4.7 Organic chemistry — revision question pack

12 specification points · notes, questions, answers and worked methods

Checked against AQA 8462 section 4.7. Review basis: the qualification registry sourced from the AQA GCSE Chemistry (8462) specification; registry verification recorded 17 July 2026.

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4.7.1.1 · Crude oil, hydrocarbons and alkanes

Explanation

Crude oil is a finite resource found in rocks. It formed from ancient biomass, mainly plankton, that was buried in mud.
Crude oil is a mixture containing many compounds; most are hydrocarbons, whose molecules contain carbon and hydrogen atoms only. Most crude-oil hydrocarbons are alkanes.
Their general formula is $\mathrm{C_nH_{2n+2}}$ , and the first four are methane, ethane, propane and butane.
To test an alkane formula, substitute its carbon count for $n$ and check the hydrogen count.
A common error is to call crude oil one compound rather than a mixture.

Worked example

Decide whether $\mathrm{C_5H_{12}}$ is an alkane.

1.Use the alkane formula $\mathrm{C_nH_{2n+2}}$ .
2.For $n=5$ , the hydrogen number is $2(5)+2=12$ .
3.The formula matches the general formula.

Answer: $\mathrm{C_5H_{12}}$ is an alkane.

Common mistakes

Don't fall into the trap of calling crude oil a single compound instead of a mixture of many compounds.
Don't fall into the trap of using $\mathrm{C_nH_{2n}}$ for an alkane instead of $\mathrm{C_nH_{2n+2}}$ .

Exam tip

When asked to identify an alkane, substitute the carbon number into $\mathrm{C_nH_{2n+2}}$ and check every atom.

Tier 1 · Easy

A molecule has the formula $\mathrm{C_3H_8}$ . State why it is a hydrocarbon and show that it fits the alkane general formula.
[2 marks]
Total for this question: 2
State why crude oil is described as a finite resource.
[1 mark]
Total for this question: 1

Tier 2 · Standard

An alkane molecule contains seven carbon atoms. Determine its molecular formula and explain how you obtained the hydrogen count.
[3 marks]
Total for this question: 3
Describe how crude oil formed from ancient biomass and explain why crude oil is a mixture.
[4 marks]
Total for this question: 4
A molecular model is represented as $\mathrm{CH_3-CH_2-CH_3}$ . Identify the alkane, give its molecular formula and calculate how many atoms the molecule contains altogether.
[3 marks]
Total for this question: 3

Tier 3 · Hard

An alkane has relative molecular mass $58$ . Using $A_r(\mathrm{C})=12$ and $A_r(\mathrm{H})=1$ , determine its molecular formula and name it.
[4 marks]
Total for this question: 4
A sample contains $\mathrm{C_{17}H_{36}}$ , $\mathrm{C_{17}H_{34}}$ and $\mathrm{C_{17}H_{32}}$ . Identify the alkane and explain why all three substances are hydrocarbons but only one is an alkane.
[4 marks]
Total for this question: 4
A sample contains two methane molecules, $\mathrm{CH_4}$ , and three ethane molecules, $\mathrm{C_2H_6}$ . Determine the total numbers of carbon and hydrogen atoms, then explain why the sample is a mixture of hydrocarbons.
[4 marks]
Total for this question: 4
An unknown alkane X has eight more hydrogen atoms than carbon atoms in each molecule. Use the alkane general formula to determine the numbers of carbon and hydrogen atoms, give the molecular formula of X and explain why X is a hydrocarbon.
[4 marks]
Total for this question: 4
A report makes three claims: crude oil is one alkane; it formed from ancient biomass consisting mainly of plankton buried in mud; and new crude oil forms quickly enough to replace what people extract. Evaluate all three claims.
[5 marks]
Total for this question: 5

4.7.1.2 · Fractional distillation and petrochemicals

Explanation

Fractional distillation separates crude oil into fractions; each fraction contains hydrocarbons with similar numbers of carbon atoms and similar boiling points. Crude oil is heated so that most of it vaporises.
The column is hot at the bottom and cooler towards the top, so vapours condense at different heights.
Large molecules with higher boiling points condense lower in the column; smaller molecules with lower boiling points rise further before condensing.
Fractions provide fuels and petrochemical feedstock for solvents, lubricants, polymers and detergents.
Fractional distillation separates molecules; it does not crack them.

A fractionating column with a temperature gradient and condensation outlets for different molecular sizes.

Worked example

Explain why a small hydrocarbon is collected higher in a fractionating column than a large hydrocarbon.

1.Small hydrocarbons have lower boiling points.
2.They remain as vapour further up the column as the temperature falls.
3.They condense higher up when the temperature becomes low enough.

Answer: Its lower boiling point makes it condense in the cooler, upper part of the column.

Common mistakes

Don't fall into the trap of saying fractions separate because they have different densities rather than different boiling points.
Don't fall into the trap of reversing the column, so small molecules are said to condense low down where it is hottest.

Exam tip

For an “explain fractional distillation” question, follow one vapour from evaporation to condensation at the level below its boiling point.

Tier 1 · Easy

What is meant by a fraction obtained from crude oil, and which physical property allows the fractions to be separated?
[2 marks]
Total for this question: 2
State the change of state when crude oil is heated before entering the column and the change of state that forms a liquid fraction.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Hydrocarbon A has a boiling point of $75\,^\circ\mathrm{C}$ and hydrocarbon B has a boiling point of $210\,^\circ\mathrm{C}$ . Explain which one condenses lower in a fractionating column.
[3 marks]
Total for this question: 3
A fraction has a boiling range of $160$ to $200\,^\circ\mathrm{C}$ . Predict which outlet, at $250\,^\circ\mathrm{C}$ , $180\,^\circ\mathrm{C}$ or $90\,^\circ\mathrm{C}$ , collects this fraction and explain the choice.
[3 marks]
Total for this question: 3
From petrol, kerosene, polymers and detergents, identify the two fuels obtained from crude-oil fractions and the two types of useful material made by the petrochemical industry.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Describe how a heated sample of crude oil is separated in a fractionating column, then explain why the collected fractions are valuable even when they are not used directly as fuels.
[5 marks]
Total for this question: 5
Explain why fractional distillation changes the composition of each collected fraction without making new hydrocarbon molecules. Include the temperature gradient and one petrochemical use in the answer.
[5 marks]
Total for this question: 5
Evaluate the claim: ‘A fraction collected from crude oil is one pure hydrocarbon because it leaves the column through one outlet.’
[5 marks]
Total for this question: 5
A fractionating column has lower, middle and upper outlets at $210$ , $140$ and $65\,^\circ\mathrm{C}$ . Fractions X, Y and Z have boiling ranges of $55$ – $75$ , $195$ – $225$ and $125$ – $155\,^\circ\mathrm{C}$ , respectively. Match each fraction to an outlet and determine which fraction contains the largest molecules.
[5 marks]
Total for this question: 5
During a fault, a fractionating column remains at $180\,^\circ\mathrm{C}$ from bottom to top. Samples from all its outlets have nearly the same composition. Explain the poor separation and describe how restoring the correct temperature pattern separates the hydrocarbons without changing their molecules.
[5 marks]
Total for this question: 5

4.7.1.3 · Properties of hydrocarbons

Explanation

As hydrocarbon molecules become larger, their boiling points increase because the intermolecular forces between molecules become stronger. Increasing molecular size also increases viscosity and decreases flammability, influencing which fractions make convenient fuels.
Complete combustion releases energy and oxidises the carbon and hydrogen, producing carbon dioxide and water.
Balance combustion equations in the order carbon, hydrogen, then oxygen.
Do not confuse incomplete combustion products with the specified products of complete combustion.
In exam answers, connect each trend explicitly to increasing molecular size and distinguish complete combustion from incomplete combustion before choosing products.

Worked example

State the products when propane, $\mathrm{C_3H_8}$ , burns completely and write the balanced equation.

1.Complete combustion forms carbon dioxide and water.
2.Balance carbon to give $3\mathrm{CO_2}$ and hydrogen to give $4\mathrm{H_2O}$ .
3.Balance oxygen with $5\mathrm{O_2}$ .

Answer: $\mathrm{C_3H_8+5O_2\rightarrow3CO_2+4H_2O}$ .

Common mistakes

Don't fall into the trap of stating that larger hydrocarbons are more flammable, when flammability decreases as molecular size increases.
Don't fall into the trap of writing carbon monoxide or carbon as a product when the question specifies complete combustion.

Exam tip

In a trend question, state the direction for all three properties: boiling point and viscosity increase, while flammability decreases.

Tier 1 · Easy

State how boiling point and viscosity change as the molecules in a homologous series of hydrocarbons become larger.
[2 marks]
Total for this question: 2
State how the flammability of hydrocarbons changes as their molecules become larger.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Complete and balance the equation for the complete combustion of propane: $\mathrm{C_3H_8 + O_2 \rightarrow CO_2 + H_2O}$ .
[2 marks]
Total for this question: 2
The complete-combustion equation is $\mathrm{2X + 22O_2 \rightarrow 14CO_2 + 16H_2O}$ . Determine the molecular formula of hydrocarbon X and explain the atom counts used.
[3 marks]
Total for this question: 3
Complete the statements for the complete combustion of a hydrocarbon: identify the element oxidised to carbon dioxide, identify the element oxidised to water and state the energy change experienced by the surroundings.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Fuel P is $\mathrm{C_5H_{12}}$ and fuel Q is $\mathrm{C_{15}H_{32}}$ . Compare their boiling points, viscosities and flammabilities, and write a balanced equation for the complete combustion of Q.
[5 marks]
Total for this question: 5
Hydrocarbons R, S and T contain 5, 12 and 24 carbon atoms per molecule, respectively. Choose the most suitable hydrocarbon for a thick road-surfacing material and justify the choice using three property trends.
[4 marks]
Total for this question: 4
Hydrocarbons A, B and C have increasing molecular size. Their boiling points are $40$ , $120$ and $210\,^\circ\mathrm{C}$ ; their viscosities are low, medium and low; and their flammabilities are high, medium and low. Identify the anomalous result, predict the corrected trend and explain how the other results support the stated order.
[4 marks]
Total for this question: 4
Equal masses of hydrocarbons P and Q are tested under the same conditions. After ten minutes in open dishes, P has lost $70\%$ of its mass and Q has lost $5\%$ . Equal volumes take $4$ s and $35$ s to flow through the same tube, respectively, and P ignites more readily. Determine which hydrocarbon has the smaller molecules and justify the decision using all three observations.
[5 marks]
Total for this question: 5
A fuel must stay mostly liquid at $80\,^\circ\mathrm{C}$ but flow through a narrow wick. Fuel A has boiling point $60\,^\circ\mathrm{C}$ and low viscosity; B has boiling point $125\,^\circ\mathrm{C}$ and medium viscosity; C has boiling point $230\,^\circ\mathrm{C}$ and high viscosity. Choose the best fuel, reject the other two and compare the likely molecular sizes and flammabilities.
[5 marks]
Total for this question: 5

4.7.1.4 · Cracking and alkenes

Explanation

Cracking breaks long-chain hydrocarbons into smaller, more useful molecules because demand for short-chain fuels is high. In catalytic cracking, a hydrocarbon is vaporised and passed over a hot catalyst; in steam cracking, hydrocarbon vapour is mixed with steam and heated strongly.
Cracking produces a mixture that includes shorter alkanes and alkenes.
Equations must conserve the number of carbon and hydrogen atoms.
Alkenes decolourise bromine water from orange to colourless and are useful for polymers and other chemicals.
Heating alone does not identify an alkene.

