3.3 Organic chemistry — revision question pack

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

Checked against AQA 7405 section 3.3. Review basis: the qualification registry sourced from the AQA A-level Chemistry (7405) specification; registry verification recorded 11 July 2026.

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Answer all questions in the spaces provided.

3.3.1.1 · Nomenclature

Explanation

Organic compounds may be represented by empirical, molecular, general, structural, displayed or skeletal formulae, each showing different information.
Members of a homologous series share a functional group and general formula, show similar chemical properties and differ successively by $\mathrm{CH_2}$ .
IUPAC naming selects the longest chain or ring containing the principal functional group, numbers it to give the required lowest locants, and identifies substituents alphabetically with multiplicative prefixes where needed.
Students must name structures and draw structures from names for chains and rings containing up to six carbon atoms each.
Every unlabelled skeletal line end and vertex represents carbon.

Worked example

Name $\mathrm{CH_3CH(CH_3)CH_2CH_2OH}$ using IUPAC rules.

1.Choose the four-carbon chain containing the alcohol group.
2.Number from the alcohol end, placing $\mathrm{-OH}$ on carbon 1.
3.The methyl substituent is on carbon 3.

Answer: 3-methylbutan-1-ol.

Common mistakes

Don't choose the visually longest horizontal chain instead of the longest chain containing the principal functional group.
Don't number from the substituent end and give the alcohol group a higher locant.
Don't miss a carbon atom at an unlabelled end of a skeletal formula.

Exam tip

For a name-from-structure question, mark the parent chain and numbering before assembling substituent prefixes and locants.

Tier 1 · Easy

Give the IUPAC name of CH₃CH(CH₃)CH₂CH₂OH.
[1 mark]
Total for this question: 1
A student names CH₃CH₂CH(CH₃)CH₂CH₃ as 2-ethylbutane. Give the correct IUPAC name and explain the error.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Write a condensed structural formula for 3-ethyl-2-methylhexane.
[1 mark]
Total for this question: 1
An acyclic saturated alcohol has molecular formula C₆H₁₄O. Its longest chain containing the OH group has five carbon atoms. Numbering gives the OH group locant 2, and the only branch is a methyl group two carbon atoms further along the chain from the carbon bearing the OH group. Give its IUPAC name and a condensed structural formula.
[2 marks]
Total for this question: 2
A cyclohexane ring bears an OH group on carbon 1, an ethyl group on carbon 3 and a methyl group on carbon 4. Give its IUPAC name and explain how the numbering and prefix order are chosen.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A compound has condensed formula CH₃C(CH₃)₂CH₂CH(OH)CH₃. Give its IUPAC name and molecular formula.
[2 marks]
Total for this question: 2
A student names CH₂=CHCH(CH₃)CH(OH)CH₃ as 3-methylpent-1-en-4-ol. Give the correct IUPAC name and explain the numbering priority used.
[3 marks]
Total for this question: 3
Draw or describe an unambiguous structure for 2-ethyl-3,3-dimethylcyclopentan-1-ol. Then determine its molecular formula and explain why the ring carbon bearing OH must be carbon 1.
[5 marks]
Total for this question: 5
Hex-3-ene has condensed structural formula CH₃CH₂CH=CHCH₂CH₃. Deduce its molecular formula, empirical formula and homologous-series general formula. Explain one limitation of the molecular formula and one different limitation of the empirical formula.
[5 marks]
Total for this question: 5
A skeletal formula consists of a five-segment zig-zag chain with a one-segment branch from the second internal vertex when counted from the left. No carbon or hydrogen symbols are shown. Give the molecular formula and IUPAC name, and state how many carbon and how many hydrogen atoms are represented but not drawn as element symbols.
[4 marks]
Total for this question: 4

3.3.1.2 · Reaction mechanisms

Explanation

A reaction mechanism explains bond making and breaking through a sequence of species and intermediates. For electron-pair mechanisms, a curly arrow starts at the source of a pair: a lone pair or covalent bond.
It ends at the atom or bond receiving that pair. Heterolytic bond breaking is shown by an arrow from the bond to the atom taking both electrons.
Structures, charges, lone pairs and every required intermediate must be unambiguous.
A radical contains an unpaired electron shown by a dot.
Free-radical mechanisms use balanced step equations and radical dots; electron-pair curly arrows are not required and must not be mixed into that notation.

Worked example

State the two curly arrows for hydroxide attacking $\mathrm{CH_3CH_2I}$ .

1.Draw an arrow from an oxygen lone pair on $\mathrm{OH^-}$ to the carbon bonded to iodine.
2.Draw an arrow from the C–I bond to iodine as the bond breaks.

Answer: $\mathrm{CH_3CH_2OH}$ and $\mathrm{I^-}$ form.

Common mistakes

Don't start a curly arrow at a positive charge instead of at the electron pair that moves.
Don't draw the bond-breaking arrow from iodine towards the C–I bond rather than from the bond to iodine.
Don't use electron-pair curly arrows in a free-radical mechanism instead of radical dots and balanced equations.

Exam tip

For every curly arrow, check that its tail touches a lone pair or bond and its head shows the destination of that electron pair.

Tier 1 · Easy

In the first step of attack by :NH₃ on a positive carbon centre, state where the bond-forming curly arrow starts and ends.
[2 marks]
Total for this question: 2
A student uses a full curly arrow to represent the movement of one electron from a chlorine radical. Explain why this notation is wrong and state how AQA radical mechanisms represent an unpaired electron.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Write the radical propagation step in which a methyl radical reacts with chlorine, and explain the required dot-and-arrow notation.
[3 marks]
Total for this question: 3
During heterolytic fission of H–Br, a student draws a curly arrow from Br^δ− towards the H–Br bond. Give the correct arrow origin and destination, and state the ions formed.
[3 marks]
Total for this question: 3
A student says that atoms travel along full curly arrows during an organic mechanism. Explain why this is wrong. State exactly what the tail and head of a full curly arrow represent.
[2 marks]
Total for this question: 2

Tier 3 · Hard

A student draws hydroxide attacking CH₃CH₂I with one curly arrow from carbon to oxygen and a second from iodine to the C–I bond. Correct both arrows and identify the electron source in each case.
[4 marks]
Total for this question: 4
A carbocation CH₃CH⁺CH₂CH₃ reacts with water. A student draws an arrow from C⁺ to O and shows a neutral alcohol immediately after bond formation. Give the corrected bond-forming arrow, the charge on the first intermediate and the electron-pair movement involving a second water molecule that forms the neutral alcohol.
[4 marks]
Total for this question: 4
A proposed electron-pair step shows CN⁻ reacting with R–CH₂⁺. The student draws an arrow from the positive charge to CN⁻, leaves CN negatively charged after the new bond forms, and states that carbon remains positively charged. Give the corrected arrow and both formal charges, explaining the carbocation carbon's valence before and after attack.
[4 marks]
Total for this question: 4
Ethane reacts with bromine under UV light. Write the initiation step, both propagation steps forming bromoethane, two different termination steps and the overall equation, showing every radical with its dot.
[6 marks]
Total for this question: 6

3.3.1.3 · Isomerism

Explanation

Structural isomers have the same molecular formula but different structural formulae. Chain isomers differ in carbon skeleton, position isomers move the same functional group or multiple bond, and functional-group isomers contain different functional groups.
Stereoisomers share a structural formula but differ in spatial arrangement.
E–Z isomerism arises from restricted rotation about planar C=C and requires two different groups on each double-bonded carbon.
Apply Cahn–Ingold–Prelog priorities separately at each carbon by comparing atomic numbers, proceeding outward only after a tie.
Higher-priority groups on the same side give Z; those on opposite sides give E.

CIP higher-priority groups are on the same side in Z and opposite sides in E.

Worked example

In an alkene, the higher-priority groups at the two C=C carbons lie on opposite sides. Assign the descriptor and justify it.

1.Use CIP rules to identify the higher-priority group at each double-bonded carbon.
2.Compare the positions of those two groups across the planar double bond.

Answer: The isomer is E because the higher-priority groups are opposite.

Common mistakes

Don't call chain isomers stereoisomers even though their atom connectivity differs.
Don't assign E or Z when one double-bonded carbon carries two identical groups.
Don't judge CIP priority by apparent group size instead of atomic number at the first point of difference.

Exam tip

For an E–Z assignment, mark the higher-priority substituent at each alkene carbon before comparing sides.

Tier 1 · Easy

Butan-1-ol and butan-2-ol have the same molecular formula. State their type of isomerism.
[1 mark]
Total for this question: 1
A student assigns an E descriptor to CH₂=C(Cl)CH₃. Explain why neither E nor Z can be assigned.
[2 marks]
Total for this question: 2

Tier 2 · Standard

At the left carbon of a C=C bond the groups are H and CH₃; at the right carbon they are CH₃ and CH₂CH₃. The left CH₃ and right CH₂CH₃ groups are drawn on opposite sides. Assign E or Z and justify your choice using CIP priorities.
[3 marks]
Total for this question: 3
Draw the structures of (E)- and (Z)-1-bromo-1-chloropropene, showing the geometry at the C=C bond, and state which CIP comparison distinguishes them.
[3 marks]
Total for this question: 3
Starting from pentan-2-one, draw or give a condensed formula for one position isomer, one chain isomer and one functional-group isomer. State the molecular formula common to all four compounds.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Draw or give condensed structural formulae for every aldehyde and ketone with molecular formula C₄H₈O. Name each one and classify the relationship between the aldehydes and the ketone.
[5 marks]
Total for this question: 5
A student claims that an acyclic alkene with formula C₅H₁₀ can have the carbon skeleton of 2-methylbutane and show E–Z isomerism. Determine whether the claim is possible and justify your conclusion by considering the permitted positions of the C=C bond.
[4 marks]
Total for this question: 4
A student claims the following four formulae represent four different compounds: CH₃CH₂CH₂CH₂OH, HOCH₂CH₂CH₂CH₃, CH₃CH(OH)CH₂CH₃ and CH₃CH₂OCH₂CH₃. Identify the duplicate, classify each of the two other genuinely different compounds relative to butan-1-ol, give one omitted chain isomer, and justify using molecular formula and connectivity.
[5 marks]
Total for this question: 5
An alkene is drawn with CH₂Cl above and CH₂OH below the left C=C carbon. CHO is above and CH₂OH is below the right C=C carbon. Determine its E–Z descriptor and justify every Cahn–Ingold–Prelog comparison, including the treatment of C=O.
[5 marks]
Total for this question: 5
Determine how many different E–Z stereoisomers hexa-2,4-diene has. List them using full descriptors and explain why assigning E or Z independently to both C=C bonds initially appears to give one extra arrangement.
[5 marks]
Total for this question: 5

3.3.2.1 · Fractional distillation of crude oil

Explanation

Alkanes are saturated hydrocarbons containing only carbon–carbon single bonds. Petroleum is a mixture consisting mainly of alkane hydrocarbons and is separated by fractional distillation.
Crude oil is vaporised and enters a fractionating column with a hot bottom and cooler top. Repeated condensation separates fractions by boiling-point range.
Longer molecules contain more electrons and have larger contact surfaces, so stronger London forces give higher boiling points and condensation lower in the column.
Shorter hydrocarbons remain gaseous to cooler, higher levels.
The process is physical: intermolecular attractions are overcome and re-formed, while covalent bonds, molecular formulae and compounds remain unchanged.

A fractionating column is hot at the bottom and cool at the top, so hydrocarbons condense at different heights.

Worked example

Explain why a $\mathrm{C_5}$ alkane condenses higher in a fractionating column than a $\mathrm{C_{15}}$ alkane.

1.The $\mathrm{C_5}$ molecule has fewer electrons and weaker London forces.
2.Its lower boiling point allows it to rise into a cooler region before condensing.

Answer: The shorter alkane is collected nearer the top of the column.

Common mistakes

Don't state that C–C bonds break during fractional distillation, confusing it with cracking.
Don't place long-chain, high-boiling hydrocarbons at the cool top of the column.
Don't call each collected fraction a pure compound rather than a boiling-range mixture.

Exam tip

For a column-position explanation, link chain length to electron number, London-force strength, boiling point and condensation height.

Tier 1 · Easy

State the physical property used to separate crude oil into fractions and name the separation process.
[2 marks]
Total for this question: 2
A student describes a crude-oil fraction as one pure alkane with a fixed boiling point. Give the correct meaning of a fraction.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Explain why a C₅ alkane is collected nearer the top of a fractionating column than a C₁₅ alkane.
[3 marks]
Total for this question: 3
A refinery must separate crude oil without changing the molecular structures of its components. Choose fractional distillation or catalytic cracking, and justify the choice in terms of bonds and physical properties.
[3 marks]
Total for this question: 3
A fractionating column has collection zones for boiling ranges below $350\,\mathrm{K}$ , from $350$ to $450\,\mathrm{K}$ and above $450\,\mathrm{K}$ . Three hydrocarbons have boiling points $325\,\mathrm{K}$ , $392\,\mathrm{K}$ and $478\,\mathrm{K}$ . Identify the collection zone for each hydrocarbon and state which zone is highest in the column.
[2 marks]
Total for this question: 2

Tier 3 · Hard

A vapour mixture contains C₇H₁₆, C₁₂H₂₆ and C₁₈H₃₈. Predict their order of condensation from highest to lowest in the column, and explain why separating them does not change any molecular formula.
[5 marks]
Total for this question: 5
A student states that a vapour condenses at an outlet because the local temperature is above its boiling range and that C–C bonds release energy as they reform. Give two corrections to this explanation.
[4 marks]
Total for this question: 4
Explain how packing in a fractionating column improves separation of hydrocarbons with fairly close boiling points. Predict the effect on the composition of the collected fractions if most of the packing is removed.
[4 marks]
Total for this question: 4

3.3.2.2 · Modification of alkanes by cracking

Explanation

Cracking breaks carbon–carbon bonds in long-chain alkanes to form smaller, more useful molecules.
Thermal cracking uses high temperature and high pressure and produces a high proportion of alkenes; its mechanism is not required.
Catalytic cracking uses high temperature, slight pressure and a zeolite catalyst, producing mainly motor fuels and aromatic hydrocarbons; its mechanism is also not required.
Every cracking equation must conserve carbon and hydrogen and commonly contains an alkane plus one or more unsaturated products.
The economic reason is to convert less-demanded heavy fractions into shorter fuels and alkene or aromatic feedstocks whose demand exceeds their direct supply from crude oil.

Worked example

Complete $\mathrm{C_{12}H_{26}\rightarrow C_8H_{18}+X}$ and identify the product type of $X$ .

1.Subtract eight product carbons from twelve reactant carbons, leaving four.
2.Subtract eighteen product hydrogens from twenty-six, leaving eight.

Answer: $X=\mathrm{C_4H_8}$ , an alkene.

Common mistakes

Don't use high pressure for catalytic cracking instead of the specified slight pressure.
Don't say thermal cracking mainly produces aromatic motor fuels rather than a high proportion of alkenes.
Don't balance a cracking equation by changing subscripts inside a molecular formula.

Exam tip

For a compare-conditions question, pair thermal high pressure with alkene production and catalytic slight pressure with zeolite and motor fuels.

Tier 1 · Easy

Complete and balance this cracking equation: $\mathrm{C_{12}H_{26}\rightarrow C_8H_{18}+X}$ .
[1 mark]
Total for this question: 1
A proposed cracking equation is $\mathrm{C_{14}H_{30}\rightarrow C_8H_{18}+C_4H_8+C_2H_6}$ . The two-carbon product is wrong. Give its correct formula and justify it by atom conservation.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A refinery cracks C₁₅H₃₂ to one molecule of C₈H₁₈, one of C₃H₆ and ethene only. Determine the number of ethene molecules formed, then explain the economic purpose of this conversion.
[3 marks]
Total for this question: 3
A refinery needs a high proportion of alkene feedstock rather than mainly motor fuels and aromatic hydrocarbons. Choose thermal or catalytic cracking, and give the pressure condition and product evidence supporting the choice.
[3 marks]
Total for this question: 3
Cracking one C₁₂H₂₆ molecule gives exactly two acyclic products. Explain, using hydrogen balance, why at least one product must be an alkene, and give one balanced equation for this cracking.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Compare the conditions and main product emphasis of thermal cracking with catalytic cracking.
[4 marks]
Total for this question: 4
One molecule of $\mathrm{C_{16}H_{34}}$ cracks to exactly three molecules: one alkane and two identical propene molecules. Deduce the alkane and write the balanced equation. A refinery wants a profitable process and has buyers for both motor fuels and propene. Evaluate whether the balanced equation alone is enough to decide between thermal and catalytic cracking: compare their pressure conditions and product emphasis, and state what further economic information is needed.
[6 marks]
Total for this question: 6
An unnamed C₁₈ alkane cracks according to $\mathrm{C_{18}H_{38}\rightarrow C_{10}H_{22}+2C_4H_8}$ . A $2.00\,\mathrm{kg}$ feed is $72.0\%$ cracked. Calculate the mass of C₄H₈ formed. Use $M_r(\mathrm{C_{18}H_{38}})=254$ and $M_r(\mathrm{C_4H_8})=56.0$ . Give your answer to 3 significant figures.
[4 marks]
Total for this question: 4
An acyclic alkane cracks as shown: $\mathrm{C_{17}H_{36}\rightarrow C_6H_{14}+C_4H_{10}+X}$ . Compound X is acyclic, contains only carbon and hydrogen, and has only C–C single and double bonds. Deduce the molecular formula of X and the number of C=C bonds in one molecule of X. Explain how the hydrogen count fixes your answer.
[4 marks]
Total for this question: 4

3.3.2.3 · Combustion of alkanes

Explanation

Alkanes are used as fuels. Complete combustion in excess oxygen forms carbon dioxide and water; incomplete combustion with limited oxygen forms carbon monoxide and/or carbon as well as water.
Internal-combustion engines also produce $\mathrm{NO_x}$ and release unburned hydrocarbons. Catalytic converters remove gaseous pollutants by converting carbon monoxide, nitrogen oxides and hydrocarbons into less harmful products.
Sulfur-containing hydrocarbons form sulfur dioxide, an acidic air pollutant.
Flue-gas desulfurisation uses basic calcium oxide or calcium carbonate to neutralise and remove $\mathrm{SO_2}$ .
Combustion equations must conserve atoms using coefficients without altering the fuel formula.

Worked example

Calculate the minimum mass of $\mathrm{CaCO_3}$ needed to remove $1.60\,\mathrm{kg}$ of $\mathrm{SO_2}$ in a $1{:}1$ reaction. Use $M_r(\mathrm{SO_2})=64.1$ and $M_r(\mathrm{CaCO_3})=100.1$ .

1.Moles $\mathrm{SO_2}=1600/64.1=24.96\,\mathrm{mol}$ .
2.The $1{:}1$ ratio requires $24.96\,\mathrm{mol}$ of $\mathrm{CaCO_3}$ .
3.Mass $=24.96\times100.1=2498.5\,\mathrm{g}$ .

Answer: $2.50\,\mathrm{kg}$ of $\mathrm{CaCO_3}$ .

Common mistakes

Don't state that incomplete combustion forms only carbon monoxide and omit possible carbon particles.
Don't claim catalytic converters remove sulfur dioxide, which is treated in flue gas with calcium compounds.
Don't convert kilograms inconsistently when using molar mass in $\mathrm{g\,mol^{-1}}$ .

Exam tip

For a pollutant-removal calculation, show the pollutant moles and equation ratio before converting the reagent moles to mass.

Tier 1 · Easy

Write the balanced equation for the complete combustion of propane.
[1 mark]
Total for this question: 1
A student writes $\mathrm{C_4H_{10}+4.5O_2\rightarrow4CO+5H_2O}$ for incomplete combustion. Rewrite the equation using the smallest whole-number coefficients.
[2 marks]
Total for this question: 2

Tier 2 · Standard

In a catalytic converter, carbon monoxide reacts with nitrogen monoxide. Write a balanced equation and state which pollutant is oxidised and which is reduced.
[3 marks]
Total for this question: 3
Two gas streams require treatment. Stream P from a power station contains sulfur dioxide; stream Q from a petrol engine contains carbon monoxide and nitrogen monoxide. Match calcium carbonate and a catalytic converter to the correct stream, and state the chemical purpose of each treatment.
[4 marks]
Total for this question: 4
Write a balanced equation for the complete combustion of an alkane with general formula $\mathrm{C_nH_{2n+2}}$ . Your oxygen coefficient may be fractional.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A flue gas contains 1.60 kg of SO₂. Calculate the minimum mass of CaCO₃ needed for complete removal using $\mathrm{CaCO_3+SO_2\rightarrow CaSO_3+CO_2}$ . Use $M_r(\mathrm{SO_2})=64.1$ and $M_r(\mathrm{CaCO_3})=100.1$ .
[3 marks]
Total for this question: 3
A student makes three claims about alkane fuels: carbon monoxide forms in excess oxygen; nitrogen oxides come from nitrogen compounds in the fuel; and catalytic converters remove sulfur dioxide from power-station flue gases. Correct all three claims.
[5 marks]
Total for this question: 5
At the same temperature and pressure, $40.0\,\mathrm{cm^3}$ of ethane burns in $110\,\mathrm{cm^3}$ of oxygen. All of the oxygen reacts. The only carbon-containing products are CO₂ and CO, and the water remains as vapour. Calculate the volumes of CO₂ and CO formed and write one balanced equation representing this product ratio.
[5 marks]
Total for this question: 5
A $2.50\,\mathrm{kg}$ fuel sample contains $0.480\%$ sulfur by mass. All the sulfur forms SO₂, which is removed by $\mathrm{CaO+SO_2\rightarrow CaSO_3}$ . The calcium oxide solid is $82.0\%$ pure by mass. Calculate the minimum mass of this solid needed. Use $A_r(\mathrm{S})=32.1$ and $M_r(\mathrm{CaO})=56.1$ . Give the final answer to 3 significant figures.
[5 marks]
Total for this question: 5
An exhaust sample entering a catalytic converter contains $0.240\,\mathrm{mol}$ CO and $0.180\,\mathrm{mol}$ NO. Assume they react only by $\mathrm{2CO+2NO\rightarrow2CO_2+N_2}$ and that the reaction goes to completion. Identify the limiting pollutant and calculate the amounts of unreacted pollutant, CO₂ and N₂ leaving the converter.
[6 marks]
Total for this question: 6

3.3.2.4 · Chlorination of alkanes

Explanation

Methane reacts with chlorine by free-radical substitution under ultraviolet radiation. Initiation is homolytic Cl–Cl fission: $\mathrm{Cl_2\xrightarrow{UV}2Cl\mathbin{\bullet}}$ .
Propagation sustains the chain because one radical is consumed and another formed: $\mathrm{Cl\mathbin{\bullet}+CH_4\rightarrow HCl+CH_3\mathbin{\bullet}}$ , then $\mathrm{CH_3\mathbin{\bullet}+Cl_2\rightarrow CH_3Cl+Cl\mathbin{\bullet}}$ .
Termination removes radicals when any two combine; possible products include chlorine, chloromethane and ethane.
All steps must balance atoms and show radical dots; electron-pair curly arrows are not required.
Continued irradiation causes further substitution, so chlorination produces a mixture of increasingly chlorinated products rather than pure chloromethane.

Initiation creates chlorine radicals; propagation regenerates a radical and sustains the chain.

Worked example

Write the initiation and both propagation equations for chlorination of methane.

1.Initiation: $\mathrm{Cl_2\xrightarrow{UV}2Cl\mathbin{\bullet}}$ .
2.Hydrogen abstraction: $\mathrm{Cl\mathbin{\bullet}+CH_4\rightarrow HCl+CH_3\mathbin{\bullet}}$ .
3.Radical regeneration: $\mathrm{CH_3\mathbin{\bullet}+Cl_2\rightarrow CH_3Cl+Cl\mathbin{\bullet}}$ .

Answer: The propagation pair has overall reaction $\mathrm{CH_4+Cl_2\rightarrow CH_3Cl+HCl}$ .

Common mistakes

Don't use heterolytic fission in initiation and form $\mathrm{Cl^+}$ and $\mathrm{Cl^-}$ instead of two radicals.
Don't call a step propagation when it consumes radicals without producing another radical.
Don't show chloromethane as the only possible product despite further substitution under UV.

Exam tip

For a named free-radical mechanism, label initiation, propagation and termination and keep every radical dot visible.

Tier 1 · Easy

Write the initiation step for methane chlorination and state the bond-fission type.
[2 marks]
Total for this question: 2
A student states that homolytic fission of $\mathrm{Cl_2}$ forms $\mathrm{Cl^+}$ and $\mathrm{Cl^-}$ . Give the correct products and explain how the bonding pair divides.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Write both propagation equations that convert methane and chlorine into chloromethane during free-radical substitution.
[2 marks]
Total for this question: 2
A mechanism contains these two equations: $\mathrm{R\mathbin{\bullet}+Cl_2\rightarrow RCl+Cl\mathbin{\bullet}}$ and $\mathrm{R\mathbin{\bullet}+Cl\mathbin{\bullet}\rightarrow RCl}$ . Identify which is propagation and which is termination, and justify each choice.
[4 marks]
Total for this question: 4
A refinery wants to maximise chloromethane rather than more highly chlorinated products during methane chlorination. State which reactant should be used in large excess, explain how this changes the collision opportunities for chlorine radicals, and explain why product separation is still required.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Give three different termination equations available in methane chlorination, and explain why prolonged irradiation lowers the purity of chloromethane.
[4 marks]
Total for this question: 4
Monochlorination of 2-methylpropane gives two structural organic products. Explain why, name both products, and write the two propagation equations for the pathway forming 2-chloro-2-methylpropane. Show every radical with a dot; do not use full curly arrows.
[6 marks]
Total for this question: 6
Methane is irradiated with a large excess of chlorine until every hydrogen atom has been substituted. Write the overall equation for forming tetrachloromethane, state how many substitution cycles are required per methane molecule, and explain why this overall equation does not represent a single step in the mechanism.
[5 marks]
Total for this question: 5
C₂H₅Cl and C₂H₄Cl₂ are detected in the product mixture of methane chlorination. Deduce the two radicals combining to form each, write both termination equations, write the equation producing ^•CH₂Cl, and explain why the two combination steps are not propagation steps.
[4 marks]
Total for this question: 4

3.3.3.1 · Nucleophilic substitution

Explanation

Halogenoalkanes contain polar C–X bonds, leaving carbon electron-deficient. Nucleophiles donate an electron pair to this carbon.
Required substitutions use $\mathrm{OH^-}$ , $\mathrm{CN^-}$ and $\mathrm{NH_3}$ to form alcohols, nitriles and amines. For primary halogenoalkanes with hydroxide or cyanide, a curly arrow runs from the nucleophile's lone pair to carbon while another runs from C–X to X.
Cyanide attacks through carbon.
Ammonia first forms $\mathrm{RNH_3^+}$ ; another ammonia removes a proton.
Hydrolysis rate follows C–X bond enthalpy: iodoalkanes react faster than bromoalkanes, then chloroalkanes, despite the opposite bond-polarity trend.

Worked example

Outline the mechanism for $\mathrm{CH_3CH_2Br}$ reacting with $\mathrm{CN^-}$ and name the product.

1.Draw a curly arrow from the carbon lone pair of $\mathrm{CN^-}$ to the carbon bonded to Br.
2.Draw a curly arrow from the C–Br bond to bromine.
3.Include the nitrile carbon in the parent-chain count.

Answer: $\mathrm{CH_3CH_2CN}$ , propanenitrile, and $\mathrm{Br^-}$ form.

Common mistakes

Don't draw cyanide attacking through nitrogen instead of through carbon.
Don't show ammonia substitution ending at $\mathrm{RNH_3^+}$ and omit deprotonation by a second ammonia molecule.
Don't predict chloroalkanes hydrolyse fastest because C–Cl is most polar, ignoring bond enthalpy.

Exam tip

For a nucleophilic-substitution mechanism, show the nucleophile lone pair, both curly arrows and the leaving halide charge.

Tier 1 · Easy

Name the organic product when 1-bromopropane undergoes nucleophilic substitution with CN⁻.
[1 mark]
Total for this question: 1
A halogenoalkane is converted by nucleophilic substitution into a nitrile containing one more carbon atom. Identify the nucleophile and state which atom of it bonds to the halogenoalkane.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Outline the one-step mechanism for CH₃CH₂CH₂Br reacting with OH⁻, including both curly arrows and the products.
[3 marks]
Total for this question: 3
A student shows CN⁻ reacting with 2-iodopropane but starts the bond-forming arrow at nitrogen and the bond-breaking arrow at iodine. Give both corrected arrow origins and destinations, and name the organic product.
[4 marks]
Total for this question: 4
The same bromoalkane gives butan-2-ol with aqueous hydroxide ions and butan-2-amine with excess ammonia. Deduce the bromoalkane, identify the attacking atom and lone pair in each reaction, and explain why neither reaction changes the carbon count.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A labelled set contains 1-chlorobutane (A), 1-bromobutane (B) and 1-iodobutane (C). Each is hydrolysed at the same concentration and temperature. Predict the rate order and explain why bond enthalpy, rather than C–X bond polarity, controls it.
[3 marks]
Total for this question: 3
Three halogenoalkanes have the same carbon skeleton. In an ethanolic silver nitrate hydrolysis test, P gives a yellow precipitate first, Q gives a white precipitate last, and R gives a cream precipitate between these times. Identify the halogen in P, Q and R, and explain the rate order.
[6 marks]
Total for this question: 6
The target molecule is 3-methylbutanenitrile, NCCH₂CH(CH₃)CH₃. Deduce a bromoalkane that gives this product in one nucleophilic-substitution step with CN⁻, and describe both curly arrows including their exact origins and destinations.
[5 marks]
Total for this question: 5
1-Bromobutane reacts with excess ammonia to form butan-1-amine. Describe the two-stage mechanism, including the curly arrows, the charged organic intermediate and the final ionic by-product.
[4 marks]
Total for this question: 4
18.40 g of 1-iodobutane ( $M_r=184.0$ ) is heated under reflux with excess ethanolic KCN; 5.45 g of pentanenitrile ( $M_r=83.0$ ) is obtained. Write the equation, explain why the product has five carbon atoms, and calculate the percentage yield.
[5 marks]
Total for this question: 5

3.3.3.2 · Elimination

Explanation

A halogenoalkane can undergo substitution and elimination concurrently with potassium hydroxide.
In aqueous conditions, $\mathrm{OH^-}$ mainly acts as a nucleophile and replaces halide; hot ethanolic conditions favour its role as a base and alkene formation.
The elimination mechanism has three simultaneous electron-pair movements: an oxygen lone pair accepts a hydrogen from the carbon adjacent to C–X, the C–H bond pair forms C=C, and the C–X bond pair moves to X.
Unsymmetrical substrates may eliminate from either adjacent carbon and form more than one structural alkene; any E–Z forms are then considered separately.

Worked example

2-bromobutane is heated with ethanolic potassium hydroxide. Name the structural alkene products and state hydroxide's role.

1.Removal of H from carbon 1 forms but-1-ene.
2.Removal of H from carbon 3 forms but-2-ene.
3. $\mathrm{OH^-}$ accepts a proton, so it acts as a base.

Answer: But-1-ene and but-2-ene form by elimination.

Common mistakes

Don't call hydroxide a nucleophile in the elimination mechanism instead of a base.
Don't omit the arrow from the C–H bond to the adjacent C–C bond, so no C=C is formed.
Don't give only the reagent KOH and omit aqueous or ethanolic conditions when distinguishing pathways.

Exam tip

For an elimination mechanism, account for all three arrows: base to H, C–H to C–C, and C–X to X.

