Use hydroxide’s role and the reaction conditions to distinguish alcohol formation from alkene formation, then connect equations to a preparation and yield calculation.
Same reagent, different role
With warm aqueous hydroxide, substitution to an alcohol is favoured. Hot ethanolic hydroxide favours elimination to an alkene. Both pathways can compete, and the substrate’s structure also affects the product mixture. Avoid saying that one solvent makes the other pathway impossible.
In substitution OH⁻ donates a pair to carbon and acts as a nucleophile. In elimination it accepts a proton from a carbon next to the one carrying X and acts as a base. This neighbouring carbon is often called the β-carbon.
The three arrows of elimination
For 2-bromopropane, show a C–H bond on either neighbouring CH₃ group. Draw an arrow from a hydroxide oxygen lone pair to that H; from the C–H bond to the bond joining the two carbons, creating C=C; and from the C–Br bond to Br. Draw propene, water and Br⁻ as products.
All three arrows move electron pairs. Hydrogen is removed from the adjacent carbon, not from the same carbon as Br. There must be a suitable adjacent C–H bond: chloromethane cannot form an alkene by this ordinary elimination.
Diagram placeholder
Elimination from 2-bromopropane
Labels to include:
- explicit neighbouring C–H
- OH⁻ lone pair to H
- C–H bond pair forms C=C
- C–Br pair goes to Br
- propene + H₂O + Br⁻
Draw the carbon bearing Br beside an adjacent carbon with an explicit H. Place the three arrows in the reactant drawing. The double bond must join these same two carbons. Hydroxide gains H to become water; bromine leaves with a negative charge.
Find every distinct neighbouring position
2-Bromobutane has two different adjacent carbon positions. Removing H from carbon 1 gives but-1-ene; removing H from carbon 3 gives but-2-ene. But-2-ene can then have E and Z arrangements. Count two structural alkene products, or three including E/Z forms, depending on what the question asks.
Work systematically: locate C–X, mark each neighbouring carbon with H, form each possible C=C, then remove duplicates caused by symmetry. Do not move the carbon skeleton.
Purification and yield in a chloroalkane preparation
A chloroalkane may be prepared from an alcohol using a specified chlorinating reagent or concentrated hydrochloric acid where suitable. Follow the question’s procedure: the apparatus, catalyst and heating requirements depend on the alcohol and reagent.
A separating funnel separates immiscible aqueous and organic layers. Identify the product layer using density information or the given procedure; the organic layer is not always the top layer. Remove the stopper before draining and vent safely when pressure develops. Washing removes specified soluble impurities; a suitable anhydrous drying agent removes residual water. Distillation then collects the required boiling range.
A washed sample is not necessarily dry, and a dry sample is not necessarily pure. Choose each step to remove a particular impurity. Use appropriate electrical heating for flammable liquids.
Quick checks
Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official AQA mark allocations.
Q1. What reagent and conditions favour propene from 2-bromopropane?Show answer
Heat with ethanolic KOH or NaOH, commonly under reflux. Hydroxide acts as a base in elimination.
Q2. Describe the arrow that creates C=C in this elimination.Show answer
It begins at the C–H bond on an adjacent carbon and ends between that carbon and the carbon bearing Br. The C–H electron pair becomes the additional bonding pair in C=C.
Q3. List the structural alkene products from 2-chlorobutane and state which shows E/Z.Show answer
But-1-ene and but-2-ene. But-2-ene shows E/Z because each double-bond carbon carries two different groups; but-1-ene has two H atoms at one end.
Q4. Why does changing water to ethanol not prove that the organic product is pure alkene?Show answer
Conditions alter the relative importance of competing substitution and elimination. Substitution can still occur, and elimination itself may give isomers.
Q5. A 1:1 preparation starts with 0.0500 mol and gives 3.70 g product of Mᵣ 92.5. Calculate percentage yield.Show answer
Theoretical mass = 0.0500 × 92.5 = 4.625 g. Yield = 3.70/4.625 × 100 = 80.0%.
Sources
Sources and examiner guidance (reviewed 2 October 2026)
- Chemrevise: Halogenoalkanes — Coverage checklist, pp1–5; original explanations and questions below.
- AQA 7405 specification — 3.3.3.1–3.3.3.3.
- AQA June 2022 AS Paper 2 mark scheme — Q05.3 p22: calculate yield using amounts and stoichiometry.
- AQA June 2022 AS Paper 2 examiner report — Q05.3 p4: synthesis yield context.
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