Worked example

Balance the cracking equation $\mathrm{C_{10}H_{22}\rightarrow C_6H_{14}+C_xH_y}$ and identify the second product type.

1.Conserve carbon: $x=10-6=4$ .
2.Conserve hydrogen: $y=22-14=8$ .
3. $\mathrm{C_4H_8}$ fits $\mathrm{C_nH_{2n}}$ , so it is an alkene.

Answer: $\mathrm{C_{10}H_{22}\rightarrow C_6H_{14}+C_4H_8}$ ; the second product is an alkene.

Common mistakes

Don't fall into the trap of describing fractional distillation as cracking, even though distillation only separates molecules.
Don't fall into the trap of giving orange to colourless as evidence for an alkane rather than for an alkene.

Exam tip

In a cracking equation, count carbon and hydrogen atoms on both sides before naming the alkene product.

Tier 1 · Easy

A colourless hydrocarbon rapidly turns orange bromine water colourless. What type of hydrocarbon is present?
[1 mark]
Total for this question: 1
State the two homologous series normally represented among the products of cracking.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Complete the cracking equation $\mathrm{C_{12}H_{26} \rightarrow C_8H_{18} + X}$ and identify the homologous series of X.
[3 marks]
Total for this question: 3
Classify cracking as thermal decomposition and explain how it differs from fractional distillation.
[3 marks]
Total for this question: 3
Describe the conditions used for catalytic cracking of a long-chain hydrocarbon and state the change that occurs to its molecules.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A refinery converts $\mathrm{C_{15}H_{32}}$ into $\mathrm{C_9H_{20}}$ and one other product. Give the other product, describe catalytic cracking, and explain two reasons the process is useful.
[5 marks]
Total for this question: 5
Describe steam cracking of a long-chain hydrocarbon, explain why a refinery uses the process, and give a test showing that one product is an alkene.
[5 marks]
Total for this question: 5
Compare catalytic cracking with steam cracking. Give the conditions that distinguish the two methods, state one feature their products have in common and explain why refineries crack long-chain hydrocarbons.
[5 marks]
Total for this question: 5
Cracking one molecule of $\mathrm{C_{13}H_{28}}$ forms one molecule of $\mathrm{C_7H_{16}}$ and two identical alkene molecules X. Determine the molecular formula and name of X, and show that carbon and hydrogen atoms are balanced.
[5 marks]
Total for this question: 5
Three colourless hydrocarbon samples from a cracking process are tested with bromine water. The bromine water remains orange with G but becomes colourless with H and J. State every conclusion justified about G, H and J, and explain two limitations of using this test alone.
[5 marks]
Total for this question: 5

4.7.2.1 · Structure and formulae of alkenes (chemistry only)

Explanation

Alkenes are hydrocarbons containing a carbon-carbon double bond, C=C; this double bond makes them unsaturated.
Their homologous-series formula is $\mathrm{C_nH_{2n}}$ , so an alkene has two fewer hydrogen atoms than the alkane with the same carbon count.
The first four members are ethene, propene, butene and pentene, and each may be shown by a molecular or fully displayed structural formula.
When recognising an alkene, check both the double bond and the formula.
A common error is to apply the alkane formula $\mathrm{C_nH_{2n+2}}$ .

Worked example

Show that $\mathrm{C_4H_8}$ can be an alkene and name the relevant member of the series.

1.Use $\mathrm{C_nH_{2n}}$ .
2.For $n=4$ , $2n=8$ , so the formula matches.
3.The four-carbon alkene is butene.

Answer: $\mathrm{C_4H_8}$ matches the alkene formula and is butene.

Common mistakes

Don't fall into the trap of using the alkane formula $\mathrm{C_nH_{2n+2}}$ for an alkene.
Don't fall into the trap of calling a molecule unsaturated without showing or identifying its carbon–carbon double bond.

Exam tip

For “recognise an alkene”, check both $\mathrm{C_nH_{2n}}$ and the displayed C=C bond.

Tier 1 · Easy

Use the general formula of the alkenes to give the molecular formula of pentene.
[1 mark]
Total for this question: 1
State the bond that makes an alkene unsaturated and give the alkene general formula.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A displayed molecule contains three carbon atoms, one C=C bond and six hydrogen atoms. Name the molecule and explain why it is unsaturated.
[2 marks]
Total for this question: 2
A student labels $\mathrm{C_5H_{10}}$ as the fully displayed formula of pentene. Explain why the label is incorrect and state what a correct fully displayed formula must show.
[3 marks]
Total for this question: 3
Complete the alkene sequence ethene, propene, ___, pentene. Give the molecular formula of the missing member and state the functional group common to every member of the sequence.
[3 marks]
Total for this question: 3

Tier 3 · Hard

An alkene has relative molecular mass $56$ . Using $A_r(\mathrm{C})=12$ and $A_r(\mathrm{H})=1$ , determine its formula and name, then give the formula of the alkane with the same number of carbon atoms.
[5 marks]
Total for this question: 5
Compare pentane and pentene by giving both molecular formulae, the difference in hydrogen count and the structural feature responsible for the difference.
[4 marks]
Total for this question: 4
A table gives ethene as $\mathrm{C_2H_4}$ , propene as $\mathrm{C_3H_8}$ and butene as $\mathrm{C_4H_8}$ . Identify the incorrect entry, write its corrected formula, explain the check used and predict the molecular formula of pentene.
[4 marks]
Total for this question: 4
One molecule of an alkene contains 18 atoms altogether. Use the alkene general formula to determine its carbon and hydrogen counts, write its molecular formula and give the molecular formula of the alkane with the same carbon count.
[4 marks]
Total for this question: 4
A student is told only that an unknown hydrocarbon has twice as many hydrogen atoms as carbon atoms. The student claims this proves the substance is butene. Evaluate the claim, state what the ratio supports and identify the extra structural or numerical evidence needed to distinguish among the first four alkenes.
[5 marks]
Total for this question: 5

4.7.2.2 · Reactions of alkenes (chemistry only)

Explanation

Alkene reactions are governed by the C=C functional group: addition places atoms across the double bond and leaves a single carbon-carbon bond.
Hydrogen adds in the presence of a nickel catalyst to form an alkane; steam adds with a phosphoric acid catalyst to form an alcohol.
Chlorine, bromine and iodine add across C=C to form saturated dihalogeno compounds; bromine water changes from orange to colourless.
Alkenes combust like other hydrocarbons but often burn with smoky flames in air because combustion is incomplete.
Do not leave C=C in an addition product.

Bromine adds across an alkene carbon–carbon double bond to form a saturated product.

Worked example

Describe the product when ethene reacts with bromine.

1.Identify the C=C functional group in ethene.
2.Open the double bond to make C–C.
3.Attach one bromine atom to each carbon.

Answer: 1,2-dibromoethane forms and bromine is decolourised.

Common mistakes

Don't fall into the trap of leaving the C=C double bond in the displayed product after an addition reaction.
Don't fall into the trap of naming bromine-water decolourisation but giving the colour change as colourless to orange.

Exam tip

For an addition product, replace C=C by C–C and attach one incoming atom or group to each carbon.

Tier 1 · Easy

Ethene is shaken with bromine water. State the observed colour change and the type of reaction.
[2 marks]
Total for this question: 2
State why alkenes are more reactive than alkanes and name the reaction type typical of alkenes.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Draw the fully displayed structural formula of the product when ethene reacts with hydrogen. Name the product and state the catalyst used.
[3 marks]
Total for this question: 3
Propene reacts with chlorine. Give the molecular formula of the product and describe the change in carbon-carbon bonding.
[3 marks]
Total for this question: 3
Complete the combustion equation $\mathrm{C_2H_4 + O_2 \rightarrow CO_2 + H_2O}$ using the smallest whole-number coefficients, then show that the number of oxygen atoms is conserved.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A hydrocarbon burns with a smoky flame and decolourises bromine water. Explain both observations, then describe how it can be converted into an alcohol and name the product when the hydrocarbon is ethene.
[6 marks]
Total for this question: 6
Evaluate these claims about ethene: bromination leaves C=C unchanged; hydrogenation produces ethanol using nickel; hydration produces ethane using phosphoric acid. Correct every error and explain the common bonding change.
[6 marks]
Total for this question: 6
Separate samples of ethene form $\mathrm{C_2H_6}$ , $\mathrm{C_2H_5OH}$ and $\mathrm{C_2H_4I_2}$ . Identify the reagent added to make each product, give the catalyst used for hydrogenation and for hydration, and explain the bonding change common to all three reactions.
[6 marks]
Total for this question: 6
An addition product has the condensed structure $\mathrm{CH_3-CHI-CH_2I}$ . Work backwards to identify the starting alkene and reagent, then explain the carbon-carbon bonding change and why the product is saturated.
[5 marks]
Total for this question: 5
Ethene is reacted completely with hydrogen over nickel to make product P. P and an untreated ethene control are then shaken separately with orange bromine water. Determine the formula of P, predict both observations and explain how the first reaction changes the evidence in the bromine-water test.
[5 marks]
Total for this question: 5

4.7.2.3 · Alcohols (chemistry only)

Explanation

Alcohols contain the -OH functional group; the first four are methanol, ethanol, propanol and butanol. They burn in air, react with sodium and are oxidised to carboxylic acids.
Their solubility in water decreases as the carbon chain becomes longer.
Ethanol is made by fermenting a sugar solution with yeast under warm conditions without oxygen; fermentation also produces carbon dioxide.
Alcohols are used as fuels and solvents.
Balanced equations are required for their combustion, but not for their other reactions in this specification.

Worked example

State the conditions needed to produce ethanol by fermentation of sugar solution.

1.Add yeast to the sugar solution.
2.Keep the mixture warm.
3.Exclude oxygen so fermentation occurs.

Answer: Yeast, warm conditions and no oxygen.

Common mistakes

Don't fall into the trap of saying fermentation needs oxygen, when yeast ferments sugar without oxygen.
Don't fall into the trap of assuming all four alcohols are equally soluble in water instead of recognising that solubility decreases with chain length.

Exam tip

For fermentation conditions, state yeast, a warm temperature and absence of oxygen.

