Cambridge AS & A Level Chemistry 9701/22 is the May/June 2026 Paper 2 (AS Level Structured Questions), variant 2. It lasts 1 hour 15 minutes, is worth 60 marks and tests the AS content. It had six compulsory questions, and the three organic questions carried 32 of the 60 marks. Question 1, on oxygen and its compounds, was the biggest single question at 14 marks. 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.

9701/22 at a glance

Question	Topic	Marks	Main command word
1	Oxygen: 2p orbitals, Li₂O bonding, oxides and chlorides with water, LiNO₃, the Haber process	14	State, identify, explain
2	Protons and electrons between charged plates; nuclear charge and atomic radius	6	Identify, explain
3	NaCl and NaBr with concentrated sulfuric acid	8	Identify, explain
4	Alcohols: chlorination, dehydration, dichromate oxidation and a structure to deduce	10	Identify, calculate, suggest
5	Free radical substitution of hexane	9	Suggest, complete
6	An unsaturated dicarboxylic acid: stereoisomers, hydration, mass spectrum, a distinguishing test	13	Suggest, explain
	Total	60

Question-by-question walkthrough

Question 1: oxygen and its compounds (14 marks)

• Electrons and orbitals (2 × 1 mark): an oxygen atom has four electrons in its 2p sub-shell. The highest-energy electron is in a 2p orbital, and you had to name it and draw its dumbbell shape. “p” alone was ignored, and the drawing needed a node at the centre. Shading, axes and orientation did not matter.
• Dot-and-cross diagram of Li₂O (2 marks): one mark for the correct charges (+ on each lithium ion, 2− on the oxide ion) and one for everything else. That means two Li⁺ ions with only their inner pair of electrons, and an O²⁻ ion with a full outer shell of eight, two of which came from lithium, all in brackets. Drawing it as covalent capped the answer at 1.
• Oxides and chlorides with water (4 × 1 mark): the Period 3 oxide that behaves like Li₂O is sodium oxide, and the name was required, since formulas were ignored. The equation is Li₂O + H₂O → 2LiOH. The solid Period 3 chlorides that give an acidic solution are phosphorus(V) chloride, PCl₅, and aluminium chloride, AlCl₃ (Al₂Cl₆ was accepted), one mark each. Silicon chloride and sodium chloride were rejected, and magnesium chloride and sulfur chlorides were ignored.
• Decomposing LiNO₃ (1 mark): the brown gas is nitrogen dioxide, NO₂. “Nitrogen oxide” on its own was ignored.
• Haber process conditions (2 marks): high pressure (about 200 atm) and an iron catalyst. Numbers without units were ignored, nickel and promoters were ignored, and platinum or vanadium oxide was rejected.
• Rate above 450 °C (3 marks): the initial rate increases, because more particles have energy equal to or greater than the activation energy, so there is a greater frequency of effective collisions. “More collisions” or “more effective collisions” without the idea of per unit time was ignored, and so was anything about yield or Le Chatelier. “Particles have more activation energy” was rejected.

Question 2: atomic structure (6 marks)

Protons, neutrons and electrons left a source at the same speed and passed between charged plates. One curve bent sharply towards the positive plate and the other bent gently towards the negative plate.

• Identify the particles (1 mark): the sharply bent curve x is the electron and the gently bent curve y is the proton. Both were needed for the mark.
• Why y bends less (1 mark): protons have a much larger mass. A higher mass-to-charge ratio or greater momentum was also accepted. “Protons are larger” and kinetic energy were ignored, and “protons have greater charge density” was rejected.
• Testing the hypothesis (2 × 1 mark): the table gave carbon (atom A, 0.077 nm), chlorine (atom B, 0.099 nm) and silicon (atom C, 0.117 nm). A and B, or A and C, support the idea that a higher nuclear charge means a larger atom. B and C disprove it, because chlorine has more protons than silicon but is smaller.
• Why A is smaller than C (2 marks): carbon has one fewer electron shell (two, not three), so there is less shielding, and the nucleus attracts the outer electrons more strongly. The reverse argument for C was accepted: its extra shell outweighs its extra protons. If you gave numbers of shells they had to be right, and “distance from the nucleus” alone did not score the first mark.

