OxfordAQA A-Level Chemistry: Organic synthesis (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 questions below are on section 3.3.14 Organic synthesis in the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams from May/June 2020 onwards. Every question is International A-level only, because the whole section is A2 organic chemistry, examined in Unit 4; Unit 5 covers all content. For atom economy, use the 3.1.2.5 formula.

Before starting, work through the study guide and the revision notes. Course links: the OxfordAQA A-Level Chemistry hub, the printable 9620 checklist and the free 10-minute diagnostics.

Data for this set. Relative atomic masses: C = 12.0, H = 1.0, Br = 79.9, Cl = 35.5, K = 39.1, O = 16.0, N = 14.0.

Questions

1. A fruit-flavour ester used to be made in a reactor with methylbenzene as the solvent. The manufacturer now uses the liquid alcohol reactant in excess and adds no other solvent. Explain two advantages of the new process. [2]

2. A pharmaceutical company makes a hydroxynitrile intermediate by reacting an aldehyde with potassium cyanide, followed by dilute acid.

(a) Explain why adding the dilute acid is a particular hazard. [1] (b) Give two reasons why process designers try to avoid hazardous starting materials. [2]

3. But-1-ene can be made by heating 1-bromobutane with potassium hydroxide in ethanol:

C₄H₉Br + KOH → C₄H₈ + KBr + H₂O

Calculate the percentage atom economy for making but-1-ene. [2]

4. State the reagent or reagents, and any condition or catalyst, for each one-step conversion.

(a) Methylbenzene to (chloromethyl)benzene. [1] (b) Benzene to ethylbenzene. [2] (c) Nitrobenzene to phenylamine. [1]

5. Chlorocyclopentane is converted into compound C:

chlorocyclopentane → (NaOH(aq), warm) → A → (acidified K₂Cr₂O₇, reflux) → B → (KCN, then dilute acid) → C

(a) Identify A, B and C. [3] (b) State whether C is optically active. Give a reason. [1]

6. A four-step synthesis has step yields of 90%, 75%, 68% and 85%. Every step has a 1 : 1 mole ratio.

(a) A chemist starts with 0.150 mol of starting material. The final product has Mr = 151.0. Calculate the mass of final product. Give your answer to 3 significant figures. [3] (b) A two-step route to the same product has step yields of 70% and 60%. State, with a reason, which route gives more product from the same starting amount. [1]

7. Samples of chlorobenzene, C₆H₅Cl, and (chloromethyl)benzene, C₆H₅CH₂Cl, are each warmed with aqueous sodium hydroxide. Only one of them forms an alcohol. Identify which, and explain the difference in reactivity. [3]

8. Phenylmethanamine, C₆H₅CH₂NH₂, can be made from methylbenzene in two steps.

(a) Give the reagent and conditions for each step, and name the intermediate. [4] (b) The second step is C₆H₅CH₂Cl + 2NH₃ → C₆H₅CH₂NH₂ + NH₄Cl. Calculate the percentage atom economy of this step. [2]

9. Hexane-1,6-diamine, H₂N(CH₂)₆NH₂, is a monomer for nylon 6,6.

(a) Devise a two-step synthesis of hexane-1,6-diamine from 1,4-dibromobutane, Br(CH₂)₄Br. For each step give the reagent and conditions, and give the structure of the organic product. [4] (b) Name the other monomer used to make nylon 6,6 and state the type of polymerisation. [2]

10. 2-Methylpropan-1-amine, (CH₃)₂CHCH₂NH₂, is to be made from propan-2-ol, (CH₃)₂CHOH.

(a) Devise a four-step synthesis. Name each organic product, and give the reagent and conditions for every step. [8] (b) Explain why the route cannot be shortened by converting propan-2-ol directly into 2-bromopropane. [1]

11. A cationic surfactant is made in two steps.

Step 1: methylbenzene is converted into (chloromethyl)benzene. Step 2: C₆H₅CH₂Cl + CH₃(CH₂)₁₅N(CH₃)₂ → [C₆H₅CH₂N(CH₃)₂(CH₂)₁₅CH₃]⁺Cl⁻

(a) Write an equation for step 1 and state the condition needed. [2] (b) Calculate the percentage atom economy of step 1. [2] (c) Name the type of reaction in step 2, and state its percentage atom economy with a reason. [2] (d) Suggest two features of this process that make it less green, and explain each. [2]

Answers

1. No energy is needed to separate and recover a solvent from the product, for example by distillation [1]. No flammable, toxic solvent vapour can escape, and there is no solvent waste to dispose of [1]. [2] Examiner insight: Each point should give a consequence; “it is greener” or “it is cheaper” with no reason attached repeats the question.

2. (a) Acid reacts with cyanide ions to release hydrogen cyanide, a very toxic gas [1]. (b) Hazardous substances put workers and the surroundings at risk if there is a leak or accident [1]. They also need costly controls such as sealed storage, protective equipment and emergency plans [1]. [3] Examiner insight: In (a), naming the gas HCN is what earns credit; “it is dangerous” does not say why.

3. Sum of Mr of reactants = 136.9 + 56.1 = 193.0 [1]; % atom economy = 56.0 ÷ 193.0 × 100 = 29.0% [1]. [2] Examiner insight: Only reactants go in the denominator; adding the masses of KBr and water instead is a common slip that gives a different answer.

