Cambridge AS & A Level Chemistry 9701/42 is the May/June 2026 Paper 4 (A Level Structured Questions), variant 2. It lasts 2 hours, is worth 100 marks and is sat by every candidate taking the full A Level. It had nine compulsory questions, and the first, on transition elements, carried 20 marks, five of them for a single titration calculation. This guide goes through the paper question by question and shows what the official mark scheme gave marks for. The paper and its mark scheme are free on the Tutopiya Learning Portal.

For an overview of all five 9701 papers from this series, see A Level Chemistry 9701 May/June 2026 past papers: what came up and how it was marked.

9701/42 at a glance

Question	Topic	Marks	Main command word
1	Transition elements: chromium complexes, acidity of Fe³⁺(aq), silver(I), ethanedioate titration	20	Explain, complete, calculate
2	Group 2: carbonate decomposition, thermal stability, sulfate solubility, Ksp	10	Suggest, explain, calculate
3	Kinetics: rate equation from initial rates, propanone bromination mechanism, heterogeneous catalysis	10	Deduce, calculate, describe
4	Partition coefficient of ethanedioic acid	6	State, calculate, suggest
5	Electrode potentials, feasibility, the standard cell, E°cell from ΔG°	11	Define, explain, draw
6	Phenol, addition and condensation polymers, reduction, synthesis of aspirin	14	Draw, suggest
7	Alanine: optical activity, isoelectric point, tripeptide, synthesis	11	Describe, draw, suggest
8	NMR of an amide, alkaline hydrolysis	8	Predict, explain, draw
9	Basicity of amines, nitration and reduction, an azo dye	10	State, explain, suggest
	Total	100

Question-by-question walkthrough

Question 1: transition elements (20 marks)

• Two other typical properties (1 mark): they act as catalysts and form coloured compounds. Both were needed for the mark. High melting points and densities were ignored.
• Why oxidation states vary (1 mark): the 3d and 4s sub-shells are close in energy. Saying “shells” for orbitals was ignored.
• Cr³⁺ configuration (1 mark): [Ar] with three unpaired electrons in separate 3d boxes and an empty 4s box.
• Bond between H₂O and Cr³⁺ (1 mark): dative covalent, or coordinate.
• Why complex ions form (1 mark): they have vacant d orbitals that are energetically accessible, so they can accept lone pairs from ligands to form dative bonds. “Partially filled d orbitals” was ignored.
• Making [Cr(H₂O)₃(OH)₃] (1 mark): sodium hydroxide (ammonia also accepted), with the reaction type: deprotonation, acid-base, precipitation or ligand exchange.
• With excess ammonia (2 marks): [Cr(H₂O)₃(OH)₃] + 6NH₃ → [Cr(NH₃)₆]³⁺ + 3H₂O + 3OH⁻ (1), and the type is ligand exchange or substitution (1).
• Stereoisomers (2 marks): two 3D octahedral drawings. In one isomer, a pair of water ligands and a pair of hydroxide ligands sit at 180° to each other; in the other, all three of each type sit at 90°.
• Why Fe³⁺(aq) is acidic (2 marks): the small, highly charged Fe³⁺ ion polarises and weakens the O–H bonds in its water ligands (1), so a ligand loses H⁺: [Fe(H₂O)₆]³⁺ → [Fe(H₂O)₅OH]²⁺ + H⁺ (1). An equation with water forming H₃O⁺ was also accepted.
• Fe²⁺ against Fe³⁺ (1 mark): the Fe²⁺ solution has the higher pH, because Fe³⁺ has a higher charge density. Both halves were needed.
• Why silver(I) salts are colourless (2 marks): Ag⁺ has a full d sub-shell (d¹⁰) (1), so no electron can be promoted between d orbitals and no visible light is absorbed (1). “No d-d splitting” or “no energy gap” contradicted the answer.
• The titration (5 marks): a 5.00 g mixture of sodium ethanedioate, ethanedioic acid dihydrate and an inert solid was made up to 500 cm³, and 25.0 cm³ portions were titrated two ways.
  • Manganate(VII): 0.0250 × 0.02180 = 5.45 × 10⁻⁴ mol MnO₄⁻, so 2.5 × that = 1.3625 × 10⁻³ mol of ethanedioate in total.
  • Hydroxide: 0.150 × 0.01360 = 2.04 × 10⁻³ mol OH⁻, so half of that = 1.02 × 10⁻³ mol of ethanedioic acid.
  • Those four mole steps scored 1 mark for any two and 2 marks for all four.
  • Sodium ethanedioate in 25 cm³ = 1.3625 × 10⁻³ − 1.02 × 10⁻³ = 3.425 × 10⁻⁴ mol (1).
  • In 500 cm³: × 20 = 6.85 × 10⁻³ mol, × 134.0 = 0.918 g (1).
  • Percentage = 0.918 ÷ 5.00 × 100 = 18.4% (1). Scaling up to 500 cm³ first, or working in masses, was equally fine, and error carried forward ran through every step.