Question 3: sodium halides and concentrated sulfuric acid (8 marks)

The table described the observations: sodium chloride gave fizzing and steamy acidic fumes, while sodium bromide also gave a brown vapour and a pungent gas.

• Equation for reaction 1 (1 mark): H₂SO₄ + NaCl → NaHSO₄ + HCl. The version making Na₂SO₄ and the ionic forms were also accepted.
• Why it is a Brønsted–Lowry reaction (2 marks): H₂SO₄ is the acid and Cl⁻ is the base (one mark), and H₂SO₄ donates a proton to Cl⁻ (one mark). A general definition of acids and bases scored nothing, and calling HCl the base was ignored.
• Species in reaction 2 (3 marks): steamy fumes HBr, brown vapour Br₂, pungent gas SO₂. If you gave both a formula and a name, the formula was marked.
• Why the brown vapour forms (1 mark): bromide ions (or HBr) are oxidised by the sulfuric acid; saying the sulfuric acid is reduced by bromide was equally good. “Bromine is oxidised” was rejected, because bromine is the product, and “displacement” was ignored.
• Why there is no pungent gas with NaCl (1 mark): sulfuric acid is not a strong enough oxidising agent to oxidise chloride, or chloride is not a strong enough reducing agent to reduce it. Saying bromide is the stronger reducing agent also scored. Bond energy, electronegativity and general reactivity arguments were ignored.

Question 4: alcohols and dichromate oxidation (10 marks)

The starting alcohol D was 2-methylpropan-2-ol, drawn in skeletal form.

• Reagent G for the chloro compound (1 mark): PCl₅, PCl₃, SOCl₂, or concentrated hydrochloric acid or HCl gas. Dilute or aqueous HCl was ignored, and so was NaCl with concentrated acid, because the question asked for a single reagent. PCl₅(aq) was rejected.
• Role of Al₂O₃ (1 mark): a catalyst or a dehydrating agent. “Acid” was ignored and “redox” was rejected.
• Structure of F (1 mark): methylpropene, H₂C=C(CH₃)₂, in any unambiguous form. The molecular formula C₄H₈ alone was ignored.
• Colour change (1 mark): orange to green, both colours needed. Any shade of orange or green was fine.
• Amount of dichromate (1 mark): 10.0 ÷ 1000 × 0.500 = 0.005 mol (5 × 10⁻³ mol).
• Amount of X (1 mark): the Mr of C₅H₁₂O₄ is 136, so 0.255 ÷ 136 = 1.875 × 10⁻³ mol. The scheme accepted 1.88 × 10⁻³ or 1.9 × 10⁻³, but rejected 1.87 × 10⁻³ (wrongly rounded), 1.880 × 10⁻³ and 1.90 × 10⁻³ (false precision) and 2.0 × 10⁻³.
• Balancing the equation (2 marks): W and dichromate reacted in a 0.0012 : 0.0020 ratio, which is 3 : 5. So the equation is 3C₅H₁₂O₄ + 5Cr₂O₇²⁻ + 40H⁺ → 3C₅H₆O₆ + 10Cr³⁺ + 29H₂O. One mark for the 5, and one for the whole equation.
• Structure of W (2 marks): W is an isomer of X with four OH groups, two of them primary. Its oxidation product Z gives an orange precipitate with 2,4-DNPH but no change with Fehling’s reagent, so Z has a ketone but no aldehyde. That fits two primary OH groups (oxidised to COOH), one secondary OH (oxidised to C=O) and one tertiary OH (unchanged). One answer was HOCH₂CH(OH)C(OH)(CH₃)CH₂OH, and the scheme also accepted (HOCH₂)₂C(OH)CH(OH)CH₃. A structure that was right apart from a connectivity slip, such as a C–H–O bond, scored 1.

