OxfordAQA A-Level Chemistry: Halogenoalkanes (9620) – 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.

The ten questions below test section 3.3.3 Halogenoalkanes, points 3.3.3.1 (nucleophilic substitution) and 3.3.3.2 (elimination), as set out in the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, used for exams from May/June 2020. This is International AS content. Expect it on the Unit 2 paper, Organic 1 and Physical 1; the Unit 5 practical and synoptic paper may revisit it.

If the mechanisms are new to you, start with the halogenoalkanes study guide or the shorter halogenoalkanes notes. For planning, use the OxfordAQA 9620 hub, the 9620 topic checklist and our free diagnostic quizzes. Relative atomic masses to use: Br = 79.9, C = 12.0, Cl = 35.5, H = 1.0, K = 39.1, N = 14.0, O = 16.0.

Questions

1. Here are five species: NH₄⁺, CN⁻, CH₄, NH₃, H⁺.

(a) Define the term nucleophile. [1] (b) Identify the two species in the list that act as nucleophiles towards halogenoalkanes, and give the feature they share. [2]

2. Nucleophiles react with a bromoalkane at one particular carbon atom. Say which one, and account for this in terms of how the bonding electrons are shared. [2]

3. 1-Bromo-2,3-dimethylbutane, BrCH₂CH(CH₃)CH(CH₃)CH₃, is heated under reflux with potassium cyanide in aqueous ethanol.

(a) Name the organic product. [1] (b) Using curly arrows, outline how this substitution takes place. [3] (c) A 4.947 g sample of 1-bromo-2,3-dimethylbutane gives 2.53 g of the product. Calculate the percentage yield. [3]

4. Chlorocyclohexane, C₆H₁₁Cl, reacts with potassium hydroxide.

(a) Outline the mechanism for the reaction in which hydroxide ions act as a nucleophile. [3] (b) Calculate the percentage atom economy for forming cyclohexanol, C₆H₁₁OH, by substitution and for forming cyclohexene, C₆H₁₀, by elimination. Both reactions use one KOH per C₆H₁₁Cl; the inorganic products are KCl, and KCl + H₂O respectively. [3] (c) State the conditions that would increase the proportion of cyclohexene. [1]

5. 3-Iodopentane, CH₃CH₂CHICH₂CH₃, is sealed in a tube with an ethanolic solution of ammonia, in excess, and heated.

(a) Outline the mechanism for the formation of the primary amine. [4] (b) Write an overall equation for the reaction. [1] (c) Explain why ammonia is used in excess. [1]

6. Data: mean bond enthalpy of C–Cl = 339 kJ mol⁻¹; of C–I = 240 kJ mol⁻¹.

A student states: “Chlorocyclopentane will react with hydroxide ions faster than iodocyclopentane, because the C–Cl bond is more polar.” Evaluate this statement. [3]

7. A student describes the mechanism for 1-bromo-3,3-dimethylbutane reacting with hydroxide ions:

“An arrow goes from the hydrogen atom of OH⁻ to the carbon bonded to bromine. A second arrow goes from the bromine atom to that carbon. The products are 3,3-dimethylbutan-1-ol and Br⁻.”

Identify three errors or omissions in the account of the mechanism. [3]

8. 2-Bromo-2-methylbutane, (CH₃)₂CBrCH₂CH₃, is heated with potassium hydroxide dissolved in ethanol.

(a) Explain the role of the hydroxide ion when an alkene forms, and how this differs from its role when an alcohol forms. [2] (b) Name the two alkenes that can form. [2] (c) Outline the mechanism for the formation of one of these alkenes. [3] (d) 12.07 g of 2-bromo-2-methylbutane gives 3.36 g of the mixture of alkenes, C₅H₁₀. Calculate the percentage yield of alkenes. [3] (e) Suggest one reason, other than mechanical losses, why the yield of alkenes is below 100%. [1]

9. 3-Chloro-3-methylpentane, CH₃CH₂C(CH₃)ClCH₂CH₃, reacts with potassium hydroxide. Substitution and elimination happen concurrently.

