Elimination from halogenoalkanesAQA A-Level Chemistry: Revision notes
Section 1
Two reactions in competition
A halogenoalkane such as 2-bromopropane, CH₃CHBrCH₃, reacts with hydroxide ions in two ways at the same time. In nucleophilic substitution the Br is replaced by OH to give an alcohol. In elimination the elements of HBr are removed to give an alkene.
The two reactions are concurrent: both take place in the same flask, and the conditions decide which product predominates. The hydroxide ion can do both jobs because it has lone pairs of electrons and can accept a proton.
Section 2
Hydroxide as nucleophile and as base
A nucleophile is an electron pair donor. A base is a proton (H⁺) acceptor. Hydroxide ions are both.
- As a nucleophile, OH⁻ uses a lone pair to attack the δ+ carbon of the polar C–Br bond. This is substitution and gives propan-2-ol.
- As a base, OH⁻ uses a lone pair to take an H⁺ from a carbon atom adjacent to the C–Br carbon. This is elimination and gives propene.
The C–Br bond is polar because bromine is more electronegative than carbon, so the carbon is δ+.
Do not say the hydroxide ion is an electrophile, or that it removes a hydrogen atom. It is a nucleophile or a base, and it removes H⁺.
Section 3
Mechanism of elimination
The mechanism of elimination from 2-bromopropane with OH⁻ has three electron-pair movements shown by curly arrows:
- a lone pair on OH⁻ forms a bond to a hydrogen on a carbon atom next to the C–Br carbon
- the electron pair of that C–H bond moves to form the C=C double bond
- the C–Br bond breaks heterolytically, both electrons going to Br, forming Br⁻
The products are propene, water and bromide ions: CH₃CHBrCH₃ + KOH → CH₃CH=CH₂ + KBr + H₂O.
Curly arrows start from a lone pair or a bond and point to where the pair ends up. Draw the arrow from the OH⁻ lone pair to the H, not from the H.
Section 4
Mechanism of substitution
In nucleophilic substitution, a lone pair on OH⁻ attacks the δ+ carbon. As the C–OH bond forms, the C–Br bond breaks heterolytically and both electrons go to Br⁻.
Products: propan-2-ol and bromide ions: CH₃CHBrCH₃ + OH⁻ → CH₃CH(OH)CH₃ + Br⁻.
The key difference is where the hydroxide ion attacks: the carbon bonded to Br for substitution, but a hydrogen on the adjacent carbon for elimination.
Section 5
Conditions decide the product
- Substitution is favoured by aqueous potassium hydroxide and gentle warming.
- Elimination is favoured by potassium hydroxide dissolved in ethanol and a higher temperature (heating under reflux).
Each set of conditions gives a mixture, with one product predominating. The alkene formed is unsaturated and decolourises bromine water, which is a test for the elimination product.
Elimination and substitution are not exclusive. Say that conditions favour one reaction, not that the other does not occur.
Must Know
- Elimination: OH⁻ acts as a base, removing H⁺ from a carbon adjacent to C–Br
- Substitution: OH⁻ acts as a nucleophile
- Elimination conditions: ethanolic KOH, hot
- Substitution conditions: aqueous KOH, warm
- Elimination of 2-bromopropane gives propene, KBr and H₂O
- The two reactions are concurrent
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Elimination from halogenoalkanes
- Hydroxide ions, OH⁻, can react with a halogenoalkane in two different ways. Each oxygen atom in OH⁻ has lone pairs of electrons, and the ion can also accept a proton. Both reactions start with the polar carbon–halogen bond.Explain how the role of the hydroxide ion in substitution differs from its role in elimination.2 marks
- A student warms 2-bromopropane, CH₃CHBrCH₃, with aqueous potassium hydroxide and obtains mainly propan-2-ol. She repeats the experiment using a solution of potassium hydroxide in ethanol and heats the mixture strongly. This time the main product is a gas that decolourises bromine water.Write an equation for the formation of the gas in the second experiment, using structural formulae for the organic species.2 marks
- A technician needs to make propene from 2-bromopropane. A competing reaction, which gives propan-2-ol, takes place at the same time under some conditions, so the technician must choose the reagent and conditions with care.Describe, in words, the mechanism for the elimination of HBr from 2-bromopropane by hydroxide ions, referring to the movement of electron pairs.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).