Question 5: free radical substitution of hexane (9 marks)

• Naming the products (2 marks): CH₃(CH₂)₅Cl is 1-chlorohexane and CH₃(CH₂)₃CHClCH₃ is 2-chlorohexane. A missing hyphen was fine. Numbering from the wrong end (6- and 5-chlorohexane), or one consistent spelling error in both names, scored 1.
• What a free radical is (1 mark): a species with one or more unpaired electrons. “It is an unpaired electron” was rejected, and “lone”, “single” or “odd” electrons were ignored.
• Bond breaking (1 mark): homolytic fission.
• Two propagation steps to 2-chlorohexane (2 marks): first a chlorine radical removes a hydrogen from carbon 2: CH₃(CH₂)₄CH₃ + Cl• → CH₃(CH₂)₃ĊHCH₃ + HCl. Then that radical reacts with Cl₂: CH₃(CH₂)₃ĊHCH₃ + Cl₂ → CH₃(CH₂)₃CHClCH₃ + Cl•. The marks were independent, and writing the radical as C₆H₁₃ was ignored in the first step because it does not show which carbon lost the hydrogen.
• Propagation then termination to dodecane (2 marks): Cl• removes a hydrogen from an end carbon to give CH₃(CH₂)₄CH₂• and HCl, then two of these radicals join to form CH₃(CH₂)₁₀CH₃.
• Radical stability (1 mark): the factor is the number of alkyl (methyl) groups attached to the carbon with the unpaired electron. “Inductive effect” alone was ignored, and so was restating the trend (tertiary more stable than secondary), because the question asked for the factor.

Question 6: an unsaturated dicarboxylic acid (13 marks)

Compound P was butenedioic acid drawn as its trans isomer.

• Draw the stereoisomer (1 mark): the cis isomer, with both COOH groups on the same side of the C=C. Drawing the trans form again scored nothing.
• Why P shows stereoisomerism (2 marks): restricted rotation about the C=C (the π bond), and two different groups on each carbon of the C=C. You had to say C=C at least once: “double bond” alone was ignored because the molecule also has C=O bonds. Saying P has a chiral carbon was rejected.
• Converting P to Q (2 marks): adding water across the C=C, so the reagent is steam and the condition is an H₃PO₄ catalyst (sulfuric acid was also accepted). Heating under reflux and pressure were ignored.
• The pair of stereoisomers of Q (2 marks): Q has one chiral carbon carrying COOH, OH, H and CH₂COOH. One mark for those four groups around the carbon drawn in 3D, and one for its mirror image. Two in-plane bonds drawn opposite each other were rejected.
• Which peak confirms P (2 marks): the peak at m/e = 116, the molecular ion, because P’s Mr is 116 (R, butanedioic acid, has Mr 118 and S, the alkene diol, 88). Citing the peaks at 99 or 71 was ignored, and any other peak was rejected.
• Fragments (2 marks): m/e = 45 is COOH⁺, and m/e = 99 is HOOCCH=CHCO⁺, which is P after losing OH. Leaving out the positive charges on both capped the answer at 1.
• Telling R from S (2 marks): the scheme gave several routes, with one mark for the reagent and one for the result. Bromine water goes from orange to colourless with S (the alkene) but not R. Sodium carbonate fizzes with R (the acid) but not S. Acidified dichromate turns green, or acidified manganate(VII) is decolourised, with S only. The observation only scored with a valid reagent. A “red” observation was rejected, and so was the iodoform test.

Where the marks are on 9701/22

• Organic chemistry carried 32 of the 60 marks: Questions 4, 5 and 6. If the AS organic reactions are weak, this paper exposes it more than any other topic.
• Recall carried 20 marks: the “state”, “name” and “identify” parts, such as the Haber conditions in 1(e)(i), the three species in 3(b) and the reagent and conditions in 6(b)(i).
• “Explain” parts carried 13 marks: 1(e)(ii), 2(b)(iii), 3(a)(ii), 3(c), 3(d), 6(a)(ii) and 6(c)(i). Most are one point per mark, so each linked step needs its own clause.
• “Suggest” parts carried 12 marks, mostly organic: radical steps in 5(b)(iii) and (iv), the structure of W, fragment formulae and the distinguishing test.
• Drawing and completing structures carried 9 marks: the 2p orbital, the dot-and-cross diagram, F, the cis isomer, the 3D enantiomers and the names in Table 5.1.
• Equations and calculations carried 6 marks: three equations (4 marks) and two one-step mole calculations (2 marks).

