A Level Chemistry (9701) topical questions with answers
Every A Level Chemistry topic, with free past-paper-style questions and full mark schemes. Pick a chapter to practise its questions, or generate a complete 9701-format paper in seconds, from easy warm-ups to A*-level "brutal" papers.
Paper 4 · A Level Structured Questions (A Level · 100 marks · 2h); Paper 5 · Planning, Analysis & Evaluation (A Level · 30 marks · 1h 15m).
Sample A Level Chemistry questions
A mixed set from across the syllabus. Questions marked A* challenge come from the hardest bank.
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]
Show answers and mark scheme
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 · 11 marksA* challenge
A weak acid HA has Ka = 1.74 × 10−5 mol dm−3.
(a)
Calculate the pH of a 0.100 mol dm−3 solution of HA. State one assumption.
[3]
(b)
A buffer contains 0.02 mol HA and 0.015 mol NaA in 1.00 dm3. Calculate its pH, and explain how it resists a change in pH when a small amount of acid is added.
[5]
(c)
A sparingly soluble salt MX has Ksp = 1.8 × 10−10 mol2 dm−6. Calculate its solubility in water and in 0.10 mol dm−3 NaX.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
2.88; [H+] = [A−] and dissociation is negligible
[H+] = √(Kac) C1 2.88 A1 assumption stated B1
3
(b)
4.63; A− + H+ → HA, large reservoir of A−
[H+] = Ka[HA]/[A−] C1 4.63 A1 added H+ removed by A−B1 equation A− + H+ → HA B1 ratio [HA]/[A−] changes little B1
5
(c)
1.34 × 10−5, 1.80 × 10−9
s = √KspC1 1.34 × 10−5A1 common ion: s = Ksp ÷ 0.10 = 1.80 × 10−9A1
3
Question 3Structured · 11 marks
The reaction between peroxodisulfate ions and iodide ions is slow but is catalysed by Fe²⁺ or Fe³⁺ ions. S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂
(a)
Explain why the uncatalysed reaction has a high activation energy.
[2]
(b)
Write two equations to show how Fe²⁺ catalyses the reaction.
[2]
(c)
Classify this catalysis as homogeneous or heterogeneous and explain why transition metal ions are good catalysts.
[2]
(d)
Describe a colorimetric method to follow the rate of formation of iodine.
[3]
(e)
In a clock reaction, a small fixed amount of sodium thiosulfate and some starch are added to the mixture. Explain why the time for the blue-black colour to appear can be used as a measure of initial rate.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
both reactants negative; repel
both reactants are negative ions B1 they repel each other so collisions with enough energy are rare B1
homogeneous (same phase) B1 transition metal ions have variable oxidation states so they can accept and donate electrons readily B1
2
(d)
colorimeter with filter; absorbance measured at intervals; calibration curve to convert to concentration
use a colorimeter with a suitable (blue) filter B1 record absorbance at regular time intervals B1 use a calibration curve to convert absorbance into [I₂] B1
3
(e)
only a small, fixed amount of reaction occurs before the colour appears, so concentrations stay close to initial values; rate ∝ 1/t
only a small fraction of the reactants is used before the colour appears, so concentrations are still close to their initial values (required) B1 the same fixed amount of iodine is formed each time, so rate ∝ 1/t B1
2
Question 4Structured · 11 marks
This question is about planning organic syntheses.
(a)
Suggest a three-step synthesis of 2-hydroxypropanoic acid from ethanol. Give reagents and conditions for each step and the intermediates.
[6]
(b)
Explain why the product is optically inactive.