Question 2: Group 2 and solubility (10 marks)

• Curly arrows (1 mark): one arrow from an O⁻ to the C–O bond, and one from the breaking C–O bond to the other oxygen, which leaves as O²⁻. Both arrows were needed.
• Thermal stability (2 marks): PbCO₃ (most stable) > CaCO₃ > CuCO₃ (least stable), with no error carried forward for the order. The explanation: a larger cation has a lower charge density, so it polarises the carbonate ion less and the C–O bonds are weakened less. Linking larger radius to a stronger ionic bond contradicted the answer.
• Sulfate solubility (3 marks): both lattice enthalpy and hydration enthalpy become less exothermic down the group (1), hydration enthalpy changes by more (1), so the enthalpy change of solution becomes less exothermic or more endothermic (1). Leaving out the Δ or the H was penalised once.
• Ksp expression and units (2 marks): Ksp = [Mg²⁺][OH⁻]² (1), units mol³ dm⁻⁹ (1).
• [OH⁻] from pH 10.35 (1 mark): [H⁺] = 10⁻¹⁰·³⁵ = 4.47 × 10⁻¹¹, so [OH⁻] = 1.00 × 10⁻¹⁴ ÷ 4.47 × 10⁻¹¹ = 2.24 × 10⁻⁴ mol dm⁻³.
• Ksp (1 mark): [Mg²⁺] is half of [OH⁻], so Ksp = 0.5 × 2.24 × 10⁻⁴ × (2.24 × 10⁻⁴)² = 5.61 × 10⁻¹². An expression without the square earned no error carried forward here.

Question 3: reaction kinetics (10 marks)

Ethanal reacts with methanol under acid conditions, and Table 3.1 gave initial rates for four experiments.

• Rate equation (4 marks): each order had to be linked to named experiments.
  • Experiments 1 and 2: [CH₃OH] × 1.5 gives rate × 1.5, so first order in CH₃OH (1).
  • Experiments 3 and 4: [H⁺] × 4/3 gives rate × 4/3, so first order in H⁺ (1).
  • Experiments 2 and 3: [CH₃CHO] × 0.5, [CH₃OH] × 2 and [H⁺] × 3 give rate × 3, so first order in CH₃CHO (1). Other pairs were accepted if worked through correctly.
  • rate = k[CH₃CHO][CH₃OH][H⁺] (1).
• Units of k (1 mark): mol⁻² dm⁶ s⁻¹.
• Rate of experiment 5 (1 mark): k = 1.25, so 1.25 × 0.250³ = 0.0195 mol dm⁻³ s⁻¹.
• Bromination of propanone (2 marks): step 1, propanone gains H⁺ on the oxygen to give the protonated ketone; step 3, the enol reacts with Br₂ to give CH₃COCH₂Br and HBr. Curly arrows were ignored, even if wrong.
• Heterogeneous catalysis (2 marks): reactants adsorb onto the catalyst surface, bonds in the reactants weaken, and products desorb. “Absorbed” and “lowers the activation energy” were ignored.

