Cambridge Chemistry 9701 Paper 1, Paper 2 and Paper 4: How to Practise

The same content can require different work in the three papers. Paper 1 demands rapid decisions, Paper 2 requires visible reasoning, and Paper 4 connects several AS and A Level areas.

The three papers at a glance

Three papers, three kinds of exam work

1

Paper 1

Decision

• 40 questions
• Rapid elimination
• Why the distractor was wrong

2

Paper 2

Built AS answer

• Calculation route
• Short, precise explanation
• Visible intermediate steps

3

Paper 4

Connection

• AS base + A Level content
• Several topics
• Longer problem solving

These are not three different kinds of chemistry. A mole calculation may appear as rapid proportional reasoning in Paper 1, a complete calculation in Paper 2 and one step within a multi-topic Paper 4 problem. Practice must therefore develop knowledge and response format together.

Use the separate Paper 5 data and planning guide for that paper.

Paper 1: record more than the correct letter

After each question, record the decisive chemical relationship, why your chosen distractor was attractive, whether the answer was secure or an informed guess, and whether knowledge or time caused the error.

Answer key — separate from the task sheet

1. A catalyst at equilibrium

C is correct: the catalyst provides a lower-activation-energy pathway in both directions. At equilibrium the two rates remain equal, and composition is unchanged at constant temperature.

• A: reaching equilibrium faster does not increase K_c; for a given reaction K_c depends on temperature.
• B: K_c does not decrease either; a catalyst does not shift equilibrium towards reactants.
• D: this ignores catalysis of the reverse reaction. A system already at equilibrium has no sustained net change.

If C was a guess, check that you can distinguish reaction rate from equilibrium composition.

During correction, the correct letter is not enough. Write the sentence that eliminated each alternative. Use short timed blocks to measure decision time and the source of delay. A practised skip-and-return rule is more useful than forcing one universal number of seconds onto every item.

Paper 2: a short answer must still be complete

A useful explanation frame is claim → chemical or particle-level reason → consequence → answer to the question. Sentences need not be long, but the decisive relationship must be visible.

Initial answer

“Magnesium oxide has stronger ionic bonds.”

What is missing?

The ion charges and the attraction–energy–melting-point link are unstated.

Improved answer

“The MgO lattice contains Mg²⁺ and O²⁻ ions, whereas the NaCl lattice contains Na⁺ and Cl⁻ ions. The greater ionic charges produce stronger electrostatic attraction between oppositely charged ions in MgO. More energy is required to overcome these attractions, so MgO has the higher melting point.”

Why it works

The improved response uses a common basis of comparison, identifies charge, states the direction of attraction and connects it to energy. “Stronger” now has a defined meaning.

Make the calculation auditable

1. Write the equation or starting relationship.
2. Record data in suitable units.
3. Show the stoichiometric or mathematical connection.
4. Preserve at least one intermediate value.
5. Round at the end, attach a unit and check the order of magnitude.

First write the volume as 0.0250 dm³. Then n(NaOH) = cV = 0.200 × 0.0250 = 0.00500 mol. The equation HCl + NaOH → NaCl + H₂O gives a 1:1 amount ratio, so n(HCl) = 0.00500 mol. The unit conversion, relationship and mole ratio can each be checked.

When an answer is wrong, do not edit only the final value. Identify the first of these five lines that failed. That turns the calculation into a specific training task.

Paper 4: do not keep AS and A Level content in separate boxes

Paper 4 is based on A Level syllabus content and requires knowledge of AS content. Topics 23–37 cannot be treated as an isolated second book. Practise links: equilibrium with energetics; rate with experimental data; organic transformation with spectra; mechanism with reaction conditions.

The reasoning network in a Paper 4 question

Identify and transform an unknown organic compound

1. 1Structural information from a spectrum
2. 2Identify the functional group
3. 3Choose reagent and conditions
4. 4Justify the mechanism or product

For an unknown organic compound, spectral evidence may first provide structural information, which supports a functional-group identification, then reagent and condition selection, then justification of a product or mechanism. If each link exists only as a memorised pair, a new context can break the chain.

After correcting a Paper 4 question, draw its topic connections. Replace “organic chemistry” with “I recognised an O–H absorption but did not connect the group to the required oxidation reagent.”

A complete Paper 4 organic task

Answer key — separate from the task sheet

1. Paper 4: from evidence to an amide

X is methyl propanoate, CH₃CH₂C(=O)OCH₃. Its formula and molecular ion of 88 are consistent with an ester; IR supports a carbonyl without a broad O–H band. The 3:2:3 integrals account for eight protons.

In CH₃CH₂–, CH₃ is a triplet from its two neighbouring protons, and CH₂ a quartet from three. CH₂ next to C=O appears near 2.3 ppm. The OCH₃ group is a three-proton singlet near 3.7 ppm; simple n+1 splitting is not applied through oxygen here. Ethyl ethanoate would instead have an OCH₂ quartet around 4 ppm and an acetyl-CH₃ singlet around 2 ppm, inconsistent with this pattern.

CH₃CH₂COOCH₃ + NaOH → CH₃CH₂COONa + CH₃OH

Products are sodium propanoate and methanol. The 1:1 ratio gives n(NaOH) = 0.0200 mol. Acidification gives propanoic acid: CH₃CH₂COO⁻ + H⁺ → CH₃CH₂COOH.

CH₃CH₂COOH + SOCl₂ → CH₃CH₂COCl + SO₂ + HCl

CH₃CH₂COCl + 2NH₃ → CH₃CH₂CONH₂ + NH₄Cl

The intermediate is propanoyl chloride; the final product is propanamide. The last reaction is nucleophilic addition–elimination: an NH₃ lone pair attacks the carbonyl carbon and the C=O π pair moves to oxygen. The tetrahedral intermediate reforms C=O and expels Cl⁻; another NH₃ accepts a proton. Curly arrows begin at electron pairs and charges must be shown. Proton transfer explains the requirement for two NH₃.

Self-review: did you justify identity from evidence, check AS stoichiometry and supply every reagent–condition–product–mechanism link? This is a learning answer, not an official mark scheme.

Choose a weekly practice frame

This adjustable illustration is neither official Cambridge advice nor a complete weekly workload.

An AS-only student is not assigned mandatory Paper 4 practice. For full A Level, Paper 4 must not displace AS foundations or practical preparation.

An illustrative four-week cycle

1. Diagnose: use a short mixed block; record which decisions required help and which errors appeared only under time.
2. Rebuild: pair an explanation with two fresh calculations addressing the main gap; retry independently two days later.
3. Focus and transfer: apply the method in a different representation on papers relevant to your route; check units and chemical justification.
4. Mix, then time and review: add timing only for a secure skill. Compare work with week one and choose the next cycle’s target.

What should the error log contain?

Do not record only “equilibrium” or “organic chemistry”. Record the first observable divergence:

• “I did not derive the exponents in K_c from the stoichiometric coefficients.”
• “My curly arrow did not begin at the electron pair.”
• “The explanation did not connect structure back to the property asked.”
• “I did not convert cm³ to dm³ before using cV.”

Attach a short repair task and dated retry to every entry. Use the Paper 3 and Paper 5 skills guide for the parallel practical system. The Chemistry 9701 overview places these papers within the AS and A Level routes.