Tier 1 · Easy

State the organic product of eliminating HBr from 2-bromopropane and name the role of OH⁻.
[2 marks]
Total for this question: 2
A student names the alkene formed by elimination from 1-bromobutane as but-3-ene. Give the correct name and state the naming error.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Two samples of 1-bromopropane are heated separately with aqueous KOH and ethanolic KOH. Give the principal organic product in each case and explain the different roles played by OH⁻.
[4 marks]
Total for this question: 4
A student states that elimination from 2-bromopentane gives only two alkene products. Give all the alkene products, including E–Z forms, and explain why pent-1-ene has no E–Z pair.
[4 marks]
Total for this question: 4
Explain why 1-bromo-2,2-dimethylpropane, BrCH₂C(CH₃)₃, cannot undergo elimination to form an alkene. Identify the structural requirement that is missing.
[3 marks]
Total for this question: 3

Tier 3 · Hard

2-bromobutane is heated with ethanolic potassium hydroxide. Name the two structurally isomeric alkene products, without counting E/Z forms separately, and state the three electron-pair movements in the elimination step.
[5 marks]
Total for this question: 5
A C₄H₉Br isomer gives one alkene on heating with ethanolic KOH. That alkene has no E–Z isomerism, and aqueous KOH converts the halogenoalkane into a tertiary alcohol. Deduce the halogenoalkane and both organic products. Explain how the evidence fixes the structure.
[5 marks]
Total for this question: 5
A $0.120\,\mathrm{mol}$ sample of 2-bromopropane reacts completely with hydroxide ions and forms only propene and propan-2-ol. The amount of propene is $0.0780\,\mathrm{mol}$ . Calculate the percentage following elimination and substitution, then explain the different role of OH⁻ in each pathway.
[4 marks]
Total for this question: 4
2-Bromopropane reacts with KOH by two different pathways depending on the conditions. Give the complete curly-arrow mechanism for both pathways, state which set of conditions favours each pathway, and identify the role of OH⁻ in each.
[6 marks]
Total for this question: 6

3.3.3.3 · Ozone depletion

Explanation

Ozone in the upper atmosphere is beneficial because it absorbs ultraviolet radiation. UV also breaks C–Cl bonds in chlorofluorocarbons and releases chlorine radicals.
These catalyse ozone decomposition through $\mathrm{Cl\mathbin{\bullet}+O_3\rightarrow ClO\mathbin{\bullet}+O_2}$ and $\mathrm{ClO\mathbin{\bullet}+O_3\rightarrow2O_2+Cl\mathbin{\bullet}}$ .
Adding the steps gives $\mathrm{2O_3\rightarrow3O_2}$ ; $\mathrm{Cl\mathbin{\bullet}}$ is regenerated and $\mathrm{ClO\mathbin{\bullet}}$ is an intermediate, so one chlorine radical destroys many ozone molecules.
This catalytic amplification explains why small CFC concentrations matter.
Consistent evidence from different research groups supported legislation banning CFC uses as solvents and refrigerants, while chemists developed chlorine-free replacements.

The chlorine-radical cycle regenerates its catalyst while converting ozone to oxygen.

Worked example

Combine the two chlorine-radical steps and identify catalyst and intermediate.

1.Add the two equations and cancel $\mathrm{Cl\mathbin{\bullet}}$ from both sides.
2.Cancel $\mathrm{ClO\mathbin{\bullet}}$ , which is formed then consumed.

Answer: $\mathrm{2O_3\rightarrow3O_2}$ ; catalyst $\mathrm{Cl\mathbin{\bullet}}$ ; intermediate $\mathrm{ClO\mathbin{\bullet}}$ .

Common mistakes

Don't describe all atmospheric ozone as harmful and omit its upper-atmosphere UV protection.
Don't call $\mathrm{ClO\mathbin{\bullet}}$ the catalyst even though it is formed and consumed.
Don't state that each chlorine radical destroys only one ozone molecule despite radical regeneration.

Exam tip

For a catalytic-cycle question, add the steps and use cancellation to distinguish the regenerated catalyst from the intermediate.

Tier 1 · Easy

State why ozone in the upper atmosphere is beneficial.
[1 mark]
Total for this question: 1
A student says that chlorine radicals are used up permanently when they react with ozone. Give the feature of the ozone-depletion cycle that corrects this claim.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Combine the two chlorine-radical ozone steps to obtain the overall equation, and identify the catalyst and intermediate.
[3 marks]
Total for this question: 3
A student claims that CFC molecules readily break down in the lower atmosphere and directly collide with ozone. Correct this account of how a CFC begins to deplete ozone.
[3 marks]
Total for this question: 3
Explain why agreement between independent research groups strengthened the case for restricting CFC uses, and state the key molecular feature required of a replacement refrigerant to avoid chlorine-radical ozone depletion.
[3 marks]
Total for this question: 3

Tier 3 · Hard

For the CFC CF₂Cl₂, write a UV photodissociation equation that releases a chlorine radical, then explain why a low concentration of this CFC can cause extensive ozone loss.
[4 marks]
Total for this question: 4
Two sealed samples initially contain the same amount of ozone. A very small amount of chlorine radicals is added to sample B. Ozone disappears much faster from B, while the chlorine-radical amount returns close to its initial value after each reaction cycle. Explain how these observations provide evidence for a catalytic chain involving $\mathrm{ClO\mathbin{\bullet}}$ .
[5 marks]
Total for this question: 5
A sample contains $3.00\times10^{-9}\,\mathrm{mol}$ of chlorine radicals. Each radical completes $8.00\times10^4$ full ozone-depletion cycles before termination. Using the overall change $\mathrm{2O_3\rightarrow3O_2}$ per cycle, calculate the mass of ozone decomposed and explain the catalytic amplification. Use $M_r(\mathrm{O_3})=48.0$ .
[5 marks]
Total for this question: 5
Researchers report four observations: UV irradiation of a CFC releases chlorine atoms; an otherwise identical dark control does not; upper-atmosphere ClO abundance rises where ozone abundance falls; and independent laboratories reproduce the photodissociation result. Evaluate how this evidence supports a chlorine-radical mechanism and a restriction on CFC use.
[5 marks]
Total for this question: 5

3.3.4.1 · Structure, bonding and reactivity

Explanation

Alkenes are unsaturated hydrocarbons containing a carbon–carbon double covalent bond.
The double covalent bond is a centre of high electron density.
This electron-rich region attracts electrophiles, which accept an electron pair.
During electrophilic addition, an electron pair from the double bond forms a new bond to the electrophile; further bond formation gives a saturated product in which C=C has become C–C.
Reactivity explanations must connect electrophile attraction and electron-pair donation to this high electron density.

The alkene double bond creates high electron density above and below the carbon framework.

Worked example

Explain why ethene reacts with an electrophile more readily than ethane.

1.Ethene's C=C is a centre of high electron density.
2.The electron-deficient electrophile is attracted and accepts a pair from the double bond.
3.Ethane lacks this exposed double-bond electron density.

Answer: Ethene undergoes electrophilic addition because of its electron-rich C=C bond.

Common mistakes

Don't define an alkene merely as a hydrocarbon with fewer hydrogens instead of identifying C=C.
Don't say an electrophile donates an electron pair to the double bond rather than accepts one.
Don't claim the entire C=C bond disappears in addition instead of becoming a C–C single bond.

Exam tip

For an alkene-reactivity explanation, use the linked phrases 'high electron density', 'attracts electrophile' and 'electron-pair donation'.

Tier 1 · Easy

Define an unsaturated hydrocarbon and state the feature that makes ethene unsaturated.
[2 marks]
Total for this question: 2
State which electrons in an alkene C=C bond are donated to an electrophile and where a bond-forming curly arrow must start.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Ethene and ethane both contain carbon–carbon bonding. State the additional bonding feature in ethene and explain how it changes electron density and reactivity.
[3 marks]
Total for this question: 3
A student states that an alkene C=C is a single covalent bond and a centre of low electron density, so it attracts nucleophiles. Give three corrections.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Explain, using electron density and bond changes, why an alkene reacts readily with an electrophile to form an addition product.
[4 marks]
Total for this question: 4
Ethene rapidly decolourises bromine in the dark, whereas ethane does not. A student attributes the difference to ethene having weaker intermolecular forces. Give the correct bonding explanation for the evidence.
[4 marks]
Total for this question: 4

3.3.4.2 · Addition reactions of alkenes

Explanation

Alkenes undergo electrophilic addition with $\mathrm{HBr}$ , $\mathrm{H_2SO_4}$ and $\mathrm{Br_2}$ . Bromine water tests for unsaturation by changing orange to colourless as bromine adds across C=C.
With HBr or sulfuric acid, the double-bond pair attacks $\mathrm{H^+}$ while the reagent bond breaks; the resulting anion attacks the carbocation.
The electron-rich alkene induces a dipole in bromine before attack and bromide completes addition.
Unsymmetrical alkenes can give major and minor products.
Their proportions are explained by carbocation stability: tertiary is more stable than secondary, which is more stable than primary, so the pathway through the more stable intermediate dominates.

Electrophilic addition proceeds through the more stable carbocation before nucleophilic attack.

Worked example

Propene reacts with HBr. Name the major product and explain its formation.

1.The C=C pair attacks H while the H–Br bond pair moves to Br.
2.Proton addition to the end carbon forms the more stable secondary carbocation.
3. $\mathrm{Br^-}$ attacks the positive middle carbon.

Answer: 2-bromopropane is the major product.

Common mistakes

Don't draw the first curly arrow from $\mathrm{H^+}$ to C=C instead of from the electron-rich bond.
Don't use permanent partial charges on bromine before the alkene induces its dipole.
Don't explain the major product by a memorised orientation rule without comparing carbocation stabilities.

Exam tip

For a major-product mechanism, draw both possible carbocations and label each primary, secondary or tertiary.

Tier 1 · Easy

State the observation when bromine water is shaken with cyclohexene and name the organic product.
[2 marks]
Total for this question: 2
A student claims that adding Br₂ to but-2-ene forms 2-bromobutane and leaves the C=C bond unchanged. Give the correct organic product and bond change.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Outline the electrophilic-addition mechanism that gives the major product when propene reacts with HBr. Include the two curly arrows, intermediate and product.
[4 marks]
Total for this question: 4
A student draws addition of Br₂ to an alkene with an arrow from Br^δ+ to C=C and an arrow from Br^δ− to the Br–Br bond. Give the two corrected first-step arrow origins and destinations, and state the charged species formed before the final attack.
[5 marks]
Total for this question: 5
But-2-ene reacts with HBr. A student expects two structural isomers because H can add to either alkene carbon. Explain why only 2-bromobutane forms as a structural isomer and identify the carbocation class.
[3 marks]
Total for this question: 3

Tier 3 · Hard

2-methylbut-2-ene reacts with HBr. Name the major and minor structural products and explain their relative amounts by comparing the carbocation intermediates.
[4 marks]
Total for this question: 4
An unknown alkene reacts with HBr and gives 2-bromo-4-methylpentane as the observed major product. The two candidate carbocations are CH₃CH⁺CH₂CH(CH₃)CH₃ and ⁺CH₂CH₂CH₂CH(CH₃)CH₃. Deduce which alkene was used, identify which carbocation leads to the major product and describe every curly arrow in that pathway, including each origin and destination.
[7 marks]
Total for this question: 7
Addition of Br₂ to an unknown acyclic alkene gives 1,2-dibromo-2-methylpropane. Deduce the alkene. The alkene then reacts with concentrated H₂SO₄. Give the structure of the major organic product and explain its formation by comparing the two possible carbocations and describing every curly arrow.
[6 marks]
Total for this question: 6
2-Methylpent-1-ene gives a single structural product with Br₂ but two with HBr. Give the Br₂ product and both HBr products, and explain, without reference to carbocation stability, why the bromine reaction can give only one structural product.
[4 marks]
Total for this question: 4

3.3.4.3 · Addition polymers

Explanation

Addition polymerisation joins many alkene or substituted-alkene monomers after their C=C bonds open. To derive a repeating unit, replace C=C with C–C, retain every substituent on its original carbon, enclose the smallest repeated section in brackets and draw a continuation bond through each bracket.
The reverse process identifies the monomer by restoring C=C between backbone carbons.
IUPAC names use poly(monomer).
Polyalkenes are unreactive because their backbones contain strong, non-polar C–C and C–H bonds; chains attract through London forces.
Rigid PVC has structural uses, while a plasticiser separates chains and increases movement to make flexible PVC.

Addition polymerisation opens the alkene double bond while retaining substituents on their original carbon.

Worked example

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

1.Open the C=C bond to form a C–C backbone.
2.Keep $\mathrm{CH_3}$ attached to the same second carbon.
3.Bracket the two-carbon unit and extend one bond through each bracket.

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

Common mistakes

Don't leave a C=C bond inside the addition-polymer repeating unit.
Don't move a substituent onto the wrong backbone carbon when opening the monomer double bond.
Don't draw end groups inside the brackets instead of continuation bonds through both sides.

Exam tip

For a polymer-to-monomer question, isolate two adjacent backbone carbons and restore the C=C without moving substituents.

Tier 1 · Easy

2-Methylpropene has the structure CH₂=C(CH₃)₂. Give the displayed repeat unit of its addition polymer.
[2 marks]
Total for this question: 2
A student draws the repeat unit of poly(tetrafluoroethene) as [-CF₂=CF₂-]_n. Identify the error and give the correct repeat unit.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A polymer segment is -CH₂-CH(CN)-CH₂-CH(CN)-. Deduce the monomer, name the addition polymer and state the strongest intermolecular force between its chains.
[4 marks]
Total for this question: 4
A two-monomer addition polymer contains the backbone fragments -CH₂-CH(CH₃)- and -CH₂-CHCl-. Deduce the two monomers and explain why changing their relative amounts can change the polymer's properties without changing its carbon backbone type.
[4 marks]
Total for this question: 4
A sample of poly(pent-1-ene) has an average relative molecular mass of $4.20\times10^5$ . Write its repeat unit and calculate its average number of repeat units. Use $A_{\mathrm r}(\mathrm C)=12.0$ and $A_{\mathrm r}(\mathrm H)=1.0$ .
[3 marks]
Total for this question: 3

Tier 3 · Hard

Unplasticised PVC is rigid, whereas PVC containing a molecular plasticiser bends more readily. Explain both observations using the structure of poly(chloroethene), intermolecular forces and chain movement. Also explain why neither sample is readily hydrolysed.
[6 marks]
Total for this question: 6
An addition polymer has repeat unit [-CH₂-C(CH₃)(COOCH₂CH₃)-]_n. Deduce and name its monomer. Explain why alkaline hydrolysis can change its side groups but does not split its main chain.
[6 marks]
Total for this question: 6
A representative random-copolymer section contains four -CF₂-CF₂- units and six -CH₂-CH₂- units. Deduce the two monomers, give their unit ratio, and calculate the percentage by mass of fluorine within these ten repeat units. Use $A_{\mathrm r}(\mathrm C)=12.0$ , $A_{\mathrm r}(\mathrm H)=1.0$ and $A_{\mathrm r}(\mathrm F)=19.0$ .
[6 marks]
Total for this question: 6
Compare the intermolecular forces between chains in poly(ethene), poly(propene) and poly(chloroethene), PVC. Explain how increasing the chain length of a polyalkene affects the strength of its intermolecular forces and its softening behaviour.
[5 marks]
Total for this question: 5

3.3.5.1 · Alcohol production

Explanation

Alcohols are produced industrially by acid-catalysed hydration of alkenes; ethene and steam form ethanol through electrophilic addition, with the acid regenerated. Ethanol is also produced by yeast fermentation of glucose under warm, aqueous, anaerobic conditions chosen to maintain enzyme activity and prevent unwanted oxidation.
Fractional distillation separates ethanol from the dilute fermentation mixture.
A biofuel comes from recently living material.
Fermentation appears carbon neutral because crop photosynthesis removes the carbon dioxide later released on combustion, but farming, fertiliser manufacture, processing and transport use energy.
Land use, food competition and other environmental or ethical effects must therefore be included in decisions.

Worked example

Explain why fermentation uses a warm temperature and anaerobic conditions.

1.A warm temperature gives useful enzyme-controlled rate without denaturing yeast enzymes.
2.Anaerobic conditions prevent ethanol being oxidised and direct glucose metabolism towards ethanol.

Answer: The conditions maximise ethanol production while preserving enzyme activity.

Common mistakes

Don't call fermented ethanol fully carbon neutral while ignoring agricultural, processing and transport emissions.
Don't use a temperature high enough to denature yeast enzymes.
Don't describe hydration as fermentation instead of acid-catalysed addition of steam to an alkene.

Exam tip

For a biofuel discussion, balance the carbon-cycle argument against at least one lifecycle emission and one land-use or ethical issue.

Tier 1 · Easy

Write the equation for fermentation of glucose to ethanol and state two conditions that keep the yeast working effectively.
[3 marks]
Total for this question: 3
A glucose fermenter is kept anaerobic but produces no carbon dioxide after its temperature reaches 65 °C. Identify the process error and state one change that should restore ethanol production.
[3 marks]
Total for this question: 3

Tier 2 · Standard

Propene is converted into propan-2-ol using steam and an acid catalyst. Outline the three mechanistic stages and explain why the catalyst is unchanged overall.
[4 marks]
Total for this question: 4
A student proposes making ethanol from ethene by adding water at room temperature with no catalyst. The industrial process operates at about 300 °C and 6-7 MPa. Identify the errors and give the correct reagent and catalyst, write the reaction equation, and explain the roles of the catalyst and high pressure.
[5 marks]
Total for this question: 5
Compare the atom economy for ethanol in hydration, C₂H₄ + H₂O → C₂H₅OH, and fermentation, C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂. Calculate both values and explain why atom economy alone does not determine the more sustainable route. Use $M_{\mathrm r}(\text{glucose})=180$ and $M_{\mathrm r}(\text{ethanol})=46.0$ .
[4 marks]
Total for this question: 4

Tier 3 · Hard

A fermenter produces $92.0\,\text{kg}$ of ethanol. Complete combustion releases all of its carbon as CO₂. Calculate the CO₂ mass using C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O, then assess whether describing the fuel as carbon neutral is justified.
[6 marks]
Total for this question: 6
A fermenter processes $18.0\,\text{kg}$ of glucose. Its ethanol yield is 80.0% of the theoretical yield from C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂. Calculate the ethanol mass produced. Then evaluate one advantage and one disadvantage of this route compared with hydration of ethene.
[6 marks]
Total for this question: 6
Equal glucose mixtures are fermented anaerobically for 48 h. At 25 °C the initial CO₂ rate is 1.8 units and the final ethanol concentration is 7.2%; at 35 °C the values are 3.4 units and 10.8%; at 50 °C they are 4.1 units and 2.3%. Explain what both measurements show, state the operating temperature, explain why using only the largest initial rate would be unsound, and state why every run was kept anaerobic.
[5 marks]
Total for this question: 5
A hydration plant feeds $100\,\mathrm{mol}$ of ethene with excess steam. On each pass, 40.0% of the ethene entering the reactor forms ethanol. After the first pass, 90.0% of the unreacted ethene is recovered and sent through the reactor once more. Calculate the total amount and mass of ethanol formed, and state why the single-pass conversion cannot reach 100%. Use $M_{\mathrm r}(\text{ethanol})=46.0$ .
[6 marks]
Total for this question: 6
For C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g), $\Delta H=-46\,\mathrm{kJ\,mol^{-1}}$ . Explain how lowering temperature affects equilibrium yield, $K_p$ and rate; state the effect of increasing pressure on yield and rate; state why a catalyst changes neither $K_p$ nor the equilibrium composition; and justify why industrial conditions are a compromise.
[6 marks]
Total for this question: 6

3.3.5.2 · Oxidation of alcohols

Explanation

Alcohols are classified as primary, secondary or tertiary by the number of carbon groups attached to the carbon bearing $\mathrm{-OH}$ .
Acidified potassium dichromate(VI) oxidises a primary alcohol first to an aldehyde and then to a carboxylic acid, while a secondary alcohol forms a ketone and a tertiary alcohol is not easily oxidised.
Distil the aldehyde as it forms to prevent further oxidation; heat under reflux with excess oxidant to obtain the acid.
Equations may use $\mathrm{[O]}$ .
Tollens' reagent gives an aldehyde a silver mirror and Fehling's solution gives a brick-red precipitate; ketones give neither observation.

Worked example

State the product and apparatus choice when propan-1-ol is oxidised to propanal.

1.Use acidified potassium dichromate(VI) and warm the primary alcohol.
2.Distil propanal as it forms so it leaves the oxidising mixture.

Answer: $\mathrm{CH_3CH_2CH_2OH+[O]\rightarrow CH_3CH_2CHO+H_2O}$ ; use distillation.

Common mistakes

Don't use reflux when preparing an aldehyde and allow further oxidation to the carboxylic acid.
Don't state that a tertiary alcohol readily oxidises to a ketone.
Don't claim ketones give a silver mirror with Tollens' reagent.

Exam tip

For a primary-alcohol preparation question, link distillation to immediate aldehyde removal or reflux to complete oxidation.

Tier 1 · Easy

Butan-2-ol is warmed with acidified potassium dichromate(VI). Name the organic product and state the colour change.
[2 marks]
Total for this question: 2
A student states that oxidation of propan-2-ol produces propanal. Identify the error and give the correct product, explaining the student's classification error.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Give the reagent, apparatus choice and an equation using [O] for converting 2-methylpropan-1-ol into 2-methylpropanal without producing much 2-methylpropanoic acid. State the aldehyde's results with Tollens' reagent and Fehling's solution.
[6 marks]
Total for this question: 6
Two alcohols are available. The first is 3-methylbutan-1-ol; the second is 2-methylbutan-2-ol. Identify the alcohol and apparatus needed to prepare 3-methylbutanal using acidified potassium dichromate(VI). Explain both choices and state what happens if the other alcohol is heated with the oxidant.
[5 marks]
Total for this question: 5
Excess butan-2-ol is oxidised by $0.0300\,\mathrm{mol}$ of dichromate(VI) ions. The redox stoichiometry is three alcohol molecules per dichromate ion. Calculate the maximum mass of butanone formed and state the inorganic colour change. Use $M_{\mathrm r}(\text{butanone})=72.0$ .
[4 marks]
Total for this question: 4

Tier 3 · Hard

Three C₄H₁₀O alcohols are butan-1-ol, butan-2-ol and 2-methylpropan-2-ol. Predict the organic result when each is heated under reflux with excess acidified dichromate(VI), and give a chemical test that distinguishes the two oxidation products.
[6 marks]
Total for this question: 6
Design an apparatus sequence to prepare and purify 3-methylbutan-2-one from 3-methylbutan-2-ol using acidified potassium dichromate(VI). Justify when reflux and distillation are used, giving a chemical reason rather than only a boiling-point statement.
[5 marks]
Total for this question: 5
An alcohol X has formula C₅H₁₂O, is secondary and contains one chiral carbon. Oxidation gives a ketone Y that has five carbon environments in its ¹³C NMR spectrum and gives no silver mirror with Tollens' reagent. Deduce X and Y and explain how every item of evidence supports the pair.
[5 marks]
Total for this question: 5
A $4.00\,\mathrm g$ sample of butan-1-ol is heated under reflux with excess acidified potassium dichromate(VI). Neutralising all the isolated acidic product uses $42.50\,\mathrm{cm^3}$ of NaOH(aq), concentration $1.00\,\mathrm{mol\,dm^{-3}}$ . Calculate the percentage yield of butanoic acid and explain the purpose of reflux. Use $M_{\mathrm r}(\text{butan-1-ol})=74.0$ and a 1:1 acid-to-NaOH ratio.
[5 marks]
Total for this question: 5
Propan-2-ol is oxidised by acidified dichromate(VI). Deduce the organic oxidation half-equation and combine it with $\mathrm{Cr_2O_7^{2-}+14H^++6e^-\rightarrow2Cr^{3+}+7H_2O}$ . Give the overall ionic equation, the alcohol-to-dichromate mole ratio and the colour change.
[6 marks]
Total for this question: 6

3.3.5.3 · Elimination

Explanation

Alcohols form alkenes by acid-catalysed elimination of water, also called dehydration. Protonation converts the poor $\mathrm{-OH}$ leaving group into water.
After water leaves, an electron pair from a C–H bond on an adjacent carbon forms C=C while $\mathrm{H^+}$ is lost and the acid catalyst is regenerated.
The electron-pair arrow must start at the bond supplying electrons.
An alcohol with suitable hydrogen atoms on both adjacent carbons can produce more than one structural alkene; any E–Z isomers are considered separately.
Alkenes made this way can supply addition-polymer monomers without deriving them directly from crude oil.

Worked example

Name the structural alkene products formed by acid-catalysed dehydration of butan-2-ol.

1.Removal of H from carbon 1 forms a double bond between carbons 1 and 2.
2.Removal of H from carbon 3 forms a double bond between carbons 2 and 3.

Answer: But-1-ene and but-2-ene.

Common mistakes

Don't remove a hydrogen from a carbon not adjacent to the carbon bearing the leaving group.
Don't draw a curly arrow from the proton towards the C–H bond instead of from the bond.
Don't forget that the acid catalyst is regenerated after elimination.

Exam tip

For an elimination-products question, inspect both carbons adjacent to the alcohol carbon before listing possible alkenes.

Tier 1 · Easy

Give the organic product and reaction type when ethanol vapour is heated with an acid catalyst and loses water.
[2 marks]
Total for this question: 2
In the final step of an acid-catalysed elimination, a student draws a curly arrow from H⁺ towards a neighbouring C-H bond. State the correct origin and destination of the electron-pair arrow that forms C=C.
[2 marks]
Total for this question: 2

Tier 2 · Standard

List all structural alkene products formed by dehydrating 3-methylpentan-3-ol and explain why more than one structural product is possible.
[4 marks]
Total for this question: 4
Choose pentan-1-ol, pentan-2-ol or 2-methylbutan-2-ol to prepare only one structural alkene by dehydration. Name the alkene and state suitable reagent and conditions. Justify why the other two alcohols are less selective.
[5 marks]
Total for this question: 5
Deduce the two structural alcohols that can each form 2-methylbut-2-ene by acid-catalysed dehydration without rearranging the carbon skeleton. Explain how the reverse of elimination identifies both.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Outline the acid-catalysed elimination mechanism that converts cyclohexanol into cyclohexene. State the origin and destination of each curly arrow and explain how the product could become a polymer feedstock without using an alkene obtained directly from crude oil.
[6 marks]
Total for this question: 6
A proposed dehydration mechanism for 2-methylbutan-2-ol shows OH⁻ leaving directly, an uncharged three-coordinate carbon intermediate, and a final curly arrow starting at H⁺. Identify and correct all three errors, state the correct intermediate charge and name the two structural alkene products.
[6 marks]
Total for this question: 6
Butan-2-ol labelled with ¹⁸O only in its OH group undergoes acid-catalysed dehydration. Predict the oxygen-containing product and its molecular-ion $m/z$ , name all the alkene products, including any E–Z isomers, and describe the electron-pair movements that account for the isotope location and catalyst regeneration. Use ¹H and ¹⁸O masses of 1 and 18.
[6 marks]
Total for this question: 6

3.3.6.1 · Identification of functional groups by test-tube reactions

Explanation

Functional groups are identified using reactions specified elsewhere in the course. Bromine water changes orange to colourless with an alkene.
A carboxylic acid effervesces with carbonate as $\mathrm{CO_2}$ forms. Warmed Tollens' reagent gives an aldehyde a silver mirror and Fehling's gives a brick-red precipitate, while ketones do not respond.
Acidified dichromate(VI) changes orange to green with oxidisable primary or secondary alcohols, so it cannot alone distinguish a primary alcohol from an aldehyde.
Required practical 6 tests alcohol, aldehyde, alkene and carboxylic acid.
Fresh portions and precise colours, precipitates or gas tests make a valid identification sequence.

Different functional groups give distinctive positive observations with selected test-tube reagents.

Worked example

Distinguish separate samples of cyclohexene and ethanoic acid using two test-tube reactions.

1.Cyclohexene decolourises bromine water from orange to colourless.
2.Ethanoic acid reacts with aqueous carbonate to effervesce.
3.The gas from the acid turns limewater milky, confirming $\mathrm{CO_2}$ .

Answer: Bromine water identifies the alkene; carbonate identifies the carboxylic acid.

Common mistakes

Don't name a reagent without stating the positive observation.
Don't use acidified dichromate alone to distinguish a primary alcohol from an aldehyde.
Don't describe bromine water as becoming clear instead of the precise orange-to-colourless change.

Exam tip

For an identification plan, use fresh portions and give reagent, conditions and observation for every sample.

Tier 1 · Easy

A colourless liquid may contain a C=C bond. State a test-tube reagent and the positive observation.
[2 marks]
Total for this question: 2
Separate fresh portions of an unknown liquid decolourise bromine water and effervesce with aqueous sodium carbonate. Deduce the two functional groups present and identify the gas.
[3 marks]
Total for this question: 3

Tier 2 · Standard

Describe two separate test-tube tests that identify which of two bottles contains ethanal and which contains ethanoic acid. Include every positive observation.
[4 marks]
Total for this question: 4
Three bottles contain cyclohexene, cyclohexanol and cyclohexanone. A student proposes bromine water followed by Tollens' reagent as a two-test sequence. Explain why this sequence cannot assign all three bottles, then state a suitable replacement for the second reagent and its two results.
[5 marks]
Total for this question: 5
In Required Practical 6, explain why each reagent is applied to a fresh portion of an unknown, why warmed Tollens' reagent should be heated in a water bath, and how a known aldehyde sample can act as a control.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Four unlabelled samples are ethanol, ethanal, ethene and ethanoic acid. Design a shortest reliable test sequence using reagents from the specification, and state the observation that assigns each sample.
[6 marks]
Total for this question: 6
An organic compound has molecular formula C₃H₄O₂. Fresh portions decolourise bromine water, effervesce with sodium carbonate solution and give no silver mirror with Tollens' reagent. Deduce its structural formula and name the compound, then write the equation for its carbonate test.
[6 marks]
Total for this question: 6
A pure liquid has formula C₄H₈O₂ and gives one gas-chromatography peak. Fresh portions do not effervesce with sodium carbonate and give no silver mirror with Tollens' reagent. After refluxing another portion with aqueous sodium hydroxide and then acidifying, a product effervesces with sodium carbonate. Evaluate whether the evidence supports one pure ester and deduce one possible structure and name for the liquid.
[6 marks]
Total for this question: 6
A $0.720\,\mathrm g$ sample of a pure C₆H₈O₄ compound produces $120\,\mathrm{cm^3}$ of CO₂ with excess aqueous sodium carbonate at room conditions. A fresh portion decolourises bromine water. Determine the number of carboxylic acid groups per molecule, identify another functional group and explain why these results do not fix one structure. Use molar gas volume $24.0\,\mathrm{dm^3\,mol^{-1}}$ and $M_{\mathrm r}=144$ .
[5 marks]
Total for this question: 5
Three unknown organic liquids A, B and C are known to be a bromoalkane, an alkene and an aldehyde, one of each. Describe how silver nitrate in ethanol, bromine water and Tollens' reagent can be used on fresh samples to identify the three liquids. Give the observation and inference for each positive result.
[6 marks]
Total for this question: 6

3.3.6.2 · Mass spectrometry

Explanation

High-resolution mass spectrometry can determine molecular formula because isotopes have precise, non-integer masses.
Candidate formulae may share a nominal $M_r$ but have different exact molecular masses.
For each candidate, multiply each isotope's precise mass by its atom count, sum all contributions and compare with the measured molecular-ion mass at the stated precision.
Carbon-12 contributes exactly $12.00000$ , while hydrogen-1 and oxygen-16 require their precise data-book values.
The molecular ion represents the intact molecule after electron loss, so its accurate $m/z$ for a singly charged ion supplies the molecular mass used in this comparison.

A high-resolution molecular-ion peak provides the precise mass used to distinguish candidate molecular formulae.

Worked example

Calculate the precise molecular mass of $\mathrm{C_3H_6O_2}$ using $^{12}\mathrm{C}=12.00000$ , $^1\mathrm{H}=1.00783$ and $^{16}\mathrm{O}=15.99491$ .

1.Carbon contribution $=3(12.00000)=36.00000$ .
2.Hydrogen contribution $=6(1.00783)=6.04698$ ; oxygen contribution $=2(15.99491)=31.98982$ .
3.Add all three contributions.

Answer: Precise molecular mass $=74.03680$ .

Common mistakes

Don't use rounded relative atomic masses and lose the exact-mass distinction between candidate formulae.
Don't forget to multiply a precise isotopic mass by the number of that atom.
Don't choose the nearest nominal integer instead of matching the measured decimal mass.

Exam tip

For an exact-mass deduction, write a separate precise-mass total for every candidate formula before comparing decimal places.