Tier 1 · Easy

Name $\mathrm{CH_3CH_2OH}$ and identify its functional group.
[2 marks]
Total for this question: 2
State how the solubility of alcohols in water changes as the carbon chain becomes longer.
[1 mark]
Total for this question: 1

Tier 2 · Standard

A student wants to produce ethanol from glucose solution by fermentation. State the biological agent, two required conditions and the other product formed.
[4 marks]
Total for this question: 4
Separate samples of propanol react with sodium and with an oxidising agent. Name the gaseous product of the first reaction and the organic product of the second, then describe the oxidising agent's role.
[3 marks]
Total for this question: 3
State two common uses of the first four alcohols and give one additional use associated with ethanol.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Ethanol is first burned completely and a separate sample is oxidised. Write the balanced combustion equation, name the oxidation product and describe the role of the oxidising agent.
[5 marks]
Total for this question: 5
A student leaves glucose solution and yeast in an open flask at $65\,^\circ\mathrm{C}$ . Explain why the ethanol yield will be poor, describe two changes that improve fermentation, and name the gaseous product.
[5 marks]
Total for this question: 5
Equal volumes of methanol and butanol are added separately to water and shaken. Predict which alcohol mixes more evenly with the water, explain the prediction using carbon-chain length and choose the better alcohol for a water-based solvent mixture.
[4 marks]
Total for this question: 4
An unknown member of the first four alcohols is oxidised to butanoic acid. Identify the alcohol, predict the gas formed when a fresh sample reacts with sodium, compare its water solubility with ethanol and explain the comparison.
[5 marks]
Total for this question: 5
Butanol and ethanol both undergo complete combustion. Balance the equation for butanol, state how many oxygen molecules are needed per ethanol molecule in its balanced equation, and explain why butanol needs more oxygen.
[4 marks]
Total for this question: 4

4.7.2.4 · Carboxylic acids (chemistry only)

Explanation

Carboxylic acids contain the -COOH functional group; the first four are methanoic, ethanoic, propanoic and butanoic acid. They dissolve in water to form acidic solutions and react with carbonates to make a salt, water and carbon dioxide.
Higher tier: carboxylic acids are weak because they only partially ionise, so an equally concentrated strong acid has a lower pH.
A carboxylic acid reacts with an alcohol to form an ester and water; ethanol with ethanoic acid forms ethyl ethanoate.
Recognise the -COOH group before naming the molecule.
A common error is to identify it as the -OH group of an alcohol.

Worked example

Predict the observation and products when ethanoic acid reacts with calcium carbonate.

1.Use the acid–carbonate reaction pattern.
2.The products are a salt, water and carbon dioxide.
3.Escaping carbon dioxide produces effervescence.

Answer: Fizzing occurs; calcium ethanoate, water and carbon dioxide form.

Common mistakes

Don't fall into the trap of identifying -COOH as the -OH functional group of an alcohol.
Don't fall into the trap of saying a weak acid is dilute rather than explaining that it only partially ionises in water (Higher tier).

Exam tip

For an acid–carbonate reaction, name all three product types and link effervescence to carbon dioxide.

Tier 1 · Easy

Name $\mathrm{CH_3COOH}$ and state the functional group that identifies it as a carboxylic acid.
[2 marks]
Total for this question: 2
Name the salt and the gas formed when propanoic acid reacts with sodium carbonate.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Dilute ethanoic acid is added to calcium carbonate. Describe the observation and name all three types of product.
[4 marks]
Total for this question: 4
An ester is named propyl methanoate. Deduce the alcohol and carboxylic acid used to make it, and state the other product.
[3 marks]
Total for this question: 3
From $\mathrm{CH_3COOH}$ , $\mathrm{C_2H_5OH}$ and $\mathrm{C_3H_7COOH}$ , identify and name the two carboxylic acids, state the functional group used to recognise them and explain why the remaining formula is not a carboxylic acid.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Two colourless liquids are ethanol and ethanoic acid. Describe a carbonate test that distinguishes them, then name the organic product when the two liquids react together.
[5 marks]
Total for this question: 5
Higher tier: Equal-concentration solutions of hydrochloric acid and ethanoic acid have different pH values. Predict which has the lower pH and explain the difference using ionisation and hydrogen-ion concentration.
[4 marks]
Total for this question: 4
Evaluate these claims about ethanoic acid: reaction with sodium carbonate makes hydrogen as the only product; reaction with ethanol makes only ethyl ethanoate; dissolving it in water makes an alkaline solution. Correct every error.
[5 marks]
Total for this question: 5
Three bottles contain methanol, methanoic acid and ethanoic acid. Carbonate added to A and B causes fizzing, while C gives no fizzing. Formula labels then show A is $\mathrm{HCOOH}$ and B is $\mathrm{CH_3COOH}$ . Identify all three liquids and explain both what the carbonate test establishes and what it cannot distinguish.
[5 marks]
Total for this question: 5

4.7.3.1 · Addition polymerisation (chemistry only)

Explanation

In addition polymerisation, many alkene monomers join to form one long-chain polymer molecule. The monomer's C=C bond opens to form C-C bonds linking neighbouring repeating units; no small molecule is produced.
The repeating unit contains the same atoms as the monomer and is shown in brackets with continuation bonds and a subscript $n$ .
To recover the monomer from a repeating unit, place C=C between the two backbone carbons.
Do not leave a double bond in the addition polymer.
Questions may require drawing the repeating unit from a displayed monomer or reversing the process, so every side group must be preserved and bonds must pass through both brackets.

General addition polymerisation of a substituted alkene into a bracketed repeating unit.

Worked example

Write the repeating unit formed from propene, $\mathrm{CH_2{=}CHCH_3}$ .

1.Use the two double-bonded carbons as the backbone.
2.Change C=C to C–C and retain the $\mathrm{CH_3}$ side group.
3.Add brackets, continuation bonds and subscript $n$ .

Answer: $[\mathrm{-CH_2-CH(CH_3)-}]_n$ .

Common mistakes

Don't fall into the trap of keeping the C=C double bond inside the addition-polymer repeating unit.
Don't fall into the trap of omitting the continuation bonds through the brackets or moving a side group to the wrong backbone carbon.

Exam tip

Copy every atom from one monomer, change C=C to C–C, then add brackets, continuation bonds and $n$ .

Tier 1 · Easy

Name the monomer used to make poly(ethene) and state the type of polymerisation.
[2 marks]
Total for this question: 2
State what happens to an alkene's C=C bond during addition polymerisation and whether a small-molecule by-product forms.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Propene, $\mathrm{CH_2{=}CHCH_3}$ , forms a polymer. Write its repeating unit and explain what happens to the carbon-carbon double bond.
[3 marks]
Total for this question: 3
A poly(chloroethene) chain contains 5000 repeating units. Determine the total numbers of carbon atoms and chlorine atoms represented by these units.
[3 marks]
Total for this question: 3
A section of poly(ethene) contains 1200 carbon atoms. Determine the number of repeating units in the section and the number of ethene monomer molecules used to form those units.
[2 marks]
Total for this question: 2

Tier 3 · Hard

A polymer has repeating unit $[\mathrm{-CH_2-CHCl-}]_n$ . Deduce the monomer formula and explain why no other product forms during polymerisation.
[4 marks]
Total for this question: 4
A student proposes $[\mathrm{CH_2{=}CHCH_3}]_n$ as the repeating unit of poly(propene). Identify two errors and write the correct repeating unit.
[5 marks]
Total for this question: 5
Structures A, B and C are $\mathrm{C_2H_6}$ , $\mathrm{CH_2{=}CH_2}$ and $[\mathrm{-CH_2-CH_2-}]_n$ . Identify the addition-polymer monomer and the polymer, explain why A cannot act as the monomer, and compare the atoms in one molecule of B with one repeating unit of C.
[5 marks]
Total for this question: 5
The monomer $\mathrm{CH_2{=}C(Cl)CH_3}$ is offered with two candidate repeating units: A is $[\mathrm{-CH_2-C(Cl)(CH_3)-}]_n$ and B is $[\mathrm{-CHCl-CH_2-CH_3-}]_n$ . Choose the correct unit and explain how the backbone and side groups expose the error in the other proposal.
[5 marks]
Total for this question: 5
One student draws a repeating unit as $[\mathrm{-CH_2-CHCl-}]_n$ and another as $[\mathrm{-CHCl-CH_2-}]_n$ . Evaluate the claim that these must be different polymers, identify their monomer and explain the atom accounting in the polymerisation.
[5 marks]
Total for this question: 5

4.7.3.2 · Condensation polymerisation (chemistry only) (HT only)

Explanation

Higher tier: condensation polymerisation requires monomers with two functional groups, allowing each monomer to join at both ends of a growing chain.
When a link forms, a small molecule such as water is usually eliminated; the polymer therefore does not contain every atom from the monomers.
A diol and a dicarboxylic acid form a polyester because -OH groups react with -COOH groups to make ester links.
Read the repeating unit across the new links and preserve the remaining carbon chains.
A common error is to treat condensation as C=C addition.

Worked example

Higher tier: explain why a diol and a dicarboxylic acid can form a polyester.

1.Each monomer has two functional groups, so it can join at both ends.
2.-OH and -COOH groups react to form ester links.
3.Each link forms with elimination of a small molecule such as water.

Answer: Repeated condensation forms a polyester with ester links and releases water.

Common mistakes

Don't fall into the trap of describing condensation polymerisation as opening a C=C bond.
Don't fall into the trap of drawing monomers with only one functional group, so they cannot join at both ends to make a chain.

Exam tip

Higher tier: identify both functional groups, the new link and the small molecule eliminated.

Tier 1 · Easy

State two features required for condensation polymerisation and name a small molecule commonly released.
[3 marks]
Total for this question: 3
Explain why ethanediol can help form a condensation polymer but ethanol cannot form a long chain by itself.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Ethanediol reacts with butanedioic acid. Identify the two functional-group types, name the polymer class and state the small molecule eliminated.
[4 marks]
Total for this question: 4
Four diol molecules and four dicarboxylic acid molecules join in one unbranched alternating chain. Determine the numbers of ester links and water molecules formed.
[3 marks]
Total for this question: 3
Explain why a condensation-polymer repeating unit does not contain all the atoms originally present in its monomers, and state the usual small molecule removed.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Use the monomers $\mathrm{HO-CH_2-CH_2-OH}$ and $\mathrm{HOOC-CH_2-CH_2-COOH}$ to write one repeating unit of the polyester and explain how its links form.
[5 marks]
Total for this question: 5
Choose which pair can make a long-chain polyester: propan-1-ol with pentanoic acid, or propane-1,3-diol with pentanedioic acid. Explain the selection in terms of functional groups, links and the by-product.
[5 marks]
Total for this question: 5
A polyester has repeating unit $[\mathrm{-O-CH_2-CH_2-O-CO-CH_2-CO-}]_n$ . Write the structural formulae of the diol and dicarboxylic acid from which it formed, identify the two functional-group types and state the by-product.
[5 marks]
Total for this question: 5
Ethanediol and hexanedioic acid form a polyester. A student proposes $[\mathrm{-O-CH_2-CH_2-O-(CH_2)_4-CO-}]_n$ as its repeating unit. Identify the missing structural feature, write a corrected repeating unit and explain how the links form.
[5 marks]
Total for this question: 5
One ethanediol unit contributes $\mathrm{C_2H_6O_2}$ and one hexanedioic acid unit contributes $\mathrm{C_6H_{10}O_4}$ before polyester links form. The resulting repeating unit is $\mathrm{C_8H_{12}O_4}$ . Determine the atoms lost, identify the small molecules they form and relate the result to the number and type of links represented.
[5 marks]
Total for this question: 5

4.7.3.3 · Amino acids (chemistry only) (HT only)

Explanation

Higher tier: an amino acid contains two different functional groups in one molecule: an amino group, -NH₂, and a carboxyl group, -COOH. Glycine is H₂NCH₂COOH.
Amino and carboxyl groups on different molecules undergo condensation to form peptide links. Each peptide link, -CONH-, forms with the loss of water, and repeated condensation produces a polypeptide.
Different amino acids can occur in the same chain, producing proteins.
Do not describe this as addition polymerisation because no C=C bond opens.
Questions may require drawing the product formed when amino acids join, so the -CONH- link and every atom in the water removed must be accounted for.

