About Halogen compounds in A Level Chemistry
Halogen compounds is one of the 18 main topics in our A Level Chemistry (Cambridge (CAIE) 9701) question bank, and about 4% of its real questions come from it. Tayyari Karo has 29 real questions on this topic from 2021 to 2026, from Paper 1, Paper 2, Paper 4; 26 are multiple choice and marked instantly, and the rest are structured questions with a model answer. By difficulty: 15 medium, 12 hard, 2 easy.
Work through the questions below, then open the explanation for each one. Redo the ones you missed a few days later.
Try 6 real questions
These are real past-paper questions. Think of your answer first, then open the explanation.
May/June 2026 · Paper 1 · Halogen compounds
A sample of 2-bromobutane is strongly heated in a concentrated ethanolic solution of sodium hydroxide. How many isomeric organic products are formed (structural isomerism and stereoisomerism taken into account)?
- 1
- 2
- 3
- 4
Show answer and explanation
Answer: C. 3
Elimination gives but-1-ene, cis-but-2-ene and trans-but-2-ene: 3 isomers.
May/June 2025 · Paper 1 · Halogen compounds
Two reactions, both with NaOH(aq) in excess under the same conditions: 1 2-bromo-2-methylbutane heated with NaOH(aq); 2 2-chloro-2-methylbutane heated with NaOH(aq). 30.18 g of 2-bromo-2-methylbutane (Mr 150.9) forms 12.32 g of product X (the tertiary alcohol, Mr 88.15). Which row is correct (percentage yield of X; relative rate of reaction)?
- 41%; 1 is faster than 2
- 41%; 2 is faster than 1
- 70%; 1 is faster than 2
- 70%; 2 is faster than 1
Show answer and explanation
Answer: C. 70%; 1 is faster than 2
Moles bromide = 30.18/150.9 = 0.200 mol, theoretical moles of alcohol X = 0.200 mol (mass 17.63 g); actual yield 12.32 g gives % yield = 12.32/17.63 × 100 ≈ 70%. The C-Br bond is weaker than C-Cl, so the bromoalkane (reaction 1) reacts faster by SN1 hydrolysis.
May/June 2024 · Paper 1 · Halogen compounds
2-bromopropane is converted to 1,2-dibromopropane in a pathway involving two reactions: 2-bromopropane → compound X (reaction 1) → 1,2-dibromopropane (reaction 2). What are the reagents and conditions for the two reactions?
- Reaction 1: heat under reflux with aqueous NaOH; Reaction 2: HBr(g) at room temperature
- Reaction 1: heat under reflux with aqueous NaOH; Reaction 2: Br2(l) at room temperature
- Reaction 1: heat under reflux with ethanolic NaOH; Reaction 2: HBr(g) at room temperature
- Reaction 1: heat under reflux with ethanolic NaOH; Reaction 2: Br2(l) at room temperature
Show answer and explanation
Answer: D. Reaction 1: heat under reflux with ethanolic NaOH; Reaction 2: Br2(l) at room temperature
Ethanolic NaOH under reflux eliminates HBr from 2-bromopropane to give propene (X); propene then undergoes electrophilic addition with liquid Br2 at room temperature to give 1,2-dibromopropane.
May/June 2023 · Paper 1 · Halogen compounds
Which row describes the solvent used and type of reaction occurring when bromoethane reacts with NaOH to form ethene?
- ethanol, elimination
- ethanol, substitution
- water, elimination
- water, substitution
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Answer: A. ethanol, elimination
Ethene is formed from bromoethane + NaOH dissolved in ethanol (not water) via an elimination reaction (heat, OH⁻ acting as a base).
May/June 2022 · Paper 1 · Halogen compounds
Ethanol can be used to make propanenitrile in two steps: CH3CH2OH →(X) CH3CH2Br →(Y) CH3CH2CN. What types of reaction are X and Y?
- free-radical substitution; electrophilic substitution
- free-radical substitution; nucleophilic substitution
- nucleophilic substitution; nucleophilic substitution
- nucleophilic substitution; electrophilic substitution
Show answer and explanation
Answer: C. nucleophilic substitution; nucleophilic substitution
Ethanol to bromoethane (e.g. with HBr or PBr3/NaBr+H2SO4) is nucleophilic substitution (Br⁻ attacks the polar C-OH carbon after protonation, or a similar mechanism); bromoethane to propanenitrile (with KCN) is also nucleophilic substitution (CN⁻ attacks the polar C-Br carbon).
May/June 2021 · Paper 1 · Halogen compounds
Which statements are correct? 1: 1,1-difluoroethane is less reactive than 1,1-dichloroethane. 2: 1,1-difluoroethane is polar. 3: The C–F bond is stronger than the C–Cl bond.
- 1, 2 and 3 are correct
- 1 and 2 only are correct
- 2 and 3 only are correct
- 1 only is correct
Show answer and explanation
Answer: A. 1, 2 and 3 are correct
The C–F bond (≈485 kJ mol⁻¹) is stronger than C–Cl (≈340 kJ mol⁻¹), making fluoroalkanes less reactive than chloroalkanes; 1,1-difluoroethane's C–F bonds are highly polar and the molecule's shape does not cancel this, so it is polar — all three correct.