Tier 1 · Easy

Calculate the precise molecular mass of C₂H₄O using ¹²C = 12.00000, ¹H = 1.00783 and ¹⁶O = 15.99491.
[2 marks]
Total for this question: 2
A student calculates the high-resolution molecular mass of C₃H₈O as 60.00000. Identify the mistake and calculate the correct value using ¹²C = 12.00000, ¹H = 1.00783 and ¹⁶O = 15.99491.
[3 marks]
Total for this question: 3

Tier 2 · Standard

A compound containing only C, H and O has a high-resolution molecular-ion mass of $74.0369$ . Choose between C₄H₁₀O and C₃H₆O₂. Use ¹²C = 12.00000, ¹H = 1.00783 and ¹⁶O = 15.99491.
[3 marks]
Total for this question: 3
An unknown containing C, H, N and O has a molecular-ion mass of 89.0477. Candidate formulae are C₃H₇NO₂ and C₄H₁₁NO. Calculate both precise masses, identify the formula and state the separation between the candidate masses. Use ¹²C = 12.00000, ¹H = 1.00783, ¹⁶O = 15.99491 and ¹⁴N = 14.00307.
[5 marks]
Total for this question: 5
Two ions both have nominal $m/z=72$ . Their formulae and calculated precise masses are C₃H₄O₂, $72.02114$ , and C₄H₈O, $72.05755$ . Explain why an instrument reporting only whole-number $m/z$ cannot distinguish them and explain, in terms of elemental composition, why the two ions have different precise masses although their nominal masses are equal.
[3 marks]
Total for this question: 3

Tier 3 · Hard

An unknown has measured molecular-ion mass $88.0526$ . Candidate formulas are C₄H₈O₂ and C₅H₁₂O. Calculate both precise masses, identify the formula and calculate the absolute error of the chosen value. Use ¹²C = 12.00000, ¹H = 1.00783 and ¹⁶O = 15.99491.
[5 marks]
Total for this question: 5
An unknown gives a molecular-ion peak at 100.0524. Candidate formulae are C₅H₈O₂ and C₆H₁₂O. Its IR spectrum has a strong C=O absorption but no broad O-H absorption, and warming with Tollens' reagent gives a silver mirror. Calculate both precise masses, select the formula and identify one functional group. Use ¹²C = 12.00000, ¹H = 1.00783 and ¹⁶O = 15.99491.
[6 marks]
Total for this question: 6
An unknown containing only C, H and O has a molecular-ion mass of $102.0681$ . Independent analysis shows that each molecule contains five carbon atoms and two oxygen atoms. Use ¹²C = 12.00000, ¹H = 1.00783 and ¹⁶O = 15.99491 to determine the molecular formula and absolute difference between its calculated and measured masses. Explain why this does not determine a unique structure.
[5 marks]
Total for this question: 5
A compound is 40.0% carbon, 6.67% hydrogen and 53.3% oxygen by mass. Its high-resolution molecular-ion mass is $90.0317$ . Determine its empirical and molecular formulae, then calculate the absolute mass difference. Use $A_{\mathrm r}(\mathrm C)=12.0$ , $A_{\mathrm r}(\mathrm H)=1.0$ , $A_{\mathrm r}(\mathrm O)=16.0$ , ¹²C = 12.00000, ¹H = 1.00783 and ¹⁶O = 15.99491.
[5 marks]
Total for this question: 5

3.3.6.3 · Infrared spectroscopy

Explanation

Molecular bonds absorb infrared radiation at characteristic wavenumbers when photon energy matches a vibrational change.
Data-book ranges identify bonds and hence possible functional groups; the fingerprint region can identify a particular molecule by comparison with a reference spectrum.
Structure deductions use both present and absent absorptions, and unexpected peaks can reveal impurities.
For example, a strong C=O absorption near $1700\,\mathrm{cm^{-1}}$ plus a very broad O–H range around $2500$ – $3000\,\mathrm{cm^{-1}}$ supports a carboxylic acid.
Infrared absorption by bonds in carbon dioxide, methane and water vapour contributes to global warming because these gases absorb outgoing terrestrial infrared radiation.

An IR spectrum is interpreted from characteristic absorptions together with missing and unexpected peaks.

Worked example

An IR spectrum has a strong absorption near $1710\,\mathrm{cm^{-1}}$ and a very broad absorption from $2500$ to $3000\,\mathrm{cm^{-1}}$ . Suggest the functional group.

1.The first absorption indicates a C=O bond.
2.The very broad lower-wavenumber absorption indicates the O–H bond of a carboxylic acid.

Answer: A carboxylic acid group is present.

Common mistakes

Don't identify a compound from one absorption without checking for supporting or missing peaks.
Don't declare a sample pure because expected peaks appear and ignore unexpected impurity absorptions.
Don't confuse wavenumber with wavelength and reverse the data-book scale.

Exam tip

For an IR structure question, cite numerical absorption ranges and state what any missing diagnostic peak rules out.

Tier 1 · Easy

An IR spectrum has a strong absorption at 1718 cm^-1. Use the Data Booklet to identify the bond indicated by this absorption.
[1 mark]
Total for this question: 1
A student says that one strong absorption at 1715 cm^-1 proves that a sample is propanone. Evaluate the claim and state what further IR evidence is needed to identify a specific compound.
[3 marks]
Total for this question: 3

Tier 2 · Standard

A liquid's IR spectrum contains a strong peak at 1740 cm^-1, no broad O-H absorption, and a fingerprint region identical to a reference spectrum of ethyl ethanoate. Deduce the compound and explain the purpose of the fingerprint comparison.
[3 marks]
Total for this question: 3
Spectrum A has a broad absorption at 3230-3550 cm^-1 and no C=O peak. After reaction, spectrum B has a strong C=O absorption near 1715 cm^-1 and no broad O-H peak. The product gives no reaction with Tollens' reagent. Deduce the change when the starting compound is propan-2-ol.
[5 marks]
Total for this question: 5
An unknown has a strong absorption near 1710 cm^-1, a very broad absorption from 2500 to 3000 cm^-1 and an absorption in the 1620-1680 cm^-1 range. Identify two functional groups and explain why these absorptions alone do not identify a unique compound.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A nominally pure propanone sample matches the propanone reference fingerprint and has a strong C=O absorption, but it also shows a broad absorption at 3230-3550 cm^-1. Suggest an impurity and explain how IR absorption by CO₂, CH₄ and water vapour contributes to global warming.
[5 marks]
Total for this question: 5
An unknown C₄H₈O liquid has a strong IR absorption at 1725 cm^-1, no broad O-H absorption and a fingerprint region identical to a reference spectrum of 2-methylpropanal. On warming with Fehling's solution it gives a brick-red precipitate. Deduce the compound and explain how each observation supports the identification.
[6 marks]
Total for this question: 6
An unknown compound has molecular formula C₂H₄O. Its IR spectrum has a strong absorption near 1720 cm^-1, no broad absorption in the 3230-3550 cm^-1 range, and no absorption in the 1000-1300 cm^-1 range. Use the Data Booklet and the molecular formula to deduce the structure and name of the compound.
[5 marks]
Total for this question: 5

3.3.7 · Optical isomerism (A-level only)

Explanation

Optical isomerism is stereoisomerism caused here by chirality at one carbon attached to four different groups.
The two enantiomers are non-superimposable mirror images and rotate plane-polarised light by equal amounts in opposite directions.
A racemic mixture contains equal amounts of both enantiomers and is optically inactive because their rotations cancel.
A chiral centre is identified by tracing all four substituents far enough to find the first difference; four bonds alone are insufficient if two groups are identical.
Enantiomer drawings must use a valid three-dimensional representation, and rotating a molecule in space does not create the other enantiomer.

A chiral carbon with four different groups forms a pair of non-superimposable mirror images.

Worked example

Determine whether carbon 2 in $\mathrm{CH_3CH(OH)CH_2CH_3}$ is chiral and explain.

1.List the four attached groups: $\mathrm{H}$ , $\mathrm{OH}$ , $\mathrm{CH_3}$ and $\mathrm{CH_2CH_3}$ .
2.All four groups differ, so the carbon is asymmetric.

Answer: Carbon 2 is a chiral centre and the molecule has two enantiomers.

Common mistakes

Don't call a tetrahedral carbon chiral without checking whether two substituents are identical.
Don't treat a rotated drawing of one molecule as its mirror-image enantiomer.
Don't say a racemic mixture does not rotate light because neither enantiomer is optically active.

Exam tip

For a chiral-centre question, list all four attached groups explicitly before drawing mirror images.

Tier 1 · Easy

Identify the chiral carbon in CH₃CH(OH)CH₂CH₃ and name the compound.
[2 marks]
Total for this question: 2
A student circles carbon 3 in 3-methylpentan-3-ol and labels it chiral. Determine whether the label is correct and justify your answer.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Describe how to draw the two enantiomers of 2-bromobutane using wedge-and-dash bonds, and state their relationship and effect on plane-polarised light.
[4 marks]
Total for this question: 4
For each structure, identify whether it contains a chiral centre and justify the decision: (A) CH₃CH(OH)CH(CH₃)₂; (B) CH₃CH(OH)CH₃; (C) CH₃COCH₂CH₃; (D) CH₃CH₂CH₂OH.
[6 marks]
Total for this question: 6
Drawing B is obtained from a wedge-and-dash drawing of butan-2-ol by swapping only H and OH at carbon 2. Drawing C is obtained from the original by swapping H with OH and also swapping CH₃ with CH₂CH₃. State the relationship of B and C to the original and justify each answer.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Ethanal and propanone each react with HCN. Explain why one product is a racemic mixture but the other is optically inactive, naming both hydroxynitrile products.
[6 marks]
Total for this question: 6
Hydrogen bromide adds to pent-1-ene through a planar secondary carbocation on the major pathway. Name the major product and explain why a polarimeter can read zero even though the product contains a chiral centre.
[6 marks]
Total for this question: 6
Pure $0.800\,\mathrm{mol}$ samples of two enantiomers rotate plane-polarised light by $+12.0^\circ$ and $-12.0^\circ$ using the same solution volume and tube. A $0.800\,\mathrm{mol}$ mixture measured under those conditions gives $+3.00^\circ$ . Assuming rotation is proportional to the difference in their amounts, calculate the amount of each enantiomer and explain why the mixture is optically active.
[5 marks]
Total for this question: 5

3.3.8 · Aldehydes and ketones (A-level only)

Explanation

Aldehydes readily oxidise to carboxylic acids and give positive Tollens' and Fehling's tests; ketones do not under those conditions.
Aqueous $\mathrm{NaBH_4}$ reduces aldehydes to primary alcohols and ketones to secondary alcohols by nucleophilic addition: $\mathrm{H^-}$ attacks the $\delta+$ carbonyl carbon, the C=O pair moves to oxygen and the alkoxide is protonated.
KCN followed by dilute acid adds HCN to form a hydroxynitrile and lengthens the carbon skeleton by one carbon.
KCN is highly toxic.
Attack on either face of a planar aldehyde or unsymmetrical ketone produces a racemate only if the new tetrahedral carbon has four different groups.

Hydride attacks the electron-deficient carbonyl carbon while the C=O electron pair moves to oxygen.

Worked example

State the organic product when propanone is reduced with aqueous $\mathrm{NaBH_4}$ and outline the first electron movement.

1. $\mathrm{H^-}$ attacks the $\delta+$ carbonyl carbon.
2.The C=O electron pair moves to oxygen before protonation.

Answer: Propan-2-ol forms by nucleophilic addition.

Common mistakes

Don't call carbonyl reduction nucleophilic addition–elimination even though no group leaves.
Don't draw hydride attacking the carbonyl oxygen instead of the $\delta+$ carbon.
Don't claim every HCN addition product is optically active without checking for four different substituents.

Exam tip

For a carbonyl mechanism, show the carbonyl dipole and start the first arrow at the nucleophile's electron pair.

Tier 1 · Easy

Butanal is treated with aqueous sodium tetrahydridoborate. Name the organic product and write an equation using [H].
[2 marks]
Total for this question: 2
A student claims that propanone gives a silver mirror because it contains a carbonyl group. Identify the error and give the correct Tollens' result, then name the product when propanone is reduced with aqueous sodium tetrahydridoborate.
[3 marks]
Total for this question: 3

Tier 2 · Standard

Two bottles contain pentanal and pentan-3-one. For Tollens' reagent and Fehling's solution, state the conditions and the observation expected with each bottle.
[4 marks]
Total for this question: 4
In a proposed sodium tetrahydridoborate mechanism for pentanal, the first curly arrow starts at the carbonyl carbon and points to H⁻, and no second arrow is shown. Identify and correct both errors, state the intermediate charge and name the final organic product.
[5 marks]
Total for this question: 5
Starting with propanal, identify reagents to make (i) propan-1-ol and (ii) 2-hydroxybutanenitrile. Name the mechanism in each case and explain the difference in carbon-chain length between the products.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Butan-2-one reacts with KCN followed by dilute acid. Name the mechanism and organic product, outline both electron-pair movements before protonation, explain the stereochemical composition and state the main KCN hazard.
[7 marks]
Total for this question: 7
A student draws nucleophilic addition of CN⁻ to pentan-3-one with an arrow from nitrogen to oxygen, leaves C=O unchanged and shows a neutral tetrahedral intermediate. Identify and correct the three errors, describe the electron-pair movements and name the final hydroxynitrile.
[7 marks]
Total for this question: 7
Ethanal is reduced using sodium tetradeuteriidoborate, NaBD₄, in ordinary water. Predict the organic product, describe every electron-pair movement through protonation, and explain why the product is CH₃CHDOH rather than CH₃CHDOD.
[5 marks]
Total for this question: 5
A $0.100\,\mathrm{mol}$ mixture contains ethanal and propanone. Excess warmed Tollens' reagent produces $5.40\,\mathrm g$ of silver. Calculate the mole percentage of each carbonyl compound and name the product formed by reducing each with aqueous NaBH₄. Use $A_{\mathrm r}(\mathrm{Ag})=108$ and two moles of Ag per mole of aldehyde.
[5 marks]
Total for this question: 5
Write the overall equation for the reaction of 2-methylpropanal with HCN and name the organic product. State the reagent combination used in the laboratory, explain why HCN gas is not used directly, and explain how those reagents supply the species needed.
[5 marks]
Total for this question: 5

3.3.9.1 · Carboxylic acids and esters (A-level only)

Explanation

Carboxylic acids are weak acids but liberate $\mathrm{CO_2}$ from carbonates. With alcohols and an acid catalyst they form esters in reversible condensation; esters are used as solvents, plasticisers, perfumes and flavourings.
Acid hydrolysis gives a carboxylic acid and alcohol, whereas alkaline hydrolysis gives a carboxylate salt and alcohol.
Vegetable oils and animal fats are triesters of propane-1,2,3-triol.
Their alkaline hydrolysis produces glycerol and soap, the salts of long-chain carboxylic acids.
Biodiesel is a mixture of long-chain methyl esters made by reacting vegetable oil with methanol in the presence of a catalyst.

Worked example

Name the ester formed from propan-1-ol and ethanoic acid and write its formation equation.

1.The alcohol supplies the propyl group and the acid supplies ethanoate.
2.Condensation removes water from the alcohol and acid.

Answer: Propyl ethanoate: $\mathrm{CH_3COOH+CH_3CH_2CH_2OH\rightleftharpoons CH_3COOCH_2CH_2CH_3+H_2O}$ .

Common mistakes

Don't name an ester from the wrong sides of the linkage, swapping alcohol-derived alkyl and acid-derived alkanoate.
Don't state that alkaline ester hydrolysis produces a carboxylic acid rather than its salt.
Don't call biodiesel glycerol instead of a mixture of long-chain methyl esters.

Exam tip

For ester reactions, identify the acyl and alkoxy sides before naming products or predicting hydrolysis.

Tier 1 · Easy

Name the ester made from propan-1-ol and ethanoic acid, state the catalyst and write the equation using condensed structures.
[3 marks]
Total for this question: 3
Methanol reacts with butanoic acid. A student names the ester butyl methanoate. Give the correct name and explain how the two parts of an ester name are assigned.
[3 marks]
Total for this question: 3

Tier 2 · Standard

Methyl propanoate is heated with excess aqueous sodium hydroxide. Name both organic products, write the equation and explain why the carboxylic-acid product is not isolated directly from this mixture.
[4 marks]
Total for this question: 4
Plan the preparation of propyl ethanoate from propan-1-ol and ethanoic acid. State the catalyst and explain why the mixture is first heated under reflux but the ester is later collected by distillation.
[5 marks]
Total for this question: 5
An esterification mixture is allowed to reach equilibrium. Predict and explain the effect on ester yield of (i) using excess alcohol, (ii) removing water as it forms and (iii) increasing catalyst concentration while keeping mixture composition and temperature fixed.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A triglyceride has $M_r=890$ . Calculate the minimum mass of NaOH needed to hydrolyse $17.8\,\text{g}$ completely, then state the two types of organic product. Use $M_r(\mathrm{NaOH})=40.0$ .
[5 marks]
Total for this question: 5
Hydrolysis of an ester produces compounds A and B. A has formula C₂H₆O and, after controlled oxidation and immediate distillation, gives a silver-mirror-positive product. B has formula C₃H₆O₂ and effervesces with sodium carbonate. Deduce A, B and the ester, and write an equation for acid hydrolysis.
[7 marks]
Total for this question: 7
An unnamed oil triester has $M_{\mathrm r}=884$ and reacts as one mole of triester with three moles of methanol to form one mole of glycerol and three moles of a methyl ester with $M_{\mathrm r}=296$ . From $44.2\,\mathrm g$ of triester, calculate the theoretical glycerol mass and the biodiesel mass recovered at 75.0% of its theoretical value. Use $M_{\mathrm r}(\text{glycerol})=92.0$ .
[6 marks]
Total for this question: 6
3-Methylbutyl ethanoate, CH₃COOCH₂CH₂CH(CH₃)₂, is used as a banana flavouring. Deduce and name the alcohol and the carboxylic acid needed, write the esterification equation, state one other common use of esters, and explain why the acid effervesces with aqueous sodium carbonate but the ester does not.
[6 marks]
Total for this question: 6

3.3.9.2 · Acylation (A-level only)

Explanation

Acid anhydrides, acyl chlorides and amides contain acyl-derived structures.
Acyl chlorides and acid anhydrides react with water, alcohols, ammonia and primary amines by nucleophilic addition–elimination, producing carboxylic acids, esters or amides as appropriate.
The nucleophile attacks the $\delta+$ acyl carbon, the C=O pair moves to oxygen, then C=O reforms as the leaving group departs; proton transfer completes the product.
Ethanoic anhydride is preferred to ethanoyl chloride for industrial aspirin manufacture because it is safer to handle and produces ethanoic acid rather than corrosive HCl.
Required practical 10 prepares and purifies a solid and a liquid and tests purity.

Worked example

Give the products when ethanoyl chloride reacts with excess ethylamine.

1.Ethylamine attacks the acyl carbon and chloride leaves after the tetrahedral intermediate.
2.A second ethylamine molecule accepts the released proton and HCl equivalent.

Answer: N-ethylethanamide and ethylammonium chloride form.

Common mistakes

Don't call the mechanism simple nucleophilic substitution and omit the tetrahedral addition intermediate.
Don't show ammonia or a primary amine producing an ester instead of an amide.
Don't claim ethanoyl chloride is industrially preferred for aspirin because it forms HCl.

Exam tip

For an acyl-chloride mechanism, show addition to C=O, re-formation of C=O, leaving-group departure and proton transfer.

Tier 1 · Easy

Ethanoyl chloride is added to ethanol. Name both products and state one visible observation.
[3 marks]
Total for this question: 3
A student states that propanoyl chloride reacts with methanol to form propyl methanoate and water. Identify the two errors and give the correct products, then state one observation.
[3 marks]
Total for this question: 3

Tier 2 · Standard

Propanoyl chloride reacts with excess methylamine. Name the amide and salt formed, and outline the addition-elimination steps.
[5 marks]
Total for this question: 5
Choose propanoic acid or propanoyl chloride to acylate propan-1-ol rapidly at room temperature. Name the ester, give the other product and explain the chemical reason for your choice.
[5 marks]
Total for this question: 5
Propanoic anhydride reacts with ethanol. Name both organic products, write a condensed equation and name the mechanism. State why HCl is not a product.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Aspirin manufacture can use ethanoic anhydride or ethanoyl chloride as the acylating agent. The desired aspirin has $M_r=180$ ; the anhydride route also makes ethanoic acid, $M_r=60.0$ , while the chloride route also makes HCl, $M_r=36.5$ . Calculate each route's atom economy and explain why industry can still prefer the anhydride route.
[6 marks]
Total for this question: 6
A proposed mechanism for butanoyl chloride reacting with ammonia shows an arrow from the carbonyl carbon to nitrogen, no movement of the C=O pi bond, a neutral oxygen in the tetrahedral intermediate and chlorine retained in the amide. Identify and correct these four errors and give the overall equation using excess ammonia.
[8 marks]
Total for this question: 8
Ethanoic anhydride reacts with excess methylamine. Name the amide and ionic coproduct, give the overall equation, and describe the electron-pair movements in the addition-elimination mechanism.
[6 marks]
Total for this question: 6
An acyl chloride with an unbranched carbon chain has a $2.13\,\mathrm g$ sample that reacts completely with excess dry ethanol. The HCl formed is swept into water and requires $25.0\,\mathrm{cm^3}$ of $0.800\,\mathrm{mol\,dm^{-3}}$ NaOH for neutralisation. Determine the acyl chloride and write the organic reaction equation. Assume one mole of acyl chloride forms one mole of HCl.
[5 marks]
Total for this question: 5
Ethanoic anhydride and ethanoyl chloride each react with water. Write the equation for the reaction of the anhydride, and compare the rate and the observations with those of ethanoyl chloride.
[5 marks]
Total for this question: 5

3.3.10.1 · Bonding (A-level only)

Explanation

Benzene is planar and all six C–C bonds have equal length intermediate between ordinary single and double bonds.
Each carbon contributes a p electron; sideways overlap around the ring produces delocalised electron density above and below the plane.
Thermochemical evidence confirms extra stability: a theoretical cyclohexa-1,3,5-triene with three isolated C=C bonds would release three times cyclohexene's hydrogenation enthalpy, but actual benzene hydrogenation is substantially less exothermic.
The difference is delocalisation stability.
Benzene therefore undergoes substitution in preference to addition because substitution restores the delocalised ring, whereas addition would permanently disrupt it.

Benzene's p orbitals form delocalised electron density above and below a planar ring of equal bonds.

Worked example

Cyclohexene hydrogenation is $-120\,\mathrm{kJ\,mol^{-1}}$ , while benzene hydrogenation is $-208\,\mathrm{kJ\,mol^{-1}}$ . Calculate the stability difference from a three-localised-double-bond model.

1.Predicted value for three isolated C=C bonds $=3(-120)=-360\,\mathrm{kJ\,mol^{-1}}$ .
2.Compare magnitudes: $360-208=152\,\mathrm{kJ\,mol^{-1}}$ .

Answer: Benzene is $152\,\mathrm{kJ\,mol^{-1}}$ more stable than the localised model.

Common mistakes

Don't describe benzene as rapidly alternating single and double bonds instead of six equal delocalised bonds.
Don't use bond length alone but omit thermochemical evidence for extra stability.
Don't say addition is preferred because benzene contains three ordinary C=C bonds.

Exam tip

For thermochemical evidence, compare actual hydrogenation with three times the cyclohexene value and interpret the less-exothermic result.

Tier 1 · Easy

Describe the shape, carbon-carbon bond lengths and pi bonding in a benzene molecule.
[3 marks]
Total for this question: 3
A model of benzene shows p orbitals on only five of its six carbon atoms. Identify the error and state how the pi system is actually arranged.
[3 marks]
Total for this question: 3

Tier 2 · Standard

Hydrogenation of cyclohexene has $\Delta H=-121\,\text{kJ mol}^{-1}$ , while hydrogenation of benzene to cyclohexane has $\Delta H=-207\,\text{kJ mol}^{-1}$ . Calculate the extra stability of benzene relative to a ring with three isolated C=C bonds.
[3 marks]
Total for this question: 3
Cyclohexene rapidly decolourises bromine water, but benzene does not under the same conditions. Explain the different observations using their pi bonding and the relative stability of the products that addition would form.
[5 marks]
Total for this question: 5
State the numbers of carbon-carbon sigma bonds, carbon-hydrogen sigma bonds and delocalised pi electrons in one methylbenzene molecule. Explain why it has six delocalised pi electrons rather than eight.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A student claims benzene is cyclohexa-1,3,5-triene with fixed alternating bonds. Evaluate the claim using measured C-C bond lengths of $0.140\,\text{nm}$ in benzene, $0.154\,\text{nm}$ for a C-C bond and $0.134\,\text{nm}$ for a C=C bond, then use delocalisation to explain why substitution is preferred to addition.
[6 marks]
Total for this question: 6
Hydrogenation of cyclohexene has $\Delta H=-119\,\text{kJ mol}^{-1}$ and hydrogenation of benzene to cyclohexane has $\Delta H=-208\,\text{kJ mol}^{-1}$ . A student calculates $3(-119)-(-208)=-149$ and concludes that benzene is 149 kJ mol^-1 less stable than a ring with three localised C=C bonds. Identify the interpretation error and state the correct stability conclusion.
[5 marks]
Total for this question: 5
Use $\Delta_{\mathrm f}H^\circ(\text{benzene})=+49\,\mathrm{kJ\,mol^{-1}}$ , $\Delta_{\mathrm f}H^\circ(\text{cyclohexane})=-156\,\mathrm{kJ\,mol^{-1}}$ and $\Delta_{\mathrm f}H^\circ(\mathrm{H_2})=0$ to calculate the enthalpy of hydrogenating benzene to cyclohexane. A localised C=C bond releases $117\,\mathrm{kJ\,mol^{-1}}$ on hydrogenation. Use both results to calculate and explain benzene's stability relative to a three-localised-double-bond model.
[5 marks]
Total for this question: 5

3.3.10.2 · Electrophilic substitution (A-level only)

Explanation

Electrophilic attack on benzene gives monosubstitution. Ring electrons attack the electrophile to form a positive sigma complex; loss of $\mathrm{H^+}$ returns the C–H electron pair to the ring and restores delocalisation.
Nitration uses concentrated nitric and sulfuric acids. Sulfuric acid generates the nitronium ion: $\mathrm{HNO_3+H_2SO_4\rightarrow NO_2^++HSO_4^-+H_2O}$ .
Nitration is important in explosive manufacture and routes to amines.
Friedel–Crafts acylation uses an acyl chloride and anhydrous $\mathrm{AlCl_3}$ to generate an acylium electrophile, $\mathrm{RCO^+}$ , forming an aromatic ketone.
Both mechanisms must show electrophile generation and delocalisation restoration.

Worked example

State the electrophile and its formation equation for nitration of benzene.

1.Concentrated sulfuric acid protonates nitric acid and promotes water loss.
2.The attacking electrophile is the nitronium ion.

Answer: $\mathrm{NO_2^+}$ ; $\mathrm{HNO_3+H_2SO_4\rightarrow NO_2^++HSO_4^-+H_2O}$ .

Common mistakes

Don't start the first curly arrow at $\mathrm{NO_2^+}$ instead of at the benzene electron pair.
Don't omit the positive sigma complex and jump directly to nitrobenzene.
Don't use aqueous $\mathrm{AlCl_3}$ in Friedel–Crafts acylation instead of anhydrous catalyst.

Exam tip

For electrophilic substitution, show electrophile generation, sigma-complex charge and the final arrow that restores ring delocalisation.

Tier 1 · Easy

State the two reagents for nitrating benzene, name the catalyst and identify the electrophile.
[3 marks]
Total for this question: 3
In nitration of benzene, a student draws the first curly arrow from NO₂⁺ towards the ring. State the correct arrow origin and destination, and explain why.
[3 marks]
Total for this question: 3

Tier 2 · Standard

Benzene reacts with propanoyl chloride in the presence of anhydrous AlCl₃. Name the organic product, write the overall equation and identify the attacking electrophile.
[4 marks]
Total for this question: 4
Benzene reacts with 2-methylpropanoyl chloride and anhydrous AlCl₃. Write the electrophile-forming equation, identify the electrophile, name the organic product and name the reaction type.
[5 marks]
Total for this question: 5
Benzene is nitrated using concentrated nitric and sulfuric acids. 7.80 g of benzene gives 9.84 g of nitrobenzene. Write the equation for the substitution, calculate the percentage yield and suggest one reason it is below 100%.
[5 marks]
Total for this question: 5

Tier 3 · Hard

Outline the complete electrophilic-substitution mechanism for reacting benzene with ethanoyl chloride and anhydrous AlCl₃. Include electrophile generation, both ring electron movements and catalyst regeneration.
[6 marks]
Total for this question: 6
A proposed nitration mechanism uses HNO₃ as the attacking electrophile, draws the first arrow from that species to benzene, shows a neutral ring intermediate and draws the final arrow from the ring towards H. Identify and correct all four errors and state how sulfuric acid is regenerated.
[8 marks]
Total for this question: 8
A Friedel-Crafts acylation using propanoyl chloride gives mainly unreacted benzene and propanoic acid. The student used hydrated AlCl₃ and wet glassware. Explain the two chemical causes, state the practical corrections, and write the equation that should generate the electrophile.
[5 marks]
Total for this question: 5
Explain why nitration is important in organic synthesis; state how the nitro group can be reduced to give phenylamine; and state what is meant by monosubstitution of benzene, giving the product of nitrating benzene once. State another major application of nitration.
[5 marks]
Total for this question: 5

3.3.11.1 · Preparation (A-level only)

Explanation

Primary aliphatic amines can be prepared by heating a halogenoalkane with excess ethanolic ammonia or by reducing a nitrile. Excess ammonia favours primary amine because the product is itself nucleophilic and can otherwise undergo further alkylation.
Direct substitution leaves the carbon count unchanged.
Cyanide substitution followed by nitrile reduction retains the nitrile carbon, so the final amine has one additional carbon relative to the starting halogenoalkane.
Nitriles may be reduced with hydrogen and nickel or $\mathrm{LiAlH_4}$ in dry ether.
Aromatic amines are prepared by reducing nitro compounds and are used in dye manufacture.

Worked example

Give a two-step route from 1-bromopropane to butan-1-amine and explain the carbon-count change.

1.React 1-bromopropane with $\mathrm{CN^-}$ to form butanenitrile.
2.Reduce the nitrile with $\mathrm{H_2/Ni}$ or $\mathrm{LiAlH_4}$ in dry ether.

Answer: Butan-1-amine forms; the cyanide carbon adds one carbon to the chain.

Common mistakes

Don't state that direct reaction with ammonia lengthens the carbon chain by one.
Don't use excess halogenoalkane when aiming to maximise primary amine yield.
Don't reduce a nitro compound but name the aromatic product as an amide instead of an amine.

Exam tip

For an amine-synthesis route, count carbons after each step and state why excess ammonia limits further substitution.