[2]
(c)
Identify the functional groups in a compound that reacts with sodium carbonate, gives a positive iodoform test and has a broad IR peak at 3200–3600 cm−1.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
oxidise to ethanal (K₂Cr₂O₇/H⁺, distil); HCN/KCN → 2-hydroxypropanenitrile; hydrolyse with dilute acid
step 1: acidified K₂Cr₂O₇, distil to give ethanal B1 step 2: HCN with KCN catalyst B1 intermediate 2-hydroxypropanenitrile B1 step 3: heat under reflux with dilute HCl B1 product 2-hydroxypropanoic acid B1 correct intermediates named B1
6
(b)
planar carbonyl attacked equally from both sides; racemic mixture
the CN⁻ attacks the planar C=O group equally from either side B1 forming a racemic mixture B1
2
(c)
carboxylic acid; CH₃CH(OH)– group; alcohol O–H
carboxylic acid (reacts with carbonate) B1 CH₃CH(OH)– / CH₃CO– group (iodoform) B1 O–H of alcohol (3200–3600 cm−1) B1
3
Question 5Structured · 11 marks
This question is about the enthalpy change of solution of NaCl. Lattice energy = −787 kJ mol−1; enthalpy change of hydration of Na⁺ = −406 kJ mol−1; of Cl⁻ = −378 kJ mol−1.
(a)
Define enthalpy change of hydration.
[2]
(b)
Construct an energy cycle and calculate the enthalpy change of solution.
[3]
(c)
Explain why the enthalpy change of hydration of Mg²⁺ is more exothermic than that of Ca²⁺.
[2]
(d)
Define entropy and predict the sign of the entropy change when a solid ionic compound dissolves.
[2]
(e)
Explain how a salt with an endothermic enthalpy of solution can still be soluble at room temperature.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
enthalpy change when 1 mol of gaseous ions dissolves in water to give infinitely dilute solution
enthalpy change when one mole of gaseous ions B1 is dissolved in a large excess of water (infinitely dilute solution) B1
Mg²⁺ smaller, higher charge density; stronger ion–dipole attraction to water
Mg²⁺ has a smaller ionic radius (same charge) so a higher charge density B1 it attracts the δ− oxygen of water molecules more strongly B1
2
(d)
measure of disorder / number of ways of arranging; usually positive
entropy is a measure of the number of ways particles and energy can be arranged (disorder) B1 usually positive because ions are freed from the lattice B1
2
(e)
ΔS positive enough that ΔG = ΔH − TΔS negative
dissolving can be feasible if ΔG = ΔH − TΔS < 0 B1 a large positive ΔS makes −TΔS outweigh the positive ΔH B1
2
Question 6Structured · 12 marks
This question is about the proton NMR spectrum of ethyl ethanoate, CH₃COOCH₂CH₃.
(a)
State the number of different proton environments in ethyl ethanoate and their relative peak areas.
[2]
(b)
Predict the splitting pattern of each peak and explain using the n + 1 rule.
[3]
(c)
Explain why the CH₂ protons have a larger chemical shift than the CH₃ protons of the ethyl group.
[2]
(d)
Explain why tetramethylsilane (TMS) is used as a standard and why CDCl₃ is used as the solvent.
[3]
(e)
Describe how thin-layer chromatography and the Rf value could be used to check the purity of the ester.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
3; 3 : 2 : 3
three environments B1 ratio 3 : 2 : 3 B1
2
(b)
CH₃CO singlet (no adjacent H); OCH₂ quartet (3 adjacent H); CH₂CH₃ triplet (2 adjacent H)
CH₃ next to C=O: singlet (no neighbouring H) B1 CH₂: quartet (three H on adjacent carbon) B1 CH₃ of ethyl: triplet (two H on adjacent carbon) B1
3
(c)
adjacent to electronegative O, deshielded
the CH₂ is bonded to an electronegative oxygen atom B1 which withdraws electron density (deshields), so the signal is downfield B1
2
(d)
TMS: single peak, 12 equivalent H, inert, defined as 0; CDCl₃ contains no ¹H, so no solvent peak
TMS gives a single strong peak (12 equivalent H) and is inert / volatile B1 it is defined as δ = 0 B1 CDCl₃ contains no ¹H atoms so does not give a peak B1
3
(e)
single spot if pure; compare Rf with authentic sample in same solvent
pure sample gives a single spot B1 compare Rf with an authentic sample run under the same conditions B1
2
Frequently asked questions
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No. They are original questions written to match the A Level Chemistry 9701 syllabus and the style of real Cambridge papers, so you can practise beyond the past papers you have already done.
Do the Chemistry questions have answers?
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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