Worked example

Higher tier: three amino acids join into one unbranched chain. State the numbers of peptide links and water molecules formed.

1.Three molecules require two joining reactions to make one chain.
2.Each joining reaction forms one peptide link.
3.Each condensation also releases one water molecule.

Answer: Two peptide links and two water molecules.

Common mistakes

Don't fall into the trap of calling the -CONH- connection an ester link rather than a peptide link.
Don't fall into the trap of saying four amino acids form four peptide links; an unbranched chain of four has three links.

Exam tip

Higher tier: for $n$ amino acids joining into one chain, expect $n-1$ peptide links and $n-1$ water molecules.

Tier 1 · Easy

Identify the two functional groups in $\mathrm{H_2NCH_2COOH}$ and name this amino acid.
[3 marks]
Total for this question: 3
Explain why one amino acid molecule can form condensation links at both ends of a polypeptide chain.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Two glycine molecules condense. Write the structural formula of the product, identify the new link and state the other product.
[4 marks]
Total for this question: 4
A polymer contains repeated $\mathrm{-CONH-}$ links. Identify the polymer class, name the monomer type and state the small molecule released during its formation.
[3 marks]
Total for this question: 3
Three glycine molecules, each with molecular formula $\mathrm{C_2H_5NO_2}$ , form one unbranched tripeptide. Determine the molecular formula of the tripeptide.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Four amino acid molecules join to make one unbranched polypeptide molecule. Determine the number of peptide links and water molecules formed, then explain how using different amino acids can change the polymer.
[5 marks]
Total for this question: 5
Glycine, $\mathrm{H_2NCH_2COOH}$ , and $\mathrm{H_2NCH(CH_3)COOH}$ form a dipeptide. Write either possible structural formula, identify the new link and name the other product.
[5 marks]
Total for this question: 5
Compare the formation of a polypeptide with the formation of a polyester from a diol and a dicarboxylic acid. Include the monomers, the links formed, the reaction type and the by-product.
[6 marks]
Total for this question: 6
Two molecules of $\mathrm{H_2NCH(CH_3)COOH}$ condense. Candidate P is $\mathrm{H_2NCH_2CONHCH_2COOH}$ and candidate Q is $\mathrm{H_2NCH(CH_3)CONHCH(CH_3)COOH}$ . Choose the peptide product, determine its molecular formula and explain why the other candidate cannot form from the stated monomers.
[5 marks]
Total for this question: 5
Two glycine molecules, $\mathrm{C_2H_5NO_2}$ , and one amino acid molecule, $\mathrm{C_3H_7NO_2}$ , form a three-unit chain. Compare the sequences G-G-A and G-A-G by calculating their molecular formulae, peptide-link and water counts, then explain why the products need not be identical.
[6 marks]
Total for this question: 6

4.7.3.4 · DNA (deoxyribonucleic acid) and other naturally occurring polymers (chemistry only)

Explanation

DNA is a large molecule essential for life because it encodes genetic instructions for the development and functioning of organisms and viruses.
Most DNA consists of two polymer chains arranged as a double helix and built from four different monomers called nucleotides.
Proteins are polymers of amino acids, while starch and cellulose are polymers made from glucose monomers.
Match each natural polymer to its own monomer type.
A common error is to describe DNA as a protein or to say that amino acids are its monomers.

A simplified DNA double helix made from two nucleotide polymer chains.

Worked example

Match DNA, protein and starch to their monomer types.

1.DNA is built from four different nucleotides.
2.Proteins are built from amino acids.
3.Starch is built from glucose.

Answer: DNA–nucleotides; protein–amino acids; starch–glucose.

Common mistakes

Don't fall into the trap of calling DNA a protein or saying its monomers are amino acids.
Don't fall into the trap of saying starch and cellulose have different monomer types, when both are polymers of glucose.

Exam tip

In a matching question, keep the pairs distinct: DNA–nucleotides, proteins–amino acids, starch/cellulose–glucose.

Tier 1 · Easy

Name the monomer type from which DNA is made and state how many different monomers are used.
[2 marks]
Total for this question: 2
Explain why starch is described as a polymer rather than a monomer.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Match each natural polymer to its monomer type: protein, starch, cellulose and DNA.
[4 marks]
Total for this question: 4
Describe DNA using its number of polymer chains, overall shape, monomer types and biological role.
[4 marks]
Total for this question: 4
A natural polymer is broken down into glucose monomers. Explain why this evidence alone does not show that the polymer is starch, and identify the other polymer it could be.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A student claims that DNA, proteins and starch must share one monomer because all three occur naturally. Evaluate the claim using monomers and one structural feature of DNA.
[5 marks]
Total for this question: 5
A natural substance consists of two polymer chains in a double helix, built from four kinds of monomer. Deduce the substance, state its role and explain why it is not a protein or a glucose polymer.
[5 marks]
Total for this question: 5
Evaluate the claim: ‘DNA contains four kinds of nucleotide, so each DNA molecule must have four polymer chains and cannot share any structural feature with proteins or starch.’
[5 marks]
Total for this question: 5
A molecule isolated from a virus encodes genetic instructions. Analysis finds four types of nucleotide arranged in two polymer chains that form a double helix. Identify the molecule and evaluate the claim that it cannot be DNA because viruses are not living organisms.
[5 marks]
Total for this question: 5
A DNA molecule contains 12 000 nucleotides shared equally between its two polymer chains. Determine the number of nucleotides in each chain, state how many different types of nucleotide DNA contains, and explain why the number of types is independent of the chain length.
[3 marks]
Total for this question: 3

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

4.7.1.1 · Crude oil, hydrocarbons and alkanes

Tier 1 · Easy

Mark scheme for 4.7.1.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	It contains carbon and hydrogen only. For $n=3$ , $2n+2=8$ , so it fits $\mathrm{C_nH_{2n+2}}$ .	First inspect the elements: C₃H₈ contains only carbon and hydrogen, so it is a hydrocarbon. Then substitute $n=3$ into the alkane formula: $2(3)+2=8$ , matching the eight hydrogen atoms.	2
Total Question 1			2
02.1	Crude oil is used much faster than it is formed (or it cannot be replaced on a human timescale).	Crude oil takes millions of years to form from ancient biomass. Human use removes it far more quickly, so the available supply is finite.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.7.1.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	C₇H₁₆ Using $\mathrm{C_nH_{2n+2}}$ with $n=7$ gives $2n+2=16$ hydrogen atoms.	Use $\mathrm{C_nH_{2n+2}}$ with $n=7$ . The number of hydrogen atoms is $2(7)+2=16$ , so the molecular formula is C₇H₁₆.	3
Total Question 1			3
02.1	Ancient biomass, mainly plankton, was buried in mud and converted into crude oil over millions of years. Crude oil is a mixture because it contains many different compounds, most of which are hydrocarbons.	Follow the formation sequence: ancient plankton was buried by mud, then changed over a very long time. A mixture contains more than one substance, and crude oil contains many different hydrocarbon compounds.	4
Total Question 2			4
03.1	Propane C₃H₈ There are 11 atoms in total.	The three-carbon chain is propane. Counting the symbols gives three carbon atoms and eight hydrogen atoms, so the molecular formula is C₃H₈ and the total atom count is $3+8=11$ .	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.7.1.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	C₄H₁₀ Butane	Write the alkane as $\mathrm{C_nH_{2n+2}}$ . Its relative molecular mass is $12n+(2n+2)=14n+2$ . Solve $14n+2=58$ to obtain $14n=56$ and $n=4$ . The formula is therefore C₄H₁₀, which is butane.	4
Total Question 1			4
02.1	$\mathrm{C_{17}H_{36}}$ is the alkane. All three are hydrocarbons because they contain carbon and hydrogen only. Only $\mathrm{C_{17}H_{36}}$ fits $\mathrm{C_nH_{2n+2}}$ when $n=17$ .	Each formula contains only C and H, so each substance is a hydrocarbon. For an alkane with seventeen carbon atoms, $2n+2=2(17)+2=36$ ; only $\mathrm{C_{17}H_{36}}$ has the required hydrogen count.	4
Total Question 2			4
03.1	There are $2(1)+3(2)=8$ carbon atoms. There are $2(4)+3(6)=26$ hydrogen atoms. Every molecule is a hydrocarbon because it contains carbon and hydrogen only. The sample is a mixture because it contains two different compounds, methane and ethane.	Count atoms across all five molecules. Two CH₄ molecules contribute two carbon and eight hydrogen atoms; three C₂H₆ molecules contribute six carbon and eighteen hydrogen atoms. Both substances are hydrocarbons, but having two substances makes the sample a mixture.	4
Total Question 3			4
04.1	For $\mathrm{C_nH_{2n+2}}$ , the difference between the hydrogen and carbon counts is $n+2$ , so $n+2=8$ . $n=6$ , so X has six carbon atoms. X has fourteen hydrogen atoms and molecular formula C₆H₁₄. X is a hydrocarbon because it contains carbon and hydrogen atoms only.	Represent X as $\mathrm{C_nH_{2n+2}}$ . Subtract the carbon count from the hydrogen count to obtain $(2n+2)-n=n+2$ . Solving $n+2=8$ gives $n=6$ , and substitution gives $2(6)+2=14$ , so X is C₆H₁₄. Its formula contains no element other than carbon and hydrogen.	4
Total Question 4			4
05.1	The first claim is incorrect because crude oil is a mixture, not one compound. The mixture contains very many compounds, most of which are hydrocarbons including alkanes. The second claim is consistent with the formation of crude oil from mainly plankton buried in mud. The third claim is incorrect because crude oil forms over geological timescales and is used much faster than it forms. Crude oil is therefore a finite resource.	Check each statement against a different defining feature of crude oil. Its composition makes it a mixture of many compounds, its origin is ancient buried biomass consisting mainly of plankton, and its extremely slow formation compared with extraction makes the resource finite.	5
Total Question 5			5