Tier 1 · Easy

Name a reagent and a condition used to convert 1-bromopropane into propylamine while limiting further substitution.
[2 marks]
Total for this question: 2
1-Bromopropane is heated separately (i) with excess ethanolic ammonia and (ii) with ethanolic potassium cyanide, followed by reduction of the nitrile. Name the primary amine formed by each route.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A two-stage route changes bromoethane into propylamine. Give the reagent and condition for each stage and name the intermediate.
[5 marks]
Total for this question: 5
Two routes are proposed for making butan-1-amine. Route A heats 1-bromobutane with excess ethanolic ammonia. Route B heats 1-bromopropane with ethanolic potassium cyanide and then reduces the nitrile. Explain why Route B is more selective for a primary amine, but requires an additional reaction step.
[4 marks]
Total for this question: 4
A sample contains $0.0250\,\mathrm{mol}$ of 1,4-dinitrobenzene. Both nitro groups are reduced completely to benzene-1,4-diamine. Write an equation using $[\mathrm{H}]$ and calculate the amount of $[\mathrm{H}]$ required.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Describe how nitrobenzene can be converted into a pure sample of phenylamine. Include the purpose of the final alkaline treatment.
[5 marks]
Total for this question: 5
A two-step synthesis of butan-1-amine from 1-bromopropane has yields of 80.0% for nitrile formation and 84.0% for nitrile reduction. Calculate the overall percentage yield and the amount, in moles, of butan-1-amine obtained from 0.300 mol of 1-bromopropane. A direct route gives 0.150 mol of butan-1-amine from 0.300 mol of 1-bromobutane. Compare the product amounts and state one selectivity advantage of the nitrile route.
[5 marks]
Total for this question: 5
Bromoethane is heated with potassium cyanide in which the carbon is ¹³C (K¹³CN) and the nitrile product is then reduced with $\mathrm{H_2/Ni}$ . Name the intermediate and final product, locate the ¹³C atom in the final product, and predict its molecular-ion $m/z$ .
[4 marks]
Total for this question: 4
$0.0750\,\mathrm{mol}$ of butanenitrile is reduced using hydrogen over a nickel catalyst. Only 0.200 g of hydrogen is available. Use $M_r(\mathrm{H_2})=2.00$ . Write the reduction equation, identify the limiting reagent, and calculate the maximum mass of butan-1-amine formed. Use $M_r(\text{butan-1-amine})=73.0$ .
[5 marks]
Total for this question: 5
A mixture contains butanenitrile and butan-1-amine only. Its total amount is $0.150\,\mathrm{mol}$ . Complete reduction increases the sample mass by $0.360\,\mathrm{g}$ because each mole of nitrile gains four hydrogen atoms. Determine the initial amounts of both compounds and explain why the carbon skeleton is unchanged. Use $A_r(\mathrm{H})=1.00$ .
[4 marks]
Total for this question: 4

3.3.11.2 · Base properties (A-level only)

Explanation

Amines are weak Brønsted–Lowry bases because the nitrogen lone pair accepts a proton, forming an alkylammonium or arylammonium ion.
Base strength depends on lone-pair availability.
Alkyl groups push electron density towards nitrogen by the positive inductive effect, making a primary aliphatic amine's lone pair more available than ammonia's and strengthening proton acceptance.
In phenylamine, the nitrogen lone pair is delocalised into the benzene ring, so it is less available to form a bond to $\mathrm{H^+}$ and phenylamine is weaker than ammonia.
The comparison concerns electron-pair availability, not N–H bond strength.

Worked example

Order methylamine, ammonia and phenylamine by decreasing base strength and explain.

1.The methyl group pushes electron density towards nitrogen, increasing lone-pair availability.
2.Phenylamine delocalises the nitrogen lone pair into the benzene ring, decreasing availability.

Answer: Methylamine $>$ ammonia $>$ phenylamine.

Common mistakes

Don't explain amine basicity by breaking an N–H bond instead of accepting a proton at the lone pair.
Don't claim delocalisation in phenylamine makes its lone pair more available.
Don't state that alkyl groups withdraw electron density from nitrogen.

Exam tip

For a base-strength comparison, state how each substituent changes nitrogen lone-pair availability.

Tier 1 · Easy

Write an equation showing ethylamine acting as a base in water.
[2 marks]
Total for this question: 2
State the Brønsted–Lowry definition of a base and write an equation for phenylamine reacting with hydrochloric acid.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Place ethylamine, ammonia and phenylamine in decreasing order of base strength. Explain both differences in the order.
[4 marks]
Total for this question: 4
Equal-concentration aqueous solutions of methylamine, ammonia and phenylamine have pH values 11.8, 11.1 and 8.7 respectively. Deduce which pH value corresponds to each base and explain the order of base strength.
[4 marks]
Total for this question: 4
Equal-concentration solutions of propylamine and ammonia have pH values 12.05 and 11.35 respectively at the same temperature. Calculate the ratio $[\mathrm{OH^-}]_{\text{propylamine}}/[\mathrm{OH^-}]_{\text{ammonia}}$ and explain the difference.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A student claims that phenylamine should be the strongest base because its nitrogen atom is attached to a large electron-rich ring. Evaluate this claim and predict which of phenylamine and propylamine forms the greater concentration of hydroxide ions in equally concentrated aqueous solutions.
[5 marks]
Total for this question: 5
Place phenylmethylamine (C₆H₅CH₂NH₂), phenylamine and ammonia in decreasing order of base strength. Explain the effect of the CH₂ group in phenylmethylamine and the benzene ring in phenylamine on the availability of the nitrogen lone pair.
[4 marks]
Total for this question: 4
Predict the favoured direction of the proton-transfer equilibrium C₂H₅NH₂ + C₆H₅NH₃⁺ ⇌ C₂H₅NH₃⁺ + C₆H₅NH₂. Explain your answer using lone-pair availability.
[4 marks]
Total for this question: 4
Phenylamine and 4-nitrophenylamine are both primary aromatic amines. Predict which is the stronger base and explain how the 4-nitro substituent changes the availability of the amine nitrogen lone pair through the benzene ring.
[4 marks]
Total for this question: 4

3.3.11.3 · Nucleophilic properties (A-level only)

Explanation

Amines are nucleophiles because the nitrogen lone pair bonds to electron-deficient carbon. With halogenoalkanes, ammonia and amines undergo nucleophilic substitution; successive alkylations form primary, secondary and tertiary amines and finally quaternary ammonium salts.
Excess ammonia favours primary amine, whereas excess halogenoalkane favours further alkylation.
Quaternary ammonium salts with a charged hydrophilic head and long hydrophobic groups act as cationic surfactants.
Ammonia and primary amines also react with acyl chlorides and acid anhydrides by nucleophilic addition–elimination.
The mechanism attacks the acyl carbon, forms a tetrahedral intermediate, eliminates the leaving group and transfers a proton.

Worked example

State the products when ethanoyl chloride reacts with excess ammonia.

1.Ammonia attacks the acyl carbon and chloride leaves after the tetrahedral intermediate.
2.A second ammonia molecule accepts the proton and neutralises HCl.

Answer: Ethanamide and ammonium chloride form.

Common mistakes

Don't start the nucleophilic curly arrow at the electron-deficient carbon instead of at the nitrogen lone pair.
Don't assume one halogenoalkane reaction can form only a primary amine and ignore successive alkylation.
Don't omit the second ammonia or amine molecule needed to accept the proton in acylation.

Exam tip

For an amine mechanism, show the nitrogen lone pair and distinguish substitution at saturated carbon from addition–elimination at acyl carbon.

Tier 1 · Easy

State the feature of an amine molecule that allows it to act as a nucleophile.
[1 mark]
Total for this question: 1
Give the structure of the amide formed when butylamine reacts with ethanoic anhydride.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Ethylamine reacts with ethanoyl chloride. Name the organic product, state the mechanism type and describe the two essential curly arrows in the addition step.
[4 marks]
Total for this question: 4
Write an equation for ethanoyl chloride reacting with excess methylamine to form N-methylethanamide. Explain the role of the second methylamine molecule.
[4 marks]
Total for this question: 4
Write equations for ethylamine reacting separately with ethanoyl chloride and with ethanoic anhydride to form N-ethylethanamide. Identify the other product in each equation and explain why the acyl-chloride equation uses two ethylamine molecules.
[4 marks]
Total for this question: 4

Tier 3 · Hard

An excess of 1-bromobutane is heated with butylamine. Predict the final nitrogen-containing product, explain why several substitution stages can occur, and relate one structural feature of the final ion to its use in a cationic surfactant.
[5 marks]
Total for this question: 5
A student draws the addition step for ethylamine reacting with propanoyl chloride using a curly arrow from the carbonyl carbon to nitrogen and leaves the C=O bond unchanged. The student names the final product ethylpropanamide. Identify each error and give the correct arrow origins and the correct name.
[5 marks]
Total for this question: 5
Propanoyl chloride is mixed with $0.0900\,\mathrm{mol}$ of ethylamine. Initially there are $0.0400\,\mathrm{mol}$ of propanoyl chloride. Using the overall 1:2 stoichiometry, calculate the maximum amount of amide formed and the amount of ethylamine left. Name both nitrogen-containing products.
[5 marks]
Total for this question: 5
Draw the nucleophilic-substitution mechanism for methylamine reacting with bromoethane to form ethylmethylamine. Include formation of the ionic intermediate and its reaction with a second methylamine molecule.
[5 marks]
Total for this question: 5
$6.12\,\mathrm{g}$ of ethanoic anhydride is mixed with $2.95\,\mathrm{g}$ of propylamine. Determine the limiting reagent and the maximum mass of amide formed, name both organic products, and explain why the amide retains one N–H bond. Use $M_r$ values: ethanoic anhydride = 102.0, propylamine = 59.0 and amide = 101.0.
[4 marks]
Total for this question: 4

3.3.12.1 · Condensation polymers (A-level only)

Explanation

Condensation polymers form when bifunctional monomers join and eliminate small molecules.
Dicarboxylic acids with diols form polyesters; dicarboxylic acids with diamines form polyamides such as nylon 6,6 or Kevlar; amino acids also form polyamides.
Repeat units retain ester $\mathrm{-COO-}$ or amide $\mathrm{-CONH-}$ linkages and bracket bonds pass through the chain.
Reverse deduction splits linkages and restores the monomer end groups.
Polyamides form hydrogen bonds between N–H and C=O groups, while polyesters have permanent dipole attractions; these intermolecular forces help explain strong fibres, fabrics and bottles without being covalent cross-links.

Condensation of a dicarboxylic acid and diamine forms a polyamide with recurring amide links.

Worked example

State the polymer type and linkage formed from a dicarboxylic acid and a diamine.

1.Each carboxyl group reacts with an amino group and eliminates water.
2.The repeating chain contains $\mathrm{-CO-NH-}$ links.

Answer: A polyamide containing amide links forms.

Common mistakes

Don't draw an addition-polymer C–C backbone and omit the condensation linkage.
Don't recover monomers from an amide link without restoring $\mathrm{-OH}$ and H end groups.
Don't call hydrogen bonds between polyamide chains covalent cross-links.

Exam tip

For a repeat-unit question, identify both bifunctional ends and place bracket bonds through the continuing polymer chain.

Tier 1 · Easy

Name the linkage formed when a dicarboxylic acid reacts with a diamine, and name the small molecule eliminated when the acid itself is used.
[2 marks]
Total for this question: 2
The molecule HOCH₂CH₂COOH can form a condensation polymer without a second monomer. Name the polymer type, the linkage formed and the small molecule eliminated.
[3 marks]
Total for this question: 3

Tier 2 · Standard

Hexane-1,6-diamine reacts with hexanedioic acid. Give the condensed formula of the polyamide repeat unit and identify the strongest intermolecular force between its chains.
[4 marks]
Total for this question: 4
A polyamide has repeat unit [–NH–(CH₂)₄–NH–CO–CH₂–CO–]_n. Deduce the two monomers used when the polymer is made from a diamine and a dicarboxylic acid.
[4 marks]
Total for this question: 4
Four diol molecules and three dicarboxylic acid molecules form one linear alternating chain, with a diol molecule at each end. Determine the number of ester links and water molecules formed, and identify the functional group at each chain end.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A polymer contains the repeating segment [–O–CH₂CH₂–O–CO–C₆H₄–CO–]_n, with the two carbonyl groups bonded at positions 1 and 4 of the benzene ring. Deduce both monomers, classify the polymer, and explain why its chains attract one another even though it has no N–H bonds.
[6 marks]
Total for this question: 6
A polyester repeat of formula C₈H₁₂O₄ can be made from ethane-1,2-diol and either hexanedioic acid or hexanedioyl dichloride. Each repeat-forming event eliminates two H₂O molecules from the acid route or two HCl molecules from the acyl chloride route. Using A_r values H = 1.0, C = 12.0, O = 16.0 and Cl = 35.5, calculate the atom economy of each route and evaluate what the values do and do not show.
[6 marks]
Total for this question: 6
Polymer A is an aromatic polyamide with repeated benzene rings and –CONH– links. Polymer B is an aliphatic polyester with flexible –CH₂– chains and –COO– links. Other factors are comparable. Predict which polymer better resists stretching and thermal softening, and explain using chain shape and intermolecular forces.
[5 marks]
Total for this question: 5
A mixture of hexane-1,6-diamine and hexanedioic acid contains a small amount of propylamine. Explain how propylamine can become incorporated at a polymer-chain end, why it stops growth at that end, and why hexane-1,6-diamine does not stop growth after only one reaction.
[4 marks]
Total for this question: 4
A section of a polyamide chain is …–NH–(CH₂)₃–NH–CO–(CH₂)₅–CO–NH–(CH₂)₃–NH–CO–(CH₂)₅–CO–… . Write the smallest repeat unit in brackets in two equivalent ways, identify the polymer class, and explain why the two bracket choices represent the same chain.
[5 marks]
Total for this question: 5

3.3.12.2 · Biodegradability and disposal of polymers (A-level only)

Explanation

Polyalkenes have strong, non-polar C–C and C–H backbones with no readily hydrolysable links, so they are chemically inert and non-biodegradable. Polyesters and polyamides contain ester or amide bonds that can be hydrolysed, shortening chains and allowing biodegradation under suitable conditions.
Disposal choices have trade-offs. Mechanical recycling conserves feedstock and reduces landfill but requires collection, sorting and cleaning and may lower material quality.
Feedstock recycling can recover chemicals but consumes energy.
Incineration reduces waste volume and may recover energy, but releases carbon dioxide and can release toxic gases without effective controls.
Landfill occupies space and leaves persistent waste.

Worked example

Explain why a polyester can be hydrolysed but poly(ethene) cannot.

1.Polyester chains contain polar ester links susceptible to hydrolysis.
2.Poly(ethene) contains only strong non-polar C–C and C–H bonds.

Answer: Hydrolysis cleaves polyester links but has no corresponding functional group in poly(ethene).

Common mistakes

Don't claim every condensation polymer is automatically biodegradable under all conditions.
Don't say polyalkenes hydrolyse because water can attack their C–C backbone.
Don't present incineration as pollution-free because energy can be recovered.

Exam tip

For a disposal evaluation, compare at least two methods using resource use, emissions and persistence rather than listing one benefit.

Tier 1 · Easy

Explain why a polyalkene chain is much less susceptible to hydrolysis than a polyester chain.
[2 marks]
Total for this question: 2
An addition polymer has repeat unit [–CH₂–C(CH₃)(COOCH₃)–]_n. Explain why hydrolysis of its ester side groups does not split the main polymer chain.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A disposable article can be made from a polyamide or from poly(propene). Compare their likely biodegradability using their chain structures.
[4 marks]
Total for this question: 4
A polyester made only from 2-hydroxypropanoic acid has repeat unit [–O–CH(CH₃)–CO–]_n. Deduce the organic product of complete acidic hydrolysis and explain why the polymer chains become shorter.
[3 marks]
Total for this question: 3
A polyamide has repeat unit [–NH–(CH₂)₃–NH–CO–(CH₂)₂–CO–]_n. Give the two organic products of complete hydrolysis with hot aqueous sodium hydroxide and explain why the polymer chain is split.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A council is choosing between landfill, mechanical recycling and energy-recovery incineration for mixed polymer waste. Evaluate the three options and justify why sorting the waste can change the best choice.
[6 marks]
Total for this question: 6
A supplier claims that an ester-linked compostable polymer is always environmentally preferable to a recyclable polyalkene. Evaluate the claim for industrial composting, landfill and a clean recycling stream.
[6 marks]
Total for this question: 6
A polyester has repeat-unit $M_r=192$ . Complete hydrolysis of each repeat section can produce one mole of benzene-1,4-dicarboxylic acid, $M_r=166$ , and one mole of ethane-1,2-diol. A chemical-recycling process treats $9.60\,\mathrm{g}$ of polymer and recovers 78.0% of the theoretical monomer amounts. Calculate the amount of each monomer recovered and the mass of the acid. State one limitation of judging the process from this yield alone.
[6 marks]
Total for this question: 6
Complete incineration of poly(ethene), represented by C₂H₄, and a polyester, represented by C₃H₄O₂, forms carbon dioxide and water. Write the equation for each repeat formula, calculate the mass of CO₂ formed from $1.00\,\mathrm{kg}$ of each polymer, and evaluate the claim that incinerating the polyester is automatically the greener disposal choice. Use $A_r$ values H = 1.0, C = 12.0 and O = 16.0.
[5 marks]
Total for this question: 5
A chemical-recycling plant hydrolyses $24.0\,\mathrm{kg}$ of a polyester. It recovers $17.6\,\mathrm{kg}$ of reusable monomers and uses $38.0\,\mathrm{kWh}$ of energy. Mechanical recycling of the same mass gives $20.4\,\mathrm{kg}$ of usable polymer and uses $11.5\,\mathrm{kWh}$ , but the product has lower quality. Calculate both material-recovery percentages and evaluate which process is preferable.
[6 marks]
Total for this question: 6

3.3.13.1 · Amino acids (A-level only)

Explanation

Amino acids contain both an acidic carboxyl group and a basic amino group. Internal proton transfer forms a zwitterion with $\mathrm{-NH_3^+}$ and $\mathrm{-COO^-}$ groups but zero overall charge.
In acid solution, the amino acid is protonated overall and the dominant ion contains $\mathrm{-NH_3^+}$ and $\mathrm{-COOH}$ .
In alkaline solution, deprotonation gives $\mathrm{-NH_2}$ and $\mathrm{-COO^-}$ .
Structures must preserve the side chain and show every charge.
Acidic and basic properties follow from proton donation by the carboxyl group and proton acceptance by the nitrogen lone pair.

Amino-acid charge changes with conditions while the carbon skeleton and side chain remain unchanged.

Worked example

Draw or state the forms of glycine in acid and alkaline solution.

1.In acid, protonate the amino group and retain $\mathrm{-COOH}$ .
2.In alkali, retain $\mathrm{-NH_2}$ and deprotonate the carboxyl group.

Answer: Acid: $\mathrm{H_3N^+CH_2COOH}$ ; alkali: $\mathrm{H_2NCH_2COO^-}$ .

Common mistakes

Don't draw neutral $\mathrm{-NH_2}$ and $\mathrm{-COOH}$ when a zwitterion is requested.
Don't give the acid-solution form a negative carboxylate charge.
Don't change the amino-acid side chain while adding or removing protons.

Exam tip

For amino-acid ions, check the overall charge after changing the protonation state of nitrogen and oxygen groups.

Tier 1 · Easy

Define the term zwitterion and state its overall charge.
[2 marks]
Total for this question: 2
Name the two functional groups in 2-aminopropanoic acid and use them to explain why the molecule is amphoteric.
[3 marks]
Total for this question: 3

Tier 2 · Standard

Give the displayed ionic forms of alanine, CH₃CH(NH₂)COOH, in strongly acidic solution and in strongly alkaline solution.
[4 marks]
Total for this question: 4
An unbranched α-amino acid has molecular formula C₄H₉NO₂ and one chiral centre. Deduce its structure, give its systematic name and write its zwitterionic form.
[4 marks]
Total for this question: 4
Separate samples each contain $0.0350\,\mathrm{mol}$ of an amino-acid zwitterion with one –NH₃⁺ group and one –COO⁻ group. Calculate the amount of HCl needed to protonate one sample fully and the amount of NaOH needed to deprotonate the other fully. Identify the group reacting in each case.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A solution contains the zwitterion of 2-aminobutanoic acid. Write net ionic equations for its separate reactions with H⁺ and OH⁻, and identify the role of the zwitterion in each reaction.
[6 marks]
Total for this question: 6
A crystalline amino acid has a much higher melting temperature than a similar-sized amine or carboxylic acid and dissolves in water but not in hexane. Explain both observations using the structure present in the crystal.
[5 marks]
Total for this question: 5
An amino acid is fully protonated as HOOCCH₂CH(NH₃⁺)COOH. Calculate the amount of hydroxide ions needed to convert $0.0125\,\mathrm{mol}$ completely into its most deprotonated form. Give that final structure and its overall charge.
[5 marks]
Total for this question: 5
A mixture contains $0.0500\,\mathrm{mol}$ in total of glycine zwitterions, H₃N⁺CH₂COO⁻, and diamino-acid zwitterions, H₃N⁺(CH₂)₂CH(NH₂)COO⁻. Complete conversion to the ions present in excess strong acid uses $0.0800\,\mathrm{mol}$ of H⁺. Determine the initial amount of each amino acid.
[4 marks]
Total for this question: 4
A lysine-type diamino acid has neutral formula H₂N(CH₂)₄CH(NH₂)COOH. Write a zwitterionic form and the predominant ions in excess strong acid and excess strong alkali. State the overall charge of each species.
[5 marks]
Total for this question: 5

3.3.13.2 · Proteins (A-level only)

Explanation

Proteins are amino-acid sequences joined by peptide $\mathrm{-CONH-}$ links.
Primary structure is sequence; secondary structure includes $\alpha$ -helices and $\beta$ -pleated sheets maintained by hydrogen bonding; tertiary structure is overall three-dimensional folding maintained by interactions including hydrogen bonds and sulfur–sulfur bonds.
Students must draw peptides from up to three amino acids and recover amino-acid structures by peptide hydrolysis.
Amino acids are separated by thin-layer chromatography, located using ninhydrin or ultraviolet light and identified by $R_f$ , calculated as distance moved by the spot centre divided by distance moved by the solvent front from the same origin.

Protein structure progresses from amino-acid sequence through local secondary structure to overall tertiary folding.

Worked example

An amino-acid spot moves $3.6\,\mathrm{cm}$ while the solvent front moves $6.0\,\mathrm{cm}$ . Calculate $R_f$ .

1.Use distances measured from the same baseline.
2. $R_f=3.6/6.0$ .

Answer: $R_f=0.60$ with no units.

Common mistakes

Don't call amino-acid sequence the secondary structure instead of primary structure.
Don't state that peptide hydrolysis breaks sulfur–sulfur bonds rather than peptide links.
Don't measure $R_f$ from the plate edge or divide solvent distance by spot distance.

Exam tip

For a protein-structure diagram, identify sequence, local helix or sheet and overall fold before naming the stabilising interaction.

Tier 1 · Easy

State what determines the primary structure of a protein.
[1 mark]
Total for this question: 1
State the interaction that stabilises an α-helix in a protein and identify the two peptide-link groups involved.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Glycine reacts with alanine so that the carboxyl group of glycine bonds to the amino group of alanine. Give the condensed structure of the dipeptide, name the new linkage and state the other product.
[4 marks]
Total for this question: 4
One mole of the tetrapeptide Gly–Ala–Gly–Cys is completely hydrolysed. State the amount of water used and the amounts of each amino acid formed.
[4 marks]
Total for this question: 4
Complete hydrolysis of a tripeptide gives glycine, alanine and serine. Partial hydrolysis produces the ordered dipeptide fragments Gly–Ala and Ala–Ser. Deduce the tripeptide sequence and give its condensed structural formula.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A tripeptide is hydrolysed completely to glycine, cysteine and alanine. Explain what happens to its peptide links, state two interactions that can maintain tertiary structure in proteins, and calculate the $R_f$ of an amino-acid spot that moves $38\,\text{mm}$ when the solvent front moves $64\,\text{mm}$ .
[6 marks]
Total for this question: 6
The peptide H₂NCH(CH₃)CONHCH₂CONHCH(CH₂SH)COOH is completely hydrolysed. Deduce the amino-acid products, state how many peptide links are broken per peptide molecule, and explain how cysteine residues can stabilise a protein's tertiary structure.
[6 marks]
Total for this question: 6
A protein contains α-helical regions and sulfur–sulfur bridges. Predict the structural levels directly affected by (i) heating that disrupts hydrogen bonds, (ii) a reducing agent that breaks S–S bonds, and (iii) complete hydrolysis of peptide links. Explain why the primary structure survives the first two treatments but not the third.
[6 marks]
Total for this question: 6
A hydrolysed protein gives three spots on a TLC plate. The spot distances are $2.16$ , $3.82$ and $5.40\,\mathrm{cm}$ and the solvent front travels $7.20\,\mathrm{cm}$ . Standards run on the same plate have $R_f$ values glycine 0.30, alanine 0.53 and leucine 0.75. Calculate the sample values and identify the amino acids, stating one limitation of the identifications. Calculate the distance a spot of $R_f$ 0.42 would travel on this plate.
[4 marks]
Total for this question: 4
Oxidation converts pairs of cysteine side-chain –SH groups into sulfur–sulfur bridges and removes two hydrogen atoms per bridge. A protein sample loses $3.60\,\mathrm{mg}$ during this oxidation. Calculate the amount of S–S bridges formed and explain which levels of protein structure are and are not changed by bridge formation. Use $A_r(\mathrm{H})=1.00$ .
[5 marks]
Total for this question: 5

3.3.13.3 · Enzymes (A-level only)

Explanation

Enzymes are protein catalysts whose tertiary folding creates an active site with a specific three-dimensional shape and arrangement of functional groups.
The active site is stereospecific: only the substrate or drug enantiomer with the correct spatial arrangement can make the required interactions and bind effectively.
A drug may inhibit an enzyme by occupying and blocking the active site, preventing substrate binding and lowering reaction rate.
Computer modelling helps compare candidate shapes and interactions before compounds are synthesised, supporting drug design.
As catalysts, enzymes provide a lower-activation-energy route and change reaction rate; they do not change the equilibrium position.

A stereospecific active site binds only the enantiomer with the matching three-dimensional group arrangement.

Worked example

Explain why only one enantiomer of a drug may inhibit an enzyme effectively.

1.The active site has a fixed, asymmetric three-dimensional arrangement.
2.Only one enantiomer positions its groups to make all required interactions.
3.The mirror-image enantiomer cannot bind in the same way.

Answer: The active site is stereospecific.

Common mistakes

Don't say both enantiomers bind equally because they have the same structural formula.
Don't claim an inhibitor increases substrate binding while occupying the active site.
Don't state that an enzyme changes equilibrium yield rather than reaction rate.

Exam tip

For stereospecificity, link three-dimensional complementarity to the different spatial group arrangement in the two enantiomers.

Tier 1 · Easy

State why an enzyme is described as a catalyst.
[2 marks]
Total for this question: 2
Explain what is meant by saying that an enzyme active site is complementary to its substrate.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Explain why one enantiomer of a chiral substrate can bind to an enzyme active site much more strongly than the other enantiomer.
[4 marks]
Total for this question: 4
A drug molecule that binds temporarily in the active site has a similar shape and arrangement of polar groups to an enzyme's substrate. Explain how it reduces the reaction rate without changing the equilibrium position.
[4 marks]
Total for this question: 4
An enzyme is tested with and without a reversible inhibitor. At low substrate concentration the rates are 12.0 and $4.0\,\mathrm{mmol\,min^{-1}}$ respectively; at high substrate concentration they are 48.0 and $42.0\,\mathrm{mmol\,min^{-1}}$ . Explain how these results support competition for the active site.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A proposed drug resembles the transition-state shape of an enzyme's normal substrate. Explain how it could inhibit the enzyme, why stereochemistry must be considered, and how computer modelling can reduce the number of compounds synthesised.
[6 marks]
Total for this question: 6
Computer modelling predicts that one enantiomer of a candidate drug binds strongly to an enzyme active site. Evaluate why this result supports selecting that enantiomer for synthesis but does not prove that it will inhibit the enzyme in practice. Suggest two experimental checks.
[6 marks]
Total for this question: 6
A mutation replaces one charged amino-acid side chain in an enzyme active site with a non-polar side chain. Explain how this single primary-structure change could reduce catalysis, and state whether it changes the equilibrium position of the catalysed reaction.
[5 marks]
Total for this question: 5
Candidate inhibitors P and Q have the same molecular formula. P can place a positive group beside a negative active-site group while also placing a hydroxyl group beside a hydrogen-bond acceptor. Q can make only the first interaction. Deduce which candidate should bind more strongly, explain why the prediction may depend on stereochemistry, and state two limitations of selecting a drug from this model alone.
[4 marks]
Total for this question: 4
A $0.612\,\mathrm{g}$ dose of the racemate ( $M_r=255.0$ ) contains equal amounts of enantiomers A and B. Only A fits the active site. Calculate the amounts of racemate and A, explain why the dose gives the same initial inhibition as $1.20\,\mathrm{mmol}$ of pure A, and state one reason pure A may still be preferable.
[4 marks]
Total for this question: 4

3.3.13.4 · DNA (A-level only)

Explanation

A nucleotide contains phosphate bonded to 2-deoxyribose, which is bonded to adenine, cytosine, guanine or thymine. A single DNA strand is a polymer of nucleotides joined by covalent bonds between one nucleotide's phosphate and another's sugar, producing a sugar–phosphate backbone with bases attached.
Two complementary strands form a double helix.
Hydrogen bonds join bases across the strands: adenine pairs with thymine and cytosine with guanine.
Complementary pairing follows from matching hydrogen-bond donor and acceptor arrangements.
The backbone itself is covalent; hydrogen bonds act between bases and hold the two strands together rather than joining adjacent sugars and phosphates.

Complementary base pairs hydrogen-bond between two covalent sugar–phosphate backbones.

Worked example

A short DNA strand has base sequence A–C–G–T. State the complementary sequence and the interaction holding each pair.

1.Pair A with T and C with G.
2.Apply the same complementary rules along the strand.

Answer: T–G–C–A; hydrogen bonds hold complementary base pairs together.

Common mistakes

Don't pair adenine with cytosine or guanine with thymine.
Don't state that hydrogen bonds form the sugar–phosphate backbone.
Don't define a nucleotide as a base only and omit phosphate and 2-deoxyribose.

Exam tip

For a DNA-bonding question, distinguish covalent backbone bonds from hydrogen bonds between complementary bases.

Tier 1 · Easy

Name the three types of component present in a DNA nucleotide.
[3 marks]
Total for this question: 3
A student states that adenine is a DNA nucleotide. Distinguish a DNA base from a DNA nucleotide and explain why the statement is incorrect.
[2 marks]
Total for this question: 2

Tier 2 · Standard

One DNA strand contains the base sequence A–C–G–T–T–A. Give the complementary sequence and distinguish the bonding along a strand from the bonding between the strands.
[4 marks]
Total for this question: 4
In a double-stranded DNA sample, 22% of the bases are adenine. Calculate the percentage of thymine, guanine and cytosine bases.
[3 marks]
Total for this question: 3
A linear double-stranded DNA fragment contains 18 base pairs, and both 5′ ends carry a phosphate group. State the numbers of nucleotides, 2-deoxyribose units and phosphate groups. Distinguish the bonding along each backbone from the bonding between paired bases.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A short double-stranded DNA section has eight A–T base pairs and eleven C–G base pairs. It is heated until the strands separate. State the number of hydrogen bonds broken and explain why a section with a greater proportion of C–G pairs generally needs more energy to separate.
[5 marks]
Total for this question: 5
A student states that heating DNA separates its strands by hydrolysing the sugar–phosphate backbone. Evaluate the statement and explain why cooling can allow the strands to pair again.
[5 marks]
Total for this question: 5
DNA is made using ³²P-labelled phosphate and ¹⁵N-labelled thymine. The purified DNA is heated until its two strands separate without hydrolysing either strand. Predict where each label is found after heating and explain which bonds have and have not been broken.
[5 marks]
Total for this question: 5
A double-stranded DNA fragment contains 80 base pairs and 196 hydrogen bonds between its strands. Determine the numbers of A–T and C–G base pairs.
[4 marks]
Total for this question: 4
Two DNA fragments each contain 30 A–T pairs and 20 C–G pairs, but their base-pair orders differ. Evaluate whether the fragments must have the same total number of hydrogen bonds, the same base sequence and the same ease of complete strand separation. State the number of adenine bases in each fragment.
[4 marks]
Total for this question: 4

3.3.13.5 · Action of anticancer drugs (A-level only)

Explanation

Cisplatin is a square-planar Pt(II) complex used as an anticancer drug. Its cis arrangement places two replaceable chloride ligands adjacent.
In cells, ligand replacement allows platinum to form co-ordinate bonds to nitrogen atoms on guanine bases, linking sites on DNA. These platinum–DNA bonds distort the double helix and prevent normal DNA replication, so rapidly dividing cancer cells cannot reproduce successfully.
The drug is not perfectly selective: it can also disrupt DNA replication in healthy dividing cells and cause adverse effects.
Society must assess this harm against therapeutic benefit.
The new Pt–N interaction is a co-ordinate covalent bond, not hydrogen bonding.

Adjacent ligand sites on cisplatin form Pt–N bonds to guanine and cross-link DNA.

Worked example

Explain how cisplatin prevents DNA replication.

1.Chloride ligands are replaced and Pt forms co-ordinate bonds to guanine nitrogen atoms.
2.Links between nearby DNA sites distort the double helix.
3.The distorted strands cannot replicate normally.

Answer: Cisplatin cross-links DNA through ligand replacement and blocks replication.