4. (a) Chlorine in UV light [1]. (b) Chloroethane [1] with an aluminium chloride (AlCl₃) catalyst [1]. (c) Tin and concentrated hydrochloric acid (then NaOH to free the amine) [1]. [4] Examiner insight: In (a) the condition carries the meaning: AlCl₃ instead of UV light would put the chlorine on the ring, which is a different product.

5. (a) A is cyclopentanol [1]. B is cyclopentanone [1]. C is 1-hydroxycyclopentane-1-carbonitrile, the ring carbon carrying both OH and CN [1]. (b) Not optically active: going round the ring either way from that carbon gives the same sequence of atoms, so it carries two identical groups and is not a chiral centre [1]. [4] Examiner insight: A clear structure for C is usually as good as a name; check that C has six carbons, because the CN carbon is new.

6. (a) Overall fraction = 0.90 × 0.75 × 0.68 × 0.85 = 0.39015 [1]; amount = 0.150 × 0.39015 = 0.05852 mol [1]; mass = 0.05852 × 151.0 = 8.84 g [1]. (b) The two-step route: 0.70 × 0.60 = 0.42, so 42.0% overall against 39.0% [1]. [4] Examiner insight: Keep unrounded values until the last line, then round to 3 significant figures as asked.

7. (Chloromethyl)benzene forms the alcohol, phenylmethanol [1]. Its chlorine is on a side-chain sp³ carbon, so the polar C–Cl bond is attacked by OH⁻ as in a halogenoalkane [1]. In chlorobenzene a chlorine lone pair overlaps with the delocalised ring electrons, strengthening the C–Cl bond, and the electron-rich ring repels the nucleophile [1]. [3] Examiner insight: The explanation needs both compounds; describing only why the side-chain chlorine reacts leaves the comparison half done.

8. (a) Step 1: chlorine [1] in UV light gives (chloromethyl)benzene [1]. Step 2: excess ammonia [1], in ethanol, heated in a sealed container, gives phenylmethanamine [1]. (b) Sum = 126.5 + 2(17.0) = 160.5 [1]; % atom economy = 107.0 ÷ 160.5 × 100 = 66.7% [1]. [6] Examiner insight: “Excess” ammonia belongs in the answer to step 2; without it, the amine product reacts further to give secondary and tertiary amines.

9. (a) Step 1: potassium cyanide, refluxed in a mixture of ethanol and water [1], gives NC(CH₂)₄CN, hexanedinitrile [1]. Step 2: reduction, for example H₂ with a nickel catalyst or LiAlH₄ [1], gives H₂N(CH₂)₆NH₂ [1]. (b) Hexanedioic acid [1]; condensation polymerisation [1]. [6] Examiner insight: Both bromine atoms are replaced in step 1, so the chain grows from four carbons to six; a product with only one CN group cannot give the diamine.

10. (a) Step 1: concentrated sulfuric (or phosphoric) acid, heat [1] gives propene [1]. Step 2: hydrogen bromide [1] gives 2-bromopropane as the major product [1]. Step 3: KCN in aqueous ethanol, heat under reflux [1] gives 2-methylpropanenitrile [1]. Step 4: reduction, for example H₂ with a nickel catalyst [1], gives 2-methylpropan-1-amine [1]. (b) The specification contains no reaction that converts an alcohol directly into a halogenoalkane, so the OH group has to be removed by elimination and HBr added to the alkene [1]. [9] Examiner insight: Name the major product in step 2; 1-bromopropane would lead to butan-1-amine, a different compound, so the position matters.

11. (a) C₆H₅CH₃ + Cl₂ → C₆H₅CH₂Cl + HCl [1]; UV light [1]. (b) Sum = 92.0 + 71.0 = 163.0 [1]; % atom economy = 126.5 ÷ 163.0 × 100 = 77.6% [1]. (c) Nucleophilic substitution (the amine lone pair attacks the CH₂ carbon) [1]; 100%, because the quaternary ammonium salt is the only product [1]. (d) Chlorine and (chloromethyl)benzene are toxic and irritant, so storage, handling and leaks are risks [1]. Step 1 makes hydrogen chloride as a by-product, which is corrosive waste and lowers the atom economy [1]. [8] Examiner insight: In (d), two different features are needed; a hazard and a waste point score more safely than two versions of the same hazard.

Where marks are usually lost

• Writing AlCl₃ instead of UV light for side-chain chlorination of methylbenzene.
• Proposing a nucleophile to replace chlorine bonded directly to a benzene ring.
• Using steps that are not in 9620, such as steam hydration or a direct alcohol-to-halogenoalkane step.
• Leaving out “excess” ammonia when a primary amine is the target.
• Adding or averaging step yields instead of multiplying them as decimals.
• Missing the coefficient in a denominator, such as 2NH₃.
• Naming the minor product of HBr addition, then building the rest of the route on it.
• Giving two answers that are the same idea (“safer” and “less hazardous”) where two points are needed.

Next steps

• Recap the topic: revision notes or the full study guide
• Drill single reactions: polymers, halogenoalkanes, alcohols, alkenes
• See every 9620 topic on the course hub
• Tick off points on the printable checklist
• Book a free trial class to work through routes with a teacher
• Find weak areas with all free 10-minute diagnostics

Official syllabus

OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: section 3.3.14 Organic synthesis (International A-level only), with percentage atom economy from 3.1.2.5.