4.7.1.2 · Fractional distillation and petrochemicals

Tier 1 · Easy

Mark scheme for 4.7.1.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	A mixture of hydrocarbons with similar numbers of carbon atoms. Their different boiling points allow separation.	A fraction is not a single pure hydrocarbon; it is a group of hydrocarbons of similar molecular size. Their boiling points differ from those in other groups, enabling fractional distillation.	2
Total Question 1			2
02.1	Liquid hydrocarbons vaporise. Hydrocarbon vapours condense.	Heating changes the crude-oil hydrocarbons from liquid to vapour. Cooling at an outlet changes selected vapours back into liquids by condensation.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.7.1.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Hydrocarbon B condenses lower in the column. B has the higher boiling point, so it condenses at a higher temperature. The column is hotter at the base, so vapours with higher boiling points condense lower down.	The bottom of the column is hotter than the top. B condenses when the vapour cools below $210\,^\circ\mathrm{C}$ , which happens relatively low in the column. A remains a vapour until it reaches a cooler, higher region near $75\,^\circ\mathrm{C}$ .	3
Total Question 1			3
02.1	The $180\,^\circ\mathrm{C}$ outlet. This temperature lies within the fraction's boiling range, so its vapours condense there.	At $250\,^\circ\mathrm{C}$ the fraction would remain mainly as vapour, while $90\,^\circ\mathrm{C}$ is well below its boiling range. The outlet at $180\,^\circ\mathrm{C}$ is in the stated range, so the vapour condenses and is collected there.	3
Total Question 2			3
03.1	Petrol is a fuel obtained from crude oil. Kerosene is a fuel obtained from crude oil. Polymers are made using petrochemical feedstock. Detergents are made using petrochemical feedstock.	Separate the list by use. Petrol and kerosene are burned as fuels. Crude-oil fractions also supply feedstock from which the petrochemical industry makes materials including polymers and detergents.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.7.1.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	The hydrocarbons vaporise, cool as they rise, and condense at heights determined by their boiling points. Fractions are feedstock for making useful petrochemicals such as polymers, solvents, lubricants or detergents.	Heat the crude oil so that its hydrocarbons enter the column mainly as vapours. The temperature falls up the column. Each hydrocarbon condenses where the temperature becomes lower than its boiling point, so similar-sized molecules are collected together. Some fractions are processed as petrochemical feedstock, supplying carbon compounds used to manufacture materials such as polymers and solvents.	5
Total Question 1			5
02.1	The column is hot at the bottom and cooler towards the top, so hydrocarbons condense at different heights according to their boiling points. The process separates existing molecules into fractions; it does not break bonds or make new hydrocarbons. A fraction can be used as feedstock for products such as polymers, solvents, lubricants or detergents.	Different boiling points make the vapours condense at different positions along the temperature gradient. This is a physical separation, so the molecules themselves are unchanged. Some separated fractions are then used as petrochemical feedstock rather than burned directly.	5
Total Question 2			5
03.1	The claim is incorrect. A fraction is a mixture of hydrocarbons, not one pure compound. The hydrocarbons in a fraction have similar numbers of carbon atoms. They therefore have similar boiling points and condense over a similar temperature range. One outlet can collect several hydrocarbons that condense in the same region of the column.	An outlet collects molecules with a range of similar boiling points. Those molecules tend to have similar chain lengths, but they are not all identical. The material leaving one outlet is therefore a fraction containing several hydrocarbons.	5
Total Question 3			5
04.1	X is collected at the upper outlet at $65\,^\circ\mathrm{C}$ . Y is collected at the lower outlet at $210\,^\circ\mathrm{C}$ . Z is collected at the middle outlet at $140\,^\circ\mathrm{C}$ . Y contains the largest molecules. Y has the highest boiling range and therefore condenses in the hotter, lower part of the column.	Compare each boiling range with the three outlet temperatures. The $55$ – $75\,^\circ\mathrm{C}$ fraction matches the cool upper outlet, the $125$ – $155\,^\circ\mathrm{C}$ fraction matches the middle outlet and the $195$ – $225\,^\circ\mathrm{C}$ fraction matches the hot lower outlet. The highest-boiling fraction contains the largest hydrocarbon molecules.	5
Total Question 4			5
05.1	A uniform temperature gives no temperature gradient for different hydrocarbons to condense at different heights. The bottom should be hot and the temperature should decrease towards the top. High-boiling hydrocarbons then condense lower in the column. Low-boiling hydrocarbons remain as vapour for longer and condense higher in the column. This is a physical separation by boiling point, so the hydrocarbon molecules are not changed.	Fractional distillation depends on vapours encountering progressively cooler regions. At one temperature, the different boiling points cannot produce a useful sequence of condensation levels. Restoring a hot base and cooler upper region lets high-boiling substances condense first and low-boiling substances rise further, separating existing molecules into fractions.	5
Total Question 5			5

4.7.1.3 · Properties of hydrocarbons

Tier 1 · Easy

Mark scheme for 4.7.1.3 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Boiling point increases. Viscosity increases.	Recall the molecular-size trends: larger hydrocarbons have stronger intermolecular attractions, so they boil at higher temperatures and flow less readily.	2
Total Question 1			2
02.1	Flammability decreases.	Shorter-chain hydrocarbons ignite more readily. Therefore, increasing molecular size makes hydrocarbons less flammable.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.7.1.3 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$\mathrm{C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O}$	Balance the three carbon atoms with $3\mathrm{CO_2}$ and the eight hydrogen atoms with $4\mathrm{H_2O}$ . The products then contain $6+4=10$ oxygen atoms, so use $5\mathrm{O_2}$ .	2
Total Question 1			2
02.1	The products contain 14 carbon atoms and 32 hydrogen atoms, shared between 2 molecules of X, so each molecule has 7 carbon and 16 hydrogen atoms. X is C₇H₁₆.	The products contain 14 carbon atoms and 32 hydrogen atoms. These came from two molecules of X, so each X molecule contains seven carbon atoms and sixteen hydrogen atoms: C₇H₁₆.	3
Total Question 2			3
03.1	Carbon is oxidised to form carbon dioxide. Hydrogen is oxidised to form water. Energy is released to the surroundings.	Combustion adds oxygen to the elements in the fuel. Its carbon atoms end up in carbon dioxide and its hydrogen atoms end up in water. Hydrocarbon combustion transfers energy to the surroundings.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.7.1.3 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Q has the higher boiling point and viscosity but is less flammable than P. $\mathrm{C_{15}H_{32} + 23O_2 \rightarrow 15CO_2 + 16H_2O}$	Q is the larger molecule, so it has stronger intermolecular forces: its boiling point and viscosity are higher, while its flammability is lower. For combustion, use $15\mathrm{CO_2}$ for carbon and $16\mathrm{H_2O}$ for hydrogen. These products contain $30+16=46$ oxygen atoms, requiring $23\mathrm{O_2}$ .	5
Total Question 1			5
02.1	T is the most suitable. Its largest molecules give it the highest boiling point and viscosity and the lowest flammability of the three.	Road-surfacing material should be thick, not evaporate readily and not ignite easily. T has the largest molecules, so it is the most viscous, has the highest boiling point and is the least flammable.	4
Total Question 2			4
03.1	The low viscosity recorded for C is anomalous. C should have the highest viscosity, or viscosity should increase from A to C. The boiling points increase from A to C as molecular size increases. The flammabilities decrease from A to C as molecular size increases.	All three expected trends must be checked separately. Increasing molecular size raises boiling point and viscosity but lowers flammability. The boiling-point and flammability data fit those trends, whereas the final viscosity value does not.	4
Total Question 3			4
04.1	P has the smaller molecules. Its greater mass loss shows that it evaporates more readily and has a lower boiling point. Its shorter flow time shows that it has a lower viscosity. Its easier ignition shows that it is more flammable. Lower boiling point, lower viscosity and greater flammability are all trends associated with smaller hydrocarbon molecules.	Translate each observation into a property before comparing molecular size. Greater evaporation indicates a lower boiling point, a shorter flow time indicates a lower viscosity, and easier ignition indicates greater flammability. All three properties independently point to P having smaller molecules than Q.	5
Total Question 4			5
05.1	Fuel B is the best compromise because its boiling point is above $80\,^\circ\mathrm{C}$ and its viscosity is not high. A is unsuitable because its boiling point is below $80\,^\circ\mathrm{C}$ , so it would vaporise too readily. C is unsuitable because its high viscosity would make it flow poorly through the wick. Molecular size is likely to increase in the order A, B, C because boiling point and viscosity increase with molecular size. Flammability is therefore likely to decrease in the order A, B, C.	Apply both design constraints rather than choosing from one property. A flows well but boils below the operating temperature; C stays liquid but is too viscous. B meets both constraints. The paired boiling-point and viscosity data also place A as the smallest and C as the largest, so A should be most flammable and C least flammable.	5
Total Question 5			5

4.7.1.4 · Cracking and alkenes

Tier 1 · Easy

Mark scheme for 4.7.1.4 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	An alkene.	The orange-to-colourless bromine-water result is the characteristic test for the carbon-carbon double bond in an alkene.	1
Total Question 1			1
02.1	Shorter-chain alkanes. Alkenes.	Cracking splits a long-chain hydrocarbon into smaller molecules. Its product mixture normally includes a shorter alkane and at least one alkene.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.7.1.4 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	X is C₄H₈. X is an alkene.	Conserve atoms. After forming C₈H₁₈, four carbon atoms and eight hydrogen atoms remain, so X is C₄H₈. This fits $\mathrm{C_nH_{2n}}$ , so it is an alkene.	3
Total Question 1			3
02.1	Cracking is thermal decomposition because heating breaks large hydrocarbon molecules into smaller molecules. Fractional distillation only separates existing hydrocarbons by boiling point; it does not break molecules or make new substances.	Thermal decomposition uses heat to split chemical bonds, so cracking forms new, smaller molecules. Fractional distillation is a physical separation based on boiling points, leaving the hydrocarbon molecules unchanged.	3
Total Question 2			3
03.1	The long-chain hydrocarbon is vaporised. The vapour is passed over a hot catalyst. Large hydrocarbon molecules split into smaller molecules.	In catalytic cracking, the hydrocarbon is first turned into a vapour. Contact with a hot catalyst then breaks large molecules into smaller hydrocarbons, including alkanes and alkenes.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.7.1.4 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	The other product is C₆H₁₂. Vaporise the hydrocarbon and pass the vapour over a hot catalyst. It supplies high-demand short-chain fuels and alkenes used for polymers or other chemicals.	Subtract the atoms in C₉H₂₀ from C₁₅H₃₂, leaving C₆H₁₂, an alkene. For catalytic cracking, vaporise the feed and contact it with a hot catalyst. The reaction turns surplus long molecules into more flammable short-chain fuels and valuable alkene feedstock.	5
Total Question 1			5
02.1	Mix hydrocarbon vapour with steam and heat the mixture to a high temperature. Cracking converts less useful long molecules into high-demand shorter fuels and alkene feedstock. Shake the product with bromine water; an alkene changes it from orange to colourless.	Steam cracking uses vaporised hydrocarbon, steam and strong heating. It improves the match between refinery output and demand while supplying reactive alkenes for chemical manufacture. Bromine-water decolourisation identifies a C=C bond in an alkene product.	5
Total Question 2			5
03.1	Catalytic cracking passes hydrocarbon vapour over a hot catalyst. Steam cracking mixes hydrocarbon vapour with steam and heats it to a very high temperature. Both methods make smaller hydrocarbon molecules. Their products include alkenes as well as alkanes. The products meet demand for small-molecule fuels and provide alkenes for polymers or other chemicals.	The catalyst distinguishes catalytic cracking, while steam and a very high temperature distinguish steam cracking. Both processes split less useful long chains into smaller molecules, creating useful fuels and reactive alkene feedstock.	5
Total Question 3			5
04.1	After C₇H₁₆ forms, the two X molecules must contain C₆H₁₂ in total. Each X molecule is therefore C₃H₆. X is propene, an alkene, because C₃H₆ fits the general formula C_nH_2n. Carbon balances because $7+(2\times3)=13$ . Hydrogen balances because $16+(2\times6)=28$ .	Subtract C₇H₁₆ from C₁₃H₂₈ to leave C₆H₁₂ for two identical molecules. Dividing both subscripts by two gives X = C₃H₆. The completed equation is $\mathrm{C_{13}H_{28} \rightarrow C_7H_{16} + 2C_3H_6}$ . Carbon checks as $7+(2\times3)=13$ and hydrogen as $16+(2\times6)=28$ . C₇H₁₆ is an alkane because $2n+2=16$ when $n=7$ ; C₃H₆ fits C_nH_2n, so X is propene, an alkene.	5
Total Question 4			5
05.1	No alkene is detected in G because the bromine water remains orange. H contains an alkene because it decolourises bromine water. J contains an alkene because it decolourises bromine water. The test alone cannot identify which specific alkene is present in H or J. The test does not show that H or J is pure because a mixture containing an alkene would also decolourise bromine water.	Use only what the colour change demonstrates. An alkene reacts by addition across C=C and consumes orange bromine, so decolourisation detects an alkene. The same result is produced by different alkenes and by mixtures that contain an alkene, so it neither identifies a particular member nor establishes purity.	5
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4.7.2.1 · Structure and formulae of alkenes (chemistry only)

