AQA A-Level Chemistry: Organic synthesis (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 3.3.14 Organic synthesis of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams from June 2016 onwards. The section is A-level only, so every question here is A-level only. It is examined on A-level Paper 2 (organic chemistry), and A-level Paper 3 can test any content. Atom economy calculations use the formula in 3.1.2.5.

Learn the ideas first in the organic synthesis study guide and the revision notes. The course hub is AQA A-Level Chemistry, the printable checklist lists every specification point, and the free 10-minute diagnostics help you decide what to revise first.

Data for this set. Relative atomic masses: H = 1.0, C = 12.0, N = 14.0, O = 16.0, Na = 23.0, Br = 79.9.

Questions

1. Explain two reasons why chemists aim to design processes that do not require a solvent. [2]

2. Explain why chemists aim to design processes that use non-hazardous starting materials. [2]

3. Propylamine can be made from 1-bromopropane and excess ammonia:

CH₃CH₂CH₂Br + 2NH₃ → CH₃CH₂CH₂NH₂ + NH₄Br

Calculate the percentage atom economy for making propylamine. [2]

4. State the reagent or reagents needed for each one-step conversion.

(a) Butanal to butan-1-ol. [1] (b) Benzene to nitrobenzene. [2]

5. Outline a two-step synthesis of butanone from 2-bromobutane. For each step give the reagent and conditions, and name the intermediate. [4]

6. Cyclohexanone is converted into compound S by this sequence:

cyclohexanone → (NaBH₄, aqueous) → P → (concentrated H₃PO₄, heat) → Q → (HBr) → R → (KCN in aqueous ethanol, reflux) → S

Identify P, Q, R and S. [4]

7. A three-step route has step yields of 78%, 64% and 90%. Each step has a 1 : 1 mole ratio. A chemist starts with 0.250 mol of starting material. The final product has Mr = 88.0. Calculate the mass of final product obtained. Give your answer to 3 significant figures. [3]

8. Propan-2-ol can be made by two routes.

Route 1: CH₃CH=CH₂ + H₂O → CH₃CH(OH)CH₃ (steam, acid catalyst) Route 2: CH₃CH=CH₂ + HBr → CH₃CHBrCH₃, then CH₃CHBrCH₃ + NaOH → CH₃CH(OH)CH₃ + NaBr

(a) Calculate the overall percentage atom economy of route 2, taking the reactants as propene, HBr and NaOH. [2] (b) State the percentage atom economy of route 1. [1] (c) Give two other reasons why route 1 is the better route. [2]

9. Devise a three-step synthesis of N-ethylethanamide, CH₃CONHCH₂CH₃, starting from ethene. For each step give the reagent and any conditions, and name the organic product. [6]

10. Compound T is 2-hydroxy-2-methylpropanenitrile, (CH₃)₂C(OH)CN.

(a) Devise a four-step synthesis of T from 1-bromopropane. For each step give the reagent and conditions, and name the organic product. [8] (b) State whether T is optically active and give a reason. [1]

11. N-phenylethanamide, C₆H₅NHCOCH₃, can be made from benzene in three steps. The last step uses ethanoic anhydride.

(a) Devise the synthesis. For each step give the reagent and conditions, and name the organic product. [6] (b) Give two industrial advantages of using ethanoic anhydride rather than ethanoyl chloride in the last step. [2]

Answers

1. Energy is needed to separate the solvent from the product, for example by distillation [1]. The solvent becomes waste that must be disposed of or recycled, and many solvents are flammable or toxic [1]. [2] Examiner insight: “Explain two reasons” asks for two different points; “it is cheaper” with no reason why is unlikely to gain credit.

2. Hazardous substances put workers and the environment at risk during storage, transport and use [1]. Avoiding them removes the need for costly safety controls such as protective equipment and special containers [1]. [2] Examiner insight: Link each point to a consequence (risk to people, extra cost); “they are dangerous” on its own repeats the question.

3. Sum of Mr of reactants = 122.9 + 2(17.0) = 156.9 [1]; atom economy = 59.0 ÷ 156.9 × 100 = 37.6% [1]. [2] Examiner insight: The coefficient 2 in front of NH₃ must be used; leaving it out gives the wrong denominator, though a correct method with that slip may still earn the method mark.

4. (a) Sodium tetrahydridoborate(III), NaBH₄, in aqueous solution [1]. (b) Concentrated nitric acid [1] and concentrated sulfuric acid [1]. [3] Examiner insight: Write “concentrated” for both acids in (b); dilute acids do not generate the nitronium ion.

