AQA A-Level Chemistry: Halogenoalkanes (7405) – 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 cover section 3.3.3 Halogenoalkanes of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: 3.3.3.1 Nucleophilic substitution, 3.3.3.2 Elimination and 3.3.3.3 Ozone depletion. All of it is AS content, so nothing here is A-level only. The topic is examined on Paper 2 at AS and A-level, and A-level Paper 3 can test any content.

Learn the ideas first in the halogenoalkanes study guide and the halogenoalkanes revision notes. The course hub is AQA A-Level Chemistry, and the printable checklist lists every specification point. A calculator is allowed for every question. Use Ar values: H = 1.0, C = 12.0, O = 16.0, Cl = 35.5, K = 39.1, Br = 79.9.

Questions

1. Explain why the carbon atom bonded to bromine in bromoethane is attacked by nucleophiles. [2]

2. Define the term nucleophile. [1]

3. 2-chlorobutane is warmed with aqueous sodium hydroxide.

(a) Name the organic product. [1] (b) Write an equation for the reaction. [1] (c) Name the type of reaction. [1]

4. 2-bromopropane reacts with potassium cyanide.

(a) State the conditions used. [1] (b) Outline the mechanism for this reaction. [3] (c) Name the organic product. [1]

5. 1-chloropropane reacts with excess ammonia to form propylamine.

(a) Write an equation for the reaction. [1] (b) Outline the mechanism for this reaction. [4] (c) Explain why excess ammonia is used. [1]

6. A student compares the rates of hydrolysis of 1-chloropropane, 1-bromopropane and 1-iodopropane. She adds each halogenoalkane to ethanol in a separate test tube, warms the tubes in a water bath, then adds aqueous silver nitrate to each.

(a) Identify the nucleophile in these reactions. [1] (b) Predict the colour of each precipitate and the order in which they appear. [2] (c) Explain this order. [2] (d) Suggest why ethanol is used. [1] (e) Explain why all three tubes are kept in the same water bath. [1]

7. 2-bromo-2-methylpropane reacts with potassium hydroxide. Two organic products form at the same time.

(a) Name the two organic products. [2] (b) Explain the two roles of the hydroxide ion in forming these products. [2] (c) Outline the mechanism for the formation of the alkene. [3]

8. 2-bromobutane is heated under reflux with potassium hydroxide dissolved in ethanol. Three isomeric alkenes form.

(a) Name the three alkenes. [3] (b) Explain why more than one structural isomer of the alkene forms. [1] (c) Explain why two of the alkenes are stereoisomers. [1] (d) State how the conditions should change to make butan-2-ol the main product. [1]

9. A student makes 2-chloro-2-methylpropane by shaking 7.40 g of 2-methylpropan-2-ol with excess concentrated hydrochloric acid. After purification she obtains 6.29 g of product.

(a) Write an equation for the reaction. [1] (b) Calculate the percentage yield. [3] (c) After removing the aqueous layer, the student purifies the crude product. Describe the steps. [3]

10. 2-bromopropane reacts with potassium hydroxide by elimination and by substitution:

elimination:  C3H7Br + KOH → C3H6 + KBr + H2O
substitution: C3H7Br + KOH → C3H7OH + KBr

(a) Calculate the percentage atom economy for making propene by elimination. [2] (b) Calculate the percentage atom economy for making propan-2-ol by substitution. [1] (c) Suggest one reason, other than an incomplete reaction, why the yield of propene is less than 100%. [1]

11. Dichlorodifluoromethane, CCl₂F₂, was once used as a refrigerant.

(a) State why ozone in the upper atmosphere is beneficial. [1] (b) Write an equation for the formation of a chlorine atom from CCl₂F₂ in the upper atmosphere. [1] (c) Explain why a C–Cl bond breaks rather than a C–F bond. [1] (d) Write two equations to show how chlorine atoms decompose ozone. [2] (e) Explain how these equations show that chlorine atoms act as a catalyst. [2] (f) Suggest why hydrofluorocarbons (HFCs) do not destroy ozone in this way. [1]

Answers

1. Bromine is more electronegative than carbon, so the C–Br bond is polar [1]. The carbon is δ+ (electron deficient), so it attracts a nucleophile, which donates an electron pair to it [1]. [2] Examiner insight: “Bromine is electronegative” on its own is weak; compare it with carbon and say what that does to the charge on carbon.

2. An electron pair donor [1]. [1] Examiner insight: “Something attracted to positive charge” describes behaviour, not the definition; write “electron pair donor”.

3. (a) Butan-2-ol [1] (b) CH₃CHClCH₂CH₃ + NaOH → CH₃CH(OH)CH₂CH₃ + NaCl (or with OH⁻ → Cl⁻) [1] (c) Nucleophilic substitution [1] [3] Examiner insight: “Substitution” without “nucleophilic” is incomplete for (c); name both parts of the type.

4. (a) KCN in aqueous ethanol, heated under reflux [1] (b) Arrow from the lone pair on the carbon of CN⁻ to the δ+ carbon of C–Br [1]. Arrow from the C–Br bond to Br [1]. Products (CH₃)₂CHCN and Br⁻ [1]. (c) 2-methylpropanenitrile [1] [5] Examiner insight: The nitrile carbon counts in the chain: (CH₃)₂CHCN has a three-carbon chain with a methyl branch, so it is not butanenitrile.