Question 4: partition coefficient (6 marks)

• Definition (1 mark): the ratio of the concentrations of a solute in two immiscible solvents at equilibrium.
• Kpc of ethanedioic acid (3 marks): 2.70 g ÷ 90 = 0.0300 mol in total; 0.0165 × 0.200 = 0.0033 mol NaOH, so 0.00165 mol of acid in the ethoxyethane; 0.0300 − 0.00165 = 0.02835 mol in the water. Both layers are 50 cm³, so Kpc = 0.00165 ÷ 0.02835 = 0.058 (ethoxyethane over water). The reverse ratio, 17.2, was also accepted.
• Butanedioic acid (2 marks): Kpc would be higher (1), because butanedioic acid is less polar, so it is less soluble in water and more soluble in ethoxyethane (1). The comparison was needed; “it is non-polar” alone was ignored.

Question 5: electrochemistry (11 marks)

• Standard electrode potential (2 marks): the emf or voltage between a half-cell and a standard hydrogen electrode (1), under standard conditions of 298 K, 1 atm (101 kPa) and 1 mol dm⁻³ solutions (1).
• Two feasible reactions (3 marks): two of these three equations (2 marks), plus the reason that E°cell is positive (1):
  • Zn + 2Fe³⁺ → Zn²⁺ + 2Fe²⁺ (E°cell = +1.53 V)
  • NO₃⁻ + 10H⁺ + 8Fe²⁺ → NH₄⁺ + 3H₂O + 8Fe³⁺ (+0.10 V)
  • NO₃⁻ + 10H⁺ + 4Zn → NH₄⁺ + 3H₂O + 4Zn²⁺ (+1.63 V)
• Why they may not happen (1 mark): a high activation energy. “The rate is too slow” and “non-standard conditions” were ignored.
• Cell diagram (3 marks): the scheme listed nine features: a labelled salt bridge, a voltmeter, hydrogen gas with a sensible delivery system, H⁺(aq) or HCl(aq), a platinum electrode in each half-cell, a solution containing NO₃⁻, NH₄⁺ and H⁺, and a complete circuit. A battery in the circuit contradicted the answer.
• E°cell from ΔG° (2 marks): ΔG° = −nE°F, so −645 600 = −3 × E° × 96 500 (1), giving E°cell = +2.23 V (1).

Question 6: phenol and polymers (14 marks)

• Phenol with bromine water (2 marks): the orange solution decolourises and a white precipitate forms (1), and the product with excess bromine is 2,4,6-tribromophenol (1). “The orange colour disappears” scored, but “the colour disappears” without naming the colour was ignored, and fizzing contradicted the answer.
• Acidity of phenol and water (2 marks): a lone pair (p orbital) on the oxygen overlaps with the delocalised ring (1), which weakens the O–H bond or stabilises the phenoxide ion, so phenol is more acidic (1). “H⁺ is more easily lost” was ignored.
• Polymers of compound C (3 marks): C is 3-(2-hydroxyphenyl)propenoic acid. The addition polymer’s repeat unit is two carbons in the chain carrying H, the COOH group and the 2-hydroxyphenyl ring, with continuation bonds (1). The condensation polymer is a polyester: two repeat units joined through an ester link between the phenol OH of one and the COOH of the next (1), with the rest correct (1).
• Reductions (2 marks): LiAlH₄ reduces the COOH to CH₂OH and leaves the C=C; excess hydrogen with nickel saturates both the C=C and the benzene ring, leaving the COOH and OH.
• Synthesis of aspirin, F (5 marks): D is 2-methylphenol and E is 2-hydroxybenzoic acid (2). Step 1, CH₃Cl with AlCl₃; step 2, hot potassium manganate(VII); step 3, CH₃COCl (3). The scheme also credited routes that form the ethanoate ester at an earlier step, such as via phenyl ethanoate. Aqueous reagents in steps 1 or 3 contradicted the answer.