(a) Write an equation for the substitution reaction and name the organic product. [2] (b) Explain why the same reagent can give two different types of product. [2] (c) Draw or write the structures of every alkene that can form, counting stereoisomers separately, and state how many there are. [3] (d) Explain why one of the alkenes exists as E and Z isomers. [1]

10. For each equation, name the type of reaction and give the role of the inorganic reagent.

(a) C₆H₁₁Cl + OH⁻ → C₆H₁₀ + H₂O + Cl⁻ [1] (b) C₅H₉I + CN⁻ → C₅H₉CN + I⁻ [1] (c) CH₃CH₂CHICH₂CH₃ + 2NH₃ → CH₃CH₂CH(NH₂)CH₂CH₃ + NH₄I [1]

Answers

1. (a) An electron pair donor (a species that donates a lone pair to form a covalent bond) [1] (b) CN⁻ and NH₃ [1]; both have a lone pair of electrons to donate [1]. [3] Examiner insight: “Negative ion” is not the shared feature, since NH₃ is neutral; answers built around the lone pair are the ones that earn credit.

2. The carbon joined to bromine: the C–Br bonding pair is drawn towards bromine, the more electronegative atom, leaving that carbon δ+ [1]. The electron-deficient carbon attracts the lone pair of a nucleophile [1]. [2] Examiner insight: Say which atom is δ+ and why; “the bond is polar” alone does not explain the attack.

3. (a) 3,4-Dimethylpentanenitrile, NCCH₂CH(CH₃)CH(CH₃)CH₃ [1] (b) δ+ on the CH₂ carbon and δ− on Br [1]; curly arrow leaving the lone pair on cyanide’s carbon and ending at the CH₂ carbon [1]; C–Br electrons moving onto Br, with Br⁻ shown as a product [1]. (c) Mr(C₆H₁₃Br) = 164.9; n = 4.947 ÷ 164.9 = 0.0300 mol [1]; theoretical mass of C₇H₁₃N (Mr 111.0) = 0.0300 × 111.0 = 3.33 g [1]; yield = 2.53 ÷ 3.33 × 100 = 76.0% [1] [7] Examiner insight: An arrow drawn from the N of cyanide would join nitrogen to the chain, which does not make this nitrile, so it is unlikely to be credited even if the rest is right.

4. (a) Dipole on C–Cl (Cδ+, Clδ−) [1]; arrow from a lone pair on O of :OH⁻ to the carbon bonded to Cl [1]; arrow from the C–Cl bond to Cl, giving cyclohexanol and Cl⁻ [1]. (b) Total Mr of reactants = 118.5 + 56.1 = 174.6 [1]; substitution: 100.0 ÷ 174.6 × 100 = 57.3% [1]; elimination: 82.0 ÷ 174.6 × 100 = 47.0% [1] (c) KOH dissolved in ethanol, heated (rather than aqueous and warm) [1] [7] Examiner insight: Atom economy uses the total mass of all reactants in the balanced equation; dividing by the Mr of chlorocyclohexane alone would give 84.4% for cyclohexanol, which is too high.

5. (a) Nitrogen’s lone pair attacks carbon 3 while the C–I electrons move onto iodine (two arrows) [1]; intermediate CH₃CH₂CH(NH₃⁺)CH₂CH₃ with I⁻ [1]; another NH₃ uses its lone pair to take a proton from N⁺ [1]; the N–H electrons return to nitrogen, giving the amine and NH₄⁺ [1]. (b) CH₃CH₂CHICH₂CH₃ + 2NH₃ → CH₃CH₂CH(NH₂)CH₂CH₃ + NH₄I [1] (c) The amine product also has a lone pair and could attack more halogenoalkane; excess ammonia makes reaction with NH₃ much more likely [1]. [6] Examiner insight: The positive charge must be on nitrogen in the intermediate; leaving it off or placing it on carbon tends to cost the intermediate mark.

6. The statement is wrong: iodocyclopentane reacts faster [1]. The C–I bond enthalpy (240 kJ mol⁻¹) is lower than C–Cl (339 kJ mol⁻¹), so less energy is needed to break it [1]. Rate depends on breaking the C–X bond, so bond enthalpy, not polarity, is the deciding factor [1]. [3] Examiner insight: “Evaluate” needs a verdict as well as the reasoning; quote the data to support it.