5. (a) CH₃CH₂CH₂Cl + 2NH₃ → CH₃CH₂CH₂NH₂ + NH₄Cl [1] (b) Arrow from the lone pair on N of NH₃ to the δ+ carbon [1]. Arrow from the C–Cl bond to Cl [1]. Intermediate CH₃CH₂CH₂NH₃⁺ with the positive charge on N [1]. Arrow from the lone pair of a second NH₃ to an H on N⁺, and arrow from that N–H bond onto N, giving CH₃CH₂CH₂NH₂ + NH₄⁺ [1]. (c) The amine product has a lone pair and could act as a nucleophile; excess NH₃ makes further substitution less likely, so the primary amine is the main product [1]. [6] Examiner insight: An equation with only one NH₃ is unbalanced unless HCl is shown as a product; using 2NH₃ and NH₄Cl is the safe form.

6. (a) Water [1] (b) Yellow (AgI), cream (AgBr), white (AgCl) [1]; order of appearance: iodo first, then bromo, chloro last [1] (c) The C–I bond has the lowest bond enthalpy / is the weakest C–X bond [1], so it needs the least energy to break and substitution is fastest [1]. (d) The halogenoalkane does not mix with water; ethanol is a solvent for both, so the reactants mix [1]. (e) Rate depends on temperature, so temperature must be the same for a fair comparison [1]. [7] Examiner insight: An answer to (c) based on polarity or electronegativity of the halogen cannot score: the rate order is explained by C–X bond enthalpy.

7. (a) 2-methylpropan-2-ol [1] and 2-methylpropene [1] (b) As a nucleophile, OH⁻ donates a lone pair to the δ+ carbon, giving the alcohol [1]. As a base, OH⁻ accepts H⁺ from a carbon next to the C–Br carbon, giving the alkene [1]. (c) Arrow from the lone pair on O of OH⁻ to an H on a CH₃ carbon [1]. Arrow from that C–H bond to between the two carbons, forming C=C [1]. Arrow from the C–Br bond to Br; products CH₂=C(CH₃)₂, H₂O and Br⁻ [1]. [7] Examiner insight: In (c) the H removed must be on a carbon adjacent to the C–Br carbon; here the central carbon carries no H at all, so an arrow to it is wrong.

8. (a) But-1-ene [1], (E)-but-2-ene [1], (Z)-but-2-ene [1] (b) H can be removed from C1 or from C3, as both are next to the C–Br carbon (C2), so the C=C forms in two positions [1]. (c) There is restricted rotation about the C=C, and each carbon of the C=C carries two different groups (H and CH₃) [1]. (d) Use aqueous potassium hydroxide (water as solvent) and a lower temperature (warm) [1]. [6] Examiner insight: For (c), “restricted rotation” alone is not enough; also say each C=C carbon has two different groups.

9. (a) (CH₃)₃COH + HCl → (CH₃)₃CCl + H₂O [1] (b)

Mr (CH3)3COH = 74.0   n = 7.40 ÷ 74.0 = 0.100 mol
Mr (CH3)3CCl = 92.5   theoretical mass = 0.100 × 92.5 = 9.25 g
% yield = 6.29 ÷ 9.25 × 100 = 68.0%

Moles of alcohol [1]; theoretical mass of product [1]; 68.0% [1] (c) Shake with sodium hydrogencarbonate solution in a separating funnel to remove acid, releasing the pressure (CO₂), then discard the aqueous layer [1]. Dry the organic layer with an anhydrous salt such as anhydrous sodium sulfate [1]. Distil and collect the fraction boiling near the boiling point of the product [1]. [7] Examiner insight: In (b) the theoretical mass must use the Mr of the chloroalkane (92.5), not the alcohol; moles stay the same because the ratio is 1 : 1.

10. (a) Sum of Mr of reactants = 122.9 + 56.1 = 179.0 [1]; atom economy = 42.0 ÷ 179.0 × 100 = 23.5% [1] (b) 60.0 ÷ 179.0 × 100 = 33.5% [1] (c) Substitution happens at the same time, so some 2-bromopropane forms propan-2-ol instead of propene [1]. [4] Examiner insight: Atom economy uses all reactants in the balanced equation and only the desired product, so dividing by the products’ total gives the wrong answer.

11. (a) It absorbs ultraviolet radiation [1]. (b) CCl₂F₂ → •CClF₂ + Cl• [1] (c) The C–Cl bond has a lower bond enthalpy (is weaker) than the C–F bond [1]. (d) Cl• + O₃ → ClO• + O₂ [1]; ClO• + O₃ → 2O₂ + Cl• [1] (e) Cl• is used in the first step and regenerated in the second [1]. It does not appear in the overall equation, 2O₃ → 3O₂, so one Cl• can destroy many ozone molecules [1]. (f) HFCs contain no chlorine (no C–Cl bond), so no chlorine atoms are formed [1]. [8] Examiner insight: Show the radical dot on Cl and ClO in (b) and (d); missing dots make the species ambiguous and can cost the mark.

Where marks are usually lost

• Curly arrows starting from the H of OH⁻, or from a minus sign with no lone pair, instead of the lone pair on O (or on C for CN⁻, N for NH₃).
• The C–X arrow pointing from the halogen to carbon instead of from the bond to the halogen.
• The second NH₃ step left out of the ammonia mechanism.
• Nitriles named without counting the C≡N carbon.
• Reactivity explained by polarity rather than C–X bond enthalpy.
• In elimination, an H removed from the C–X carbon itself, or OH⁻ called a nucleophile.
• Aqueous and ethanolic conditions swapped between substitution and elimination.
• Ozone equations without radical dots, or with Cl• not regenerated.
• In yield calculations, the theoretical mass worked out from the wrong compound’s Mr.

Next steps

• Recap with the halogenoalkanes revision notes.
• Reread the halogenoalkanes study guide.
• See what the alkene products do next in alkenes.
• Course hub: AQA A-Level Chemistry.
• Printable checklist for every specification point.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

AQA, AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: section 3.3.3 Halogenoalkanes (3.3.3.1 to 3.3.3.3).