Question 7: alanine (11 marks)

• Optically active (1 mark): rotates the plane of plane-polarised light.
• Racemic mixture (1 mark): equal amounts of two optical isomers (enantiomers).
• Isoelectric point (1 mark): the pH at which the amino acid has no overall charge (exists as a zwitterion).
• Alanine at pH 10 (1 mark): CH₃CH(NH₂)COO⁻.
• Ala-Ala-Ala (2 marks): two displayed peptide links, with C=O shown (1), and the rest of the structure correct (1).
• Synthesis of J (5 marks): G is the N-ethanoyl derivative of alanine and H is its acyl chloride (2). Step 1, CH₃COCl; step 2, PCl₅, SOCl₂, or PCl₃ with heat; step 3, benzene with AlCl₃ (3). Adding AlCl₃ or concentrated sulfuric acid in step 1 contradicted the answer.

Question 8: NMR spectroscopy (8 marks)

Compound K was N-(2-chloroethyl)-2-methylpropanamide.

• Standard (1 mark): tetramethylsilane. Only that name scored, spelled correctly; “TMS” and the formula were ignored.
• Carbon-13 peaks (1 mark): five.
• Splitting patterns (2 marks): w (the two CH₃ groups) is a doublet, x (the CH) is a multiplet (heptet or septet accepted), and y and z (the two CH₂ groups) are triplets. “Double” or “triple” did not count.
• The extra peak in CDCl₃ (2 marks): a range of 5.0 to 12.0 ppm, the amide N–H range on the data table (1), because in D₂O the N–H proton exchanges with deuterium and disappears, which does not happen in CDCl₃ (1). A single value instead of the range did not score.
• Hydrolysis with hot NaOH (2 marks): sodium 2-methylpropanoate (the carboxylate, isolated at pH 10) and 2-aminoethanol, since the C–Cl is hydrolysed as well as the amide. Drawing O–Na as a covalent bond contradicted the answer.

Question 9: amines (10 marks)

• Basicity (4 marks): Q (phenylmethylamine) > ammonia > P (4-methylphenylamine) (1). Basicity depends on how available the nitrogen lone pair is to accept a proton (1). In P, the lone pair overlaps with the ring, so it is less available; in Q, the CH₂ group pushes electron density onto the nitrogen, so the lone pair is more available (2 from these three ideas).
• Steps from methylbenzene (2 marks): step 1, concentrated nitric and sulfuric acids; step 2, tin and concentrated hydrochloric acid. Calling tin a catalyst contradicted the answer.
• Reduction equation (1 mark): + 6[H] on the left and + 2H₂O on the right.
• Diazonium salt M (1 mark): the 4-methylbenzenediazonium ion, with the positive charge on the correct nitrogen. A covalent N–Cl bond contradicted the answer.
• Azo dye steps (2 marks): step 1, NaNO₂ and HCl (or HNO₂) at 10 °C or below; step 2, the sulfonated phenol shown in N under alkaline conditions (NaOH). Heating in step 1 contradicted the answer.

Where the marks are on 9701/42

• Organic chemistry carried 43 marks (Questions 6 to 9), physical chemistry 27 (Questions 3 to 5) and inorganic chemistry 30 (Questions 1 and 2).
• Drawing structures carried 18 marks: stereoisomers, the tribromophenol, both polymers, the reduction products, D and E, alanine at pH 10, the tripeptide, G and H, the hydrolysis products and the diazonium ion.
• Calculations carried 13 marks: the titration (5), [OH⁻] and Ksp (2), the rate of experiment 5 (1), Kpc (3) and E°cell (2). The rate equation deduction added 4 more.
• Reagents and conditions carried 10 marks across 6(e), 7(d) and 9. One list of reagents covers all of them.
• Question 1 alone is worth 20 marks. Transition element chemistry costs more on this paper than any other topic.

