OCR A Level Chemistry A: Haloalkanes (H432) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions test section 4.2.2, Haloalkanes. The source is OCR’s H432 document for A Level Chemistry A, specification Version 3.1 (dated May 2026), with first assessment in 2017. Between them the questions reach all five lettered statements of this Module 4 section, and question 5 and question 6 use the PAG7 rate comparison. As H432 is a full A level, no question is labelled AS. Questions 10 to 12 are longer: they bring in mole calculations and extended explanations.

Revise first with the haloalkanes study guide and the haloalkanes revision notes. The OCR A Level chemistry hub has the other units, and the H432 chemistry checklist lets you log what you have secured. For a broader test, take a 10-minute diagnostic.

Ar values for this set: C 12.0; Ag 107.9; H 1.0; Br 79.9; O 16.0; Cl 35.5. Other data appears in the question that needs it.

Questions

1. This question is on nucleophiles in haloalkane hydrolysis.

(a) Define the term nucleophile. [1] (b) From this list, choose the two species that can act as nucleophiles and explain your choice: H₂O, CH₄, OH⁻, H⁺. [2]

2. 2-Chloro-2-methylbutane, (CH₃)₂CClCH₂CH₃, is converted into an alcohol.

(a) State the reagent and conditions. [2] (b) Give a balanced equation, then name the alcohol formed. [2]

3. Sketch and label the mechanism for the reaction of 1-chloro-2-methylpropane, (CH₃)₂CHCH₂Cl, with hydroxide ions. Show curly arrows, relevant dipoles and lone pairs. [4]

4. Use these bond enthalpies: C–Cl 342 kJ mol⁻¹, C–Br 283 kJ mol⁻¹, C–I 232 kJ mol⁻¹.

Explain the order of the rates of hydrolysis of 1-chloro-3-methylbutane, 1-bromo-3-methylbutane and 1-iodo-3-methylbutane. [3]

5. A student compares the rates of hydrolysis of three primary haloalkanes using aqueous silver nitrate and ethanol in a water bath.

(a) Explain why ethanol is added. [1] (b) Explain why all tubes are kept in the same water bath. [1] (c) State one other variable the student must control. [1] (d) One tube gives a cream precipitate. Write an ionic equation, with state symbols, for its formation. [1] (e) Name the halogen present in the haloalkane in that tube. [1]

6. The student’s results at 55 °C are shown.

Haloalkane	Time for precipitate / s
1-chloro-2-methylpropane	2400
1-bromo-2-methylpropane	300
1-iodo-2-methylpropane	20

(a) Calculate the relative rate, 1/time, for each haloalkane. [2] (b) By what factor is the iodo compound’s rate greater than the bromo compound’s? [1] (c) State what the results show about the strength of the carbon–halogen bonds. [1]

7. In the experiment in question 6, water is the nucleophile.

(a) Write an equation for the hydrolysis of 1-bromo-2-methylpropane by water. [1] (b) State which atom in water donates the electron pair. [1]

8. 1,2-Dichloro-1,1,2,2-tetrafluoroethane, CF₂ClCF₂Cl, is a CFC.

(a) Explain why UV radiation breaks a C–Cl bond and not a C–F bond in this molecule. [1] (b) Write an equation for the formation of chlorine radicals from CF₂ClCF₂Cl. [1] (c) Write two equations to show how chlorine radicals catalyse the breakdown of ozone. [2] (d) Write the overall equation. [1]

9. Nitrogen monoxide radicals, •NO, also break down ozone. Write two equations for this process and explain why •NO is described as a catalyst. [3]

10. 5.38 g of 1-chloroheptane, C₇H₁₅Cl, is heated under reflux with excess aqueous sodium hydroxide. 3.94 g of heptan-1-ol, C₇H₁₅OH, is isolated.