Tier 1 · Easy

Mark scheme for 4.7.2.1 Tier 1 · Easy
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01.1	C₅H₁₀	Pentene contains five carbon atoms, so set $n=5$ in $\mathrm{C_nH_{2n}}$ . This gives ten hydrogen atoms and the formula C₅H₁₀.	1
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02.1	A carbon-carbon double bond, C=C. $\mathrm{C_nH_{2n}}$	Unsaturation in an alkene is identified by a C=C bond. Members of the homologous series follow $\mathrm{C_nH_{2n}}$ .	2
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Tier 2 · Standard

Mark scheme for 4.7.2.1 Tier 2 · Standard
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01.1	Propene. It is unsaturated because it contains a carbon-carbon double bond.	Three carbon atoms give the prefix prop-. A C=C bond identifies the alkene ending -ene, so the name is propene. The double bond is also the structural reason it is described as unsaturated.	2
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02.1	$\mathrm{C_5H_{10}}$ is a molecular formula because it gives only the numbers of each type of atom. A fully displayed formula must show every atom and every bond, including the C=C bond.	A molecular formula records atom totals but not their connections. A fully displayed formula draws all C-C and C-H bonds and must make the alkene's carbon-carbon double bond visible.	3
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03.1	The missing member is butene. C₄H₈ Every member contains a carbon-carbon double bond, C=C.	The names increase by one carbon atom at each step, placing butene between propene and pentene. With four carbon atoms, C_nH_2n gives C₄H₈. All alkenes in the sequence contain C=C.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.7.2.1 Tier 3 · Hard
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01.1	The alkene is C₄H₈, butene. The corresponding alkane is C₄H₁₀.	For $\mathrm{C_nH_{2n}}$ , the relative molecular mass is $12n+2n=14n$ . Solve $14n=56$ to get $n=4$ , giving C₄H₈, butene. The alkane formula is $\mathrm{C_nH_{2n+2}}$ , so with four carbons it is C₄H₁₀.	5
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02.1	Pentane is C₅H₁₂ and pentene is C₅H₁₀. Pentene has two fewer hydrogen atoms because it contains a carbon-carbon double bond instead of only single carbon-carbon bonds.	Use $\mathrm{C_nH_{2n+2}}$ for pentane and $\mathrm{C_nH_{2n}}$ for pentene with $n=5$ . The C=C bond makes pentene unsaturated and accounts for the difference of two hydrogen atoms.	4
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03.1	The propene entry is incorrect. Propene is C₃H₆. Alkenes fit C_nH_2n, so three carbon atoms require six hydrogen atoms. Pentene is C₅H₁₀.	Check each row using C_nH_2n. For $n=3$ , $2n=6$ , so propene cannot be C₃H₈. Applying the same rule with $n=5$ gives C₅H₁₀ for pentene.	4
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04.1	For $\mathrm{C_nH_{2n}}$ , the total atom count is $n+2n=3n$ , so $3n=18$ . $n=6$ , so the alkene has six carbon atoms and twelve hydrogen atoms. The alkene formula is C₆H₁₂. The alkane with six carbon atoms is C₆H₁₄.	An alkene molecule described by $\mathrm{C_nH_{2n}}$ contains $3n$ atoms in total. Solving $3n=18$ gives $n=6$ , so the formula is C₆H₁₂. Replacing the alkene rule with $\mathrm{C_nH_{2n+2}}$ gives C₆H₁₄ for the corresponding alkane.	4
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05.1	A hydrogen-to-carbon atom ratio of $2:1$ is consistent with the alkene formula $\mathrm{C_nH_{2n}}$ . The evidence supports the substance being an alkene but does not prove that it is butene. Ethene, propene, butene and pentene all have the same $2:1$ hydrogen-to-carbon ratio. The number of carbon atoms, or the complete molecular formula, would identify which member is present. A displayed formula showing C=C would also confirm the alkene functional group and reveal the carbon count.	Convert the verbal ratio into the series relationship H $=2n$ when C $=n$ . That relationship is shared by every member, so it identifies a pattern rather than one chain length. A carbon count or full formula is required to choose the member, while a displayed C=C bond supplies structural confirmation.	5
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4.7.2.2 · Reactions of alkenes (chemistry only)

Tier 1 · Easy

Mark scheme for 4.7.2.2 Tier 1 · Easy
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01.1	Orange to colourless. Addition reaction.	Bromine adds across ethene's C=C bond. Its orange colour disappears as bromine is consumed, so the reaction is addition.	2
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02.1	Alkenes contain a reactive carbon-carbon double bond. Addition reaction.	The C=C functional group makes an alkene more reactive than an alkane, which has only single carbon-carbon bonds. Reagents add across the double bond in addition reactions.	2
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Tier 2 · Standard

Mark scheme for 4.7.2.2 Tier 2 · Standard
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01.1	$\begin{array}{c}\quad \mathrm{H}\qquad\mathrm{H}\\ \quad\vert\qquad\vert\\ \mathrm{H-C-C-H}\\ \quad\vert\qquad\vert\\ \quad \mathrm{H}\qquad\mathrm{H}\end{array}$ Ethane. Nickel catalyst.	Add one hydrogen atom to each carbon of C=C and replace the double bond with a single C-C bond. Draw every C-H bond: each carbon in ethane has three hydrogen atoms and one bond to the other carbon. The hydrogenation uses a nickel catalyst.	3
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02.1	C₃H₆Cl₂ The C=C double bond becomes a C-C single bond as one chlorine atom bonds to each of the two carbon atoms.	Addition uses one chlorine molecule without removing atoms from propene, giving C₃H₆Cl₂. The double bond opens and is replaced by a single carbon-carbon bond.	3
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03.1	A coefficient of 2 before both CO₂ and H₂O balances carbon and hydrogen. A coefficient of 3 before O₂ balances oxygen, giving $\mathrm{C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O}$ . The completed equation has six oxygen atoms on each side.	Balance two carbon atoms with $2\mathrm{CO_2}$ and four hydrogen atoms with $2\mathrm{H_2O}$ . The products contain six oxygen atoms, requiring $3\mathrm{O_2}$ . The left side then also contains six oxygen atoms.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.7.2.2 Tier 3 · Hard
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01.1	It is an alkene: incomplete combustion can make a smoky flame, and bromine adds across C=C. React it with steam using a phosphoric acid catalyst; ethene forms ethanol.	A smoky flame indicates incomplete combustion, which alkenes tend to show in air. Decolourising bromine water confirms a reactive C=C bond because bromine adds across it. Hydration adds water across the double bond: pass ethene with steam over a phosphoric acid catalyst to produce ethanol.	6
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02.1	Bromination changes C=C to C-C and forms 1,2-dibromoethane. Hydrogenation uses nickel but produces ethane, not ethanol. Hydration uses phosphoric acid but produces ethanol, not ethane. All three are addition reactions in which atoms add across C=C and a C-C single bond remains.	Match each reagent to its correct reaction: bromine gives the dibromo compound, hydrogen over nickel gives the alkane, and steam over phosphoric acid gives the alcohol. In each case the double bond opens during addition.	6
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03.1	Hydrogen forms C₂H₆. A nickel catalyst is used for hydrogenation. Steam forms C₂H₅OH. A phosphoric acid catalyst is used for hydration. Iodine forms C₂H₄I₂. In every reaction, atoms add across C=C and it becomes a single carbon-carbon bond.	Work backwards from the atoms gained by ethene. Adding H₂ gives ethane and requires nickel; adding water as steam gives ethanol and uses phosphoric acid; adding I₂ gives C₂H₄I₂. Each is an addition reaction across the double bond.	6
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04.1	The starting alkene is propene, $\mathrm{CH_3-CH{=}CH_2}$ . The reagent is iodine, $\mathrm{I_2}$ . One iodine atom has added to each carbon that was in the C=C bond. The carbon-carbon double bond has become a carbon-carbon single bond. The product is saturated because it contains no carbon-carbon double bond.	Locate the adjacent carbon atoms carrying iodine in the product. Remove one iodine from each and restore a double bond between those carbons, giving $\mathrm{CH_3-CH{=}CH_2}$ . Forward addition of I₂ then explains both C-I bonds and the replacement of C=C by C-C.	5
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05.1	P is ethane, C₂H₆. Hydrogen adds across C=C over nickel, changing it to C-C. The bromine water mixed with P remains orange. The bromine water mixed with untreated ethene becomes colourless. Hydrogenation removes the reactive C=C functional group, so P cannot undergo the bromine addition reaction.	Complete hydrogenation adds H₂ to C₂H₄, producing C₂H₆. Ethane has only a carbon-carbon single bond and leaves bromine water orange. The control retains C=C, so it reacts with bromine and removes the orange colour.	5
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4.7.2.3 · Alcohols (chemistry only)

Tier 1 · Easy

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01.1	Ethanol. The -OH functional group.	The molecule contains two carbon atoms, giving the prefix eth-. The -OH group gives the alcohol ending -anol, so the compound is ethanol.	2
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02.1	Solubility decreases.	The shorter alcohols mix more readily with water. Increasing the length of the carbon chain reduces their solubility.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.7.2.3 Tier 2 · Standard
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01.1	Yeast. Warm conditions and absence of oxygen. Carbon dioxide.	Add yeast to the glucose solution and keep it warm so its enzymes work. Exclude oxygen so fermentation rather than aerobic respiration occurs. The sugar is converted into ethanol and carbon dioxide.	4
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02.1	Sodium produces hydrogen gas. Oxidation produces propanoic acid. The oxidising agent supplies oxygen to propanol or removes hydrogen from it.	Alcohols react with sodium to release hydrogen. Oxidising a primary alcohol with three carbon atoms produces the corresponding three-carbon carboxylic acid, propanoic acid; oxidation is gain of oxygen or loss of hydrogen.	3
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03.1	Alcohols can be used as fuels. Alcohols can be used as solvents. Ethanol is used in alcoholic drinks.	The first four alcohols burn and dissolve many substances, so fuels and solvents are standard uses. Ethanol is also the alcohol present in alcoholic drinks.	3
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Tier 3 · Hard