7. The first arrow should start from a lone pair on the oxygen, not from hydrogen [1]. The second arrow should start from the C–Br bond and point to Br, not from Br to carbon [1]. The dipole (Cδ+ and Brδ−) is missing [1]. [3] Examiner insight: The products are correct; spotting a “mistake” that is not one gains nothing, so check each claim before you criticise it.

8. (a) In elimination OH⁻ acts as a base: it removes H⁺ from a carbon neighbouring the one bonded to bromine [1]; in substitution it acts as a nucleophile, donating its lone pair to the δ+ carbon [1]. (b) 2-Methylbut-1-ene [1]; 2-methylbut-2-ene [1]. (c) For 2-methylbut-2-ene: hydroxide’s oxygen lone pair attacks a hydrogen of the CH₂ group [1]; the electrons of that C–H bond shift into the gap between C3 and C2, making the double bond [1]; the C–Br pair goes to bromine, and the products are the alkene, H₂O and Br⁻ [1]. (For 2-methylbut-1-ene, the H comes from a CH₃ carbon.) (d) n(C₅H₁₁Br) = 12.07 ÷ 150.9 = 0.0800 mol [1]; theoretical mass of C₅H₁₀ = 0.0800 × 70.0 = 5.60 g [1]; yield = 3.36 ÷ 5.60 × 100 = 60.0% [1] (e) Substitution happens at the same time, so some of the halogenoalkane becomes 2-methylbutan-2-ol instead [1]. [11] Examiner insight: In (c) the hydroxide arrow must end on a hydrogen; an arrow to carbon describes substitution, so the elimination mechanism is unlikely to gain that mark.

9. (a) CH₃CH₂C(CH₃)ClCH₂CH₃ + OH⁻ → CH₃CH₂C(CH₃)(OH)CH₂CH₃ + Cl⁻ [1]; 3-methylpentan-3-ol [1] (b) OH⁻ has a lone pair, so it can donate it to the δ+ carbon as a nucleophile (substitution) [1], or use it to accept a proton from an adjacent carbon as a base (elimination) [1]. (c) CH₂=C(CH₂CH₃)₂, 2-ethylbut-1-ene [1]; CH₃CH=C(CH₃)CH₂CH₃, 3-methylpent-2-ene, as E and Z forms [1]; three alkenes in total [1]. (d) Rotation about C=C is restricted, and in 3-methylpent-2-ene neither alkene carbon has two identical substituents (one has H and CH₃, the other CH₃ and C₂H₅) [1]. [8] Examiner insight: Taking H from either ethyl group leads to one and the same alkene, because the ethyl groups match; listing it twice suggests you have not checked for duplicates.

10. (a) Elimination; OH⁻ acts as a base [1] (b) Nucleophilic substitution; CN⁻ acts as a nucleophile [1] (c) Nucleophilic substitution; NH₃ acts as a nucleophile (the second NH₃ acts as a base) [1] [3] Examiner insight: Each mark needs both parts, the reaction type and the role; “substitution” without “nucleophilic” is weaker than the full name.

Where marks are usually lost

• Arrows starting from a hydrogen atom or from a negative charge sign rather than from a lone pair.
• Using the nitrogen end of cyanide as the nucleophile.
• Missing the second ammonia molecule and the NH₄⁺ product.
• Dividing by the halogenoalkane’s Mr alone in atom economy calculations.
• Explaining the reactivity order with polarity instead of C–X bond enthalpy.
• Removing hydrogen from the carbon that carries the halogen in an elimination.
• Counting equivalent alkenes twice, or missing E and Z forms.
• Rounding moles too early, which shifts a percentage yield in the last figure.

Next steps

• Halogenoalkanes: condensed notes
• Halogenoalkanes: full study guide
• Curly arrows and isomers (organic basics)
• OxfordAQA chemistry course hub
• Printable 9620 topic checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

Questions written against OxfordAQA’s International AS and A-level Chemistry (9620) specification, Version 5.3 (exams from May/June 2020): section 3.3.3 Halogenoalkanes, an International AS section with two points, nucleophilic substitution (3.3.3.1) and elimination (3.3.3.2).