(a) Write an equation for the reaction and state why heating under reflux is used. [2] (b) Calculate the percentage yield of heptan-1-ol. [3] (c) The aqueous layer is acidified with dilute nitric acid before aqueous silver nitrate is added. Explain why the acid is needed. [1] (d) The aqueous layer gives 5.17 g of silver chloride. Calculate the percentage of the 1-chloroheptane that was hydrolysed. [2]

11. A bromoalkane, CₙH₂ₙ₊₁Br, is fully hydrolysed. 3.018 g of the bromoalkane gives 3.756 g of AgBr after adding acidified silver nitrate.

(a) What colour is the silver halide formed? [1] (b) Calculate the relative molecular mass of the bromoalkane and find n. [4] (c) The bromoalkane is primary and unbranched. Give its name and the name of the alcohol formed with NaOH(aq). [2] (d) Predict, with a reason, whether the matching chloroalkane would be hydrolysed faster or slower. [1]

12. Bromotrifluoromethane, CF₃Br, was used in fire extinguishers.

(a) Write an equation for the action of UV on CF₃Br, and explain which bond breaks. [2] (b) Write two equations showing how the radical formed catalyses ozone breakdown. [2] (c) Explain why CFCs were widely used, why their low reactivity caused a problem, and how scientific evidence led to action. [3]

Answers

1. (a) An electron pair donor [1] (b) H₂O and OH⁻ [1]; each has a lone pair on oxygen to donate to an electron-deficient atom, while CH₄ and H⁺ have no lone pair [1] [3] Examiner insight: “A negative ion” is not a definition; water is neutral, so the answer must mention an electron pair being donated.

2. (a) Aqueous sodium hydroxide (or NaOH(aq)) [1]; heat under reflux [1] (b) (CH₃)₂CClCH₂CH₃ + NaOH → (CH₃)₂C(OH)CH₂CH₃ + NaCl [1]; 2-methylbutan-2-ol [1] [4] Examiner insight: The OH goes on carbon 2, where the Cl was; check the position number in your name matches.

3. Dipole shown: Cδ+–Clδ− [1]; :OH⁻ with lone pair, curly arrow from the lone pair to the δ+ carbon [1]; curly arrow from the C–Cl bond to Cl [1]; products (CH₃)₂CHCH₂OH and :Cl⁻ [1] [4] Examiner insight: An arrow that starts on the O atom itself, rather than on its lone pair or negative charge, does not show where the electron pair comes from.

4. The C–X bond must be broken in hydrolysis [1]. Bond enthalpy decreases C–Cl > C–Br > C–I, so C–I needs least energy to break [1]. Rate order: iodo > bromo > chloro compound [1] [3] Examiner insight: Use the data given; an answer built on polarity or electronegativity explains the wrong thing and gains little credit.

5. (a) The haloalkane is immiscible with the aqueous silver nitrate; ethanol dissolves both, so they can mix and react [1] (b) Temperature affects rate, so it must be the same in every tube [1] (c) Any one: same volume or amount of each haloalkane; same volume and concentration of AgNO₃(aq); same chain length; same volume of ethanol [1] (d) Ag⁺(aq) + Br⁻(aq) → AgBr(s) [1] (e) Bromine [1] [5] Examiner insight: For (c), “same amount” alone is vague; say which substance and whether you mean volume or concentration.

6. (a) Chloro: 1/2400 = 4.17 × 10⁻⁴ s⁻¹; bromo: 1/300 = 3.33 × 10⁻³ s⁻¹ [1]; iodo: 1/20 = 0.0500 s⁻¹ [1] (b) 300 ÷ 20 = 15 times faster [1] (c) Bond strength order is C–Cl > C–Br > C–I; the weakest bond breaks fastest [1] [4] Examiner insight: Give units of s⁻¹ for 1/time, and divide times for a ratio rather than subtracting them.

7. (a) (CH₃)₂CHCH₂Br + H₂O → (CH₃)₂CHCH₂OH + H⁺ + Br⁻ (or HBr) [1] (b) The oxygen atom (from one of its lone pairs) [1] [2] Examiner insight: Water is a reactant here, so write it in the equation; NaOH does not appear in this experiment.

