens and one halogen atom adds to each carbon from the original double bond.
The functional group changes from an alkene to a dihalogenoalkane. For example, ethene reacts with bromine to form 1,2-dibromoethane.
Key idea: The reaction is addition because atoms are added across the C=C double bond and only one organic product is formed from the alkene.
Bromine or chlorine adds across the C=C at room temperature, without UV light, and the orange-brown colour of bromine disappears.
Quick Check: Predict the Product
Work out what chlorine does to an alkene that is not shown on this page.
Why Bromine Acts as an Electrophile
The π bond in an alkene is a region of high electron density. As a Br2 molecule approaches the alkene, the electron-rich π bond repels the shared electron pair in the Br-Br bond.
This repulsion induces a dipole in the bromine molecule. One bromine atom becomes slightly positive, Brδ+, while the other becomes slightly negative, Brδ–.
The slightly positive bromine atom acts as the electrophile because it is attracted to the electron-rich π bond.
Electrophile: An electron-pair acceptor. In this reaction, the polarised bromine molecule provides Brδ+, which accepts electron density from the alkene π bond.
The π electrons repel the Br–Br bonding pair, inducing a dipole: the nearer bromine atom becomes Brδ+ and acts as the electrophile.
Quick Check: The Induced Dipole
In each round, choose the one statement that is accurate.
Bond Fission and Carbocation Formation
The Br-Br bond breaks by heterolytic fission. This means both electrons from the covalent bond go to one bromine atom, forming a bromide ion, Br–.
At the same time, the alkene π bond forms a new bond to Brδ+. One of the carbon atoms from the original double bond becomes positively charged, producing a carbocation intermediate.
Remember: A carbocation is an organic ion with a positively charged carbon atom.
The π bond polarises Br2, Brδ+ adds first, then Br– attacks the carbocation to give the dibromo product.
Quick Check: Charges, Atoms and Amounts
Answer as quickly as you can, using new alkenes each time.
The Mechanism in Three Steps
Step 1: The π bond polarises Br2
The high electron density in the C=C bond repels the Br-Br bonding pair, making the nearer bromine atom Brδ+.
Step 2: The electrophile adds
The alkene π bond donates electron density to Brδ+, forming a C-Br bond and a carbocation intermediate.
Step 3: Bromide attacks
The Br– ion donates a lone pair to the carbocation, forming a second C-Br bond and producing the dihalogenoalkane.
Quick Check: Chlorine with Cyclopentene
Drag the words and numbers into place to build the account of the reaction.
Bromine and Chlorine Give Similar Products
The same addition pattern applies to chlorine. With Cl2, the product is a dichloroalkane. With Br2, the product is a dibromoalkane.
| Halogen | Product type | Example from ethene |
|---|---|---|
| Br2 | Dibromoalkane | 1,2-dibromoethane |
| Cl2 | Dichloroalkane | 1,2-dichloroethane |
Exam focus: Do not describe this as substitution. The C=C double bond breaks and two atoms are added, so this is an addition reaction.
Quick Check: Work It Out, Then Flip
Work each answer out on paper first, then turn the card over to check your reasoning.
Quick Check: Halogen Addition from Start to Finish
Each round covers one part of the page, from the product to the mechanism. Choose the one accurate statement.
FAQs
These questions cover the halogen addition points that cost marks.
Bromine is non-polar, so how can it be an electrophile?
The π electrons repel the Br-Br bonding pair as the molecule approaches. This induces a dipole, and the nearer atom becomes Brδ+.
Is UV light needed?
No. UV light is for alkanes and free-radical substitution. Alkenes add bromine at room temperature in the dark.
What is the product of propene and bromine?
1,2-dibromopropane, CH3CHBrCH2Br. One bromine atom goes on each carbon of the original C=C bond.
How does the Br-Br bond break?
Heterolytically. Both bonding electrons go to the bromine atom further from the alkene, which leaves as Br–.
Is the product a halogenoalkane?
Yes, a dihalogenoalkane. It is saturated, with two halogen atoms on adjacent carbons.
Copyright notice: This OLS revision content, including the explanations, layout, diagrams, tables and embedded learning structure, is authored for Online Learning System by Dr. Mohammed Al-Fatah. It may not be copied, reproduced, redistributed or adapted without written permission.
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