Mark scheme for 4.7.2.3 Tier 3 · Hard
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01.1	$\mathrm{C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O}$ Ethanoic acid. The oxidising agent supplies oxygen or removes hydrogen from ethanol.	For combustion, balance carbon with $2\mathrm{CO_2}$ and hydrogen with $3\mathrm{H_2O}$ . The products then contain seven oxygen atoms, one already in ethanol, so $3\mathrm{O_2}$ supplies the other six. Oxidation converts ethanol into ethanoic acid; the oxidising agent causes this gain of oxygen or loss of hydrogen.	5
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02.1	$65\,^\circ\mathrm{C}$ can denature yeast enzymes or kill the yeast, and the open flask allows oxygen to enter. Use a warm temperature of about $30$ to $40\,^\circ\mathrm{C}$ and exclude oxygen with a suitable closure or air lock. Carbon dioxide is the gaseous product.	Fermentation requires living yeast with active enzymes, so the mixture must be warm rather than hot. Oxygen should be excluded to maintain anaerobic conditions. Under these conditions glucose forms ethanol and carbon dioxide.	5
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03.1	Methanol mixes more evenly with the water. Methanol has the shorter carbon chain. The solubility of alcohols in water decreases as carbon-chain length increases. Methanol is therefore the better choice for the water-based mixture.	Methanol has one carbon atom whereas butanol has four. Increasing carbon-chain length lowers an alcohol's solubility in water, so methanol forms the more uniform mixture and is the suitable choice here.	4
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04.1	The alcohol is butanol. Oxidation preserves the four-carbon chain and produces butanoic acid. Reaction with sodium produces hydrogen gas. Butanol is less soluble in water than ethanol. Alcohol solubility decreases as the carbon chain becomes longer, and butanol has the longer chain.	Work backwards from the four-carbon carboxylic acid to the corresponding four-carbon alcohol, butanol. Apply the general sodium reaction to predict hydrogen. Then compare chain lengths: butanol has four carbon atoms and ethanol has two, so butanol is less soluble in water.	5
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05.1	$\mathrm{C_4H_9OH + 6O_2 \rightarrow 4CO_2 + 5H_2O}$ . Carbon is balanced at $4=4$ and hydrogen at $10=10$ ; oxygen is balanced because $1+12=8+5=13$ . Ethanol needs three oxygen molecules: $\mathrm{C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O}$ . Butanol needs more oxygen because each molecule has more carbon and hydrogen atoms to oxidise than ethanol.	Balance four carbon atoms with 4CO₂ and ten hydrogen atoms with 5H₂O. The products then contain 13 oxygen atoms, so the one oxygen atom in butanol plus 6O₂ supplies 13. Ethanol balances with 3O₂. Butanol has four carbon and ten hydrogen atoms, compared with two carbon and six hydrogen atoms in ethanol, so more oxygen is required for complete combustion.	4
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4.7.2.4 · Carboxylic acids (chemistry only)

Tier 1 · Easy

Mark scheme for 4.7.2.4 Tier 1 · Easy
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01.1	Ethanoic acid. The -COOH functional group.	Count both carbon atoms, including the one in -COOH, to obtain the eth- prefix. The -COOH group gives the name ethanoic acid.	2
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02.1	Sodium propanoate. Carbon dioxide.	An acid and a carbonate form a salt, water and carbon dioxide. Propanoic acid produces a propanoate salt, so sodium carbonate gives sodium propanoate.	2
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Tier 2 · Standard

Mark scheme for 4.7.2.4 Tier 2 · Standard
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01.1	Fizzing or effervescence is seen. A calcium ethanoate salt, water and carbon dioxide are formed.	Use the general acid-carbonate reaction: acid plus carbonate produces a salt, water and carbon dioxide. The escaping carbon dioxide causes fizzing; ethanoic acid forms an ethanoate salt.	4
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02.1	Propanol. Methanoic acid. Water.	The first part of an ester name comes from the alcohol, so propyl identifies propanol. The methanoate part comes from methanoic acid. Ester formation also produces water.	3
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03.1	CH₃COOH is ethanoic acid. C₃H₇COOH is butanoic acid. Both are recognised by the –COOH functional group. C₂H₅OH is an alcohol, not a carboxylic acid.	Look for the –COOH group. CH₃COOH has two carbon atoms overall and is ethanoic acid; C₃H₇COOH has four and is butanoic acid. C₂H₅OH ends in –OH instead.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.7.2.4 Tier 3 · Hard
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01.1	Add a carbonate: ethanoic acid fizzes and releases carbon dioxide, whereas ethanol does not. Ethyl ethanoate is formed, together with water.	Add the same small amount of carbonate to separate samples. Only ethanoic acid gives effervescence because an acid-carbonate reaction releases carbon dioxide. When ethanol and ethanoic acid react, the alcohol and acid combine in a condensation reaction to make the ester ethyl ethanoate and water.	5
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02.1	Hydrochloric acid has the lower pH. Hydrochloric acid ionises completely, whereas ethanoic acid ionises only partially because it is weak. Hydrochloric acid therefore has the greater hydrogen-ion concentration.	At equal concentration, the strong acid produces more H⁺ ions because it ionises completely. The weak carboxylic acid only partially ionises, so its H⁺ concentration is lower and its pH is higher.	4
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03.1	Reaction with sodium carbonate produces carbon dioxide, not hydrogen. The other products are sodium ethanoate and water. Reaction with ethanol produces the ester ethyl ethanoate. Water is also formed in the ester-forming reaction. Ethanoic acid forms an acidic solution in water, with a pH below 7.	Use the characteristic reaction patterns. An acid and a carbonate make a salt, water and carbon dioxide. A carboxylic acid and an alcohol make an ester and water. A carboxylic acid dissolved in water is acidic rather than alkaline.	5
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04.1	C is methanol because it is the only listed liquid that does not react with carbonate to release carbon dioxide. A is methanoic acid because its formula is HCOOH. B is ethanoic acid because its formula is CH₃COOH. Fizzing establishes that A and B are acids reacting with carbonate to form carbon dioxide. The carbonate test alone cannot distinguish methanoic acid from ethanoic acid because both give the same type of observation.	Use the carbonate result first to separate the alcohol from the two carboxylic acids. Both acids effervesce because their reactions release carbon dioxide, so that test cannot name either acid. The formula labels then distinguish the one-carbon methanoic acid from the two-carbon ethanoic acid.	5
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4.7.3.1 · Addition polymerisation (chemistry only)

Tier 1 · Easy

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01.1	Ethene. Addition polymerisation.	The name inside poly(ethene) identifies the alkene monomer as ethene. Alkene monomers join by opening C=C bonds, which is addition polymerisation.	2
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02.1	The C=C double bond opens and becomes C-C bonds linking monomers. No small-molecule by-product forms.	Addition polymerisation rearranges the alkene double bonds to create links between monomers. All monomer atoms remain in the polymer, so nothing small is eliminated.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.7.3.1 Tier 2 · Standard
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01.1	$[\mathrm{-CH_2-CH(CH_3)-}]_n$ C=C becomes C-C as new bonds link the monomers.	Use the two double-bonded carbon atoms as the polymer backbone, retain the CH₃ side group, and replace C=C with C-C. Put the unit in brackets with bonds passing through the brackets: $[\mathrm{-CH_2-CH(CH_3)-}]_n$ .	3
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02.1	10 000 carbon atoms. 5000 chlorine atoms.	Each $[\mathrm{-CH_2-CHCl-}]$ repeating unit contains two carbon atoms and one chlorine atom. Therefore, $5000\times2=10\,000$ carbon atoms and $5000\times1=5000$ chlorine atoms are represented.	3
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03.1	There are 600 repeating units. 600 ethene molecules formed those units.	Each poly(ethene) repeating unit contains two carbon atoms, so $1200 \div 2=600$ repeating units. Addition polymerisation uses one ethene molecule for each repeating unit, giving 600 monomer molecules.	2
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Tier 3 · Hard

Mark scheme for 4.7.3.1 Tier 3 · Hard
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01.1	$\mathrm{CH_2{=}CHCl}$ The double bond opens and all monomer atoms remain in the polymer, so no small molecule is eliminated.	Take one pair of backbone carbons and restore a double bond between them, keeping every attached atom in place. This gives $\mathrm{CH_2{=}CHCl}$ . Addition polymerisation only rearranges the bonding around C=C, so the repeating unit has the same atoms as the monomer and there is no by-product.	4
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02.1	The C=C bond should become C-C, and only the two former double-bonded carbons form the repeating backbone while CH₃ remains a side group. $[\mathrm{-CH_2-CH(CH_3)-}]_n$	Open propene's double bond and use those two carbon atoms as the chain backbone. Keep the third carbon as a CH₃ side group, then show continuation bonds through brackets and the subscript $n$ .	5
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03.1	B, ethene, is the monomer. C is poly(ethene), the addition polymer. A cannot act as the monomer because ethane has no carbon-carbon double bond. One ethene molecule and one repeating unit contain the same two carbon and four hydrogen atoms. The C=C bond opens to form single carbon-carbon bonds in the polymer chain.	An addition-polymer monomer must contain C=C, which identifies B. Opening that double bond links many ethene molecules to give C. No atoms are lost, so each C₂H₄ monomer supplies the atoms in one –CH₂–CH₂– repeating unit.	5
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04.1	A is the correct repeating unit. The two carbon atoms that formed C=C must become the two-carbon polymer backbone. The chlorine atom remains attached to the same backbone carbon after the double bond opens. The CH₃ group also remains a side group on that backbone carbon. B incorrectly places all three carbon atoms in the continuing backbone and changes the original attachments.	Mark the two atoms on either side of C=C before opening the bond. Those two atoms repeat along the main chain, while Cl and CH₃ keep their attachments to the second carbon. Candidate A preserves that connectivity; candidate B moves the CH₃ carbon into the backbone.	5
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05.1	The claim is incorrect because the second unit is the first unit read in the opposite direction. Both drawings represent poly(chloroethene). Their monomer is chloroethene, $\mathrm{CH_2{=}CHCl}$ . Opening C=C links the same two carbon atoms into a chain while the chlorine remains attached to one of them. One repeating unit contains the same atoms as one monomer because no small molecule is produced.	Trace the chain through each pair of brackets from both directions. Reversing a repeating unit does not change which atoms are connected, so the drawings are equivalent. Restoring C=C gives chloroethene, and addition polymerisation retains every monomer atom in the repeating unit.	5
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4.7.3.2 · Condensation polymerisation (chemistry only) (HT only)