Common mistakes

Don't say cisplatin hydrogen-bonds to guanine rather than forming Pt–N co-ordinate bonds.
Don't draw the trans isomer when explaining the adjacent reactive sites of cisplatin.
Don't claim cisplatin affects only cancer cells and therefore has no adverse effects.

Exam tip

For drug action, link ligand replacement to Pt–N bonding, DNA distortion and failed replication in a causal sequence.

Tier 1 · Easy

Name the DNA base to which cisplatin bonds and identify the donor atom in that base.
[2 marks]
Total for this question: 2
State the ligand that is replaced when cisplatin acts, and name the bond formed to guanine.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Explain, using ligand replacement and DNA structure, how cisplatin can stop a cancer cell from replicating.
[4 marks]
Total for this question: 4
Cisplatin and transplatin are cis–trans isomers. Describe the relative positions of the two chloride ligands in each isomer and explain why the cis arrangement is important for disrupting DNA replication.
[4 marks]
Total for this question: 4
Cisplatin, cis-[Pt(NH₃)₂Cl₂], reacts with a guanine base, G, so that one chloride ligand is replaced by a nitrogen atom of guanine. Write an equation for this ligand replacement, name the type of bond formed between platinum and nitrogen, and state the oxidation state and co-ordination number of platinum in the product.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Cisplatin reduces tumour growth but can damage bone marrow and the digestive lining. Explain both observations and state why treatment decisions must balance benefit against adverse effects.
[5 marks]
Total for this question: 5
After equal-concentration treatments, cultured cells show relative DNA synthesis rates of 100% with no drug, 24% with cisplatin and 81% with transplatin. Evaluate the claim that these data prove the cis arrangement is the sole reason cisplatin kills cancer cells.
[6 marks]
Total for this question: 6
A hypothetical square-planar Pt(II) drug has only one replaceable ligand but can still form one Pt–N bond to guanine. Predict why it is likely to disrupt DNA replication less than cisplatin, and explain why it could still cause adverse effects.
[5 marks]
Total for this question: 5
Two equal groups of 160 patients receive either a standard treatment or a treatment that includes cisplatin. Tumour shrinkage occurs in 64 patients with the standard treatment and 104 with cisplatin. Severe adverse effects occur in 12 patients with the standard treatment and 36 with cisplatin. Calculate the response and severe-adverse-effect percentages for both groups, then evaluate whether cisplatin should automatically be used for every patient, and state why the adverse-effect rate rises even though cisplatin bonds only to guanine in DNA.
[5 marks]
Total for this question: 5
A cell sample contains $7.20\,\mathrm{nmol}$ of accessible guanine sites. Each cisplatin molecule must form two Pt–N bonds to make one DNA cross-link. If $2.80\,\mathrm{nmol}$ of cisplatin reaches the DNA and every molecule reacts twice, determine the limiting reactant, the maximum cross-link amount and the guanine amount left. Explain how the links inhibit replication.
[5 marks]
Total for this question: 5

3.3.14 · Organic synthesis (A-level only)

Explanation

An organic synthesis may use reactions in the specification in a route of up to four steps.
Retrosynthetic planning identifies the target functional group, works backwards to feasible precursors, then checks every forward reagent, condition and intermediate.
Carbon skeletons must be tracked: cyanide substitution adds one carbon, while many oxidation, reduction, addition and elimination steps preserve it unless another reactant supplies carbon.
Selective conditions such as distillation versus reflux determine products.
Sustainable processes avoid solvents where possible, choose non-hazardous starting materials, use fewer steps and achieve high percentage atom economy, reducing material use, waste, separation demand and safety risk.

Worked example

Design a two-step route from ethanol to ethyl ethanoate.

1.Oxidise ethanol using $\mathrm{K_2Cr_2O_7/H_2SO_4}$ and heat under reflux to form ethanoic acid.
2.React the ethanoic acid with ethanol using concentrated $\mathrm{H_2SO_4}$ as a catalyst to form ethyl ethanoate.

Answer: Ethanol $\rightarrow$ ethanoic acid $\rightarrow$ ethyl ethanoate.

Common mistakes

Don't miss the extra carbon introduced by cyanide and propose the wrong target chain length.
Don't list an intermediate without a reagent or condition for the next conversion.
Don't claim a longer route is greener while ignoring lower atom economy, extra solvent and additional waste.

Exam tip

For a synthesis question, show every intermediate structure and place reagent and condition over each arrow.

Tier 1 · Easy

Give two reasons, other than cost, why a chemist may prefer a high-atom-economy synthesis with fewer reaction steps.
[2 marks]
Total for this question: 2
Route A to an organic product has an overall yield of 68% and an atom economy of 55%. Route B has an overall yield of 60% and an atom economy of 82%. Identify the route that maximises isolated product from the same theoretical amount, and the route that incorporates the greater fraction of reactant atoms into the desired product.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Design a three-step synthesis of ethanoic acid starting from ethene. Give the reagent and condition for each step and name the intermediate after each of the first two steps.
[6 marks]
Total for this question: 6
A student proposes this method for preparing butyl ethanoate: heat butan-1-ol with ethanoic acid under reflux using concentrated sulfuric acid as a catalyst; cool the mixture and transfer it to a separating funnel; wash the organic layer with aqueous sodium hydroxide; add aqueous magnesium sulfate to the organic layer; then distil. Identify the two errors in the method and give a correction for each.
[4 marks]
Total for this question: 4
Route A has three successive percentage yields of 92.0%, 84.0% and 75.0% and an atom economy of 88%. Route B is one step with a 68.0% yield and a 63% atom economy. Calculate Route A's overall yield. Route A requires a solvent at every step; Route B is solvent-free. Give one reason why Route B could be considered the more sustainable process and one reason why Route A could be.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Starting from bromoethane and using no more than three reaction steps, prepare N-propylethanamide. Give structures or names of both intermediates, reagents, conditions and the reason the first step changes the carbon-chain length.
[8 marks]
Total for this question: 8
Design a three-step synthesis of propan-2-amine from propan-2-ol. Give the reagent and condition for each step and name the organic product formed in each step.
[7 marks]
Total for this question: 7
A student claims that propene can be converted to butanenitrile in two steps by adding HBr and then heating the major bromoalkane with ethanolic KCN. Track the major carbon skeleton, name the bromoalkane and nitrile, state the condition used for the cyanide substitution, and evaluate the claim.
[5 marks]
Total for this question: 5
Give a synthesis of CH₃CH₂CH(OH)CH₂NH₂ from propan-1-ol in no more than three steps. State every intermediate, reagent and condition, name the final product, and explain where the fourth carbon atom enters the route.
[5 marks]
Total for this question: 5
Give a three-step route from 2-bromobutane to 2-hydroxy-2-methylbutanenitrile. State the organic product, reagent and condition for every step, track the change in carbon number, and explain why the final product is a racemate.
[6 marks]
Total for this question: 6

3.3.15 · Nuclear magnetic resonance spectroscopy (A-level only)

Explanation

NMR reveals positions of $^{13}\mathrm{C}$ or $^1\mathrm{H}$ atoms through chemical shifts on the $\delta$ scale, which depend on molecular environment. $^{13}\mathrm{C}$ spectra are simpler and normally give one signal per distinct carbon environment. $^1\mathrm{H}$ integration gives relative numbers of equivalent protons.
For adjacent non-equivalent aliphatic protons, the $n+1$ rule gives doublet, triplet or quartet splitting for one, two or three equivalent neighbours.
Samples use deuterated solvent or $\mathrm{CCl_4}$ .
TMS is inert, volatile, gives one sharp highly shielded signal and defines $\delta=0$ .
Spectra, integration and data-book shifts combine to suggest structures.

A simplified proton NMR spectrum shows splitting patterns and the TMS reference at zero chemical shift.

Worked example

An ethyl-group signal from $\mathrm{CH_3}$ has two equivalent neighbouring protons. Predict its splitting and relative integration.

1.Apply $n+1$ with $n=2$ neighbouring protons.
2.The methyl environment contains three equivalent protons.

Answer: The $\mathrm{CH_3}$ signal is a triplet with relative integration 3.

Common mistakes

Don't count signal height instead of integrated area as the relative proton number.
Don't apply $n+1$ to the protons within the same equivalent environment.
Don't say TMS is suitable because it reacts readily with the sample.

Exam tip

For a structure deduction, make a table of chemical shift, integration and splitting before assembling fragments.

Tier 1 · Easy

Give two properties that make tetramethylsilane suitable as the standard for chemical shift in NMR spectroscopy.
[2 marks]
Total for this question: 2
A compound with molecular formula C₃H₆O gives two ¹³C NMR signals, at 30 ppm and 205 ppm. Deduce the compound and explain why it gives only two signals.
[3 marks]
Total for this question: 3

Tier 2 · Standard

A compound gives three ¹H NMR signals: a triplet with integration 3, a singlet with integration 3 and a quartet with integration 2. Explain the splitting and deduce a structure consistent with molecular formula C₄H₈O₂.
[5 marks]
Total for this question: 5
An alcohol has molecular formula C₃H₈O and two ¹³C NMR signals. Its ¹H NMR spectrum has a doublet at 1.2 ppm integrating to 6, a 1H signal at 3.9 ppm, and a broad signal integrating to 1. Deduce the structure and justify the splitting and integrations.
[5 marks]
Total for this question: 5
The ¹H NMR spectrum of 2-aminoethan-1-ol, HOCH₂CH₂NH₂, is recorded before and after shaking with D₂O. Protons bonded to oxygen or nitrogen exchange with deuterium, and deuterium gives no signal in a ¹H NMR spectrum. Predict the number and relative integrations of the proton environments that remain after the exchange, and explain which original signals disappear.
[4 marks]
Total for this question: 4

Tier 3 · Hard

An ester has molecular formula C₅H₁₀O₂ and five ¹³C NMR signals. Its ¹H NMR spectrum contains two triplets, each integrating to 3, and two quartets, each integrating to 2; one quartet is substantially further downfield than the other. Deduce the ester and justify every signal pattern.
[7 marks]
Total for this question: 7
An organic compound has molecular formula C₄H₈O₂ and a molecular-ion peak at m/z 88. Its IR trace contains a broad carboxylic-acid O–H band from 2500 to 3000 cm⁻¹ and a strong absorption at 1710 cm⁻¹. Its ¹³C NMR spectrum has signals at 19, 34 and 180 ppm. Its ¹H NMR spectrum has a 6H doublet at 1.1 ppm, a 1H signal at 2.6 ppm and a broad 1H singlet at 11.5 ppm. Deduce the structure and justify the evidence from every technique.
[8 marks]
Total for this question: 8
Explain how ¹³C and ¹H NMR can distinguish pentan-2-one from pentan-3-one without using exact chemical-shift values. State the number of ¹³C environments for each and the diagnostic ¹H pattern caused by the symmetry of pentan-3-one.
[5 marks]
Total for this question: 5
A mixture contains ethyl ethanoate and propanone only. A non-overlapping OCH₂ quartet from ethyl ethanoate has integral 3.20, and the propanone methyl singlet has integral 7.20. Determine the mole ratio and mole percentages of the two compounds. Explain why raw integral values cannot be compared without accounting for the protons producing each signal.
[5 marks]
Total for this question: 5
A compound has molecular formula C₃H₆O₂. The compound gives no effervescence with aqueous sodium carbonate. Its ¹³C NMR spectrum has three signals, including one in the ester carbonyl region. Its ¹H NMR spectrum has a 1H singlet far downfield, a 2H quartet and a 3H triplet. Deduce the structure and justify the chemical environments, integrations and splitting.
[6 marks]
Total for this question: 6

3.3.16 · Chromatography (A-level only)

Explanation

Chromatography separates and identifies mixture components through their balance between solubility in the mobile phase and retention by the stationary phase. TLC uses a solid-coated plate with solvent rising; column chromatography uses solvent moving down packed solid; gas chromatography passes carrier gas through a heated column containing solid or liquid-coated-solid stationary phase.
Greater mobile-phase solubility increases movement, while stronger stationary retention slows it. $R_f$ equals component distance divided by solvent-front distance.
$R_f$ values and retention times identify substances only by comparison with standards under identical conditions.
GC–MS adds a mass spectrum to support each separated component's identification.
Required practical 12 uses TLC.

TLC separates spots between a common baseline and solvent front so their movement ratios can be compared.

Worked example

On a TLC plate, a spot moves $4.2\,\mathrm{cm}$ and the solvent front moves $7.0\,\mathrm{cm}$ . Calculate $R_f$ .

1.Measure both distances from the same baseline.
2. $R_f=4.2/7.0$ .

Answer: $R_f=0.60$ with no units.

Common mistakes

Don't divide solvent-front distance by spot distance and obtain an impossible $R_f>1$ .
Don't compare $R_f$ values measured with different solvents as though conditions were identical.
Don't claim a matching GC retention time alone proves identity without supporting mass-spectrum evidence.

Exam tip

For a chromatogram calculation, measure to the centre of the spot from the baseline and keep $R_f$ unitless.

Tier 1 · Easy

On a TLC plate, a spot travels $4.5\,\text{cm}$ from the start line while the solvent front travels $7.5\,\text{cm}$ . Calculate the $R_f$ value.
[2 marks]
Total for this question: 2
State three requirements when setting up a TLC plate in a developing chamber so that the starting spots and solvent movement can be interpreted reliably.
[3 marks]
Total for this question: 3

Tier 2 · Standard

Two dyes are placed on the same TLC plate. Dye P is very soluble in the solvent and weakly retained by the coating; dye Q is less soluble and more strongly retained. Predict which dye has the larger $R_f$ and explain your answer.
[4 marks]
Total for this question: 4
An unknown gives one TLC spot with the same $R_f$ value as a pure reference compound in one solvent. Evaluate the identification and suggest one chromatographic improvement.
[4 marks]
Total for this question: 4
A reaction mixture is separated by column chromatography. Explain how the stationary and mobile phases separate its components, and describe how TLC can be used to decide which collected fractions contain the same pure product and may be combined.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A GC trace of a flavour mixture has peaks at $2.8$ , $4.6$ and $7.1\,\text{min}$ . Under identical conditions, reference compounds X, Y and Z have retention times $2.8$ , $4.6$ and $6.9\,\text{min}$ respectively. Explain what can be concluded, what cannot be concluded about the last peak, and how coupling the GC instrument to a mass spectrometer strengthens identification.
[6 marks]
Total for this question: 6
Three reference compounds P, Q and R are tested by TLC. In solvent A they travel 6.40, 6.64 and 6.88 cm while the solvent front travels 8.00 cm. In solvent B they travel 1.60, 4.00 and 6.40 cm with the same solvent-front distance. Calculate all six $R_f$ values, identify the better solvent for separating the references, and evaluate an identification of an unknown that travels 4.00 cm in solvent B.
[7 marks]
Total for this question: 7
A gas chromatogram has peak areas 18, 42 and 60 arbitrary units. Assume equal detector response per mole. Calculate the percentage represented by each peak. GC–MS shows that the middle peak contains two co-eluting compounds. Explain what abundance conclusion can and cannot be made for that peak and how the mass spectra reveal co-elution.
[6 marks]
Total for this question: 6
Analysis of five consecutive fractions from a column gives these masses of target and impurity: F₁, $0.04\,\mathrm{g}$ target and $0.16\,\mathrm{g}$ impurity; F₂, $0.28\,\mathrm{g}$ and $0.04\,\mathrm{g}$ ; F₃, $0.38\,\mathrm{g}$ and zero; F₄, $0.25\,\mathrm{g}$ and zero; F₅, $0.05\,\mathrm{g}$ and $0.09\,\mathrm{g}$ . The total target loaded was $1.00\,\mathrm{g}$ . Calculate the product purity and target recovery if (i) F₂–F₄ are pooled and (ii) only F₃–F₄ are pooled. Choose the pool for a specification requiring at least 98.0% purity, and suggest how collection could be changed to recover more pure target.
[6 marks]
Total for this question: 6
A laboratory must (i) check rapidly whether a non-volatile dye mixture contains a known dye, (ii) isolate gram quantities of its components, and (iii) identify volatile components in a fragrance mixture with structural evidence. Identify the most suitable chromatographic method for each task and justify each choice.
[6 marks]
Total for this question: 6

Answer key

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

3.3.1.1 · Nomenclature

Tier 1 · Easy

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01.1	3-methylbutan-1-ol	The longest chain containing the OH group has four carbon atoms, so the parent is butanol. Number from the OH end: the OH group is on carbon 1 and the methyl branch is on carbon 3, giving 3-methylbutan-1-ol.	1
02.1	3-methylpentane; the student did not choose the longest continuous carbon chain.	Trace the longest continuous chain before naming any branches. It contains five carbon atoms, so the parent is pentane. The remaining carbon is a methyl group on carbon 3, giving 3-methylpentane. A four-carbon parent is not permitted when a five-carbon chain is present.	2

Tier 2 · Standard

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01.1	CH₃CH(CH₃)CH(CH₂CH₃)CH₂CH₂CH₃	Write a six-carbon parent chain. Attach a methyl group to carbon 2 and an ethyl group to carbon 3, keeping both branch points as CH groups: CH₃CH(CH₃)CH(CH₂CH₃)CH₂CH₂CH₃.	1
02.1	4-methylpentan-2-ol CH₃CH(OH)CH₂CH(CH₃)CH₃	The principal functional group fixes the numbering direction, so the five-carbon parent is numbered to put OH on carbon 2. The methyl branch is then on carbon 4. The name is 4-methylpentan-2-ol and the condensed structure is CH₃CH(OH)CH₂CH(CH₃)CH₃.	2
03.1	3-ethyl-4-methylcyclohexan-1-ol; number from the OH-bearing carbon in the direction giving the substituents the lowest locants, and list ethyl before methyl alphabetically.	The alcohol is the principal functional group, so its carbon is carbon 1 and the suffix is cyclohexan-1-ol. Of the two possible directions around the ring, one gives substituent locants 3 and 4 while the other gives 4 and 5, so choose 3 and 4. Prefixes are then written alphabetically, ignoring locant numbers: ethyl precedes methyl. The complete name is 3-ethyl-4-methylcyclohexan-1-ol.	3

Tier 3 · Hard

Mark scheme for 3.3.1.1 Tier 3 · Hard
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01.1	4,4-dimethylpentan-2-ol C₇H₁₆O	The longest chain containing the OH group has five carbon atoms. Number from the end nearer the OH group, placing OH on carbon 2 and the two remaining methyl substituents on carbon 4: 4,4-dimethylpentan-2-ol. Counting all atoms gives seven carbons, sixteen hydrogens and one oxygen, so the molecular formula is C₇H₁₆O.	2
02.1	3-methylpent-4-en-2-ol; number from the end that gives the alcohol group the lowest locant.	The alcohol is the principal functional group and is named with the suffix -ol, so its locant takes priority over the alkene locant when the chain is numbered. Numbering from the alcohol end gives OH on carbon 2, the methyl group on carbon 3 and the double bond starting at carbon 4: 3-methylpent-4-en-2-ol.	3
03.1	A five-membered carbon ring with OH on carbon 1, ethyl on carbon 2 and two methyl groups on carbon 3. C₉H₁₈O; the alcohol is the principal functional group and receives the lowest locant.	Start with a five-carbon ring and label the OH-bearing carbon as carbon 1 because the -ol suffix fixes the numbering origin. Number toward the ethyl group so it is on carbon 2; place both methyl groups on carbon 3. The structure contains five ring carbons, two ethyl carbons and two methyl carbons, giving nine carbons. A saturated monocyclic hydrocarbon has formula C_nH_2n; replacing one C-H bond by C-OH adds oxygen without changing the total hydrogen count, so the formula is C₉H₁₈O.	5
04.1	The molecular formula is C₆H₁₂. The empirical formula is CH₂. The general formula for an acyclic alkene containing one C=C bond is C_nH_2n. The molecular formula does not show the atom connectivity or the position of the C=C bond. The empirical formula gives only the simplest whole-number ratio and does not show the actual number of atoms in one molecule.	Count six carbon atoms and twelve hydrogen atoms in the condensed structure to obtain C₆H₁₂. Dividing both subscripts by six gives the simplest ratio CH₂. Hex-3-ene is an acyclic monoalkene, so it fits C_nH_2n. C₆H₁₂ preserves the numbers of atoms but not their connections or the double-bond locant; CH₂ also loses the molecular size because it records only the simplest ratio.	5
05.1	The molecular formula is C₇H₁₆. The IUPAC name is 3-methylhexane. Seven carbon atoms are represented but not drawn as element symbols. Sixteen hydrogen atoms are represented but not drawn as element symbols.	A five-segment main zig-zag contains six carbons: two line ends and four internal vertices. The branch contributes one more carbon at its unlabelled end, giving seven carbons in total. The longest chain is hexane and the branch is methyl on carbon 3, so the name is 3-methylhexane. Completing carbon valencies gives sixteen hydrogens and formula C₇H₁₆. Skeletal notation omits the symbols for all seven carbons and all sixteen carbon-bound hydrogens here.	4

3.3.1.2 · Reaction mechanisms

Tier 1 · Easy

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01.1	It starts at the nitrogen lone pair and ends at the carbon atom where the new C–N bond forms.	A curly arrow follows an electron pair. Ammonia supplies the pair from nitrogen, so the tail is placed on the N lone pair and the arrowhead is placed at the electron-deficient carbon that receives the pair.	2
02.1	A full curly arrow represents movement of an electron pair, so it cannot represent one electron; AQA radical mechanisms use balanced equations and a dot for the unpaired electron, not curly arrows.	A full curly arrow tracks a pair of electrons and must start at a bond or lone pair, so it cannot show the movement of one electron. In AQA radical mechanisms, use balanced equations and show each unpaired electron with a radical dot; curly arrows are not required.	2

Tier 2 · Standard

Mark scheme for 3.3.1.2 Tier 2 · Standard
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01.1	$\mathrm{CH_3\mathbin{\bullet}+Cl_2\rightarrow CH_3Cl+Cl\mathbin{\bullet}}$ ; show the radical dots and no electron-pair curly arrows.	The methyl radical removes one chlorine atom from Cl₂, making chloromethane and regenerating a chlorine radical: $\mathrm{CH_3\mathbin{\bullet}+Cl_2\rightarrow CH_3Cl+Cl\mathbin{\bullet}}$ . The unpaired electron is represented by a dot on each radical. AQA does not require electron-pair curly arrows for a radical mechanism.	3
02.1	The arrow starts at the H–Br bond and ends at Br; H⁺ and Br⁻ form.	The bonding electron pair is the electron source, so the curly-arrow tail must start in the H–Br bond. Bromine receives both electrons, so the arrowhead ends on Br. The products of this heterolytic fission are H⁺ and Br⁻.	3
03.1	A full curly arrow shows movement of an electron pair, not an atom; its tail starts at the source lone pair or bond and its head shows the atom or bond receiving the pair.	Read a full curly arrow as electron-pair bookkeeping. The tail must touch the lone pair or covalent bond that supplies two electrons, and the head marks the atom or bond to which that pair moves. The atoms are represented by the structures before and after the step; they do not move along the arrow.	2

Tier 3 · Hard

Mark scheme for 3.3.1.2 Tier 3 · Hard
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01.1	Arrow 1 goes from an oxygen lone pair on OH⁻ to the carbon bonded to iodine; arrow 2 goes from the C–I bond to iodine.	The incoming bond is made using an oxygen lone pair, so the first arrow starts at :OH⁻ and points to the carbon bearing iodine. The C–I bond then breaks heterolytically, so the second arrow starts in that bond and points to iodine, which takes both bonding electrons and leaves as I⁻.	4
02.1	Draw the arrow from an oxygen lone pair on H₂O to C⁺; the first intermediate is CH₃CH(OH₂⁺)CH₂CH₃; deprotonation uses a lone pair on a second water molecule to H and an arrow from that O–H bond back to O.	Water supplies the electron pair, so the first arrow starts at an oxygen lone pair and ends at the positively charged carbon, forming a C–O bond. Oxygen then has three bonds and carries a positive charge in CH₃CH(OH₂⁺)CH₂CH₃. A second water molecule removes H⁺: draw an arrow from its oxygen lone pair to that H and another from the intermediate's O–H bond to O, giving neutral butan-2-ol.	4
03.1	The arrow starts at the carbon lone pair of CN⁻ and ends at the positive carbon; the carbocation carbon has three bonds before attack and four afterwards; both formal charges are removed, giving neutral R–CH₂–CN.	Cyanide attacks through carbon. Curly arrows follow electrons, so the tail belongs on the carbon lone pair of CN⁻, not on the electron-deficient positive charge. The arrowhead is placed at the carbocation carbon, where the new C–C bond forms. That carbon has three bonds and an incomplete valence before attack; the donated pair supplies its fourth bond. The cyanide and carbocation charges are therefore both removed in neutral R–CH₂–CN.	4
04.1	Initiation: Br₂ → 2Br• (UV light) Propagation 1: Br• + C₂H₆ → HBr + C₂H₅• Propagation 2: C₂H₅• + Br₂ → C₂H₅Br + Br• Termination: C₂H₅• + C₂H₅• → C₄H₁₀ Termination: C₂H₅• + Br• → C₂H₅Br Overall: C₂H₆ + Br₂ → C₂H₅Br + HBr	UV light homolyses the halogen bond in initiation, giving two bromine radicals. Propagation alternates: a bromine radical abstracts hydrogen from ethane to give HBr and an ethyl radical, then the ethyl radical attacks a bromine molecule to give bromoethane and regenerate the bromine radical. Termination combines any two radicals; two ethyl radicals give butane, and an ethyl radical with a bromine radical gives bromoethane. Summing the two propagation steps cancels the radicals and gives the overall equation.	6

3.3.1.3 · Isomerism

Tier 1 · Easy

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01.1	position isomerism	Both compounds have the same four-carbon skeleton and the same alcohol functional group, but the OH group occupies a different carbon. They are position isomers.	1
02.1	One carbon of the C=C bond has two identical H substituents, so the alkene does not show E–Z isomerism.	E–Z isomerism requires each double-bonded carbon to have two different substituents. The CH₂ end has H and H, so that condition fails before CIP priorities are compared.	2

Tier 2 · Standard

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01.1	E; the higher-priority group at each alkene carbon is on the opposite side.	On the left carbon, carbon in CH₃ has higher atomic number than hydrogen, so CH₃ has priority. On the right, the first atoms tie as carbon, but CH₂CH₃ next presents C,H,H whereas CH₃ presents H,H,H, so ethyl has priority. Those two higher-priority groups are opposite, hence E.	3
02.1	E: CH₃ and Br on opposite sides of the planar C=C bond (skeletal or displayed drawing) Z: CH₃ and Br on the same side of the planar C=C bond At carbon 1, Br has higher priority than Cl; at carbon 2, CH₃ has higher priority than H.	The connectivity is C(Br)(Cl)=CHCH₃; a drawing must show the two geometries by placing the higher-priority groups on opposite sides for E and the same side for Z. On carbon 1, atomic number gives Br priority over Cl. On carbon 2, the carbon of CH₃ has priority over H. The relative positions of Br and CH₃ therefore distinguish the two isomers.	3
03.1	Position: CH₃CH₂COCH₂CH₃, pentan-3-one. Chain: CH₃COCH(CH₃)CH₃, 3-methylbutan-2-one. Functional group: CH₃CH₂CH₂CH₂CHO, pentanal. Common molecular formula: C₅H₁₀O.	Move the ketone carbonyl along the same straight chain to obtain pentan-3-one. Keep a ketone but branch the carbon skeleton to obtain 3-methylbutan-2-one. Change the ketone to an aldehyde while retaining five carbons to obtain pentanal. Counting atoms in each structure gives C₅H₁₀O, so each qualifies as a structural isomer of pentan-2-one.	4

Tier 3 · Hard

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01.1	CH₃CH₂CH₂CHO, butanal (CH₃)₂CHCHO, 2-methylpropanal CH₃COCH₂CH₃, butan-2-one The aldehydes are chain isomers; either aldehyde and the ketone are functional-group isomers.	An aldehyde needs a terminal CHO group. A straight four-carbon chain gives butanal, while a branched skeleton gives 2-methylpropanal. A ketone needs an internal C=O group; with four carbons its only position is carbon 2, giving butan-2-one. Changing only the carbon skeleton gives chain isomerism, whereas changing aldehyde to ketone gives functional-group isomerism.	5
02.1	The claim is impossible; every alkene on the 2-methylbutane carbon skeleton has either a terminal CH₂ group or two identical methyl groups on one double-bonded carbon, so none shows E–Z isomerism.	Place the double bond on each distinct adjacent pair of carbon atoms in the 2-methylbutane skeleton. A terminal placement gives a CH₂ end with two identical H groups. The internal placement gives one alkene carbon two identical CH₃ groups. Each possible structure therefore fails the two-different-groups test at one carbon, so the stated combination of constraints is impossible.	4
03.1	The first two formulae are the same molecule, butan-1-ol, written from opposite ends. Butan-2-ol is a position isomer; ethoxyethane is a functional-group isomer; 2-methylpropan-1-ol is an omitted chain isomer. All have molecular formula C₄H₁₀O but different connectivity where they are genuine isomers.	Reading HOCH₂CH₂CH₂CH₃ from the opposite end gives CH₃CH₂CH₂CH₂OH, so those entries duplicate butan-1-ol. Moving OH to carbon 2 gives the position isomer butan-2-ol. Moving oxygen into a C–O–C linkage gives the ether ethoxyethane, a functional-group isomer. Branching the carbon skeleton while retaining terminal OH gives the omitted chain isomer 2-methylpropan-1-ol. Each genuine isomer keeps C₄H₁₀O but changes connectivity.	5
04.1	The atoms directly attached to each C=C carbon are all carbon, so each first comparison is tied. On the left, CH₂Cl presents Cl,H,H whereas CH₂OH presents O,H,H, so CH₂Cl has higher priority. For CHO, the C=O bond is treated as the carbon being attached to O,O,H. On the right, O,O,H outranks the O,H,H presented by CH₂OH, so CHO has higher priority. The two higher-priority groups are on the same side, so the descriptor is Z.	Apply CIP rules separately at the two alkene carbons. Each substituent begins with carbon, so compare the atoms attached to those carbons in decreasing atomic-number order. At the left, Cl,H,H outranks O,H,H because chlorine has a higher atomic number than oxygen. At the right, count the double-bonded oxygen twice: CHO presents O,O,H, which outranks O,H,H from CH₂OH at the second entry. The higher-priority groups CH₂Cl and CHO are both above the double bond, giving Z.	5
05.1	Each C=C bond has two different groups on each double-bonded carbon, so both bonds can be assigned E or Z. Independent assignment gives the four formal combinations (2E,4E), (2E,4Z), (2Z,4E) and (2Z,4Z). End-to-end symmetry makes the (2E,4Z) and (2Z,4E) drawings the same stereoisomer. There are three different stereoisomers. They are (2E,4E)-hexa-2,4-diene, (2E,4Z)-hexa-2,4-diene and (2Z,4Z)-hexa-2,4-diene.	Both double bonds satisfy the two-different-groups condition, so start with two choices at carbon 2 and two at carbon 4. This gives four formal combinations. The molecule has identical ends, so turning an (2E,4Z) drawing end-for-end produces the arrangement labelled (2Z,4E); these are not separate compounds. The two same-descriptor arrangements remain distinct from each other and from the mixed arrangement, leaving three stereoisomers: (2E,4E), (2E,4Z) and (2Z,4Z).	5

3.3.2.1 · Fractional distillation of crude oil

Tier 1 · Easy

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01.1	different boiling points; fractional distillation	The hydrocarbons vaporise and condense at different temperatures because they have different boiling points. Repeated vaporisation and condensation in a fractionating column is fractional distillation.	2
02.1	A fraction is a mixture of hydrocarbons with a similar range of boiling points.	Fractional distillation groups hydrocarbons that condense over a similar temperature interval. A collected fraction therefore contains several hydrocarbons and has a boiling range rather than being one pure compound with one boiling point.	2