5. Step 1: aqueous sodium hydroxide [1], warm; the intermediate is butan-2-ol [1]. Step 2: acidified potassium dichromate(VI) [1], heat under reflux, to give butanone [1]. [4] Examiner insight: “Aqueous” matters in step 1. KOH in ethanol favours elimination and gives butenes, which breaks the route.

6. P is cyclohexanol [1]. Q is cyclohexene [1]. R is bromocyclohexane [1]. S is the nitrile C₆H₁₁CN (cyclohexanecarbonitrile), with one more carbon than R [1]. [4] Examiner insight: A correct structure is acceptable where a name is hard; check that S has seven carbons, because the CN carbon is new.

7. Overall fraction = 0.78 × 0.64 × 0.90 = 0.44928 [1]; amount = 0.250 × 0.44928 = 0.1123 mol [1]; mass = 0.1123 × 88.0 = 9.88 g [1]. [3] Examiner insight: Keep unrounded values between steps; rounding to 0.449 and then 0.112 gives 9.86 g, which is wrong at 3 s.f. Error carried forward is usually allowed after an early slip.

8. (a) Sum of Mr of reactants = 42.0 + 80.9 + 40.0 = 162.9 [1]; atom economy = 60.0 ÷ 162.9 × 100 = 36.8% [1]. (b) 100%: addition gives a single product [1]. (c) Route 1 has one step instead of two, so the overall yield is likely to be higher and less energy and time are used [1]. Route 1 produces no sodium bromide waste and uses fewer reagents [1]. [5] Examiner insight: In (c) “higher atom economy” is not an “other” reason; two separate further points are needed.

9. Step 1: hydrogen bromide [1] gives bromoethane, CH₃CH₂Br [1]. Step 2: excess ammonia in ethanol, heated in a sealed container [1], gives ethylamine (ethanamine), CH₃CH₂NH₂ [1]. Step 3: ethanoyl chloride (or ethanoic anhydride) [1] gives N-ethylethanamide, CH₃CONHCH₂CH₃ [1]. [6] Examiner insight: “Excess” ammonia is part of the condition in step 2; without it, further substitution gives secondary and tertiary amines.

10. (a) Step 1: KOH in ethanol, hot, under reflux [1] gives propene [1]. Step 2: steam with an acid catalyst such as phosphoric acid [1] gives propan-2-ol as the major product [1]. Step 3: acidified potassium dichromate(VI), heat under reflux [1] gives propanone [1]. Step 4: KCN, followed by dilute acid [1] gives T, 2-hydroxy-2-methylpropanenitrile [1]. (b) Not optically active: the carbon bonded to OH and CN also carries two identical CH₃ groups, so it is not a chiral centre [1]. [9] Examiner insight: Step 2 must give propan-2-ol; propan-1-ol would oxidise to propanal or propanoic acid, not propanone, so naming the major product shows the route works.

11. (a) Step 1: concentrated nitric acid and concentrated sulfuric acid [1] give nitrobenzene [1]. Step 2: reduction with tin and concentrated hydrochloric acid, then sodium hydroxide [1], gives phenylamine [1]. Step 3: ethanoic anhydride [1] gives N-phenylethanamide and ethanoic acid [1]. (b) Ethanoic anhydride is cheaper [1]. It is less corrosive, reacts less vigorously with water and does not release hydrogen chloride gas [1]. [8] Examiner insight: For (b), give two distinct advantages; “safer” and “less hazardous” count as one idea unless you say why.

Where marks are usually lost

• Adding step yields instead of multiplying them.
• Missing coefficients (such as 2NH₃) from the atom economy denominator.
• Giving a reagent with no conditions, or the wrong solvent (aqueous vs ethanolic KOH).
• Forgetting “excess” ammonia for a primary amine, or “excess” oxidant and reflux for a carboxylic acid.
• Ignoring the major product when an unsymmetrical alkene reacts with steam or HBr.
• Using NaBH₄ to reduce a nitrile or a nitro compound.
• Not counting carbons: KCN is needed whenever the chain grows by one.
• Giving one vague idea (“safer”, “greener”) for two marks.

Next steps

• Organic synthesis revision notes
• Organic synthesis study guide
• Reaction practice: halogenoalkanes, alkenes, alcohols
• AQA A-Level Chemistry course hub
• Printable AQA A-Level Chemistry checklist
• 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 3.3.14 Organic synthesis (A-level only), with atom economy from 3.1.2.5.