Tier 1 · Easy

Mark scheme for 4.7.3.2 Tier 1 · Easy
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01.1	Each monomer has two functional groups. The monomers join repeatedly while a small molecule, commonly water, is released.	Two reactive ends are needed so a monomer can connect on both sides and extend the chain. Each condensation link usually forms with the elimination of water.	3
Total Question 1			3
02.1	Ethanediol has two functional groups and can form links at both ends. Ethanol has only one functional group, so it cannot continue a chain at both ends.	A monomer in a condensation chain needs two reactive sites. Ethanediol has an -OH group at each end, whereas ethanol has only one -OH group and terminates growth.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.7.3.2 Tier 2 · Standard
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01.1	Ethanediol has two -OH groups and butanedioic acid has two -COOH groups. A polyester forms and water is eliminated.	The diol supplies an -OH group at each end, while the dicarboxylic acid supplies two -COOH groups. Alcohol and carboxylic-acid groups form ester links, so the product is a polyester and each new link eliminates water.	4
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02.1	Seven ester links. Seven water molecules.	Eight separate monomer molecules require seven joins to make one unbranched chain. Each condensation join forms one ester link and eliminates one water molecule, giving seven of each.	3
Total Question 2			3
03.1	Functional groups on neighbouring monomers react to make links. Some atoms are removed from the monomers when each link forms. The removed atoms usually form water.	Condensation polymerisation joins bifunctional monomers while eliminating a small molecule. The atoms in that small molecule are not present in the polymer repeating unit; in the usual examples they leave as H₂O.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.7.3.2 Tier 3 · Hard
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01.1	$[\mathrm{-O-CH_2-CH_2-O-CO-CH_2-CH_2-CO-}]_n$ -OH and -COOH groups react to form ester links, eliminating water.	Remove H from an alcohol -OH and OH from a carboxyl -COOH at each joining point; these form water. The remaining oxygen connects to the carbonyl carbon, producing ester links. Keeping both carbon chains gives $[\mathrm{-O-CH_2-CH_2-O-CO-CH_2-CH_2-CO-}]_n$ .	5
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02.1	Propane-1,3-diol and pentanedioic acid can make the long-chain polyester. Each has two functional groups, so molecules can join at both ends to form repeated ester links. Water is eliminated when each link forms.	Propan-1-ol and pentanoic acid each have only one relevant functional group, so their reaction cannot keep extending at both ends. The diol and dicarboxylic acid are difunctional, allowing repeated ester-link formation with water eliminated.	5
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03.1	The diol is HO–CH₂–CH₂–OH. The dicarboxylic acid is HOOC–CH₂–COOH. The diol has two hydroxyl groups. The acid has two carboxyl groups. Water is the by-product.	Split the repeating unit at its ester links, –COO–. Add –OH to both ends of the two-carbon alcohol section to recover HO–CH₂–CH₂–OH, and add –OH to each carbonyl end of the acid section to recover HOOC–CH₂–COOH. Forming the ester links removes water.	5
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04.1	The proposal is missing one carbonyl group, C=O, from the dicarboxylic-acid section. A corrected unit is $[\mathrm{-O-CH_2-CH_2-O-CO-(CH_2)_4-CO-}]_n$ . The unit contains ester links, -COO-. An -OH group from the diol reacts with a -COOH group from the acid at each link. Water is eliminated as the condensation links form.	Preserve both carbonyl carbons from the two -COOH groups of hexanedioic acid. Joining the acid section to ethanediol through oxygen creates -COO- ester links at both sides of the repeating pattern. This uses the repeat-unit convention: the two links represented across one diol-diacid repeat eliminate two H₂O molecules.	5
Total Question 4			5
05.1	Before linking, the combined atom total is C₈H₁₆O₆. The repeating unit has four fewer hydrogen atoms and two fewer oxygen atoms, so H₄O₂ is lost. H₄O₂ corresponds to two water molecules. The repeating unit represents two ester links formed from -OH and -COOH groups. Losing water while the links form shows that the process is condensation polymerisation.	Add the monomer contributions to obtain C₈H₁₆O₆. Subtract the stated repeating-unit formula C₈H₁₂O₄; the difference is H₄O₂, or two H₂O molecules. This uses the repeat-unit convention: the repeating pattern spans two ester links and therefore two water losses.	5
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4.7.3.3 · Amino acids (chemistry only) (HT only)

Tier 1 · Easy

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01.1	An amino group, -NH₂. A carboxyl group, -COOH. Glycine.	Read the end groups separately: H₂N- is the amino group and -COOH is the carboxyl group. The specified amino acid with formula H₂NCH₂COOH is glycine.	3
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02.1	It contains both an amino group and a carboxyl group, so it has two reactive functional groups.	The -NH₂ group can react with a -COOH group on another molecule, while the molecule's own -COOH group can react at the other end. This allows chain growth.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.7.3.3 Tier 2 · Standard
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01.1	H₂NCH₂CONHCH₂COOH A peptide link, -CONH-, and water.	Join the -COOH group of one glycine to the -NH₂ group of the other. Remove OH from -COOH and H from -NH₂ to make water. The remaining groups form H₂NCH₂CONHCH₂COOH with a -CONH- peptide link.	4
Total Question 1			4
02.1	Polypeptide (or protein). Amino acids. Water.	The $\mathrm{-CONH-}$ group is a peptide link. Repeated peptide links make a polypeptide from amino-acid monomers, with water eliminated during each condensation.	3
Total Question 2			3
03.1	Three glycine molecules contain C₆H₁₅N₃O₆ in total. Two peptide links form, so two H₂O molecules are removed. The atoms removed have the combined formula H₄O₂. The tripeptide formula is C₆H₁₁N₃O₄.	First multiply the glycine formula by three to obtain C₆H₁₅N₃O₆. Joining three monomers in one chain creates two links and removes two water molecules, H₄O₂. Subtraction gives C₆H₁₁N₃O₄.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.7.3.3 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Three peptide links and three water molecules. Different amino acids can be arranged in different sequences, producing different polypeptides or proteins.	Joining four separate molecules into one chain requires three joining reactions. Each condensation reaction makes one peptide link and releases one water molecule, so there are three of each. Changing the amino acids or their order changes the sequence of groups along the polypeptide, giving a different protein.	5
Total Question 1			5
02.1	H₂NCH₂CONHCH(CH₃)COOH (or H₂NCH(CH₃)CONHCH₂COOH). A peptide link, -CONH-. Water.	Remove OH from a -COOH group and H from an -NH₂ group to make water. Join the remaining carbonyl carbon to nitrogen, forming -CONH-. Either amino acid can appear first, giving the two accepted orientations.	5
Total Question 2			5
03.1	A polypeptide is made from amino acid monomers. A polyester can be made from a diol and a dicarboxylic acid. Polypeptides contain peptide, –CONH–, links. Polyesters contain ester, –COO–, links. Both form by condensation polymerisation. Both usually release water when the links form.	Identify the functional groups that react in each case. Amino and carboxyl groups make peptide links, while hydroxyl and carboxyl groups make ester links. Both processes repeatedly join monomers and eliminate water, so both are condensation polymerisations.	6
Total Question 3			6
04.1	Q is the peptide product. Q contains the peptide link -CONH-. Before condensation, the two monomers contain C₆H₁₄N₂O₄ in total. Removing one H₂O gives Q the molecular formula C₆H₁₂N₂O₃. P cannot form from the stated monomers because it lacks both CH₃ side groups; water loss removes no carbon atoms.	Each monomer has formula C₃H₇NO₂, so two contain C₆H₁₄N₂O₄. Two monomers form one link, so one water molecule is lost. This gives C₆H₁₂N₂O₃. Q preserves both CH₃ side groups and contains -CONH-. P is the glycine dipeptide skeleton and has only four carbon atoms, so it cannot come from two three-carbon monomers.	5
Total Question 4			5
05.1	Before condensation, either sequence has combined formula C₇H₁₇N₃O₆. Three amino acid molecules form two peptide links in one unbranched chain. Two water molecules, H₄O₂, are eliminated. Each product therefore has molecular formula C₇H₁₃N₃O₄. The two products have the same atom totals and the same numbers of links and water molecules. They need not be identical because the amino acids occur in different orders along the chain.	Add two glycine formulae and one C₃H₇NO₂ formula to obtain C₇H₁₇N₃O₆. This uses the finite-chain convention: three monomers make two links, so subtract two H₂O molecules to get C₇H₁₃N₃O₄. Reordering does not change atom totals, but it changes the monomer sequence.	6
Total Question 5			6

4.7.3.4 · DNA (deoxyribonucleic acid) and other naturally occurring polymers (chemistry only)

Tier 1 · Easy

Mark scheme for 4.7.3.4 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Nucleotides. Four different nucleotides.	Recall that each DNA chain is a nucleotide polymer. The specification identifies four different nucleotide monomers.	2
Total Question 1			2
02.1	Starch contains many glucose monomer units joined together.	A polymer is a large molecule made from many repeating monomer units. Starch meets this definition because many glucose units are linked in each molecule.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.7.3.4 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Protein - amino acids. Starch - glucose. Cellulose - glucose. DNA - nucleotides.	Proteins are polypeptides, so their monomers are amino acids. Starch and cellulose are both glucose polymers. DNA is formed from nucleotide monomers.	4
Total Question 1			4
02.1	Most DNA contains two polymer chains arranged in a double helix. The chains are made from four different nucleotide monomers. DNA encodes genetic instructions for the development and functioning of organisms and viruses.	Link four specification facts: DNA is usually double-stranded, the strands form a double helix, four nucleotide types build them, and their sequence stores genetic instructions.	4
Total Question 2			4
03.1	The evidence is not enough to identify starch. Both starch and cellulose are polymers made from glucose monomers. The other possible polymer is cellulose.	Work backwards from the monomer but check whether it maps to only one polymer. Glucose is the monomer for both starch and cellulose, so identifying glucose cannot distinguish between those two natural polymers.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.7.3.4 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	The claim is incorrect: DNA uses four different nucleotides, proteins use amino acids, and starch uses glucose. Most DNA has two polymer chains arranged in a double helix.	Being naturally occurring does not determine a polymer's monomer. Identify each separately: nucleotides build DNA, amino acids build proteins, and glucose builds starch. DNA is further distinguished because most DNA molecules contain two polymer chains coiled into a double helix.	5
Total Question 1			5
02.1	The substance is DNA, which encodes genetic instructions. Its monomers are four different nucleotides; proteins use amino acids, while starch and cellulose use glucose.	The two-chain double helix and four nucleotide types identify DNA. Its function is to encode genetic instructions. A protein has amino-acid monomers, while starch and cellulose are built from glucose, so neither matches the evidence.	5
Total Question 2			5
03.1	The claim is incorrect. Most DNA molecules have two polymer chains, not four. The two chains form a double helix. Four refers to the different nucleotide monomers, not the number of chains. DNA, proteins and starch are all polymers, although proteins use amino acids and starch uses glucose as monomers.	Keep monomer variety separate from chain count. DNA normally has two chains arranged as a double helix, built using four nucleotide types. It shares the general feature of being a polymer with proteins and starch, but their monomers differ.	5
Total Question 3			5
04.1	The molecule is DNA. Four nucleotide monomer types are characteristic of DNA. Two polymer chains arranged as a double helix provide further structural evidence. The claim is incorrect because DNA encodes genetic instructions for viruses as well as for living organisms. Whether the source is a virus does not change the molecule's monomers or double-helix evidence.	Combine function, monomer and structure rather than relying on the source. Encoding genetic instructions, using four nucleotide types and forming a two-chain double helix all identify DNA. DNA carries genetic instructions for the development and functioning of viruses as well as organisms.	5
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05.1	Each chain contains 6 000 nucleotides. DNA contains four different types of nucleotide. Every DNA chain is built from the same four monomer types; a longer chain changes the number and order of monomers, not the number of available types.	The two chains share the total equally, so $12\,000\div2=6\,000$ nucleotides per chain. DNA uses four nucleotide monomer types regardless of chain length. Increasing the chain length adds monomers and allows a longer order or sequence, but it does not create additional types.	3
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