Tier 2 · Standard

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01.1	The C₅ alkane has weaker London forces and a lower boiling point, so it remains gaseous until it reaches the cooler upper region.	The C₁₅ molecule has more electrons and a larger contact area, so its London forces are stronger and more energy is needed to separate its molecules. It therefore has the higher boiling point and condenses in the hotter lower region. The C₅ alkane has weaker London forces, so it rises farther before condensing.	3
02.1	Use fractional distillation; it separates by boiling point using vaporisation and condensation, so intermolecular forces are overcome but covalent bonds are not broken.	The required operation is physical separation. Fractional distillation exploits different boiling ranges through repeated vaporisation and condensation. These changes overcome intermolecular attractions only. Catalytic cracking would break C–C covalent bonds and make different molecules, so it cannot meet the constraint.	3
03.1	$325\,\mathrm{K}$ : lowest-temperature zone; $392\,\mathrm{K}$ : middle zone; $478\,\mathrm{K}$ : highest-temperature zone; the lowest-temperature zone is highest.	Match each boiling point to the interval in which that component condenses: $325\,\mathrm{K}$ belongs below $350\,\mathrm{K}$ , $392\,\mathrm{K}$ lies between $350$ and $450\,\mathrm{K}$ , and $478\,\mathrm{K}$ lies above $450\,\mathrm{K}$ . Temperature decreases up the column, so the lowest-boiling range is collected highest and the highest-boiling range lowest.	2

Tier 3 · Hard

Mark scheme for 3.3.2.1 Tier 3 · Hard
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01.1	Highest to lowest: C₇H₁₆, C₁₂H₂₆, C₁₈H₃₈; separation is a physical change with no covalent bonds broken.	Boiling point increases with chain length because the number of electrons, molecular surface contact and London forces increase. C₁₈H₃₈ therefore condenses first in the hot lower region, C₁₂H₂₆ above it, and C₇H₁₆ highest. Only intermolecular separation and condensation occur, so covalent structures and molecular formulae remain unchanged.	5
02.1	Condensation occurs when the local temperature is low enough, below the relevant boiling range; covalent C–C bonds are not broken or re-formed, and only intermolecular attractions change.	A component remains gaseous while the column is hotter than its boiling range and condenses after reaching a sufficiently cool region. Fractional distillation is a physical process: molecules stay chemically unchanged, so C–C and C–H covalent bonds remain intact. Energy changes involve intermolecular attractions between molecules.	4
03.1	Packing provides a large surface for repeated condensation and vaporisation; higher-boiling components preferentially condense while lower-boiling components continue upward. Removing packing gives fewer separation stages, so boiling ranges overlap more and each fraction is less pure.	Rising vapour meets cooler liquid on the large surface supplied by the packing. Repeated condensation and vaporisation enrich the rising vapour in lower-boiling hydrocarbons and the descending liquid in higher-boiling hydrocarbons. If packing is removed, there are fewer such repeated separations. Components with close boiling points then leave together more readily, so the collected fractions contain broader, more overlapping mixtures.	4

3.3.2.2 · Modification of alkanes by cracking

Tier 1 · Easy

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01.1	$X=\mathrm{C_4H_8}$	Subtract the atoms in C₈H₁₈ from C₁₂H₂₆. The remainder has $12-8=4$ carbon atoms and $26-18=8$ hydrogen atoms, so $X=\mathrm{C_4H_8}$ , an alkene.	1
02.1	The correct two-carbon product is $\mathrm{C_2H_4}$ .	The first two products, $\mathrm{C_8H_{18}}$ and $\mathrm{C_4H_8}$ , account for twelve carbon atoms and twenty-six hydrogen atoms. The remaining product must therefore contain two carbon atoms and four hydrogen atoms, so it is $\mathrm{C_2H_4}$ . The corrected equation $\mathrm{C_{14}H_{30}\rightarrow C_8H_{18}+C_4H_8+C_2H_4}$ conserves all fourteen C and thirty H atoms.	2

Tier 2 · Standard

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01.1	$\mathrm{C_{15}H_{32}\rightarrow C_8H_{18}+C_3H_6+2C_2H_4}$ ; two ethene molecules; it converts a less-demanded long fraction into demanded fuel and alkene feedstock.	After C₈H₁₈ and C₃H₆, four carbon atoms and eight hydrogen atoms remain, corresponding to two C₂H₄ molecules. The balanced equation is $\mathrm{C_{15}H_{32}\rightarrow C_8H_{18}+C_3H_6+2C_2H_4}$ . Cracking adjusts supply to demand by making a shorter motor-fuel alkane and alkenes used as chemical feedstock from a less useful long-chain fraction.	3
02.1	Choose thermal cracking; it uses high pressure and produces a high proportion of alkenes.	Match the required product profile before matching the conditions. Thermal cracking is the alkene-rich route and uses high temperature with high pressure. Catalytic cracking uses a zeolite at slight pressure and is associated mainly with motor fuels and aromatic hydrocarbons.	3
03.1	Two saturated acyclic products containing 12 carbon atoms in total would contain 28 hydrogen atoms, not 26, so at least one product must be an alkene. $\mathrm{C_{12}H_{26}\rightarrow C_7H_{16}+C_5H_{10}}$ , where $\mathrm{C_5H_{10}}$ is an alkene such as pent-1-ene.	If both products were saturated acyclic hydrocarbons, their combined formula would follow the two-alkane hydrogen total $2(12)+4=28$ . The reactant supplies only 26 hydrogen atoms, so both products cannot be saturated and at least one must be unsaturated. One valid atom balance is $\mathrm{C_{12}H_{26}\rightarrow C_7H_{16}+C_5H_{10}}$ : it has 12 carbon atoms and 26 hydrogen atoms on each side.	3

Tier 3 · Hard

Mark scheme for 3.3.2.2 Tier 3 · Hard
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01.1	Thermal: high temperature and high pressure, with a high percentage of alkenes. Catalytic: high temperature, slight pressure and a zeolite catalyst, mainly motor fuels and aromatic hydrocarbons.	Give both condition sets and link each to its product profile. Thermal cracking is distinguished by high pressure and its alkene-rich output. Catalytic cracking is distinguished by slight pressure plus a zeolite and is directed mainly towards motor-fuel molecules and aromatics.	4
02.1	Decane, $\mathrm{C_{10}H_{22}}$ ; $\mathrm{C_{16}H_{34}\rightarrow C_{10}H_{22}+2C_3H_6}$ . The equation alone is not enough to choose the more profitable process. Thermal cracking uses high temperature and high pressure and produces a high proportion of alkenes; catalytic cracking uses high temperature, slight pressure and a zeolite catalyst and mainly produces motor fuels and aromatic hydrocarbons. The refinery must compare product demand or selling prices with operating and catalyst costs.	Two propene molecules contain six carbon atoms and twelve hydrogen atoms. Subtracting these from $\mathrm{C_{16}H_{34}}$ leaves $\mathrm{C_{10}H_{22}}$ , which is decane and fits the alkane general formula. The balanced equation is $\mathrm{C_{16}H_{34}\rightarrow C_{10}H_{22}+2C_3H_6}$ . That atom balance identifies possible products but does not specify the economically best route. Thermal cracking favours alkenes but requires high pressure; catalytic cracking uses a zeolite at slight pressure and mainly produces motor fuels and aromatics. A justified business choice therefore needs product values or demand and the relative energy, compression and catalyst costs.	6
03.1	$635\,\mathrm{g}$ of C₄H₈ to three significant figures	Convert the feed to grams and find its amount: $n(\mathrm{C_{18}H_{38}})=2000/254=7.8740157\,\mathrm{mol}$ . Only 72.0% cracks, so the reacting amount is $7.8740157\times0.720=5.6692913\,\mathrm{mol}$ . The equation forms two moles of C₄H₈ per mole cracked, giving $11.3385826\,\mathrm{mol}$ . Its mass is $11.3385826\times56.0=634.96\,\mathrm{g}$ , hence $635\,\mathrm{g}$ to three significant figures.	4
04.1	Atom conservation gives $\mathrm{X=C_7H_{12}}$ . A saturated acyclic seven-carbon hydrocarbon would have formula $\mathrm{C_7H_{16}}$ . X has four fewer hydrogen atoms, corresponding to two units of unsaturation. The stated bonding and acyclic constraints make both units C=C bonds, so X is a diene.	The named products account for ten carbon atoms and twenty-four hydrogen atoms, leaving seven carbon atoms and twelve hydrogen atoms for X: C, $17=6+4+7$ ; H, $36=14+10+12$ . The acyclic saturated reference formula is $\mathrm{C_nH_{2n+2}}$ , so a seven-carbon alkane would be $\mathrm{C_7H_{16}}$ . Each C=C bond reduces the hydrogen count by two relative to that reference. The deficit of four hydrogens therefore represents two C=C bonds. The prompt excludes rings and triple bonds, so the two unsaturation units must form a diene.	4

3.3.2.3 · Combustion of alkanes

Tier 1 · Easy

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01.1	$\mathrm{C_3H_8+5O_2\rightarrow3CO_2+4H_2O}$	Three carbon atoms require 3CO₂, and eight hydrogen atoms require 4H₂O. The products then contain ten oxygen atoms, so five O₂ molecules are needed: $\mathrm{C_3H_8+5O_2\rightarrow3CO_2+4H_2O}$ .	1
02.1	$\mathrm{2C_4H_{10}+9O_2\rightarrow8CO+10H_2O}$	The fractional-coefficient equation is atom-balanced, so multiply every coefficient by two. This gives $\mathrm{2C_4H_{10}+9O_2\rightarrow8CO+10H_2O}$ , with eight C, twenty H and eighteen O atoms on each side.	2

Tier 2 · Standard

Mark scheme for 3.3.2.3 Tier 2 · Standard
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01.1	$\mathrm{2CO+2NO\rightarrow2CO_2+N_2}$ ; CO is oxidised and NO is reduced.	Pair two CO molecules with two NO molecules to conserve C, N and O: $\mathrm{2CO+2NO\rightarrow2CO_2+N_2}$ . Carbon gains oxygen as CO becomes CO₂, so CO is oxidised. NO loses oxygen as its nitrogen forms N₂, so NO is reduced.	3
02.1	Use calcium carbonate for P to remove or neutralise SO₂; use a catalytic converter for Q to oxidise CO to CO₂ and reduce NO to N₂.	Calcium carbonate reacts with acidic sulfur dioxide in flue-gas desulfurisation, so it belongs with stream P. In stream Q, a catalytic converter couples oxidation of toxic CO to CO₂ with reduction of NO to N₂, removing both engine pollutants.	4
03.1	$\mathrm{C_nH_{2n+2}+\frac{3n+1}{2}O_2\rightarrow nCO_2+(n+1)H_2O}$ $\mathrm{2C_nH_{2n+2}+(3n+1)O_2\rightarrow 2nCO_2+(2n+2)H_2O}$	Balance carbon first to obtain $n\mathrm{CO_2}$ . The fuel has $2n+2$ hydrogen atoms, so it forms $(n+1)\mathrm{H_2O}$ . The products contain $2n+(n+1)=3n+1$ oxygen atoms, requiring $(3n+1)/2$ molecules of O₂. Therefore the general equation is $\mathrm{C_nH_{2n+2}+\frac{3n+1}{2}O_2\rightarrow nCO_2+(n+1)H_2O}$ .	3

Tier 3 · Hard

Mark scheme for 3.3.2.3 Tier 3 · Hard
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01.1	$2.50\,\mathrm{kg}$ of CaCO₃	Convert 1.60 kg to 1600 g. The amount of SO₂ is $n=1600/64.1=24.96\,\mathrm{mol}$ . The equation has a 1:1 ratio, so 24.96 mol of CaCO₃ is required. Its mass is $m=24.96\times100.1=2498.5\,\mathrm{g}=2.50\,\mathrm{kg}$ to three significant figures.	3
02.1	CO forms during incomplete combustion when oxygen is limited; nitrogen oxides form when N₂ and O₂ from air react at high engine temperatures; sulfur dioxide in flue gas is removed using calcium oxide or calcium carbonate, not a vehicle catalytic converter.	Limited oxygen prevents full oxidation of carbon to CO₂, so CO can form. Engine temperatures are high enough for nitrogen and oxygen from the air to react, producing nitrogen oxides; nitrogen need not be present in the fuel. Power-station sulfur dioxide is removed by basic calcium compounds such as CaO or CaCO₃, whereas a vehicle catalytic converter treats CO and nitrogen oxides.	5
03.1	$20.0\,\mathrm{cm^3}$ CO₂ and $60.0\,\mathrm{cm^3}$ CO $\mathrm{2C_2H_6+\frac{11}{2}O_2\rightarrow CO_2+3CO+6H_2O}$	Gas volumes at the same conditions follow mole ratios. Forty volumes of C₂H₆ contain enough carbon for 80 volumes of CO₂ plus CO, and enough hydrogen for 120 volumes of water vapour. Let the CO₂ volume be $x$ , so CO is $80-x$ . Oxygen atoms in the products correspond to $2x+(80-x)+120=x+200$ volume-units of atoms, supplied by twice the oxygen volume. Hence $2(110)=x+200$ , giving $x=20.0$ and CO $=60.0\,\mathrm{cm^3}$ . Dividing all volumes by 20 gives the stated equation.	5
04.1	The fuel contains $12.0\,\mathrm{g}$ of sulfur. The amount of sulfur, and hence SO₂, is $0.373832\,\mathrm{mol}$ . The 1:1 equation requires $0.373832\,\mathrm{mol}$ of CaO. The required pure CaO mass is $20.9720\,\mathrm{g}$ . The minimum mass of the $82.0\%$ pure solid is $25.6\,\mathrm{g}$ to three significant figures.	Convert the fuel mass to grams and apply the mass percentage: $2500\times0.00480=12.0\,\mathrm{g}$ of S. Its amount is $12.0/32.1=0.3738318\,\mathrm{mol}$ . Each sulfur atom gives one SO₂ molecule, and the removal equation is 1:1, so the same amount of CaO is needed. The pure CaO mass is $0.3738318\times56.1=20.9720\,\mathrm{g}$ . Because this is only 82.0% of the solid, divide by 0.820: $20.9720/0.820=25.5756\,\mathrm{g}$ , giving $25.6\,\mathrm{g}$ .	5
05.1	The reacting ratio CO:NO is 1:1. NO is the limiting pollutant. $0.180\,\mathrm{mol}$ of CO reacts. $0.0600\,\mathrm{mol}$ of CO remains unreacted. $0.180\,\mathrm{mol}$ of CO₂ forms. $0.0900\,\mathrm{mol}$ of N₂ forms.	The equation uses equal amounts of CO and NO. The smaller initial amount, $0.180\,\mathrm{mol}$ NO, is therefore limiting and consumes $0.180\,\mathrm{mol}$ CO. The CO remaining is $0.240-0.180=0.0600\,\mathrm{mol}$ . The coefficient ratio gives one mole of CO₂ per mole of CO consumed, so $0.180\,\mathrm{mol}$ CO₂ forms. Two moles of NO give one mole of N₂, so the nitrogen amount is $0.180/2=0.0900\,\mathrm{mol}$ .	6

3.3.2.4 · Chlorination of alkanes

Tier 1 · Easy

Mark scheme for 3.3.2.4 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$\mathrm{Cl_2\xrightarrow{UV}2Cl\mathbin{\bullet}}$ ; homolytic fission	UV radiation breaks the Cl–Cl bond so that one bonding electron goes to each chlorine atom. This homolytic fission gives $\mathrm{Cl_2\xrightarrow{UV}2Cl\mathbin{\bullet}}$ .	2
02.1	Two chlorine radicals, $\mathrm{2Cl\mathbin{\bullet}}$ ; one electron from the bond goes to each chlorine atom.	Homolytic means equal bond breaking. Each chlorine atom takes one electron from the Cl–Cl bonding pair, so two neutral radicals $\mathrm{Cl\mathbin{\bullet}}$ form rather than ions.	2

Tier 2 · Standard

Mark scheme for 3.3.2.4 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$\mathrm{Cl\mathbin{\bullet}+CH_4\rightarrow HCl+CH_3\mathbin{\bullet}}$ $\mathrm{CH_3\mathbin{\bullet}+Cl_2\rightarrow CH_3Cl+Cl\mathbin{\bullet}}$	First a chlorine radical abstracts H from methane: $\mathrm{Cl\mathbin{\bullet}+CH_4\rightarrow HCl+CH_3\mathbin{\bullet}}$ . The methyl radical then removes Cl from chlorine: $\mathrm{CH_3\mathbin{\bullet}+Cl_2\rightarrow CH_3Cl+Cl\mathbin{\bullet}}$ . The regenerated chlorine radical carries the chain forward.	2
02.1	$\mathrm{R\mathbin{\bullet}+Cl_2\rightarrow RCl+Cl\mathbin{\bullet}}$ is propagation because a radical is used and another is formed; $\mathrm{R\mathbin{\bullet}+Cl\mathbin{\bullet}\rightarrow RCl}$ is termination because two radicals combine and no radical remains.	Track radicals rather than memorising species. The first equation consumes $\mathrm{R\mathbin{\bullet}}$ but regenerates $\mathrm{Cl\mathbin{\bullet}}$ , allowing the chain to continue, so it is propagation. The second removes two radicals and produces only a stable molecule, so it terminates the chain.	4
03.1	Use methane in large excess, so chlorine radicals are more likely to collide with CH₄ than with CH₃Cl; this favours the first substitution over further substitution. Further substitution still occurs, so a mixture forms and chloromethane must be separated from the other products.	A chlorine radical can abstract H from either CH₄ or a chlorinated product. Using a large excess of methane makes encounters with CH₄ much more frequent, so formation of CH₃Cl is favoured relative to its further chlorination. This changes product proportions rather than making the reaction perfectly selective: some CH₃Cl molecules still undergo further substitution, so fractional distillation is needed to separate the product mixture.	4

Tier 3 · Hard

Mark scheme for 3.3.2.4 Tier 3 · Hard
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01.1	$\mathrm{Cl\mathbin{\bullet}+Cl\mathbin{\bullet}\rightarrow Cl_2}$ $\mathrm{CH_3\mathbin{\bullet}+Cl\mathbin{\bullet}\rightarrow CH_3Cl}$ $\mathrm{CH_3\mathbin{\bullet}+CH_3\mathbin{\bullet}\rightarrow C_2H_6}$ Chloromethane can undergo further radical substitution to form more highly chlorinated products.	Termination removes radicals by pairing them: two chlorine radicals form Cl₂, methyl plus chlorine forms CH₃Cl, and two methyl radicals form C₂H₆. Under continued UV exposure, C–H bonds remaining in CH₃Cl can also be substituted, producing CH₂Cl₂, CHCl₃ and CCl₄; the product is therefore a mixture.	4
02.1	The primary H atoms are equivalent but differ from the single tertiary H, so 1-chloro-2-methylpropane and 2-chloro-2-methylpropane form. $\mathrm{Cl\mathbin{\bullet}+(CH_3)_3CH\rightarrow HCl+(CH_3)_3C\mathbin{\bullet}}$ $\mathrm{(CH_3)_3C\mathbin{\bullet}+Cl_2\rightarrow(CH_3)_3CCl+Cl\mathbin{\bullet}}$	There are two chemically different hydrogen environments: nine equivalent primary H atoms and one tertiary H atom. Substitution at these positions gives 1-chloro-2-methylpropane and 2-chloro-2-methylpropane. For the tertiary pathway, $\mathrm{Cl\mathbin{\bullet}}$ abstracts the tertiary H to give HCl and $\mathrm{(CH_3)_3C\mathbin{\bullet}}$ . That radical then reacts with $\mathrm{Cl_2}$ , forming $\mathrm{(CH_3)_3CCl}$ and regenerating $\mathrm{Cl\mathbin{\bullet}}$ . Radical dots are explicit; electron-pair curly arrows are not used.	6
03.1	$\mathrm{CH_4+4Cl_2\rightarrow CCl_4+4HCl}$ ; four successive substitution cycles are required. Each cycle has its own radical propagation steps and replaces one C–H bond; the overall equation is their sum, not a single collision or a single step in the mechanism.	Each substitution replaces one H by Cl and also forms one HCl, so four hydrogens require four Cl₂ molecules overall: $\mathrm{CH_4+4Cl_2\rightarrow CCl_4+4HCl}$ . The organic sequence is CH₄ to CH₃Cl to CH₂Cl₂ to CHCl₃ to CCl₄. Every conversion proceeds through hydrogen abstraction and chlorine-radical regeneration. Adding those four propagation cycles cancels their radical intermediates and gives the overall equation, but no single collision breaks four C–H bonds and forms all products at once.	5
04.1	$\mathrm{CH_3\mathbin{\bullet}+\mathbin{\bullet}CH_2Cl\rightarrow C_2H_5Cl}$ $\mathrm{\mathbin{\bullet}CH_2Cl+\mathbin{\bullet}CH_2Cl\rightarrow C_2H_4Cl_2}$ $\mathrm{Cl\mathbin{\bullet}+CH_3Cl\rightarrow HCl+\mathbin{\bullet}CH_2Cl}$ Each termination equation removes two radicals without regenerating a radical, so it is not propagation.	Split each detected molecule into two plausible radical fragments. C₂H₅Cl results from methyl and chloromethyl radicals: $\mathrm{CH_3\mathbin{\bullet}+\mathbin{\bullet}CH_2Cl\rightarrow C_2H_5Cl}$ . Pairing two chloromethyl radicals gives $\mathrm{\mathbin{\bullet}CH_2Cl+\mathbin{\bullet}CH_2Cl\rightarrow C_2H_4Cl_2}$ . Chloromethyl radical is produced when a chlorine radical abstracts H from chloromethane: $\mathrm{Cl\mathbin{\bullet}+CH_3Cl\rightarrow HCl+\mathbin{\bullet}CH_2Cl}$ . That abstraction is propagation because one radical is consumed and another formed. The two combination equations are termination: each consumes two radicals and forms no radical, so neither propagates the chain.	4

3.3.3.1 · Nucleophilic substitution

Tier 1 · Easy

Mark scheme for 3.3.3.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	butanenitrile	CN⁻ replaces Br and bonds through its carbon atom. That carbon becomes part of the main chain, increasing the carbon count from three to four, so the product is butanenitrile.	1
02.1	CN⁻; its carbon atom forms the new bond.	Cyanide ion is the nucleophile that extends a carbon chain by one atom. It attacks through its carbon end, so the carbon of CN⁻ becomes the nitrile carbon in the product.	2

Tier 2 · Standard

Mark scheme for 3.3.3.1 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Arrow from an O lone pair on OH⁻ to the carbon bonded to Br; arrow from the C–Br bond to Br; products propan-1-ol and Br⁻.	Show the C–Br bond polarised toward Br. Draw one curly arrow from a lone pair on :OH⁻ to the carbon bearing Br, and at the same time a second from the C–Br bond to Br. The new C–O bond gives CH₃CH₂CH₂OH and bromide leaves as Br⁻.	3
02.1	Arrow from the carbon lone pair of CN⁻ to the carbon bonded to I; arrow from the C–I bond to I; product 2-methylpropanenitrile.	Show the C–I bond polarised C^δ+–I^δ−. The nucleophilic arrow starts at the carbon lone pair of ⁻:CN and ends at the carbon bonded to iodine. At the same time, the leaving-group arrow starts in the C–I bond and ends on I, forming I⁻. Counting the nitrile carbon in the parent gives 2-methylpropanenitrile.	4
03.1	The halogenoalkane is 2-bromobutane; the oxygen lone pair in OH⁻ and the nitrogen lone pair in NH₃ attack the carbon bonded to Br. Each nucleophile replaces Br but contains no carbon atom, so the four-carbon skeleton is retained.	Both products have the new functional group on carbon 2 of the same four-carbon chain, so replacing that group by Br identifies CH₃CHBrCH₂CH₃, 2-bromobutane. Hydroxide attacks through an oxygen lone pair; ammonia attacks through a nitrogen lone pair. In each case that atom forms the new bond to carbon while Br leaves. Because neither OH⁻ nor NH₃ contributes a carbon atom, substitution retains the original carbon skeleton.	4

Tier 3 · Hard

Mark scheme for 3.3.3.1 Tier 3 · Hard
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01.1	1-iodobutane fastest, then 1-bromobutane, then 1-chlorobutane; C–I has the lowest bond enthalpy and C–Cl the highest.	Substitution requires breaking the carbon–halogen bond. Bond enthalpy decreases from C–Cl to C–Br to C–I, so C–I breaks most readily and C–Cl least readily. Although C–Cl is the most polar of the three bonds, the observed order is iodo > bromo > chloro, showing that bond strength is the controlling factor here.	3
02.1	P contains iodine, Q contains chlorine and R contains bromine; hydrolysis is fastest for C–I and slowest for C–Cl because bond enthalpy increases from C–I to C–Br to C–Cl.	AgI is yellow, AgCl is white and AgBr is cream, so P is the iodoalkane, Q the chloroalkane and R the bromoalkane. The precipitate appears as halide ions are released by hydrolysis. The C–I bond has the lowest bond enthalpy and breaks most readily; C–Cl has the highest and breaks least readily, giving the order iodoalkane > bromoalkane > chloroalkane.	6
03.1	Use 1-bromo-2-methylpropane, CH₃CH(CH₃)CH₂Br. Draw an arrow from the carbon lone pair of ⁻:CN to the carbon bonded to Br, and an arrow from the C–Br bond to Br, forming Br⁻.	Work backwards by removing the nitrile carbon from the target and replacing its bond to the carbon skeleton with C–Br. This gives BrCH₂CH(CH₃)CH₃, named 1-bromo-2-methylpropane. In the forward reaction, the carbon lone pair of ⁻:CN attacks the carbon bearing Br, so the first arrow runs from that lone pair to the carbon. Simultaneously the C–Br bond pair moves to Br, shown by an arrow from the bond to Br, releasing Br⁻ and forming the target nitrile.	5
04.1	The first arrow starts at the nitrogen lone pair of NH₃ and ends at the carbon bonded to Br. A simultaneous arrow runs from the C–Br bond to Br, forming Br⁻. The first-stage products are CH₃CH₂CH₂CH₂NH₃⁺ and Br⁻. A second NH₃ molecule deprotonates the intermediate, forming butan-1-amine and NH₄Br.	Ammonia first acts as a nucleophile. Draw an arrow from its nitrogen lone pair to the carbon bearing Br and another from the C–Br bond to Br. Nitrogen then has four bonds, so the first-stage products are CH₃CH₂CH₂CH₂NH₃⁺ and Br⁻. A second ammonia molecule acts as a base: its nitrogen lone pair attacks H on the intermediate while that N–H bond pair returns to the intermediate's nitrogen. This forms CH₃CH₂CH₂CH₂NH₂, named butan-1-amine, plus NH₄⁺; pairing with Br⁻ gives NH₄Br.	4
05.1	$\mathrm{CH_3CH_2CH_2CH_2I+KCN\rightarrow CH_3CH_2CH_2CH_2CN+KI}$ The carbon atom in CN⁻ forms the new bond and becomes part of the product chain, so four carbons become five. The amount of 1-iodobutane is $18.40/184.0=0.1000\,\mathrm{mol}$ . The 1:1 equation gives a theoretical pentanenitrile mass of $0.1000\times83.0=8.30\,\mathrm{g}$ . Percentage yield $=(5.45/8.30)\times100=65.7\%$ to three significant figures.	Cyanide substitutes for iodide in a 1:1 reaction: $\mathrm{CH_3CH_2CH_2CH_2I+KCN\rightarrow CH_3CH_2CH_2CH_2CN+KI}$ . CN⁻ attacks through carbon, so its carbon becomes the fifth atom in the pentanenitrile chain. The starting amount is $18.40/184.0=0.1000\,\mathrm{mol}$ , giving a theoretical $0.1000\,\mathrm{mol}$ of product and a theoretical mass of $0.1000\times83.0=8.30\,\mathrm{g}$ . Therefore the percentage yield is $(5.45/8.30)\times100=65.6626\%$ , which is $65.7\%$ to three significant figures.	5

3.3.3.2 · Elimination

Tier 1 · Easy

Mark scheme for 3.3.3.2 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	propene; OH⁻ acts as a base	OH⁻ removes a hydrogen from a carbon next to the C–Br carbon while Br⁻ leaves. A C=C bond forms, giving propene, and proton acceptance identifies OH⁻ as a base.	2
02.1	but-1-ene; the C=C bond must be given the lowest possible locant.	Number the four-carbon chain from the end nearest the double bond. The C=C bond begins at carbon 1, so the product is but-1-ene. Numbering it from the other end to give locant 3 breaks the lowest-locant rule.	2

Tier 2 · Standard

Mark scheme for 3.3.3.2 Tier 2 · Standard
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01.1	Aqueous KOH gives propan-1-ol by nucleophilic substitution; ethanolic KOH gives propene by elimination. OH⁻ donates a lone pair to carbon as a nucleophile in substitution but accepts a proton as a base in elimination.	In aqueous solution, OH⁻ attacks the carbon bonded to Br and replaces Br⁻, so propan-1-ol is favoured; it acts as a nucleophile. In ethanol, OH⁻ removes a hydrogen from the adjacent carbon while the C–H electrons form C=C and Br⁻ leaves, so propene is favoured; it acts as a base.	4
02.1	pent-1-ene, (E)-pent-2-ene and (Z)-pent-2-ene; pent-1-ene has two H atoms on its terminal double-bonded carbon.	Removing H from either carbon adjacent to the C–Br carbon gives pent-1-ene or pent-2-ene. Each carbon of pent-2-ene has two different groups, so it has E and Z forms. The terminal carbon of pent-1-ene is CH₂, so it carries two identical groups and E–Z isomerism is impossible.	4
03.1	The carbon adjacent to the C–Br carbon has no hydrogen atom; elimination requires a hydrogen on an adjacent carbon so its C–H bond can supply the C=C bond.	The carbon bearing Br is CH₂. Its only adjacent carbon is C(CH₃)₃, which already has four C–C bonds and therefore no hydrogen. Hydroxide cannot remove an adjacent hydrogen, so there is no C–H electron pair available to form a double bond between those two carbons. The substrate therefore lacks the required adjacent hydrogen for elimination.	3

Tier 3 · Hard

Mark scheme for 3.3.3.2 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	but-1-ene and but-2-ene O lone pair to a neighbouring H; C–H bond to the C–C bond; C–Br bond to Br.	A hydrogen can be removed from either carbon adjacent to carbon 2. Removal from carbon 1 gives but-1-ene; removal from carbon 3 gives but-2-ene. For either route, draw an arrow from :OH⁻ to the chosen H, a second from that C–H bond into the adjacent C–C bond to make C=C, and a third from C–Br to Br.	5
02.1	2-bromo-2-methylpropane; elimination gives 2-methylpropene; substitution gives 2-methylpropan-2-ol.	A tertiary alcohol with four carbon atoms requires the halogen to be on the tertiary carbon of 2-bromo-2-methylpropane. Its three adjacent methyl groups are equivalent, so removal of H from any one gives the same alkene, 2-methylpropene. One alkene carbon is CH₂, so no E–Z pair is possible. Replacing Br by OH under aqueous conditions gives 2-methylpropan-2-ol.	5
03.1	Elimination: $65.0\%$ ; substitution: $35.0\%$ . OH⁻ is a base that accepts H⁺ in elimination and a nucleophile that donates a lone pair to carbon in substitution.	Each substrate molecule gives one molecule of either organic product. The elimination fraction is $0.0780/0.120=0.650$ , or 65.0%. The alcohol amount is $0.120-0.0780=0.0420\,\mathrm{mol}$ , so substitution is $0.0420/0.120=35.0\%$ . In elimination, OH⁻ accepts an adjacent proton while C=C forms. In substitution, it donates an oxygen lone pair to the carbon bonded to Br and replaces Br⁻.	4
04.1	Substitution arrow 1: from an oxygen lone pair of OH⁻ to carbon 2; arrow 2: from the C–Br bond to Br, forming propan-2-ol and Br⁻. Substitution is favoured by warm aqueous KOH; OH⁻ acts as a nucleophile by donating an electron pair to carbon. Elimination arrow 1: from an oxygen lone pair of OH⁻ to an H on either adjacent methyl group. Elimination arrow 2: from that C–H bond into the adjacent C–C bond, forming the C=C. Elimination arrow 3: from the C–Br bond to Br, giving propene, H₂O and Br⁻. Elimination is favoured by hot ethanolic KOH; OH⁻ acts as a base by accepting H⁺.	For aqueous hydroxide, show the two simultaneous substitution arrows: oxygen lone pair to the carbon bearing Br and C–Br bond to Br. The products are propan-2-ol and Br⁻, and hydroxide is the nucleophile. For hot ethanolic hydroxide, either adjacent methyl group is equivalent. Show all three simultaneous elimination arrows: oxygen lone pair to an adjacent H, that C–H bond pair into the C–C bond to form C=C, and the C–Br bond pair to Br. The products are propene, H₂O and Br⁻, and hydroxide is the base.	6

