Elimination Reactions of Halogenoalkanes
A concise revision guide to the elimination reactions of halogenoalkanes with ethanolic potassium hydroxide: the hydroxide ion as a base, the alkenes formed, why substitution and elimination compete, and how the solvent decides.
- 3.3.3.2i
- 3.3.3.2ii
- 3.3.3.2iii-b
What these spec points say
- 3.3.3.2i know that a halogenoalkane can undergo concurrent substitution and elimination reactions (eg 2-bromopropane with potassium hydroxide)
- 3.3.3.2ii explain the role of the reagent as both nucleophile and base in elimination reactions
- 3.3.3.2iii-b outline elimination mechanisms of halogenoalkanes
Elimination: Hydroxide Ions as a Base
Watch a hydroxide ion in hot ethanol take a proton from 2-bromopropane while the bromide ion leaves and a C=C double bond forms, then compare substitution and elimination.
© Dr. Mohammed Al-Fatah – onlinelearningsystem.net
GCSE Recap: Alkenes and Neutralisation
Four quick questions on the GCSE ideas this page builds on.
Hydroxide as a Base
The same reagent, potassium hydroxide, gives a different reaction when the solvent is changed.
Heating a halogenoalkane under reflux with potassium hydroxide dissolved in ethanol removes a hydrogen atom and the halogen from adjacent carbons and forms an alkene: CH₃CH₂CH₂Br + OH⁻ → CH₃CH=CH₂ + H₂O + Br⁻.
This is an elimination reaction, and here the hydroxide ion acts as a base: it removes a proton rather than attacking the carbon.
The hydrogen that is removed must come from a carbon next to the carbon carrying the halogen, because the double bond forms between those two carbons. If that carbon has no hydrogen, elimination cannot happen.
Key idea: Aqueous KOH: OH⁻ is a nucleophile, substitution, alcohol. Ethanolic KOH, heated: OH⁻ is a base, elimination, alkene.
Quick Check: Same Reagent, Different Solvent
Answer each quick question on the conditions and the role of the hydroxide ion.
The Elimination Mechanism
The mechanism is a single step drawn with three curly arrows.
The first arrow goes from the lone pair on the oxygen of the hydroxide ion to a hydrogen atom on the carbon next to the C–Br carbon: the base takes that proton.
The second arrow goes from the C–H bond being broken into the space between the two carbons: those electrons become the second bond of the C=C.
The third arrow goes from the C–Br bond to the bromine: the bond breaks heterolytically and a bromide ion leaves. The products are the alkene, water and a bromide ion.
Elimination and substitution compete, so a mixture of products usually forms.
The balance can be pushed either way by the solvent and the temperature: water and lower temperatures favour substitution, ethanol and higher temperatures favour elimination.
The elimination of 2-bromopropane by ethanolic potassium hydroxide, and the same reagent giving substitution in water.
Exam focus: Three arrows: lone pair to H, C–H bond into C–C, C–Br bond to Br. Label the hydroxide ion as the base.
Extension: beyond the AQA specification
Primary halogenoalkanes tend towards substitution and tertiary halogenoalkanes towards elimination.
In a tertiary compound the three alkyl groups crowd the δ+ carbon and block the nucleophile, while there are plenty of neighbouring hydrogens for the base to remove and the alkene formed is stable.
You may meet this in wider reading; AQA does not examine it.
Quick Check: Nucleophile or Base?
Write a short explanation of the two roles of the hydroxide ion.
More Than One Alkene
When the carbon carrying the halogen has two different neighbouring carbons that both carry hydrogen, the base can remove a hydrogen from either side and two structural isomers form.
2-bromobutane gives but-1-ene when the hydrogen comes from carbon 1 and but-2-ene when it comes from carbon 3.
But-2-ene has two different groups on each carbon of the double bond, so it exists as E and Z isomers, giving three alkenes in total. The more substituted alkene, but-2-ene, is usually the major product.
2-bromopropane, by contrast, gives only propene because both its neighbouring carbons are identical CH₃ groups, and 1-bromopropane gives only propene because the only neighbouring carbon is carbon 2.
| Compound | Neighbouring carbons with H | Alkenes formed |
|---|---|---|
| 1-bromopropane | C2 only | propene |
| 2-bromopropane | C1 and C3, identical | propene |
| 2-bromobutane | C1 and C3, different | but-1-ene, E-but-2-ene, Z-but-2-ene |
| 2-bromo-2-methylpropane | three identical CH₃ | 2-methylpropene |
| 1-bromo-2,2-dimethylpropane | none with H | no elimination |
Exam technique: List the carbons next to the C–X carbon, check each for hydrogen, draw one alkene per different neighbour, then check the products for E/Z isomerism.
Check each carbon next to the C–X carbon for hydrogen, count one alkene per different neighbour, then look for E/Z isomers: 2-bromobutane gives three alkenes.
Quick Check: Every Possible Alkene
Click every alkene that can form from each halogenoalkane.
Quick Check: Substitution versus Elimination
Pick the one accurate statement in each round to check the whole page.
Common Exam Points
Say
“Potassium hydroxide dissolved in ethanol, heat under reflux; the hydroxide ion acts as a base.” “H and Br are removed from adjacent carbons and a C=C bond forms.”
Do not say
“Ethanoic KOH.” “The OH⁻ attacks the carbon” (that is substitution). “Elimination of HBr” without saying which carbons lose the H and the Br.
Watch for
The solvent is the whole point of the question: if it says “in ethanol” think elimination, if it says “aqueous” think substitution, and if it says both were formed explain the mixture.
FAQs
Use these quick answers to check elimination.
What exactly does “hydroxide acts as a base” mean?
The hydroxide ion uses its lone pair to remove a hydrogen ion from a carbon next to the C–X carbon, forming water, rather than bonding to the carbon.
Why does the solvent matter so much?
In water the hydroxide ion is surrounded by water molecules and behaves as a nucleophile towards the δ+ carbon; in ethanol it acts as a stronger base and removes a proton instead.
Hot ethanolic KOH (ethanol as the solvent, heated) favours elimination; warm aqueous KOH favours substitution.
Which hydrogen is removed?
One on a carbon adjacent to the carbon carrying the halogen, because the double bond forms between those two carbons.
Do I always get a mixture of products?
Usually. Substitution and elimination compete, and an unsymmetrical secondary or tertiary compound can also give more than one alkene. Questions ask for the major product or for all the possible alkenes.
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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