A Level Chemistry: Amines — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Amines revision notes

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Section A

1. Explain why amines are basic. [2]

2. Place these in order of increasing base strength, and justify: ammonia, phenylamine, ethylamine, diethylamine. [4]

3. State the reagents and conditions for converting nitrobenzene to phenylamine. [3]

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Section B

4. Bromoethane is heated with excess ammonia in ethanol under pressure.

(a) Name the product and the mechanism. [2]

(b) Explain why an excess of ammonia is used. [3]

(c) Give the structure of the product formed if ammonia is not in excess and the reaction continues. [1]

5. Phenylamine is treated with nitrous acid at 5 °C, and the product is then reacted with phenol in alkaline solution.

(a) Name the intermediate formed with nitrous acid. [1]

(b) Explain why the temperature must be kept below 10 °C. [2]

(c) Name the final product type and explain why such compounds are strongly coloured. [3]

6. Compare the basicity of ethylamine and phenylamine, explaining the difference fully. [4]

7. Propanenitrile, CH₃CH₂CN, is reduced to a primary amine.

(a) State a suitable reagent for this reduction. [1] (b) Name the amine formed. [1] (c) Explain why this route is useful for lengthening a carbon chain, given that the nitrile itself was made from a halogenoalkane and KCN. [2]

8. Aqueous ethylamine is added to aqueous copper(II) sulfate.

(a) State the observation. [1] (b) Explain this observation, drawing a comparison with a reaction of aqueous ammonia you have met elsewhere in the course. [2]

9. Phenylamine is shaken with aqueous bromine at room temperature.

(a) State the observation. [1] (b) Name the organic product. [1] (c) Explain why phenylamine reacts with bromine water even faster than phenol does, and why no catalyst is needed. [2]

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Answers

1. The nitrogen atom has a lone pair of electrons [1] which can accept a proton [1].

2. phenylamine < ammonia < ethylamine < diethylamine [1]. Alkyl groups are electron-donating, increasing electron density on the nitrogen and making the lone pair more available [1], so more alkyl groups gives a stronger base [1]. In phenylamine the lone pair is delocalised into the benzene ring, making it much less available [1].

3. Tin and concentrated hydrochloric acid [1], heated [1], then sodium hydroxide to liberate the free amine from its salt [1]. The NaOH step is essential and routinely omitted.

4. (a) Ethylamine [1]; nucleophilic substitution [1].

(b) The ethylamine produced is itself a nucleophile [1] and can attack another bromoethane molecule [1], giving secondary, tertiary and quaternary products; excess ammonia makes further substitution statistically less likely [1].

(c) (C₂H₅)₂NH — diethylamine [1] (or triethylamine / tetraethylammonium bromide).

5. (a) A diazonium salt (benzenediazonium chloride) [1].

(b) Above about 10 °C the diazonium salt decomposes [1], releasing nitrogen gas and forming phenol instead of the desired product [1].

(c) An azo dye [1]. Coupling produces an extended delocalised electron system across both rings [1], which absorbs light in the visible region, so the compound appears strongly coloured [1].

6. Ethylamine is the stronger base [1]. Its ethyl group donates electron density to the nitrogen through the positive inductive effect, making the lone pair more available to accept a proton [1]. In phenylamine, the nitrogen lone pair overlaps with the delocalised π system of the benzene ring [1], so it is partially delocalised and much less available [1].

7. (a) LiAlH₄ in dry ether (or H₂ with a nickel catalyst) [1]. (b) Propylamine (1-aminopropane), CH₃CH₂CH₂NH₂ [1]. (c) The original halogenoalkane, CH₃CH₂Br, has only two carbons; reacting it with KCN adds a third carbon (as the nitrile carbon), giving propanenitrile [1]. Reducing the nitrile then converts that extra carbon’s triple-bonded nitrogen into a CH₂NH₂ group, so the overall two-step sequence has increased the chain length by one carbon, something a direct substitution reaction with ammonia cannot achieve [1].

8. (a) A deep blue solution forms [1]. (b) Ethylamine acts as a ligand, donating its nitrogen lone pair to the Cu²⁺ ion to form a complex ion [1], in the same way aqueous ammonia forms a deep blue complex with copper(II) ions — the amine’s lone pair behaves like ammonia’s in this respect, since both donate through nitrogen [1].

9. (a) An immediate white precipitate forms [1]. (b) 2,4,6-tribromophenylamine (2,4,6-tribromoaniline) [1]. (c) The nitrogen lone pair on the –NH₂ group is delocalised into the ring, activating it towards electrophilic substitution even more strongly than phenol’s –OH group does, so no halogen carrier is needed to polarise the Br–Br bond [1]; the ring is already electron-rich enough to do this itself, exactly as with phenol but to a greater degree [1].

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Where marks are usually lost

• Explaining basicity without reference to lone-pair availability.
• Saying phenylamine is a stronger base than ammonia.
• Omitting the NaOH step after reducing nitrobenzene.
• Forgetting the below-10 °C condition for diazotisation.
• Explaining azo dye colour without mentioning delocalisation.
• Forgetting that the nitrile route to a primary amine adds a carbon to the chain — a synthetic detail examiners specifically test when asking how to increase chain length by one carbon.
• Describing the amine–copper(II) reaction as a simple acid-base reaction rather than complex/ligand formation.

Two preparation routes to a primary amine, compared

Route	Reagents	Chain length	Product type
Halogenoalkane + excess NH₃	Ethanolic ammonia, heat, pressure	Unchanged	Primary amine (with some over-substitution risk)
Nitrile + LiAlH₄ (or H₂/Ni)	Reduction	Increased by one carbon (via the earlier KCN step)	Primary amine only

The nitrile route is the one to reach for whenever a synthesis question specifically asks for a chain one carbon longer than the starting halogenoalkane, since direct substitution with ammonia can never change the number of carbons present. For the full basicity comparison and diazotisation chemistry, see the Amines revision notes.