Reactions of Alkenes with Hydrogen Halides
A concise revision guide to the addition reaction of alkenes with hydrogen halides, including HBr and HCl, the electrophilic addition mechanism, carbocation formation and why one product may form in greater amount with unsymmetrical alkenes.
- 6C.22iii
- 6C.23
- 6C.24ii
- 6C.24iii-a
- 6C.24iii-b
What these spec points say
- 6C.22iii understand the addition reactions of alkenes with: hydrogen halides to produce halogenoalkanes
- 6C.23 understand that heterolytic bond fission of a covalent bond results in the formation of ions
- 6C.24ii understand the mechanism of the electrophilic addition reactions between alkenes and: hydrogen halides, including addition to unsymmetrical alkenes
- 6C.24iii-a use curly arrow notation in electrophilic addition mechanisms
- 6C.24iii-b know the relative stability of primary secondary and tertiary carbocation intermediates
Electrophilic Addition in Motion
Choose a step to watch HBr and then Br₂ add across the π bond of ethene, one curly arrow at a time.
© Dr. Mohammed Al-Fatah – onlinelearningsystem.net
Before You Start
This page builds on electrophilic addition and on bond polarity.
- An electrophile is an electron-pair acceptor.
- Bromine is more electronegative than hydrogen, so H-Br is polar: Hδ+-Brδ-.
- Alkenes react with halogens by electrophilic addition through a carbocation.
- Propene, CH3CH=CH2, has two different carbons in its C=C bond.
GCSE Recap: Covalent Molecules and Acids
Four quick GCSE questions on covalent bonds, acids and addition before you meet the hydrogen halides.
The Overall Reaction
Alkenes react with hydrogen halides such as hydrogen bromide, HBr, or hydrogen chloride, HCl, in an addition reaction.
The C=C double bond opens. The hydrogen atom adds to one carbon atom from the original double bond, and the halogen atom adds to the other carbon atom.
The functional group changes from an alkene to a halogenoalkane.
Key idea: This is addition because two atoms are added across the C=C double bond and the alkene becomes a saturated halogenoalkane.
Quick Check: Formula and Reaction Type
Work out the product of an addition that this page does not show.
Why HBr and HCl Act as Electrophiles
In hydrogen bromide, bromine is more electronegative than hydrogen. This means the H-Br bond is polar, with an uneven distribution of electron density.
The hydrogen atom carries a partial positive charge, Hδ+, while the bromine atom carries a partial negative charge, Brδ–.
The electron-rich π bond in the alkene is attracted to Hδ+. The hydrogen atom acts as the electrophile because it accepts electron density from the alkene π bond.
Electrophile: An electron-pair acceptor. In this reaction, Hδ+ is attracted to the electron-rich π bond and begins the electrophilic addition mechanism.
Exam focus: The electrophile is the Hδ+ end of HBr, not Br–. The π bond attacks the hydrogen atom first.
Quick Check: Which End Attacks?
Decide whether the statement about hydrogen iodide is true or false.
Watch: one alkene, two different products
But-1-ene and HBr give two different products. Both form, but one forms far more. Which one, and why?
The Electrophilic Addition Mechanism
The π bond donates electron density to Hδ+, forming a new C-H bond. At the same time, the H-Br bond breaks by heterolytic fission, producing Br–.
After H+ has added, one carbon atom from the original C=C double bond becomes positively charged. This produces a carbocation intermediate.
The bromide ion, Br–, then donates a lone pair to the carbocation, forming the final bromoalkane product.
Step 1: The π bond attacks Hδ+
The alkene π bond is electron-rich, so it is attracted to the partially positive hydrogen atom in HBr.
Step 2: H-Br breaks heterolytically
Both electrons from the H-Br bond move to bromine, forming Br– and leaving a carbocation on the organic molecule.
Step 3: Br– attacks the carbocation
The bromide ion donates a lone pair to the positively charged carbon, forming a new C-Br bond.
The π bond reacts with Hδ+ first, then Br- attacks the carbocation intermediate to form the halogenoalkane.
Quick Check: Run the Mechanism
Follow hydrogen iodide and oct-4-ene through the mechanism, one step at a time.
Unsymmetrical Alkenes Can Form More Than One Product
When an alkene is unsymmetrical, HBr or HCl can add in two different ways. This is because H+ can add to either carbon atom of the C=C double bond.
Each possible route forms a different carbocation intermediate. These carbocations can have different stabilities, so the products are not always formed in equal amounts.
Route 1
H+ adds to one carbon atom of the C=C bond, forming one possible carbocation intermediate.
Route 2
H+ adds to the other carbon atom of the C=C bond, forming a different carbocation intermediate.
Key idea: The major product usually forms from the pathway that produces the more stable carbocation intermediate.
Compare the two possible carbocations: the more stable one gives the major product.
Quick Check: Compare the Two Routes
Drag the words and numbers into place to compare the two carbocations.
Why One Product Forms in Greater Amount
A carbocation is an organic ion with a positively charged carbon atom. Carbocations are stabilised when alkyl groups are attached to the positively charged carbon.
Alkyl groups are electron-releasing. They help spread out and reduce the positive charge, making the carbocation less reactive and more stable.
Because the more stable carbocation is formed more readily, the reaction proceeds mainly through that pathway. This leads to a higher yield of the corresponding product.
| Carbocation type | Alkyl groups attached to C+ | Relative stability |
|---|---|---|
| Primary | One alkyl group | Less stable |
| Secondary | Two alkyl groups | More stable |
| Tertiary | Three alkyl groups | Most stable |
Exam focus: Explain the major product by comparing carbocation stability, not just by stating which product forms.
Alkyl groups stabilise carbocations by releasing electron density towards the positively charged carbon.
More alkyl groups on C⁺ release more electron density, so tertiary carbocations are the most stable.
HBr and Propene: Major and Minor Products
Choose a step to watch HBr add to propene by two routes, compare the two carbocations and see why 2-bromopropane is the major product.
© Dr. Mohammed Al-Fatah – onlinelearningsystem.net
Quick Check: Explain the Major Product
Write a short explanation, then compare it with the mark points and the model answer.
Quick Check: Pick the Accurate Statement
In each round, choose the one statement that is accurate.
FAQs
These questions cover the hydrogen halide points that cost marks.
Which atom of HBr does the π bond attack?
The hydrogen atom. Bromine is more electronegative, so hydrogen carries δ+ and accepts the electron pair.
Why is 2-bromopropane the major product from propene?
It forms through a secondary carbocation, which is more stable than the primary one because it has two electron-releasing alkyl groups.
Does but-2-ene give two products with HBr?
No. It is symmetrical, so both routes give the same secondary carbocation and only 2-bromobutane forms.
Does the bromide ion choose which carbon to attack?
No. The choice is made when H+ adds. Br– then attacks whichever carbocation formed.
Is a tertiary carbocation more stable than a secondary one?
Yes. Three alkyl groups release more electron density and spread the positive charge further.
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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