How to self-mark 9701/42 with the mark scheme

1. Learn the abbreviations first. A slash separates alternatives, A means accept, R means reject, I means ignore, CON means a contradiction that cancels the mark, DEP means the mark depends on an earlier one, ecf means error carried forward, ora means “or reverse argument” and u/c means unconditional.
2. Mark long calculations step by step. In 1(h), give yourself the method marks even if the final percentage is wrong, and carry your own numbers forward.
3. Check every order against named experiments. In 3(a)(i), an order without the experiments that show it did not score.
4. Look for contradictions. “No d-d splitting”, a battery in the cell, fizzing with bromine water, aqueous reagents in synthesis steps and covalent O–Na or N–Cl bonds all cancelled marks on this paper.
5. Check how a structure is drawn. Polymer repeat units need continuation bonds, the condensation polymer needs the ester link displayed, and the tripeptide needs both C=O groups displayed.
6. Check the exact terms. Tetramethylsilane in full, “multiplet” or “septet” rather than “multiple”, “adsorb” rather than “absorb”.

Sit 9701/42 as a timed mock

The paper allows 120 minutes for 100 marks, a little over a minute per mark. A workable plan:

Block	Marks	Time
Question 1	20	23 minutes
Question 2	10	11 minutes
Question 3	10	11 minutes
Question 4	6	7 minutes
Question 5	11	13 minutes
Question 6	14	16 minutes
Question 7	11	13 minutes
Question 8	8	9 minutes
Question 9	10	11 minutes
Check calculations, charges and structures	–	6 minutes

Lay out the titration in 1(h) as labelled mole steps, so that a slip early on costs one mark rather than five.

Other May/June 2026 9701 papers

• 9701 Paper 12 (Multiple Choice)
• 9701 Paper 22 (AS Level Structured Questions)
• 9701 Paper 32 (Advanced Practical Skills)
• 9701 Paper 52 (Planning, Analysis and Evaluation)

Practise and download

• Download 9701/42 and its mark scheme on the Learning Portal →
• All 9701 May/June 2026 papers: overview and marking patterns
• Try a free A Level Chemistry quiz

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Frequently asked questions

What topics were on 9701 Paper 42 May/June 2026? +

Nine questions: chromium and iron complexes with an ethanedioate titration (20 marks), Group 2 carbonates, sulfates and Ksp (10), the rate equation for ethanal with methanol (10), partition coefficients (6), electrode potentials (11), phenol, polymers and aspirin synthesis (14), alanine and peptides (11), NMR (8), and amines and an azo dye (10).

How many marks is 9701/42 and how long is it? +

It is out of 100 marks and lasts 2 hours. Question 1, on transition elements, is the biggest at 20 marks.

What was the answer to the titration in 9701/42 Question 1(h)? +

The percentage by mass of sodium ethanedioate was 18.4%. The mixture contained 0.02725 mol of ethanedioate in total and 0.0204 mol of ethanedioic acid in 500 cm³, so 0.00685 mol (0.918 g) was sodium ethanedioate out of 5.00 g.

What was Ksp for magnesium hydroxide on 9701/42? +

A pH of 10.35 gives [OH⁻] = 2.24 × 10⁻⁴ mol dm⁻³, so Ksp = 0.5 × 2.24 × 10⁻⁴ × (2.24 × 10⁻⁴)² = 5.61 × 10⁻¹² mol³ dm⁻⁹.

What was the rate equation on 9701/42 2026? +

Rate = k[CH₃CHO][CH₃OH][H⁺], first order in each. k is 1.25 mol⁻² dm⁶ s⁻¹, so experiment 5 with every concentration at 0.250 mol dm⁻³ has an initial rate of 0.0195 mol dm⁻³ s⁻¹.

Where can I download 9701/42 May/June 2026 and its mark scheme? +

Both are free to download and print on the Tutopiya Learning Portal, on the Cambridge AS & A Level Chemistry 9701 past papers page.

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