3.3.3.3 · Ozone depletion

Tier 1 · Easy

Mark scheme for 3.3.3.3 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	It absorbs ultraviolet radiation.	The protective role is absorption of incoming UV radiation, reducing the amount that reaches organisms at Earth's surface.	1
02.1	The chlorine radical is regenerated in a later step, so it acts as a catalyst.	Although $\mathrm{Cl\mathbin{\bullet}}$ is consumed in the first step, it is formed again in the next step. It is therefore unchanged overall and can repeat the cycle.	1

Tier 2 · Standard

Mark scheme for 3.3.3.3 Tier 2 · Standard
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01.1	$\mathrm{2O_3\rightarrow3O_2}$ ; catalyst Cl $\mathbin{\bullet}$ ; intermediate ClO $\mathbin{\bullet}$ .	Add $\mathrm{Cl\mathbin{\bullet}+O_3\rightarrow ClO\mathbin{\bullet}+O_2}$ and $\mathrm{ClO\mathbin{\bullet}+O_3\rightarrow2O_2+Cl\mathbin{\bullet}}$ . Cancel Cl $\mathbin{\bullet}$ and ClO $\mathbin{\bullet}$ to leave $\mathrm{2O_3\rightarrow3O_2}$ . Chlorine radical is used then regenerated, so it is the catalyst; ClO radical is made then consumed, so it is an intermediate.	3
02.1	CFCs are sufficiently stable to reach the upper atmosphere; ultraviolet radiation then causes homolytic C–Cl bond fission and releases chlorine radicals, which catalyse ozone decomposition.	The persistence of CFCs allows them to reach the upper atmosphere without reacting extensively lower down. Higher-energy UV radiation breaks a C–Cl bond homolytically, producing a $\mathrm{Cl\mathbin{\bullet}}$ radical. It is this radical, not an unchanged CFC molecule, that enters the catalytic ozone-depletion cycle.	3
03.1	Independent consistent results make the causal evidence more reliable and less likely to be due to one method or group; a replacement should contain no chlorine that UV can release as Cl radicals.	When separate groups using different measurements obtain compatible evidence, systematic error in one investigation is less likely to explain the conclusion. That reproducibility supports legislation because it strengthens the link between CFCs and ozone loss. A replacement designed to avoid this mechanism must not contain chlorine that upper-atmosphere UV can release as Cl $\mathbin{\bullet}$ radicals.	3

Tier 3 · Hard

Mark scheme for 3.3.3.3 Tier 3 · Hard
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01.1	$\mathrm{CF_2Cl_2\xrightarrow{UV}CF_2Cl\mathbin{\bullet}+Cl\mathbin{\bullet}}$ The chlorine radical is regenerated in a catalytic cycle and can decompose many ozone molecules.	UV light causes homolytic breaking of one C–Cl bond: $\mathrm{CF_2Cl_2\xrightarrow{UV}CF_2Cl\mathbin{\bullet}+Cl\mathbin{\bullet}}$ . The released Cl $\mathbin{\bullet}$ reacts with O₃ to form ClO $\mathbin{\bullet}$ , which reacts with another O₃ and reforms Cl $\mathbin{\bullet}$ . Because the radical is regenerated rather than used up overall, it repeats the cycle and a small CFC amount can remove much more ozone.	4
02.1	$\mathrm{Cl\mathbin{\bullet}+O_3\rightarrow ClO\mathbin{\bullet}+O_2}$ ; $\mathrm{ClO\mathbin{\bullet}+O_3\rightarrow Cl\mathbin{\bullet}+2O_2}$ . Regeneration explains why a small radical amount removes much ozone; $\mathrm{ClO\mathbin{\bullet}}$ is an intermediate.	The first step is $\mathrm{Cl\mathbin{\bullet}+O_3\rightarrow ClO\mathbin{\bullet}+O_2}$ . The second is $\mathrm{ClO\mathbin{\bullet}+O_3\rightarrow Cl\mathbin{\bullet}+2O_2}$ . Chlorine radical is consumed and then regenerated, matching the observation that its amount returns while ozone continues to fall. It is therefore the catalyst, while $\mathrm{ClO\mathbin{\bullet}}$ is formed then consumed and is the intermediate. The overall change is $\mathrm{2O_3\rightarrow3O_2}$ .	5
03.1	$2.30\times10^{-2}\,\mathrm{g}$ of O₃ to three significant figures Cl $\mathbin{\bullet}$ is regenerated in the second step, so one radical repeats the cycle many times; ClO $\mathbin{\bullet}$ is the intermediate.	One cycle destroys two O₃ molecules per chlorine radical. The ozone amount is therefore $3.00\times10^{-9}\times8.00\times10^4\times2=4.80\times10^{-4}\,\mathrm{mol}$ . Its mass is $4.80\times10^{-4}\times48.0=2.304\times10^{-2}\,\mathrm{g}$ , or $2.30\times10^{-2}\,\mathrm{g}$ to three significant figures. Cl $\mathbin{\bullet}$ is regenerated at the end of each cycle, while ClO $\mathbin{\bullet}$ is formed then consumed, so one radical can destroy many ozone molecules before a termination event removes it.	5
04.1	The irradiated-versus-dark comparison supports UV photodissociation as the source of chlorine radicals rather than spontaneous breakdown. Detection of ClO where ozone is lost supports its predicted role as an intermediate in the catalytic cycle. The inverse ClO–ozone relationship is consistent with the mechanism, although correlation alone would not prove causation. Independent reproduction makes a method-specific error or result unique to one research group less likely. The agreement of a controlled photochemical result, a mechanistic intermediate and atmospheric observations provides a strong evidence base for restricting CFC emissions.	Use each observation for the claim it can test. The dark control isolates UV as the cause of C–Cl fission. ClO $\mathbin{\bullet}$ is formed when Cl $\mathbin{\bullet}$ reacts with O₃, so finding ClO where ozone falls is a specific mechanistic prediction, though the field correlation alone cannot establish the whole causal chain. Repetition by independent laboratories reduces the chance of one apparatus or group producing the result. The converging controlled, mechanistic and atmospheric evidence is therefore much stronger than any observation alone and supports precautionary restrictions on CFC use.	5

3.3.4.1 · Structure, bonding and reactivity

Tier 1 · Easy

Mark scheme for 3.3.4.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	A hydrocarbon containing a carbon–carbon multiple bond; ethene contains a C=C bond.	A hydrocarbon contains only carbon and hydrogen. It is unsaturated when it contains a C=C or another carbon–carbon multiple bond; ethene has one C=C double bond.	2
02.1	The electron pair from the C=C bond is donated; the curly arrow starts at the C=C bond.	The double bond is the electron-rich region. Its electron pair supplies the new bond to the electrophile, so the arrow tail must touch the C=C bond rather than a carbon atom or the electrophile.	2

Tier 2 · Standard

Mark scheme for 3.3.4.1 Tier 2 · Standard
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01.1	Ethene contains a C=C double covalent bond; it is a centre of high electron density that attracts electrophiles.	Ethane has only a C–C single bond, whereas ethene has a C=C double covalent bond. The double bond is a centre of high electron density, so electron-deficient electrophiles are attracted to it and ethene undergoes addition reactions.	3
02.1	C=C is a double covalent bond; it is a centre of high electron density; it attracts electrophiles.	Alkenes contain a carbon–carbon double covalent bond. This bond is a centre of high electron density, not low electron density. Electron-pair acceptors are attracted to that region, so the attacking species are electrophiles rather than nucleophiles.	3

Tier 3 · Hard

Mark scheme for 3.3.4.1 Tier 3 · Hard
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01.1	The electron-rich C=C attracts an electron-pair acceptor; an electron pair from the double bond forms a new covalent bond, and a second new bond completes addition as C=C becomes C–C.	The C=C bond is a centre of high electron density. An electrophile is attracted to and accepts an electron pair from the double bond, forming a new covalent bond. Subsequent bonding at the other alkene carbon gives two new bonds to the attacking species overall, while the carbon–carbon bond in the saturated product is single.	4
02.1	Ethene has an electron-rich C=C bond that polarises and attracts Br₂; the C=C electron pair forms one C–Br bond, then Br⁻ forms the second. Ethane has no C=C electron-rich centre and does not undergo this electrophilic addition.	The observation concerns chemical reactivity, not boiling or intermolecular attractions. Electron density in the alkene C=C induces a dipole in approaching Br₂. The C=C electron pair forms one C–Br bond to Br^δ+; Br⁻ then forms the second C–Br bond. Ethane contains only C–C and C–H single bonds and lacks this high-electron-density reaction centre.	4

3.3.4.2 · Addition reactions of alkenes

Tier 1 · Easy

Mark scheme for 3.3.4.2 Tier 1 · Easy
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01.1	orange to colourless; 1,2-dibromocyclohexane	Br₂ adds across the ring's C=C bond, consuming coloured bromine and forming a dibromo compound. The observation is orange to colourless and the product has Br on the two formerly double-bonded carbons: 1,2-dibromocyclohexane.	2
02.1	2,3-dibromobutane; C=C becomes C–C as one Br bonds to each alkene carbon.	Addition uses both atoms from Br₂. One bromine atom bonds to each carbon that was in the double bond, and the carbon–carbon bond in the saturated product is single. The product is therefore 2,3-dibromobutane, not a monobromoalkane.	2

Tier 2 · Standard

Mark scheme for 3.3.4.2 Tier 2 · Standard
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01.1	Arrow from C=C to H in H–Br and from H–Br to Br; secondary CH₃C⁺HCH₃ intermediate; arrow from :Br⁻ to C⁺; product 2-bromopropane.	Draw a curly arrow from the C=C bond to H and another from the H–Br bond to Br. Attach H to the end carbon so the positive charge is on the middle carbon, giving the more stable secondary carbocation CH₃C⁺HCH₃. Then draw an arrow from a lone pair on Br⁻ to C⁺, forming 2-bromopropane.	4
02.1	Arrow from C=C to Br^δ+; arrow from the Br–Br bond to Br^δ−; a positively charged carbon intermediate and Br⁻ form.	Show Br₂ with an induced Br^δ+–Br^δ− dipole. The alkene electron pair moves from the C=C bond to Br^δ+, so that is the first arrow. The Br–Br bonding pair moves from the bond to Br^δ−, giving Br⁻ and a carbocation bearing the first Br substituent. A lone pair on Br⁻ then attacks the C⁺ centre to form the dibromo product.	5
03.1	The two carbons of but-2-ene's symmetrical C=C have equivalent surroundings, so either protonation direction gives the same carbocation and hence the same structural isomer, 2-bromobutane. The carbocation is secondary.	But-2-ene is symmetrical: each alkene carbon is bonded to H and CH₃. Adding H to either end therefore produces the same secondary carbocation at the other carbon, and bromide attack gives the single connectivity CH₃CHBrCH₂CH₃, named 2-bromobutane. Only one structural isomer is possible because both protonation directions converge on the same intermediate.	3

Tier 3 · Hard

Mark scheme for 3.3.4.2 Tier 3 · Hard
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01.1	Major: 2-bromo-2-methylbutane, formed through a tertiary carbocation. Minor: 2-bromo-3-methylbutane, formed through a secondary carbocation.	Proton addition to carbon 3 places the positive charge on carbon 2, which is bonded to three carbon groups and is therefore tertiary. Br⁻ attack gives 2-bromo-2-methylbutane, the major product. Proton addition to carbon 2 instead places the charge on carbon 3, a secondary carbocation; Br⁻ attack gives 2-bromo-3-methylbutane, the minor product. The tertiary intermediate is more stable, so its pathway is favoured.	4
02.1	The alkene was 4-methylpent-1-ene. The secondary carbocation CH₃CH⁺CH₂CH(CH₃)CH₃ leads to the observed major product; the alternative carbocation is primary. Major-pathway arrows: from the C=C bond to H in H–Br, from the H–Br bond to Br, then from a lone pair on Br⁻ to the positively charged carbon.	Both candidate cations preserve a six-carbon skeleton and can arise only by protonating the terminal double bond of 4-methylpent-1-ene. For the favoured route, draw an arrow from the C=C bond to H in H^δ+–Br^δ− so H bonds to carbon 1. Draw a second arrow from the H–Br bond to Br, forming Br⁻ and the secondary carbocation CH₃CH⁺CH₂CH(CH₃)CH₃. Then draw an arrow from a lone pair on Br⁻ to carbon 2, forming 2-bromo-4-methylpentane. Protonation in the opposite orientation would produce the primary carbocation ⁺CH₂CH₂CH₂CH(CH₃)CH₃. The secondary intermediate is more stable, so its product is major.	7
03.1	The alkene is 2-methylpropene, (CH₃)₂C=CH₂. The major product is (CH₃)₃C–OSO₂OH, formed through the tertiary carbocation (CH₃)₃C⁺; the alternative pathway has a primary carbocation, (CH₃)₂CHCH₂⁺. Arrows run from C=C to H of H₂SO₄, from the O–H bond to O, and from an oxygen lone pair of HSO₄⁻ to the positively charged carbon.	Remove one Br from each adjacent brominated carbon and restore the double bond between carbons 1 and 2. This gives (CH₃)₂C=CH₂, 2-methylpropene. For the major sulfuric-acid pathway, draw an arrow from C=C to H of H₂SO₄ so H bonds to the CH₂ carbon, and an arrow from the acid's O–H bond to O. This forms HSO₄⁻ and the tertiary carbocation (CH₃)₃C⁺. The opposite protonation would form the primary carbocation (CH₃)₂CHCH₂⁺. The tertiary carbocation is more stable, so it predominates; an arrow from a lone pair on the negatively charged oxygen of HSO₄⁻ to the positive carbon gives (CH₃)₃C–OSO₂OH as the major product.	6
04.1	Br₂ gives 1,2-dibromo-2-methylpentane. HBr gives 2-bromo-2-methylpentane. HBr also gives 1-bromo-2-methylpentane. The two atoms added from Br₂ are identical, so reversing their orientation across the two alkene carbons gives the same structural product.	Across CH₂=C(CH₃)CH₂CH₂CH₃, Br₂ places one Br on each alkene carbon, giving BrCH₂CBr(CH₃)CH₂CH₂CH₃, named 1,2-dibromo-2-methylpentane. HBr can place Br on carbon 2 to give 2-bromo-2-methylpentane or on carbon 1 to give 1-bromo-2-methylpentane. Br₂ cannot create a second connectivity by reversing orientation because the two added atoms are identical: either assignment still places Br on both carbons 1 and 2.	4

3.3.4.3 · Addition polymers

Tier 1 · Easy

Mark scheme for 3.3.4.3 Tier 1 · Easy
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01.1	[-CH₂-C(CH₃)₂-]_n, with a backbone bond passing through each bracket.	Open the monomer's C=C bond to make the two-carbon C-C backbone. Keep both CH₃ groups attached to the same second carbon, then bracket the unit and continue the backbone bonds through the brackets.	2
02.1	The C=C bond should have opened during addition polymerisation. The correct repeat unit is [-CF₂-CF₂-]_n, with a continuation bond through each bracket.	Addition polymerisation converts the monomer pi bond into two new single bonds to neighbouring repeat units. Retain both fluorine atoms on each backbone carbon, but replace C=C by C-C.	2

Tier 2 · Standard

Mark scheme for 3.3.4.3 Tier 2 · Standard
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01.1	Monomer: CH₂=CHCN. Poly(prop-2-enenitrile), commonly poly(acrylonitrile). Permanent dipole-dipole attractions between polar C≡N groups.	Identify the repeating pair -CH₂-CH(CN)- and restore a C=C bond between those two backbone carbons. The monomer is therefore CH₂=CHCN. Name the polymer by placing the systematic monomer name in poly(...); the polar nitrile bonds give permanent dipole-dipole attraction.	4
02.1	The monomers are propene, CH₂=CHCH₃, and chloroethene, CH₂=CHCl. Both monomers form a saturated C-C addition-polymer backbone. Changing the proportion of polar C-Cl groups changes the permanent dipole-dipole attractions between chains, so chain movement and physical properties can change.	Restore a C=C bond separately across each pair of backbone carbons without moving its substituent. Then distinguish the unchanged saturated backbone from the variable side groups that control attractions between chains.	4
03.1	The repeat unit is [-CH₂-CH(CH₂CH₂CH₃)-]_n. $M_{\mathrm r}$ of C₅H₁₀ is $70.0$ , so the average number of repeat units is $(4.20\times10^5)/70.0=6.00\times10^3$ .	Open the C=C bond of pent-1-ene without moving its propyl substituent. Addition polymerisation loses no atoms, so one repeat unit has the monomer formula C₅H₁₀ and $M_{\mathrm r}=5(12.0)+10(1.0)=70.0$ . Divide the polymer's average relative molecular mass by this value to obtain $6000$ repeat units.	3

Tier 3 · Hard

Mark scheme for 3.3.4.3 Tier 3 · Hard
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01.1	C-Cl bonds make neighbouring PVC chains polar, so permanent dipole-dipole attractions act between them. In unplasticised PVC these attractions hold chains close together and restrict movement, making the material rigid. Plasticiser molecules fit between chains, increase their separation and weaken the total attraction between polymer chains, so the chains slide more easily. The polymer backbone contains strong C-C bonds and no hydrolysable functional group, so water does not readily split the chains.	Link the polar C-Cl bonds first to attractions between chains, then link those attractions to restricted chain movement. A plasticiser separates the chains and reduces those effective attractions, increasing flexibility. Finally inspect the backbone: addition polymerisation leaves only robust carbon-carbon links, not ester or amide links that water could hydrolyse.	6
02.1	The monomer is CH₂=C(CH₃)COOCH₂CH₃, ethyl 2-methylprop-2-enoate. Hydroxide ions can hydrolyse the ester side groups, producing carboxylate groups and ethanol. The main chain contains C-C bonds formed by opening the monomer C=C bond and contains no ester link, so ester hydrolysis does not cut the backbone.	Restore the monomer double bond between the two backbone carbons and retain both substituents on the same carbon. Locate the ester in the pendant group rather than the backbone; hydrolysis therefore changes that group while leaving the carbon-carbon chain continuous.	6
03.1	The monomers are tetrafluoroethene, CF₂=CF₂, and ethene, CH₂=CH₂. Their unit ratio in the stated sample is $4:6=2:3$ . The four fluorinated units contain fluorine mass $4(4)(19.0)=304$ . The total relative mass of the ten units is $4[2(12.0)+4(19.0)]+6[2(12.0)+4(1.0)]=568$ , so fluorine is $(304/568)\times100=53.5\%$ by mass.	Restore C=C separately across each two-carbon backbone unit to identify the monomers. Count units directly for the composition ratio. A C₂F₄ unit has relative mass $100$ and a C₂H₄ unit has relative mass $28$ ; use all ten units in the denominator but only the sixteen fluorine atoms in the numerator.	6
04.1	Poly(ethene) and poly(propene) are non-polar, so the attractions between their chains are London forces. These London forces arise from temporary dipoles inducing dipoles in neighbouring chains. PVC has polar C-Cl bonds, so permanent dipole-dipole attractions act between its chains in addition to London forces. A longer polyalkene chain contains more electrons and has more points of contact with neighbouring chains, so its London forces are stronger. More energy is therefore needed to separate longer chains, so a longer-chain sample softens at a higher temperature than a shorter-chain sample of the same polymer.	Classify the hydrocarbon polyalkenes as non-polar, then identify the permanent bond dipoles introduced by C-Cl bonds in PVC. For the chain-length trend, link a larger electron cloud and greater chain contact to stronger London forces, then connect stronger attractions to the extra thermal energy required for softening.	5

3.3.5.1 · Alcohol production

Tier 1 · Easy

Mark scheme for 3.3.5.1 Tier 1 · Easy
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01.1	C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂. Use yeast at about 30-40 °C and exclude oxygen; an aqueous glucose solution is also required.	Balance six carbon atoms by making two ethanol and two carbon dioxide molecules; this also balances hydrogen and oxygen. A moderate temperature keeps yeast enzymes active without denaturing them, while anaerobic conditions favour fermentation rather than aerobic respiration.	3
02.1	The temperature is too high, so yeast enzymes have denatured and fermentation has stopped. Use a fresh yeast culture and maintain a warm temperature of about 30-40 °C under anaerobic conditions.	Use the absence of carbon dioxide as evidence that fermentation is not occurring. Anaerobic conditions are already suitable, so diagnose the excessive temperature and restore active enzymes rather than adding oxygen.	3

Tier 2 · Standard

Mark scheme for 3.3.5.1 Tier 2 · Standard
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01.1	The propene pi bond accepts H⁺, forming the more stable secondary carbocation. A water lone pair attacks the carbocation to make a protonated alcohol. Loss of H⁺ forms propan-2-ol and regenerates the acid catalyst.	Use the electron-rich C=C bond to attack H⁺ in the first step, placing the positive charge on the middle carbon. Water then donates a lone pair to that carbon. Deprotonation gives CH₃CH(OH)CH₃; because the H⁺ consumed first is released last, it is catalytic.	4
02.1	React ethene with steam using a phosphoric acid catalyst. The acid supplies a lower-activation-energy catalytic route and is regenerated. The reaction is C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g). Higher pressure favours the side with fewer gas molecules, increasing the equilibrium yield of ethanol.	Replace liquid water by steam and add the acid catalyst; the temperature and pressure are supplied rather than recalled. Use the gaseous equilibrium stoichiometry to justify pressure, and keep the catalyst explanation kinetic rather than claiming that it changes equilibrium yield.	5
03.1	Hydration has $100\%$ atom economy because ethanol is the only product and all reactant atoms enter it. Fermentation has ethanol atom economy $[2(46.0)/180]\times100=51.1\%$ . The remaining glucose atoms form the CO₂ coproduct, lowering atom economy for the desired ethanol. A sustainability comparison must also include whether the feedstock is renewable, energy for reaction and purification, lifecycle emissions and land use; high atom economy alone is not decisive.	For each balanced equation, divide the formula mass of desired ethanol by the total formula mass of reactants, including the coefficient of two in fermentation. Then keep this material-efficiency metric separate from renewability, process energy and wider environmental effects.	4

Tier 3 · Hard

Mark scheme for 3.3.5.1 Tier 3 · Hard
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01.1	$176\,\text{kg}$ of CO₂. Growing the glucose crop can remove the same carbon from atmospheric CO₂, so the combustion carbon can form a short carbon cycle. The description is not fully justified if cultivation, fertiliser, distillation or transport uses fossil energy; land and food-crop competition also matter.	The amount of ethanol is $92.0/46.0=2.00\,\text{kmol}$ . The equation gives twice as much CO₂, so $n(\mathrm{CO_2})=4.00\,\text{kmol}$ and $m=4.00\times44.0=176\,\text{kg}$ . Carbon neutrality concerns the whole life cycle, not only this combustion equation.	6
02.1	$7.36\,\text{kg}$ of ethanol. Fermentation can use renewable plant-derived glucose and operates at a low temperature. It is slow or batch-based and produces dilute ethanol that requires energy-intensive fractional distillation; land use and lifecycle emissions also weaken a carbon-neutrality claim.	$n(\text{glucose})=18000/180=100\,\text{mol}$ . The theoretical ethanol amount is $200\,\text{mol}$ , with mass $200\times46.0=9.20\,\text{kg}$ . Applying the 80.0% yield gives $9.20\times0.800=7.36\,\text{kg}$ . Compare the whole processes, not only their feedstocks.	6
03.1	Choose 35 °C because it gives the greatest final ethanol concentration, so it gives the best sustained production of the three temperatures. The larger initial CO₂ rate at 50 °C shows that reactions are initially faster at the higher temperature. The low final ethanol concentration at 50 °C indicates that yeast enzymes denature or yeast cells become inactive during the run. At 25 °C the enzymes remain active but reactions are slower, so less ethanol is formed in 48 h. All runs are anaerobic, preventing oxygen from diverting glucose metabolism away from ethanol production or promoting oxidation of ethanol.	Separate an early rate measurement from the accumulated yield after 48 h. The 50 °C run begins fastest but does not maintain activity, whereas 35 °C combines a useful enzyme-controlled rate with the highest final ethanol concentration. The controlled anaerobic condition is necessary in every comparison.	5
04.1	The first pass forms $100(0.400)=40.0\,\mathrm{mol}$ of ethanol. $60.0\,\mathrm{mol}$ of ethene remains after the first pass. The recovered amount is $60.0(0.900)=54.0\,\mathrm{mol}$ , and its second pass forms $54.0(0.400)=21.6\,\mathrm{mol}$ of ethanol. The two passes form $40.0+21.6=61.6\,\mathrm{mol}$ of ethanol in total. The ethanol mass is $61.6\times46.0=2.83\times10^3\,\mathrm g$ , or $2.83\,\mathrm{kg}$ . The hydration is reversible and reaches equilibrium, so ethene is not fully converted in one pass.	Apply the stated fractional conversion to the fresh feed, not to the original feed again on the second pass. Recover only 90.0% of the first-pass remainder, apply the same reactor conversion to that recovered stream, then combine the two ethanol amounts before converting to mass. The reversible hydration establishes equilibrium before all the ethene can react in one pass.	6
05.1	Lowering temperature favours the exothermic forward reaction, so the equilibrium ethanol yield and $K_p$ increase. Lowering temperature reduces reaction rate because fewer collisions have enough energy to react. Increasing pressure shifts equilibrium towards the one-mole gas side, so the equilibrium ethanol yield increases. Increasing pressure raises gas concentrations, so collisions are more frequent and the reaction rate increases. A catalyst provides an alternative route with a lower activation energy and increases the rates of both directions, so equilibrium is established faster; it does not change the relative stability of reactants and products, so changes neither $K_p$ nor the equilibrium composition. Industry balances a useful rate and yield against the energy demand and equipment cost or hazard of high pressure.	Treat equilibrium position, equilibrium constant and kinetics separately. Use the negative enthalpy to predict the temperature effect and the two-gas-moles-to-one stoichiometry for pressure. A catalyst changes the route and time to equilibrium only. The chosen plant conditions must therefore trade the thermodynamic benefits against rate, compression cost and safe containment.	6

3.3.5.2 · Oxidation of alcohols

Tier 1 · Easy

Mark scheme for 3.3.5.2 Tier 1 · Easy
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01.1	Butanone; the solution changes from orange to green.	Butan-2-ol is secondary because its OH-bearing carbon is bonded to two other carbons. Oxidation therefore forms the ketone butanone, while dichromate(VI) ions are reduced from orange to green chromium(III) ions.	2
02.1	The product is propanone. Propan-2-ol is a secondary alcohol, so oxidation forms a ketone; propanal would come from the primary alcohol propan-1-ol.	Count the carbon groups attached to the carbon bearing OH. Two carbon groups make this alcohol secondary, and removing hydrogen from that carbon and from O-H produces a C=O group within the chain.	2

Tier 2 · Standard

Mark scheme for 3.3.5.2 Tier 2 · Standard
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01.1	Use acidified potassium dichromate(VI), warm gently and distil the aldehyde as it forms. (CH₃)₂CHCH₂OH + [O] → (CH₃)₂CHCHO + H₂O. On warming, Tollens' reagent gives a silver mirror and Fehling's solution gives a brick-red precipitate.	The starting compound is a primary alcohol, so one oxidation step gives an aldehyde. Gentle heating supplies the reaction rate, and immediate distillation removes the volatile aldehyde from contact with oxidant before it undergoes the second oxidation step. The isolated aldehyde reduces Tollens' silver ions and Fehling's copper(II) ions, giving the stated positive observations.	6
02.1	Identify 3-methylbutan-1-ol because it is a primary alcohol and can be oxidised to 3-methylbutanal. Warm and distil the aldehyde as it forms so that it leaves the oxidising mixture before further oxidation to 3-methylbutanoic acid. 2-Methylbutan-2-ol is a tertiary alcohol and is not readily oxidised by acidified potassium dichromate(VI), so the mixture stays orange and no carbonyl product forms.	Classify the two alcohols from the number of carbon groups attached to the OH-bearing carbon. Select the primary alcohol, then remove its aldehyde promptly by distillation; the tertiary isomer cannot supply the required carbonyl product by this oxidation.	5
03.1	$n(\text{butanone})=3(0.0300)=0.0900\,\mathrm{mol}$ because one secondary-alcohol molecule gives one ketone molecule. $m=0.0900\times72.0=6.48\,\mathrm g$ . The dichromate(VI) mixture changes from orange to green as chromium(III) ions form.	Use the supplied redox ratio before the organic 1:1 ratio: $0.0300$ mol of dichromate oxidises $0.0900$ mol of alcohol. Butan-2-ol is secondary, so each oxidised molecule forms one molecule of butanone and is not carried on to a carboxylic acid by this reagent.	4

Tier 3 · Hard

Mark scheme for 3.3.5.2 Tier 3 · Hard
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01.1	Butan-1-ol gives butanoic acid; butan-2-ol gives butanone; 2-methylpropan-2-ol shows no easy oxidation. Add sodium carbonate: butanoic acid effervesces because CO₂ forms, whereas butanone does not react. Tollens' reagent would not distinguish these final products because neither gives a positive aldehyde result.	Classify the alcohols by counting carbon groups on the OH-bearing carbon: primary, secondary and tertiary respectively. Reflux with excess oxidant carries the primary alcohol to the acid and the secondary alcohol to the ketone. Then choose a test for the acid functional group, such as carbonate and effervescence, rather than an aldehyde test.	6
02.1	First heat the secondary alcohol with acidified potassium dichromate(VI) under reflux so reactants are not lost and oxidation can approach completion. A secondary alcohol forms a ketone, which is not readily oxidised further by this reagent, so keeping the product in the hot mixture does not convert it into a carboxylic acid. After reaction, rearrange the apparatus for distillation and collect the ketone over its boiling range, leaving less volatile reagents in the flask.	Use reflux for reaction completion, based on the secondary-alcohol oxidation limit, and distillation afterward for isolation. These are two successive apparatus functions, not alternative names for the same heating arrangement.	5
03.1	X is pentan-2-ol and Y is pentan-2-one. Carbon 2 in pentan-2-ol is bonded to H, OH, CH₃ and CH₂CH₂CH₃, so it is chiral and the alcohol is secondary. Oxidation of a secondary alcohol forms a ketone without changing the carbon skeleton. Pentan-2-one has five non-equivalent carbon environments, consistent with five ¹³C signals. A ketone is not readily oxidised by Tollens' reagent, so it gives no silver mirror.	Start with the oxidation class: a secondary alcohol must produce a ketone. Test the possible five-carbon secondary alcohols against the chirality evidence, then retain the unbranched skeleton because its ketone has five distinct carbon positions. Use the negative Tollens result as confirmation of ketone rather than aldehyde chemistry.	5
04.1	The starting amount of butan-1-ol is $4.00/74.0=0.054054\,\mathrm{mol}$ . Complete oxidation has 1:1 organic stoichiometry, so the theoretical butanoic acid amount is $0.054054\,\mathrm{mol}$ . The titre shows that $n(\text{butanoic acid})=1.00(42.50/1000)=0.04250\,\mathrm{mol}$ was isolated. The percentage yield is $(0.04250/0.054054)\times100=78.6\%$ . Reflux permits prolonged heating without loss of volatile material, allowing the primary alcohol and intermediate aldehyde to remain with excess oxidant until carboxylic acid forms.	Convert the alcohol mass to moles and use the unchanged carbon skeleton to obtain the theoretical acid amount. The monoprotic acid reacts one-to-one with NaOH, so the titre measures the isolated acid amount directly. Divide actual moles by theoretical moles. Reflux is chosen because the target is the fully oxidised acid rather than an aldehyde removed by distillation.	5
05.1	The organic product is propanone, CH₃COCH₃. The oxidation half-equation is $\mathrm{CH_3CH(OH)CH_3\rightarrow CH_3COCH_3+2H^++2e^-}$ . Multiplying the organic half-equation by three supplies the six electrons consumed by one dichromate ion. The overall equation is $\mathrm{3CH_3CH(OH)CH_3+Cr_2O_7^{2-}+8H^+\rightarrow3CH_3COCH_3+2Cr^{3+}+7H_2O}$ . The alcohol-to-dichromate mole ratio is $3:1$ . The mixture changes from orange dichromate(VI) to green chromium(III).	Oxidising a secondary alcohol to a ketone removes two hydrogen atoms, so place two H⁺ and two electrons on the product side of the organic half-equation. Multiply that half-equation by three, add the supplied reduction half-equation and cancel six electrons and six of the fourteen H⁺ ions. The coefficients then give the mole ratio, while chromium changes from oxidation state +6 to +3.	6