How to self-mark 9701/22 with the mark scheme

1. Learn the abbreviations first. A means accept, R reject, I ignore, ecf error carried forward, BOD benefit of the doubt, ora or reverse argument, owtte or words to that effect, and M1, M2 the separate marking points. An underlined word must appear in your answer.
2. Mark each M point separately. In 1(e)(ii), the increase in rate, the activation energy point and the frequency of effective collisions each earn their own mark.
3. Read the guidance column. It holds the reject and ignore notes, such as “inductive effect” in 5(c) and “double bond” without C=C in 6(a)(ii). An answer that sounds right can still be on the ignore list.
4. Apply the list rule. For “state two” in 1(e)(i), mark only your first two answers, and treat two answers that contradict each other as one wrong answer.
5. Check significant figures and units. In 4(b)(iii) a wrongly rounded or over-precise value lost the mark, and a missing or wrong unit normally costs the final calculation mark.
6. Use ecf where it applies. In 5(b)(iii), a wrong radical in the first step can still earn the second step mark if you use it correctly.

Sit 9701/22 as a timed mock

The paper allows 75 minutes for 60 marks, about 1¼ minutes per mark. A workable plan:

Block	Marks	Time
Question 1	14	17 minutes
Question 2	6	7 minutes
Question 3	8	9 minutes
Question 4	10	12 minutes
Question 5	9	11 minutes
Question 6	13	15 minutes
Check equations, structures and significant figures	–	4 minutes

In Question 4, write the 3 : 5 mole ratio down before you balance the equation, and sketch Z before you draw W: working backwards from the oxidation product is quicker than guessing.

Other May/June 2026 9701 papers

• 9701 Paper 12 (Multiple Choice)
• 9701 Paper 32 (Advanced Practical Skills)
• 9701 Paper 42 (A Level Structured Questions)
• 9701 Paper 52 (Planning, Analysis and Evaluation)

Practise and download

• Download 9701/22 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

Need help with a 9701 topic? Book a free trial lesson with an A Level Chemistry tutor →

Frequently asked questions

What topics were on 9701 Paper 22 May/June 2026? +

Six questions: oxygen and its compounds with the Haber process; protons and electrons in an electric field and atomic radius; sodium halides with concentrated sulfuric acid; alcohols with a dichromate calculation and a structure to deduce; free radical substitution of hexane; and an unsaturated dicarboxylic acid with stereoisomers and a mass spectrum.

How many marks is 9701/22 and how long is it? +

It is out of 60 marks and lasts 1 hour 15 minutes. All six questions are compulsory. Question 1, on oxygen and its compounds, is the biggest at 14 marks.

What was the answer to the moles calculation in 9701/22 Question 4? +

The Mr of C₅H₁₂O₄ is 136, so 0.255 g is 0.255 ÷ 136 = 1.875 × 10⁻³ mol. The scheme accepted 1.88 × 10⁻³ or 1.9 × 10⁻³, but rejected 1.87 × 10⁻³, 1.880 × 10⁻³ and 2.0 × 10⁻³.

How much of 9701/22 was organic chemistry? +

32 of the 60 marks. Questions 4, 5 and 6 covered alcohols, free radical substitution of hexane, and stereoisomerism and mass spectrometry of an unsaturated dicarboxylic acid.

Is 9701/22 an AS or A Level paper? +

It tests the AS content. AS candidates sit it with Papers 1 and 3, and full A Level candidates take Papers 4 and 5 as well.

Where can I download 9701/22 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.

Ready to Excel in Your Studies?

Get personalised help from Tutopiya's expert tutors. Whether it's IGCSE, IB, A-Levels, or any other curriculum — we match you with the perfect tutor and your first session is free.

Book Your Free Trial