8. (a) C–Cl has a lower bond enthalpy than C–F, so less energy is needed to break it [1] (b) CF₂ClCF₂Cl → CF₂ClCF₂• + •Cl [1] (c) •Cl + O₃ → •ClO + O₂ [1]; •ClO + O → •Cl + O₂ [1] (d) O₃ + O → 2O₂ [1] [5] Examiner insight: Every radical needs its dot, including the organic fragment in (b).

9. •NO + O₃ → •NO₂ + O₂ [1]; •NO₂ + O → •NO + O₂ [1]; •NO is regenerated in the second step, so it is not used up overall [1] [3] Examiner insight: Built by analogy with the chlorine cycle, each equation must still balance atoms on both sides.

10. (a) C₇H₁₅Cl + NaOH → C₇H₁₅OH + NaCl [1]; reflux allows long heating without losing the volatile organic compounds [1] (b) Mr C₇H₁₅Cl = 134.5, n = 5.38 ÷ 134.5 = 0.0400 mol [1]; theoretical mass = 0.0400 × 116.0 = 4.64 g [1]; yield = 3.94 ÷ 4.64 × 100 = 84.9% [1] (c) To remove excess OH⁻, which would otherwise form a brown precipitate with Ag⁺ and hide the result [1] (d) Mr AgCl = 143.4, n(AgCl) = 5.17 ÷ 143.4 = 0.03605 mol [1]; 0.03605 ÷ 0.0400 × 100 = 90.1% [1] [8] Examiner insight: Work out Mr of heptan-1-ol (116.0) from the formula; using the haloalkane’s Mr for the product is a frequent slip.

11. (a) Cream [1] (b) Mr AgBr = 187.8; n(AgBr) = 3.756 ÷ 187.8 = 0.0200 mol [1]; n(bromoalkane) = 0.0200 mol (one Br each) [1]; Mr = 3.018 ÷ 0.0200 = 150.9 [1]; 14.0n + 1.0 + 79.9 = 150.9, so n = 5 [1] (c) 1-bromopentane [1]; pentan-1-ol [1] (d) Slower, because the C–Cl bond enthalpy is greater than C–Br [1] [8] Examiner insight: Show the link from n(AgBr) to n(bromoalkane); stating Mr = 150.9 without the 1 : 1 step leaves the logic unclear.

12. (a) CF₃Br → •CF₃ + •Br [1]; C–Br is the weakest bond in the molecule, much weaker than C–F [1] (b) •Br + O₃ → •BrO + O₂ [1]; •BrO + O → •Br + O₂ [1] (c) CFCs are non-toxic, non-flammable and unreactive, so they suited refrigerants, aerosols and solvents [1]; their low reactivity means they survive long enough to reach the stratosphere, where UV releases chlorine radicals [1]; evidence of ozone loss was accepted and governments agreed to phase out CFCs (Montreal Protocol, 1987), leading to chlorine-free replacements [1] [7] Examiner insight: In (c), each of the three ideas needs its own clear sentence; a list of uses alone answers only the first part.

Where marks are usually lost

• Defining a nucleophile without the words “electron pair donor”.
• Drawing the first curly arrow from carbon to oxygen instead of from the lone pair to carbon.
• Omitting the Cδ+–Xδ− dipole or the charge on the halide ion.
• Explaining rate trends with polarity rather than bond enthalpy.
• Writing NaOH in the equation for hydrolysis by water.
• Confusing cream (AgBr) with yellow (AgI).
• Breaking C–F in a CFC or halon radical equation.
• Missing dots on radicals, or leaving the catalyst in the overall equation.
• Using the wrong Mr for the product in yield calculations.
• Subtracting times instead of dividing them when comparing rates.

Next steps

• Recap with the haloalkanes revision notes
• Reread the full explanations in the haloalkanes study guide
• Sharpen calculations with the amount of substance practice questions
• See every unit on the OCR A Level chemistry hub
• Track outcomes on the H432 chemistry checklist
• Book a free trial class
• Check other topics with a free 10-minute diagnostic

Official syllabus

Written against the OCR H432 document, A Level Chemistry A, in specification Version 3.1 (May 2026, first assessment 2017): section 4.2.2 Haloalkanes, statements (a) to (e).