Eliminate hydrogen halide to form an alkene
Use ethanolic NaOH and heat to remove HX from a halogenoalkane and form a C=C bond.
AS Level; Cambridge International Chemistry 9701, 2025–2027 syllabus. Use the named reagents, solvents, temperatures and pressure conditions; begin curly arrows at a bond or lone pair and distinguish typical mechanisms from absolute rules. Written practice is self-checked
Before you begin
- Use covalent formulas, nucleophiles, electrophiles, homolytic and heterolytic fission, and the alkene and alcohol pathways in Topics 13–14.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What product class forms when HX is eliminated from a halogenoalkane?
What you will learn
- Recognise elimination of HX
- Use ethanolic NaOH and heat
- Distinguish an alkene from the aqueous substitution product
Elimination forms a double bond
A halogenoalkane heated with NaOH in ethanol can lose X and a hydrogen on an adjacent carbon. A C=C bond forms between those carbons.
Bromoethane gives ethene: C₂H₅Br + NaOH → C₂H₄ + NaBr + H₂O, with ethanolic NaOH and heat.
Aqueous and ethanolic solvents favor different outcomes
Aqueous NaOH favors substitution to an alcohol; ethanolic NaOH with heat favors elimination to an alkene.
The alkene has one fewer H and no halogen. Check the neighboring carbon from which the hydrogen is removed and balance the coproducts.
Pause and explain
Which condition favors the syllabus alkene-forming route from bromoethane?
Worked example
Write the elimination equation for 2-bromopropane and state the conditions.
Show the worked solution
- Remove Br from carbon 2 and H from an adjacent carbon.
- Form a C=C between the two carbons.
- Show sodium bromide and water as coproducts.
Answer CH₃CHBrCH₃ + NaOH → CH₃CH=CH₂ + NaBr + H₂O; ethanolic NaOH, heat.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Write the equation for eliminating HBr from 1-bromopropane with ethanolic NaOH.
Give me a hint
Remove Br from C1 and H from neighboring C2.
Compare my reasoning
- Keep the three-carbon skeleton.
- Form propene between C1 and C2.
- Add NaBr and H₂O to balance the equation.
CH₃CH₂CH₂Br + NaOH → CH₃CH=CH₂ + NaBr + H₂O; ethanol, heat.
Look for these in your work
- Skeleton preserved
- C=C positioned
- Coproducts and conditions given
A learner uses aqueous NaOH to make propene from 2-bromopropane. Correct the conditions and identify the competing product.
Give me a hint
Compare solvent-dependent substitution and elimination.
Compare my reasoning
- Aqueous NaOH favors replacement of Br by OH.
- That product is propan-2-ol.
- For propene, use ethanolic NaOH and heat.
Use ethanolic NaOH and heat for elimination to propene; aqueous NaOH favors substitution to propan-2-ol.
Look for these in your work
- Aqueous route named
- Product identified
- Elimination condition corrected
A halogenoalkane has no hydrogen on a carbon adjacent to the C–Br carbon. Can ordinary HX elimination form a C=C at that site? Explain.
Give me a hint
Elimination requires a hydrogen and leaving group on adjacent carbons.
Compare my reasoning
- The Br-bearing carbon loses the leaving group.
- A neighboring carbon must lose H to make the second bond.
- Without an adjacent H, that elimination route cannot form the C=C at that site.
No. The adjacent hydrogen required to form the second bond is absent.
Look for these in your work
- Bond changes stated
- Adjacent atom requirement recognized
- Conclusion follows
Common mistakes
- Using aqueous NaOH for the stated alkene route
- Replacing Br with OH instead of eliminating
- Forming C=C without removing an adjacent H
What can you explain now?
What product class forms when HX is eliminated from a halogenoalkane?
Compare with the explanation
An alkene
Elimination forms a C=C bond.
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Make it stick
How do the solvent, product and bond change identify elimination rather than substitution?
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