A Level Chemistry Paper 4 · A Level Structured Questions practice papers
A Level · 100 marks · 2h. Each generated paper follows the 9701 Paper 4 · A Level Structured Questions format, with the same number and style of questions as the real exam, and comes with a Cambridge-style mark scheme.
Sample Paper 4 · A Level Structured Questions questions
Question 1Structured · 11 marksA* challenge
Compound Y, C3H6O2, has the following proton NMR spectrum: δ 1.2 (3H, triplet), δ 2.4 (2H, quartet), δ 11.7 (1H, singlet). The peak at δ 11.7 disappears when D2O is added.
(a)
Identify Y, explaining each feature of the spectrum.
[5]
(b)
An isomer of Y, Z, shows only two singlets in its proton NMR spectrum. Identify Z and give the ratio of the peak areas.
[2]
(c)
How many peaks would be seen in the 13C NMR spectrum of each of Y and Z?
[2]
(d)
In thin-layer chromatography of a mixture, a spot travels 3.6 cm and the solvent front 8.0 cm. Calculate the Rf value and explain why spots are often viewed under UV light.
[2]
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Part
Answer and marking points
Marks
(a)
propanoic acid CH3CH2COOH; triplet: CH3 next to CH2; quartet: CH2 next to CH3; δ 11.7 COOH proton; D2O exchange confirms O–H
propanoic acid B1 triplet explained by n + 1 rule B1 quartet explained B1 δ 11.7 is carboxylic acid O–H B1 disappears with D2O (exchangeable proton) B1
5
(b)
methyl ethanoate CH3COOCH3; 1 : 1
methyl ethanoate B1 1 : 1 B1
2
(c)
Y 3, Z 3
3 B1 3 B1
2
(d)
0.45; many organic compounds are colourless; they fluoresce/absorb UV
0.45 B1 colourless compounds made visible B1
2
Question 2Structured · 10 marksA* challenge
The initial rate of the reaction A + B → products is measured at constant temperature.
experiment
[A] / mol dm−3
[B] / mol dm−3
initial rate / mol dm−3 s−1
1
0.10
0.10
1.2 × 10−4
2
0.20
0.10
4.8 × 10−4
3
0.10
0.20
1.2 × 10−4
Table 2.1
(a)
Deduce the order with respect to A and to B. Explain your reasoning.
[3]
(b)
Write the rate equation and calculate the rate constant, with units.
[3]
(c)
Calculate the initial rate when [A] = 0.30 mol dm−3 and [B] = 0.30 mol dm−3.
[1]
(d)
Suggest a two-step mechanism consistent with your rate equation, identifying the rate-determining step.
[2]
(e)
For a different first-order reaction, k = 3.5 × 10−4 s−1. Calculate its half-life.
[1]
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Part
Answer and marking points
Marks
(a)
order 2 w.r.t. A, 0 w.r.t. B
A doubled (exp 1→2): rate × 4, order 2 B1 B doubled (exp 1→3): rate × 1, order 0 B1 reasoning shown B1
3
(b)
rate = k[A]2; k = 0.012 mol−1 dm3 s−1
correct rate equation B1 0.012 A1 correct units B1
3
(c)
1.1 × 10−3
1.1 × 10−3B1
1
(d)
slow step involves the species in the rate equation in those numbers; fast second step completes the reaction
rate-determining step contains 2 A only B1 second (fast) step gives the overall equation B1
2
(e)
1980 s
0.693 ÷ k = 1980 s B1
1
Question 3Structured · 12 marks
An alloy containing copper is dissolved in nitric acid; the excess acid is removed and the solution neutralised. Excess potassium iodide is added: 2Cu²⁺ + 4I⁻ → 2CuI + I₂. The iodine produced requires 24.80 cm3 of 0.0500 mol dm−3 sodium thiosulfate: I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻. The mass of alloy used was 0.150 g.
(a)
State the electronic configuration of Cu, Cu⁺ and Cu²⁺.
