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A Level Chemistry: Halogen compounds, questions with answers

Practise 29 real questions on Halogen compounds from 2021 to 2026. See the answer and explanation as soon as you answer.

29real questions
26multiple choice
11exam sessions

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. 1
  2. 2
  3. 3
  4. 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)?

  1. 41%; 1 is faster than 2
  2. 41%; 2 is faster than 1
  3. 70%; 1 is faster than 2
  4. 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?

  1. Reaction 1: heat under reflux with aqueous NaOH; Reaction 2: HBr(g) at room temperature
  2. Reaction 1: heat under reflux with aqueous NaOH; Reaction 2: Br2(l) at room temperature
  3. Reaction 1: heat under reflux with ethanolic NaOH; Reaction 2: HBr(g) at room temperature
  4. 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?

  1. ethanol, elimination
  2. ethanol, substitution
  3. water, elimination
  4. water, substitution
Show answer and explanation

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?

  1. free-radical substitution; electrophilic substitution
  2. free-radical substitution; nucleophilic substitution
  3. nucleophilic substitution; nucleophilic substitution
  4. 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. 1, 2 and 3 are correct
  2. 1 and 2 only are correct
  3. 2 and 3 only are correct
  4. 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.

Sessions with Halogen compounds questions

SessionPaper 1Paper 2Paper 4
May/June 20262––
May/June 20253––
October/November 202521–
May/June 20243––
October/November 20243––
May/June 20232––
October/November 20231–1
May/June 20223–1
October/November 20221––
May/June 20212––
October/November 20214––

Common questions

What does the Halogen compounds topic cover on this page?

29 real A Level Chemistry (9701) questions on Halogen compounds from 2021 to 2026, from Paper 1, Paper 2, Paper 4. Each one shows the year, session and paper it came from.

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