3.5.5Physical Chemistry
Elimination Reactions of Alcohols — AQA AS Chemistry Revision Notes & Topical Questions
5 min read 5 parts Lesson 86 of 102
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Part 1 of 5Dehydration of Alcohols
Reaction type
- Reaction type Elimination removes a small molecule, here water, from the alcohol to leave a C=C bond.
Reagents
- Reagents Hot concentrated sulfuric acid or concentrated phosphoric(V) acid acts as the dehydrating agent and catalyst.
Alternative
- Alternative Alcohol vapour passed over heated aluminium oxide also loses water to form an alkene.
Ethanol
- Ethanol C₂H₅OH → CH₂=CH₂ + H₂O gives ethene, which can be collected over water.
Propan-2-ol
- Propan-2-ol CH₃CH(OH)CH₃ → CH₃CH=CH₂ + H₂O gives propene.
Part 2 of 5Acid-Catalysed Mechanism
1
A lone pair on the oxygen of the alcohol is donated to an H⁺ ion from the acid, giving a protonated alcohol, R–OH₂⁺.
2
The C–O bond breaks heterolytically, with both electrons going to oxygen, so a water molecule leaves.
3
This forms a carbocation on the carbon that held the –OH group.
4
A C–H bond on the adjacent carbon breaks, and its electrons move to form the C=C double bond.
5
The H⁺ lost in this step is released, so the acid is regenerated and acts as a catalyst.
Part 3 of 5Possible Products
Possible Products
Symmetrical cases
- Symmetrical cases Ethanol and propan-2-ol each give only one alkene, because the neighbouring carbons are equivalent.
Unsymmetrical cases
- Unsymmetrical cases Butan-2-ol can lose a hydrogen from either side, giving but-1-ene and but-2-ene.
Geometric isomers
- Geometric isomers But-2-ene forms as both E and Z isomers, so three alkene products are possible from butan-2-ol.
Use of the alkene
- Use of the alkene Alkenes made from plant-derived ethanol can form addition polymers without using crude oil.
Part 4 of 5Two Different Eliminations
Two Different Eliminations
Alcohol
- Alcohol Acid-catalysed: H⁺ attacks the oxygen, and water is the leaving group.
Haloalkane
- Haloalkane Base-induced: hot ethanolic hydroxide removes H and X, and HX is lost.
Reagent role
- Reagent role In the haloalkane case OH⁻ acts as a base, whereas in the alcohol case the acid acts as a proton donor.
Drawing marks
- Drawing marks Draw every C–H bond that breaks explicitly, and do not mix the two mechanisms.
| Feature | Alcohol dehydration | Haloalkane elimination |
|---|---|---|
| Reagent | Conc. H₂SO₄ or H₃PO₄, hot | Ethanolic NaOH or KOH, heat |
| Small molecule lost | H₂O | HBr, HCl or HI |
| Role of reagent | Acid catalyst | Base |
| Product | Alkene | Alkene |
Part 5 of 5Exam Checklist & Tips
3.5.5 – Key Points for the Exam
- Elimination A reaction in which a small molecule is removed from a larger one, leaving a double bond.
- Dehydration Removal of water from a compound.
- Carbocation An ion with a positively charged carbon atom that has only three bonds.
- Catalyst A substance that speeds up a reaction and is regenerated at the end of the reaction.
Common Mistakes
- Drawing OH⁻ attacking a hydrogen in the alcohol mechanism; the first step is protonation of the –OH oxygen.
- Leaving out the H atom on the bond that breaks and drawing it in skeletal shorthand.
- Stating that the acid is used up; it is regenerated.
- Forgetting that an unsymmetrical alcohol can give more than one alkene.
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