UNIT 2 CHEMISTRY · PEARSON AAQ · E1.7.5
Elimination Reactions of Halogenoalkanes
FREE CHEMISTRY REVISION · TOPIC 60
Elimination Reactions of Halogenoalkanes
Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · E1.7.5
1. What you need to know
Predict, name and draw the alkene formed when a halogenoalkane is heated with ethanolic sodium hydroxide. Write a balanced equation, state the conditions, and distinguish elimination from the aqueous substitution route. This lesson covers Unit 2 E1.7.5; mechanisms and practical procedures are not required.
2. Elimination removes groups from adjacent carbons
In an elimination reaction, a small molecule is removed from an organic molecule and a carbon–carbon double bond forms. For the reaction here, remove a halogen atom from one carbon and a hydrogen atom from an adjacent carbon. The two carbons become joined by C=C. The carbon skeleton is retained.
The overall organic change is loss of HX, where X is a halogen. With sodium hydroxide, the complete equation gives an alkene, a sodium halide and water. Do not write HX as an additional product in that same complete equation.
CH₃CH₂Br + NaOH → CH₂=CH₂ + NaBr + H₂O
Bromoethane gives ethene. The hydrogen and bromine removed from the organic compound are on neighbouring carbons. Each carbon in ethene has four bonds when its C=C double bond counts as two.
3. Reagent and conditions
Use ethanolic sodium hydroxide and heat, commonly heating under reflux. Ethanolic means the hydroxide is supplied in ethanol. Hot, concentrated ethanolic hydroxide favours elimination. Ethanol is the solvent; it is not added to the carbon chain or consumed in the overall equation.
Heating alone is not enough to identify the pathway: aqueous hydroxide substitution is also commonly heated. Always state the solvent. Reaction conditions favour a pathway rather than guaranteeing one pure product; substitution can compete. No apparatus diagram or detailed technique is needed for this specification.
| Conditions | Main transformation taught here | Organic product |
|---|---|---|
| NaOH in ethanol + heat | Elimination: lose H and X from adjacent carbons | Alkene |
| NaOH in water + heat | Substitution: replace X with OH | Alcohol |
CH₃CH₂Br + NaOH(aq) → CH₃CH₂OH + NaBr
For elimination, water is a product. For this substitution equation, the OH group enters the organic product. Never use the alcohol product with the stated ethanolic elimination conditions.
4. Predict and name the product
- Locate the carbon bonded to the halogen.
- Find each adjacent carbon that carries at least one H.
- Remove X and one such H, then make the bond between those carbons a double bond.
- Keep every carbon and check four bonds per carbon.
- Name the longest chain containing C=C and number it to give the double bond the lowest locant. Check whether different adjacent positions give distinct products.
Worked example 1 — bromoethane
CH₃CH₂Br + NaOH → CH₂=CH₂ + NaBr + H₂O
Heat with ethanolic NaOH. Only one adjacent carbon position is available, so the alkene is ethene. Atom check: C₂H₆BrNaO on each side; alkene H₄ plus water H₂ accounts for all six hydrogens.
Worked example 2 — 1-bromopropane
CH₃CH₂CH₂Br + NaOH → CH₃CH=CH₂ + NaBr + H₂O
Remove Br from the end carbon and H from the neighbouring CH₂. The alkene is propene, CH₂=CHCH₃. These two formula orientations represent the same product; numbering from the double-bond end avoids an unnecessarily high locant.
Worked example 3 — 1-chlorobutane
CH₃CH₂CH₂CH₂Cl + NaOH → CH₃CH₂CH=CH₂ + NaCl + H₂O
Heat with ethanolic NaOH. The product is but-1-ene, also written CH₂=CHCH₂CH₃. It has four carbons and formula C₄H₈. But-2-ene would put the double bond away from the original C–Cl position and is not this direct elimination product.
Worked example 4 — 2-bromopropane
CH₃CHBrCH₃ + NaOH → CH₂=CHCH₃ + NaBr + H₂O
The carbons on either side of the C–Br carbon are equivalent. Removing a hydrogen from either end gives propene. Do not count the same product twice simply because its structure is drawn from the opposite end.
5. When more than one alkene can form
For an unsymmetrical halogenoalkane, different adjacent carbons can give different double-bond positions. Start by drawing all valid connections; a balanced molecular formula alone does not identify them.
Worked example 5 — 2-bromobutane: two positions
CH₃CHBrCH₂CH₃ + NaOH → CH₂=CHCH₂CH₃ + NaBr + H₂O
CH₃CHBrCH₂CH₃ + NaOH → CH₃CH=CHCH₃ + NaBr + H₂O
These are alternative equations, each for one route, giving but-1-ene or but-2-ene. Do not put both alkenes on the right of one equation with a coefficient of one for each. But-2-ene also has E and Z forms; but-1-ene does not, because its terminal C has two identical H atoms. Product proportions are not required here.
6. Check the structural requirement
Worked example 6 — no adjacent hydrogen
Chloromethane, CH₃Cl, has no neighbouring carbon. It cannot form an alkene by the adjacent-carbon elimination described here. Do not invent a second carbon or remove H and Cl from the same carbon to draw C=C.
For a simple acyclic monohalogenoalkane, CₙH₂ₙ₊₁X gives CₙH₂ₙ where a suitable adjacent H exists. This formula check supports the structure; it does not replace drawing the double-bond position.
Common misconceptions
- Ethanolic means ethanoic acid: it means ethanol is the solvent.
- OH replaces Br in elimination: that describes substitution; elimination forms C=C.
- H and X leave the same carbon: they must be on adjacent carbons for this route.
- Every halogenoalkane gives only one alkene: different adjacent positions may give distinct products.
- The sodium halide is the only other product: include water in the full NaOH elimination equation.
- Heat alone identifies elimination: specify ethanolic hydroxide.
Examination tips
- Write ethanolic NaOH and heat above or alongside your equation.
- Mark the original C–X position before drawing C=C.
- Use a structural formula and the lowest double-bond locant.
- Check all carbons still have four bonds and conserve C, H, X, Na and O.
- Show alternative products as separate balanced equations.
- Explain the overall change without curly arrows or mechanisms.
Check your understanding
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