Free Amines, amides & amino acids questions for A Level Chemistry (9701), part of Inorganic & organic. Each question has the answer and the marking points an examiner would look for.
Generate a full Amines, Amides & Amino Acids worksheet →All Inorganic & organic questionsPhenylamine is a weaker base than ethylamine.
Explain the difference in basicity.
| Part | Answer and marking points | Marks |
|---|---|---|
| lone pair delocalised into ring in phenylamine lone pair on N in phenylamine is delocalised into the benzene ring B1 less available to accept a proton B1 | 2 |
Ammonia, ethylamine and phenylamine are all bases.
Place them in order of increasing base strength and explain the order.
Describe how phenylamine is made from benzene in two steps, giving reagents and conditions.
Draw the structure of the zwitterion of aminoethanoic acid, and explain why amino acids have high melting points.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | phenylamine < ammonia < ethylamine; ethyl group donates electrons increasing lone pair availability; in phenylamine lone pair delocalised into ring correct order B1 basicity depends on availability of N lone pair B1 alkyl group electron-donating B1 lone pair delocalised into benzene ring B1 | 4 |
| (b) | conc HNO3/conc H2SO4, 55 °C → nitrobenzene; Sn/conc HCl, heat, then NaOH nitration reagents B1 temperature below 55 °C B1 Sn and concentrated HCl, heat B1 then NaOH(aq) B1 | 4 |
| (c) | H3N+CH2COO−; strong ionic attractions between zwitterions correct zwitterion B1 ionic attraction between zwitterions B1 | 2 |
This question is about amides and peptide hydrolysis.
Write an equation for the formation of N-methylethanamide from ethanoyl chloride and methylamine.
Explain why amides are not basic, unlike amines.
Describe the hydrolysis of a dipeptide using acid and using alkali, stating the products in each case.
Explain how the isoelectric point is used to separate amino acids by electrophoresis.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | CH₃COCl + 2CH₃NH₂ → CH₃CONHCH₃ + CH₃NH₃⁺Cl⁻ correct amide B1 second methylamine (or HCl) as by-product B1 | 2 |
| (b) | N lone pair delocalised into C=O; not available to accept proton the lone pair on nitrogen is delocalised over the C=O group B1 so it is not available to accept a proton B1 | 2 |
| (c) | acid: heat with HCl(aq) → amino acids as NH₃⁺ cations; alkali: heat with NaOH → carboxylate salts with NH₂ acid: heat under reflux with dilute HCl B1 products are protonated amino acids (–NH₃⁺, –COOH) B1 alkali: heat with aqueous NaOH B1 products are sodium salts (–NH₂, –COO⁻Na⁺) B1 | 4 |
| (d) | at buffer pH above pI negative, moves to anode; at pI does not move at its isoelectric point an amino acid has no net charge and does not move B1 at a pH above/below the pI it is negatively/positively charged and moves to the anode/cathode B1 | 2 |
This question is about the base strength of nitrogen compounds.
Put ammonia, ethylamine and phenylamine in order of base strength, strongest first.
Explain the position of each compound in your order.
Write an equation for the reaction of ethylamine with hydrochloric acid and name the product.
Describe a chemical test to distinguish phenylamine from ethylamine.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | ethylamine > ammonia > phenylamine ethylamine > ammonia > phenylamine B1 | 1 |
| (b) | ethyl donates electrons; ammonia reference; phenyl delocalises lone pair ethyl group is electron-donating, increasing lone-pair availability B1 ammonia has no donating group B1 in phenylamine the lone pair overlaps with the π system of the ring B1 making it less available to accept a proton B1 | 4 |
| (c) | C₂H₅NH₂ + HCl → C₂H₅NH₃⁺Cl⁻; ethylammonium chloride C₂H₅NH₂ + HCl → C₂H₅NH₃Cl B1 ethylammonium chloride B1 | 2 |
| (d) | bromine water gives white precipitate with phenylamine add aqueous bromine: phenylamine decolourises it and gives a white precipitate (2,4,6-tribromophenylamine) B1 no precipitate with ethylamine B1 | 2 |
Table 5.1 shows the pKa values of the conjugate acids of three bases.
| base | ammonia | ethylamine | phenylamine |
|---|---|---|---|
| pKa of conjugate acid | 9.25 | 10.8 | 4.6 |
Use Table 5.1 to put the three bases in order of base strength, strongest first.
Explain why ethylamine is a stronger base than ammonia.
Explain why phenylamine is a much weaker base than ammonia.
3.72 g of phenylamine reacts with excess ethanoyl chloride. Write the equation and calculate the mass of N-phenylethanamide formed if the yield is 72%. (Mr: phenylamine 93; N-phenylethanamide 135)
Describe how phenylamine is made from nitrobenzene.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | ethylamine > ammonia > phenylamine ethylamine > ammonia > phenylamine B1 | 1 |
| (b) | ethyl group electron-donating; N lone pair more available the ethyl group is electron-donating (positive inductive effect) B1 increasing the electron density on N, so the lone pair accepts a proton more readily B1 | 2 |
| (c) | lone pair delocalised into ring; less available the lone pair on N is delocalised into the benzene ring B1 so it is less available to accept a proton B1 | 2 |
| (d) | 3.89 g C₆H₅NH₂ + CH₃COCl → C₆H₅NHCOCH₃ + HCl B1 n = 3.72/93 = 0.0400 mol C1 × 135 × 0.72 C1 3.89 g A1 | 4 |
| (e) | Sn and conc HCl, heat; then NaOH tin and concentrated hydrochloric acid, heated under reflux B1 then add NaOH(aq) to release the free amine B1 | 2 |
Amino acids exist as zwitterions.
Draw the zwitterion of glycine, H₂NCH₂COOH.
Explain why amino acids have high melting points.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | H₃N⁺CH₂COO⁻ H₃N⁺CH₂COO⁻ B1 | 1 |
| (b) | ionic attractions between zwitterions zwitterions have strong ionic attractions between them B2 | 2 |
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