[3]
(b)
Explain why CuI is white whereas Cu²⁺ compounds are coloured.
[2]
(c)
Calculate the percentage by mass of copper in the alloy.
[4]
(d)
State the indicator used and the colour at the end-point, and explain why it is added near the end-point only.
Cu⁺ has a full 3d subshell B1 no electron can be promoted between split d orbitals, so no visible light is absorbed B1
2
(c)
52.5%
moles S₂O₃²⁻ = 1.24 × 10−3M1 moles I₂ = 6.20 × 10−4, moles Cu²⁺ = 1.24 × 10−3M1 mass Cu = 0.0787 g M1 52.5% A1
4
(d)
starch; blue-black to colourless (white ppt); added late to avoid starch–iodine complex not releasing iodine
starch B1 blue-black to colourless (white suspension of CuI) B1 added when the solution is pale yellow, because at high [I₂] a large amount of iodine is bound in the starch complex and released only slowly, so the end-point is not sharp B1
3
Question 4Structured · 12 marks
This question is about methylbenzene and its reactions.
(a)
Methylbenzene reacts with chlorine in two different ways depending on conditions. State the conditions and the organic product of each.
[4]
(b)
Explain why substitution in the ring occurs mainly at the 2- and 4-positions.
[2]
(c)
Methylbenzene is heated with alkaline KMnO₄, then acidified. Name the product and explain why benzene does not react.
[2]
(d)
Describe the mechanism of the Friedel–Crafts acylation of benzene with ethanoyl chloride and AlCl₃.
[4]
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Part
Answer and marking points
Marks
(a)
UV light: (chloromethyl)benzene, side-chain substitution; AlCl₃ catalyst: 2- and 4-chloromethylbenzene, ring substitution
UV light / heat B1 (chloromethyl)benzene C₆H₅CH₂Cl B1 AlCl₃ / FeCl₃ catalyst, room temperature B1 2-chloro- and 4-chloromethylbenzene B1
4
(b)
methyl group electron-donating; activates 2,4 positions, stabilises intermediate
the methyl group donates electrons into the ring B1 it increases electron density at the 2- and 4-positions, stabilising the intermediate B1
2
(c)
benzoic acid; benzene has no side chain, ring stable
benzoic acid B1 benzene has no alkyl side chain and the delocalised ring resists oxidation B1
CH₃COCl + AlCl₃ → CH₃CO⁺ + AlCl₄⁻ B1 curly arrow from ring to CH₃CO⁺ M1 intermediate with positive charge and broken delocalisation A1 loss of H⁺ restores ring, AlCl₃ + HCl regenerated A1
4
Question 5Structured · 3 marks
Phenol is more reactive than benzene.
Describe the reaction of phenol with bromine water and explain the difference from benzene.
[3]
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Part
Answer and marking points
Marks
white precipitate 2,4,6-tribromophenol; lone pair donated into ring
decolourises and gives white precipitate B1 2,4,6-tribromophenol B1 lone pair on O delocalised into ring increases electron density B1
3
Question 6Structured · 12 marks
This question is about proteins and polyamides.
(a)
Describe the primary, secondary and tertiary structure of a protein.
[4]
(b)
Explain why heating a protein changes its shape but not its primary structure.
[2]
(c)
Compare the structure of a protein with that of nylon-6,6.
[2]
(d)
Explain why poly(lactic acid) is described as biodegradable and why this is an advantage.
[2]
(e)
Draw the repeat unit of the polymer formed from 2-hydroxypropanoic acid, CH₃CH(OH)COOH, and name the type of polymerisation.
[2]
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Part
Answer and marking points
Marks
(a)
sequence of amino acids; H bonding α-helix / β-sheet; folding by R-group interactions
primary: sequence of amino acids joined by peptide bonds B1 secondary: α-helix or β-pleated sheet held by hydrogen bonds between C=O and N–H B1 tertiary: overall folding held by R-group interactions B1 e.g. disulfide bridges, ionic bonds, hydrogen bonds, van der Waals B1
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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in