A Level Chemistry: Multi-Step Synthesis Routes — 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 available free from your board.

Related: Multi-Step Synthesis Routes revision notes

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Section A

1. State the reagents and conditions to convert propan-1-ol to (a) propanal, (b) propanoic acid. [4]

2. Name the three reagents that add a carbon atom to a chain. [3]

3. State the reagent and conditions to convert a nitrile to a primary amine, and explain why NaBH₄ cannot be used. [3]

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Section B

4. Devise a synthesis of butanoic acid starting from 1-bromopropane.

(a) State the number of carbon atoms in each and explain what this tells you about the route. [2]

(b) Give the reagents and conditions for each step. [4]

(c) Name the mechanism of the first step. [1]

5. Ethyl ethanoate can be made from ethanol by two different routes.

(a) Route 1 uses ethanoic acid directly. State the reagents and conditions, and give one disadvantage. [3]

(b) Route 2 converts the acid to ethanoyl chloride first. State the reagent for that conversion, and explain why this route gives a better yield. [3]

6. 1-Bromopropane is heated with NaOH(aq) in one experiment and with NaOH in ethanol in another. Explain why the two give different products, naming both products. [3]

7. Benzene is converted to phenylethanone (C₆H₅COCH₃) in one step, and methylbenzene is converted to benzoic acid in a separate reaction.

(a) Name the reagents and mechanism for converting benzene to phenylethanone. [3]

(b) State the reagent and conditions for oxidising methylbenzene’s side chain to give benzoic acid. [2]

8. Benzene is nitrated to form nitrobenzene.

(a) State the reagents and temperature. [3]

(b) Explain why the temperature must be carefully controlled, and what happens if it is exceeded. [2]

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Answers

1. (a) K₂Cr₂O₇/H₂SO₄ [1], warm and distil off the product immediately [1]. (b) K₂Cr₂O₇/H₂SO₄ [1], heat under reflux [1]. Distil vs reflux is the single most commonly dropped mark in the whole topic.

2. KCN in ethanol with a halogenoalkane, which adds exactly one carbon (the nitrile carbon) [1]; HCN with an aldehyde or ketone, also adding exactly one carbon [1]; Friedel–Crafts alkylation or acylation with RCl or RCOCl and AlCl₃, which adds however many carbons are in the R group chosen — not necessarily just one [1].

3. LiAlH₄ in dry ether (or H₂/Ni) [1]. NaBH₄ is not a powerful enough reducing agent [1] — it reduces aldehydes and ketones only, not nitriles [1].

4. (a) 1-bromopropane has 3 carbons, butanoic acid has 4 [1]. A route that adds exactly one carbon is therefore needed — the nitrile route (KCN then hydrolysis) is the standard method taught at this level for adding a single carbon to an aliphatic chain [1].

(b) Step 1: KCN in ethanol, heat under reflux [1] → butanenitrile [1]. Step 2: dilute HCl(aq), heat under reflux [1] → butanoic acid [1]. Ethanolic KCN is essential — aqueous conditions would give the alcohol instead.

(c) Nucleophilic substitution [1].

5. (a) Ethanoic acid + ethanol, concentrated H₂SO₄ catalyst, heat under reflux [1]. Disadvantage: the reaction is reversible, so the equilibrium yield is limited [1] and the reaction is slow [1].

(b) SOCl₂ [1]. The reaction of an acyl chloride with an alcohol is fast, irreversible and near-quantitative at room temperature [1], so the yield is much higher than the equilibrium-limited esterification [1].

6. NaOH(aq) and heat gives nucleophilic substitution → propan-1-ol [1]. NaOH in ethanol and heat gives elimination of HBr → propene [1]. The solvent determines whether OH⁻ acts mainly as a nucleophile or as a base [1].

7. (a) Ethanoyl chloride (CH₃COCl) with an AlCl₃ catalyst [1] — this is Friedel–Crafts acylation [1], an electrophilic substitution [1]. (b) Hot alkaline potassium manganate(VII) (KMnO₄ with NaOH(aq)) [1], then dilute acid (e.g. dilute H₂SO₄), which converts the benzoate ion formed in the alkaline mixture into benzoic acid [1]. The side chain is oxidised all the way down to a single –COOH group on the ring, regardless of how long the chain was, provided the carbon attached to the ring carries at least one hydrogen; a side chain with no such hydrogen, such as the tert-butyl group in tert-butylbenzene, is not oxidised this way.

8. (a) Concentrated nitric acid with concentrated sulfuric acid (as catalyst) [1], at a temperature between 25 °C and 60 °C [1], generating the electrophile NO₂⁺ [1]. (b) Above about 60 °C, a second nitro group is introduced (dinitration), giving a mixture of dinitrobenzene isomers as well as the desired mononitrobenzene, reducing the yield of the intended product [1]; keeping the temperature within the 25–60 °C range ensures the reaction stops at a single substitution [1].

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Where marks are usually lost

• Writing “oxidise” without naming both the reagent and the conditions.
• Refluxing where distillation is needed, so the aldehyde over-oxidises.
• Using aqueous rather than ethanolic KCN.
• Routes that gain or lose carbons with no reaction accounting for it.
• Using NaBH₄ on a nitrile or carboxylic acid.
• Forgetting AlCl₃ as the catalyst in a Friedel–Crafts reaction, or calling it addition rather than electrophilic substitution.
• Forgetting the risk of dinitration if the nitration temperature is not kept within 25–60 °C.

Bringing aliphatic and aromatic routes together

A synthesis question can require moving between an aliphatic starting material and an aromatic ring in the same route, so it is worth holding the aromatic-specific reactions in mind alongside the aliphatic interconversion ladder above: nitration (conc. HNO₃/H₂SO₄, 25–60 °C) introduces a nitro group that can then be reduced to an amine (Sn/conc. HCl, then NaOH); Friedel–Crafts alkylation (RCl/AlCl₃) and acylation (RCOCl/AlCl₃) attach an alkyl or acyl group directly to the ring; and oxidising an alkyl side chain with hot alkaline KMnO₄, then dilute acid, converts it all the way to a –COOH group regardless of its original length, which is a distinct reaction from the aliphatic alcohol oxidation ladder covered earlier. For the full reagent table and worked routes, see the Multi-Step Synthesis Routes revision notes.