3.3.5.3 · Elimination

Tier 1 · Easy

Mark scheme for 3.3.5.3 Tier 1 · Easy
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01.1	Ethene; acid-catalysed elimination (dehydration). CH₃CH₂OH → CH₂=CH₂ + H₂O.	Remove OH from one carbon and H from the adjacent carbon, then place a double bond between those two carbons. One ethanol molecule therefore gives ethene and water.	2
02.1	The curly arrow starts at the neighbouring C-H bond. It ends between that carbon and the positively charged carbon, forming the C=C pi bond as H⁺ is lost.	A curly arrow follows an electron pair. The electrons supplied by the C-H bond become the new pi bond, so an arrow cannot start at the proton.	2

Tier 2 · Standard

Mark scheme for 3.3.5.3 Tier 2 · Standard
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01.1	2-Ethylbut-1-ene and 3-methylpent-2-ene. A beta hydrogen can be removed from an end ethyl group or from the methyl substituent; the two ethyl groups are equivalent, so removal from either one gives the same 3-methylpent-2-ene structure.	Mark the carbon bearing OH, then inspect every adjacent carbon for a removable H. Forming the double bond towards either equivalent ethyl group gives 3-methylpent-2-ene; forming it towards the methyl group gives CH₂=C(CH₂CH₃)₂, named 2-ethylbut-1-ene from the longest chain containing C=C.	4
02.1	Choose pentan-1-ol to make pent-1-ene. Heat with concentrated sulfuric acid or concentrated phosphoric acid. Only carbon 2 is adjacent to the OH-bearing carbon in pentan-1-ol, so elimination gives one C=C position. Pentan-2-ol and 2-methylbutan-2-ol have hydrogen-bearing carbons on two different adjacent positions, so each can form two structural alkenes.	For each candidate, inspect every carbon adjacent to the carbon bearing OH. The terminal alcohol has only one possible adjacent carbon from which elimination can form a double bond.	5
03.1	One precursor is 2-methylbutan-2-ol, (CH₃)₂C(OH)CH₂CH₃. The other is 3-methylbutan-2-ol, CH₃CH(OH)CH(CH₃)CH₃. Reverse dehydration by adding H and OH across the C=C bond in the first orientation to place OH on one alkene carbon, then in the second orientation to place OH on the other. Each alcohol has an H on the carbon adjacent to its OH-bearing carbon, so loss of H₂O can restore the same C=C bond.	Work backwards without moving carbon groups. Add the elements of water across the double bond in both possible orientations, then name each resulting alcohol from the longest chain containing OH.	4

Tier 3 · Hard

Mark scheme for 3.3.5.3 Tier 3 · Hard
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01.1	An oxygen lone pair attacks H⁺, producing protonated cyclohexanol; the acid O-H bond returns to the acid oxygen if the acid is drawn explicitly. The C-O bond pair moves to oxygen so H₂O leaves. The adjacent C-H bond pair forms the C=C bond as H⁺ is lost, regenerating the acid catalyst; no base-attack arrow is drawn. Cyclohexene or another alkene made by dehydrating a fermentation-derived alcohol can undergo addition polymerisation, replacing a crude-oil-derived monomer route.	First turn poor leaving group OH into neutral water by protonation. Next break C-O heterolytically, keeping that electron pair on oxygen. Finally start the curly arrow at an adjacent C-H bond and end it between the two carbons while H⁺ leaves. Because H⁺ is returned, the sequence is catalytic; coupling alcohol production from biomass with dehydration supplies an alkene feedstock.	6
02.1	The OH group must first be protonated so that neutral H₂O, not OH⁻, can leave. Loss of water gives a positively charged tertiary carbocation at carbon 2. The final electron-pair arrow starts at an adjacent C-H bond and ends between the adjacent carbon and carbon 2 to form C=C; H⁺ is regenerated. The structural products are 2-methylbut-1-ene and 2-methylbut-2-ene.	Track charge after each electron-pair movement. Protonation makes a viable neutral leaving group, heterolytic C-O bond breaking leaves a carbocation, and electrons from either distinct adjacent C-H position can restore a four-bond carbon by forming either alkene.	6
03.1	The labelled oxygen leaves in H₂¹⁸O, whose singly charged molecular ion has $m/z=2(1)+18=20$ . The products are but-1-ene, (E)-but-2-ene and (Z)-but-2-ene. A lone pair on the labelled alcohol oxygen attacks H⁺, making the labelled OH group into a neutral water leaving group. The C-O bond pair moves to oxygen, so H₂¹⁸O leaves and the organic intermediate contains no oxygen. An electron pair from a C-H bond on either adjacent carbon forms C=C to give the possible alkene positions. H⁺ is lost in this final step, regenerating the acid catalyst.	Track the labelled atom rather than only balancing the overall reaction. Protonation and heterolytic C-O cleavage carry the alcohol oxygen into water; subsequent electron movement occurs entirely in the carbon skeleton. Inspect both adjacent carbons for alkene positions, apply E/Z to but-2-ene and account for return of H⁺.	6

3.3.6.1 · Identification of functional groups by test-tube reactions

Tier 1 · Easy

Mark scheme for 3.3.6.1 Tier 1 · Easy
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01.1	Shake with bromine water; its orange colour is discharged to colourless.	Use bromine water at room temperature. An alkene adds bromine across C=C, consuming coloured Br₂, so record the specific change orange to colourless.	2
02.1	A carbon-carbon double bond is present because bromine water is decolourised. A carboxylic acid group is present because carbonate produces effervescence. The gas is carbon dioxide.	Treat each positive result independently: bromine adds across C=C, while an acid reacts with carbonate to release CO₂. Fresh portions prevent the first reagent from changing the second test mixture.	3

Tier 2 · Standard

Mark scheme for 3.3.6.1 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Add sodium carbonate to fresh portions: ethanoic acid effervesces and ethanal does not; the gas is CO₂. Warm fresh portions with Tollens' reagent: ethanal forms a silver mirror, while ethanoic acid gives no silver mirror.	Use carbonate to test acidity and Tollens' reagent to test an aldehyde. The two independent positive results cross-check the assignment instead of relying only on the absence of a reaction.	4
02.1	Test each sample with bromine water: cyclohexene decolourises it from orange to colourless; the other two do not. Tollens' reagent gives no reaction with both cyclohexanol and cyclohexanone, so it cannot distinguish them. Instead, warm fresh portions of the remaining two samples with acidified potassium dichromate(VI). Cyclohexanol changes the dichromate solution from orange to green; cyclohexanone gives no change. Also accept sodium: effervescence with cyclohexanol only; or acidified potassium manganate(VII): purple to colourless with cyclohexanol only.	Audit the proposed negative results before accepting the sequence. After the alkene is identified, choose a reagent that reacts differently with a secondary alcohol and ketone, and use fresh portions so the first test cannot contaminate the second.	5
03.1	Fresh portions prevent one reagent reacting with, diluting or contaminating the sample before a later test. A water bath gives controlled gentle heating and avoids heating the Tollens mixture directly with a flame. A known aldehyde should give a silver mirror under the same conditions. Its positive result shows that the freshly prepared Tollens' reagent and heating procedure work, so a negative unknown result is meaningful.	Treat validity as part of identification. Independent portions keep the chemical tests independent; controlled warming supplies the condition for the aldehyde reaction; a positive control separates a genuine negative result from failure of the reagent or procedure.	4

Tier 3 · Hard

Mark scheme for 3.3.6.1 Tier 3 · Hard
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01.1	Test fresh portions with sodium carbonate: the sample that effervesces CO₂ is ethanoic acid. Warm fresh portions of the other samples with Tollens' reagent: the silver-mirror sample is ethanal. Shake fresh portions of the remaining two with bromine water: orange to colourless identifies ethene. The remaining sample is ethanol; warming it with acidified dichromate confirms this by an orange-to-green change.	Remove the acid first with the selective carbonate test, then remove the aldehyde with Tollens' reagent. Bromine water distinguishes the remaining alkene from alcohol. The final dichromate result is a positive confirmation of ethanol rather than an assignment by elimination alone; fresh portions prevent one reagent changing the next result.	6
02.1	The structural formula is CH₂=CHCOOH, prop-2-enoic acid. Bromine-water decolourisation identifies C=C and carbonate effervescence identifies COOH; the negative Tollens result is consistent with no aldehyde group. 2CH₂=CHCOOH + Na₂CO₃ → 2CH₂=CHCOONa + H₂O + CO₂.	Combine both positive tests within the supplied formula rather than assigning only one group. One carboxylic acid group uses two oxygen atoms, and the remaining unsaturation is a C=C bond. Balance the acid-carbonate equation using two acid molecules per carbonate ion.	6
03.1	The single gas-chromatography peak supports one component, so the observations are consistent with one pure compound rather than a mixture. No initial carbonate effervescence shows that the liquid is not a carboxylic acid. The negative Tollens' test shows that an aldehyde group is not present. Aqueous sodium hydroxide hydrolyses an ester; acidification converts the carboxylate product into a carboxylic acid, which then effervesces with sodium carbonate. One possible structure consistent with C₄H₈O₂ is CH₃COOCH₂CH₃, ethyl ethanoate. The evidence identifies an ester but does not distinguish every ester isomer, so it supports rather than uniquely proves ethyl ethanoate.	Use the single chromatographic peak to assess purity, then compare the tests before and after hydrolysis. The original liquid is neither an acid nor an aldehyde, but alkaline hydrolysis followed by acidification generates a carbonate-positive carboxylic acid. This supports an ester such as ethyl ethanoate, while the lack of isomer-specific evidence prevents a unique identification.	6
04.1	The compound amount is $0.720/144=5.00\times10^{-3}\,\mathrm{mol}$ . The carbon dioxide amount is $(120/1000)/24.0=5.00\times10^{-3}\,\mathrm{mol}$ . The 1:1 molecular-to-CO₂ ratio means that each molecule supplies two acidic protons, so it contains two COOH groups. Bromine-water decolourisation identifies a C=C bond. Different carbon skeletons, C=C positions or E-Z arrangements can contain two COOH groups and one C=C bond, so further evidence is required for a unique structure.	Convert both sample mass and gas volume to amounts. The carbonate equation consumes two carboxylic acid groups for every CO₂ molecule formed. Since one mole of compound gives one mole of gas, each molecule contains two acid groups. Treat the fresh-portion bromine result independently, then distinguish functional-group identification from complete structural identification.	5
05.1	Warm fresh samples with silver nitrate in ethanol; the bromoalkane gives a cream precipitate. The precipitate is silver bromide, so this result identifies the compound containing a C-Br bond as the halogenoalkane. Shake fresh samples with orange bromine water; the alkene rapidly decolourises it to colourless. Decolourisation shows that bromine has added across a C=C bond, so this result identifies the alkene. Warm fresh samples with Tollens' reagent; the aldehyde forms a silver mirror, or a grey precipitate of silver. Reduction of Tollens' reagent identifies the aldehyde group, so the liquid giving this result is the aldehyde.	Use a fresh portion for each reagent. Match the silver-halide precipitate to the halogenoalkane, bromine-water decolourisation to addition at C=C and silver formation with Tollens' reagent to oxidation of an aldehyde. State both what is seen and what functional group that observation supports.	6

3.3.6.2 · Mass spectrometry

Tier 1 · Easy

Mark scheme for 3.3.6.2 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$44.02623$	Add the isotope contributions: $2(12.00000)+4(1.00783)+15.99491=44.02623$ . Keep the precision supplied because rounding to $44$ would discard the evidence used for formula identification.	2
02.1	The student used whole-number relative masses for H and O instead of the exact isotopic masses. The precise molecular mass is 60.05755.	Calculate $3(12.00000)+8(1.00783)+15.99491=36.00000+8.06264+15.99491=60.05755$ .	3

Tier 2 · Standard

Mark scheme for 3.3.6.2 Tier 2 · Standard
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01.1	C₃H₆O₂; its calculated mass is $74.03680$ , compared with $74.07321$ for C₄H₁₀O.	For C₄H₁₀O, calculate $4(12.00000)+10(1.00783)+15.99491=74.07321$ . For C₃H₆O₂, calculate $3(12.00000)+6(1.00783)+2(15.99491)=74.03680$ . The second value differs from the measurement by only $0.00010$ .	3
02.1	C₃H₇NO₂ has precise mass 89.04770; C₄H₁₁NO has precise mass 89.08411. The formula is C₃H₇NO₂. The candidate masses are separated by 89.08411 - 89.04770 = 0.03641.	For C₃H₇NO₂, calculate $3(12.00000)+7(1.00783)+14.00307+2(15.99491)=89.04770$ . For C₄H₁₁NO, calculate $4(12.00000)+11(1.00783)+14.00307+15.99491=89.08411$ . The measured mass selects the first formula, and the 0.03641 separation exceeds 0.03.	5
03.1	Both values round to nominal $m/z=72$ , so a whole-number measurement gives the same result for both ions. Only ¹²C has an exactly integer precise mass; replacing CH₄ by O changes the precise total by $0.03641$ while leaving the nominal mass unchanged. High-resolution (precise) mass measurement therefore distinguishes the two molecular formulae.	Compare the candidates after rounding both to whole numbers — they coincide, so nominal mass cannot separate them. The formulae differ by CH₄ versus O, and because the precise masses of ¹H and ¹⁶O are not exact integers this swap shifts the precise mass by $0.03641$ without changing the nominal mass.	3

Tier 3 · Hard

Mark scheme for 3.3.6.2 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	C₄H₈O₂: $88.05246$ ; C₅H₁₂O: $88.08887$ . The formula is C₄H₈O₂, with absolute mass error $\|88.05260-88.05246\|=0.00014$ .	Calculate $4(12.00000)+8(1.00783)+2(15.99491)=88.05246$ and $5(12.00000)+12(1.00783)+15.99491=88.08887$ . The first candidate is much closer to $88.0526$ . Subtract without attaching a sign because the question asks for absolute error: $0.00014$ .	5
02.1	C₅H₈O₂: 100.05246; C₆H₁₂O: 100.08887. The formula is C₅H₈O₂; its calculated mass differs from the measured value by only 0.00006. Aldehyde is one functional group present: it accounts for C=O, the absent O-H band and the positive Tollens result.	Calculate $5(12.00000)+8(1.00783)+2(15.99491)=100.05246$ and $6(12.00000)+12(1.00783)+15.99491=100.08887$ . Use the mass to choose the formula, then combine both spectroscopic and test-tube evidence to assign an aldehyde rather than claiming the whole structure is uniquely known.	6
03.1	The five C and two O atoms contribute $5(12.00000)+2(15.99491)=91.98982$ . The remaining mass is $102.06810-91.98982=10.07828$ , corresponding to $10.07828/1.00783\approx10$ H atoms. The molecular formula is C₅H₁₀O₂. Its calculated precise mass is $102.06812$ , with absolute difference $0.00002$ from the measurement. Different structural or functional-group isomers can share one molecular formula, so other evidence such as spectra or reactions is needed.	Subtract the known carbon and oxygen contributions before dividing the residual exact mass by the hydrogen isotope mass. Recalculate the complete formula as a check, then distinguish formula determination from structural identification.	5
04.1	For a $100\,\mathrm g$ sample: C $=40.0/12.0=3.333\,\mathrm{mol}$ , H $=6.67/1.0=6.67\,\mathrm{mol}$ and O $=53.3/16.0=3.331\,\mathrm{mol}$ , giving the ratio $1:2:1$ . The empirical formula is CH₂O. The precise empirical-unit mass is $12.00000+2(1.00783)+15.99491=30.01057$ . $90.0317/30.01057\approx3$ , so the molecular formula is C₃H₆O₃. Its calculated precise mass is $90.03171$ , giving an absolute difference of $0.00001$ from the measured mass.	Assume a 100 g sample and convert each percentage to moles using ordinary relative atomic masses. Reduce the mole ratio to obtain the empirical formula. Then switch to precise isotopic masses, calculate one empirical unit's exact mass and compare it with the molecular-ion measurement to obtain the integer multiplier and final exact-mass check.	5

3.3.6.3 · Infrared spectroscopy

Tier 1 · Easy

Mark scheme for 3.3.6.3 Tier 1 · Easy
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01.1	A C=O bond.	Locate 1718 cm^-1 in the carbonyl range in the Data Booklet. This establishes the presence of C=O but does not by itself distinguish an aldehyde, ketone, acid or ester.	1
02.1	The absorption supports the presence of a C=O bond but does not by itself prove propanone. Many aldehydes, ketones, acids and esters absorb in the carbonyl region. The fingerprint region must match a reference spectrum of propanone, with other diagnostic absorptions also considered.	Use a characteristic absorption to identify a bond or group, not a unique molecule. Identification requires agreement of the fingerprint region with an authentic reference spectrum.	3

Tier 2 · Standard

Mark scheme for 3.3.6.3 Tier 2 · Standard
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01.1	The compound is ethyl ethanoate. The C=O peak and absence of O-H support an ester, while the matching fingerprint region identifies the particular ester rather than only its functional group.	First use the diagnostic region: C=O is present and an alcohol or acid O-H is absent. Several esters could fit those facts, so compare the complex fingerprint pattern; an identical pattern under the same conditions provides the molecule-specific match.	3
02.1	Spectrum A shows the alcohol O-H group of propan-2-ol. Spectrum B shows that O-H has been lost and a carbonyl group has formed. The product is propanone; the negative Tollens result supports a ketone rather than an aldehyde. The change is oxidation of a secondary alcohol to a ketone.	Compare the spectra rather than reading either alone. The disappearing alcohol band and appearing carbonyl band show oxidation, and the Tollens result fixes the carbonyl class.	5
03.1	The strong C=O absorption together with the very broad 2500-3000 cm^-1 O-H absorption identifies a carboxylic acid group. The 1620-1680 cm^-1 absorption indicates a C=C bond. More than one compound or positional isomer can contain both groups. A matching fingerprint region against a reference spectrum, or further chemical evidence, is needed for a specific identity.	Interpret the carbonyl and broad acid O-H absorptions together rather than as unrelated peaks. Add the alkene evidence, but stop at functional groups because characteristic ranges are shared by many molecules; molecular identification needs the fingerprint pattern or another independent constraint.	4

Tier 3 · Hard

Mark scheme for 3.3.6.3 Tier 3 · Hard
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01.1	A possible impurity is propan-2-ol or water because the extra broad absorption indicates O-H. Bonds in molecules such as CO₂, CH₄ and H₂O absorb outgoing infrared radiation at matching vibrational frequencies. Energy is redistributed by re-emission and collisions, so less energy escapes directly to space and the atmosphere warms.	Treat the unexpected O-H band as impurity evidence while the fingerprint still supports propanone as the main component. For the climate explanation, connect the characteristic bond vibrations of greenhouse gases to absorption of terrestrial IR and then to redistribution and retention of energy in the atmosphere.	5
02.1	The compound is 2-methylpropanal, (CH₃)₂CHCHO. The 1725 cm^-1 absorption identifies C=O and the absence of broad O-H excludes an alcohol or carboxylic acid. The matching fingerprint region identifies the specific compound. The brick-red precipitate is the positive Fehling's result expected from an aldehyde.	Use the molecular formula and functional-group region to narrow the class, the fingerprint match for molecular identity, and the independent test-tube result to confirm aldehyde reactivity.	6
03.1	The strong absorption near 1720 cm^-1 identifies a C=O bond. The absence of a broad 3230-3550 cm^-1 absorption shows that an alcohol O-H group is not present. No 1000-1300 cm^-1 C-O absorption excludes an isomer containing a C-O single bond, so the oxygen is present only as C=O. A ketone carbonyl carbon must be bonded to two carbon groups, which is impossible for a molecule containing only two carbon atoms. The structure is CH₃CHO, ethanal; this contains C=O and has molecular formula C₂H₄O.	Use the strong absorption to place C=O, the missing broad band to exclude an alcohol, and the absent 1000-1300 cm^-1 band to exclude any C-O single bond. Then use the two-carbon formula to distinguish the carbonyl classes: a ketone would require at least three carbon atoms, so the carbonyl must be terminal and the structure is ethanal.	5

3.3.7 · Optical isomerism (A-level only)

Tier 1 · Easy

Mark scheme for 3.3.7 Tier 1 · Easy
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01.1	Carbon 2 is chiral; the compound is butan-2-ol.	Carbon 2 is bonded to H, OH, CH₃ and CH₂CH₃, which are four different groups. No other carbon in the structure meets that test.	2
02.1	The label is incorrect: carbon 3 is not chiral. It is attached to two identical ethyl groups, as well as OH and CH₃, so it does not have four different groups.	List all four attachments to the proposed centre explicitly. Two matching CH₂CH₃ groups are enough to rule out chirality.	2

Tier 2 · Standard

Mark scheme for 3.3.7 Tier 2 · Standard
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01.1	At carbon 2, draw CH₃ and CH₂CH₃ in the plane, with Br on a solid wedge and H on a dashed bond; draw the partner with Br dashed and H wedged. They are non-superimposable mirror images and rotate plane-polarised light equally in opposite directions.	Keep the same four groups and reverse the three-dimensional arrangement at the only chiral centre. Swapping the wedge and dash for Br and H produces the mirror configuration; do not change the structural formula or move a group to another carbon.	4
02.1	A contains a chiral centre at carbon 2: it is bonded to H, OH, CH₃ and CH(CH₃)₂, four different groups. B does not: its central carbon is bonded to two identical CH₃ groups. C does not: the carbonyl carbon is trigonal planar rather than a tetrahedral carbon bonded to four different groups, and its other carbons are CH₃ or CH₂. D does not: the carbon bonded to OH is CH₂ and therefore has two identical H atoms.	Inspect every plausible carbon rather than only the OH-bearing carbon. A has one tetrahedral carbon with four distinct attachments; B fails through duplicate carbon groups, C through carbonyl geometry, and D through duplicate hydrogen atoms.	6
03.1	B is the other enantiomer of the original. One interchange of two groups reverses the three-dimensional arrangement at the chiral carbon, giving a non-superimposable mirror image. C represents the same enantiomer as the original. Two interchanges restore the original handed arrangement; C can be rotated in space to superimpose on the original.	Keep all four groups attached to the same carbon and track only their spatial arrangement. One pairwise exchange inverts a single chiral centre, whereas a second exchange reverses that inversion. Rotation may change how a structure looks on the page but not its handedness.	4

Tier 3 · Hard

Mark scheme for 3.3.7 Tier 3 · Hard
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01.1	Ethanal forms 2-hydroxypropanenitrile, whose central carbon is attached to H, OH, CN and CH₃, so it is chiral. The planar ethanal carbonyl is attacked equally from either face, forming equal amounts of the two enantiomers; their rotations cancel in a racemate. Propanone forms 2-hydroxy-2-methylpropanenitrile, whose central carbon has two identical CH₃ groups, so it has no chiral centre and is optically inactive.	Write the four substituents on the former carbonyl carbon for each product. Ethanal gives four different groups and its planar starting group exposes two equally likely faces. Propanone retains two identical methyl groups, so even attack from opposite faces cannot create a pair of enantiomers.	6
02.1	The major product is 2-bromopentane. Its carbon 2 is bonded to Br, H, CH₃ and CH₂CH₂CH₃, so it is chiral. The secondary carbocation is trigonal planar, so Br⁻ can attack either face with equal probability. A racemic mixture of two enantiomers forms; their equal and opposite rotations cancel, so the mixture is optically inactive.	Apply the familiar planar-intermediate principle to an unfamiliar electrophilic-addition context. Optical inactivity of a sample does not prove that every molecule in it is achiral.	6
03.1	The observed fraction of the pure positive rotation is $3.00/12.0=0.250$ . Therefore $n_+-n_-=0.250(0.800)=0.200\,\mathrm{mol}$ , while $n_++n_-=0.800\,\mathrm{mol}$ . $n_+=0.500\,\mathrm{mol}$ and $n_-=0.300\,\mathrm{mol}$ . The rotations cancel only for equal amounts. Here the positive enantiomer is in excess, leaving a net positive rotation.	Convert the measured rotation into the fractional excess that would produce it under the same conditions. Solve the difference equation together with the total-amount equation, then connect the unequal amounts to incomplete cancellation of equal and opposite rotations.	5

3.3.8 · Aldehydes and ketones (A-level only)

Tier 1 · Easy

Mark scheme for 3.3.8 Tier 1 · Easy
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01.1	Butan-1-ol; CH₃CH₂CH₂CHO + 2[H] → CH₃CH₂CH₂CH₂OH.	Reduction adds two hydrogen equivalents across C=O. Because the starting carbonyl is terminal, the product has a terminal CH₂OH group and is the primary alcohol butan-1-ol.	2
02.1	Propanone gives no silver mirror with Tollens' reagent because it is a ketone, not an aldehyde. Reduction with aqueous sodium tetrahydridoborate forms propan-2-ol.	Do not treat every carbonyl compound as an aldehyde. Tollens' reagent distinguishes aldehydes from ketones, while reduction of a ketone converts C=O into the secondary-alcohol group CHOH.	3

Tier 2 · Standard

Mark scheme for 3.3.8 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Warm with Tollens' reagent: pentanal gives a silver mirror; pentan-3-one shows no change. Warm with Fehling's solution: pentanal gives a brick-red precipitate; pentan-3-one leaves the blue solution unchanged.	Pentanal is an aldehyde and is oxidised by both mild oxidising reagents. Pentan-3-one is a ketone and is not oxidised in either test, so each fresh-portion test assigns the same bottle independently.	4
02.1	The first arrow must start at the electron pair on H⁻ and end at the carbonyl carbon. A second arrow must start at the C=O pi bond and end at oxygen. The tetrahedral intermediate has an O⁻ group, which is then protonated. The product is pentan-1-ol.	Follow electron pairs from donor to acceptor. Hydride supplies the attacking pair, while the carbonyl pi electrons move onto oxygen to keep carbon within an octet; protonation then gives the alcohol.	5
03.1	Use aqueous sodium tetrahydridoborate, NaBH₄, to make propan-1-ol by nucleophilic addition. Use KCN followed by dilute acid to make 2-hydroxybutanenitrile by nucleophilic addition. Hydride adds no carbon atom, so the alcohol retains three carbons. CN⁻ forms a C-C bond through its carbon atom, and its nitrile carbon becomes part of the product chain, increasing the carbon count to four.	Match each target group to the nucleophile. Hydride converts C=O into CHOH without changing the carbon skeleton; cyanide attacks through carbon and remains as C≡N, so the product contains one additional carbon.	4

Tier 3 · Hard

Mark scheme for 3.3.8 Tier 3 · Hard
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01.1	Nucleophilic addition forms 2-hydroxy-2-methylbutanenitrile. A lone pair on the carbon of CN^- attacks the delta-positive carbonyl carbon while the C=O pi pair moves to oxygen; the resulting O^- is then protonated. The carbonyl group is planar, so attack occurs equally from either face. The product carbon is bonded to OH, CN, CH₃ and CH₂CH₃, producing a racemic mixture of two enantiomers. KCN is highly toxic or poisonous.	Use CN^- as the electron-pair donor and the carbonyl carbon as the electron-pair acceptor. Moving the pi pair to oxygen prevents carbon from exceeding an octet. Protonation gives the hydroxynitrile; checking its four substituents confirms a new chiral centre and equal attack on the two planar faces explains the racemate. Treat cyanide as acutely toxic throughout preparation and disposal.	7
02.1	The attacking electron pair is on the carbon end of CN⁻, and its arrow ends at the carbonyl carbon, not oxygen. A second arrow moves the C=O pi electron pair onto oxygen as the C-C bond forms. The tetrahedral intermediate contains O⁻; it is not neutral. An oxygen lone pair then attacks H⁺ from dilute acid to form OH. The product is 2-ethyl-2-hydroxybutanenitrile.	Preserve the carbon octet by moving the carbonyl pi pair to oxygen during attack. For naming, include the nitrile carbon as carbon 1 of the longest chain; the former carbonyl carbon is carbon 2 and bears both hydroxy and ethyl substituents.	7
03.1	D⁻ attacks the delta-positive carbonyl carbon and the C=O pi electron pair moves to oxygen. This gives CH₃CH(D)O⁻ as the tetrahedral alkoxide intermediate. An oxygen lone pair attacks H in water and the water O-H bond pair returns to its oxygen, producing CH₃CH(D)OH. The D supplied as D⁻ therefore becomes bonded to the carbonyl carbon. Ordinary water supplies H to the oxygen, not D, so the O-H group forms and the product is CH₃CHDOH rather than CH₃CHDOD.	Follow each isotope through the two stages. D⁻ is the nucleophile and bonds to carbon as the C=O pi pair moves to oxygen. The resulting alkoxide is then protonated by ordinary water, so H bonds to oxygen.	5
04.1	The silver amount is $5.40/108=0.0500\,\mathrm{mol}$ . The ethanal amount is $0.0500/2=0.0250\,\mathrm{mol}$ . The propanone amount is $0.100-0.0250=0.0750\,\mathrm{mol}$ . The mixture is $25.0\%$ ethanal and $75.0\%$ propanone by moles. NaBH₄ reduces ethanal to ethanol and propanone to propan-2-ol.	Only the aldehyde reduces Tollens' reagent, so use the silver amount and the supplied two-to-one ratio to find ethanal. Subtract from the total mixture amount for propanone and convert both to percentages. For the reduction products, convert the terminal aldehyde carbonyl to a primary alcohol and the internal ketone carbonyl to a secondary alcohol.	5
05.1	(CH₃)₂CHCHO + HCN → (CH₃)₂CHCH(OH)CN. The product is 2-hydroxy-3-methylbutanenitrile. Use KCN followed by dilute acid. HCN is a toxic gas, so it is not used directly. KCN supplies CN⁻ for nucleophilic attack, and the dilute acid supplies H⁺ to protonate the intermediate.	Add HCN across the C=O group without losing any atoms: CN bonds to the carbonyl carbon and H converts the oxygen into OH. Include the nitrile carbon in the longest parent chain and number it carbon 1, giving 2-hydroxy-3-methylbutanenitrile. In practice, KCN supplies the cyanide nucleophile and dilute acid supplies the proton while avoiding direct use of toxic gaseous HCN.	5

3.3.9.1 · Carboxylic acids and esters (A-level only)

Tier 1 · Easy

Mark scheme for 3.3.9.1 Tier 1 · Easy
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01.1	Propyl ethanoate; concentrated sulfuric acid catalyst. CH₃COOH + CH₃CH₂CH₂OH ⇌ CH₃COOCH₂CH₂CH₃ + H₂O.	Take propyl from the alcohol and ethanoate from the acid. Join the alcohol oxygen to the acid's acyl carbon, remove water, and show a reversible arrow because esterification establishes an equilibrium.	3
02.1	The ester is methyl butanoate. The alcohol supplies the alkyl part, methyl, and the carboxylic acid supplies the alkanoate part, butanoate.	Read an ester name as alkyl alkanoate. Track the group attached to oxygen back to the alcohol and the carbonyl-containing portion back to the acid.	3

Tier 2 · Standard

Mark scheme for 3.3.9.1 Tier 2 · Standard
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01.1	Products: methanol and sodium propanoate. CH₃CH₂COOCH₃ + NaOH → CH₃CH₂COONa + CH₃OH. Alkaline conditions deprotonate the acid product to its carboxylate salt; acidification is needed to obtain propanoic acid.	Split the ester at the acyl C-O bond. The alkoxy fragment becomes methanol, while the acyl fragment is present as propanoate in excess alkali. Add dilute acid after hydrolysis if the neutral carboxylic acid is required.	4
02.1	Heat propan-1-ol with ethanoic acid and concentrated sulfuric acid catalyst under reflux. Reflux allows prolonged heating while condensing volatile reactants back into the flask, so they remain available to reach equilibrium. After reaction, distillation separates the volatile ester from the reaction mixture over its boiling range. The product name is propyl ethanoate because propyl comes from the alcohol and ethanoate from the acid.	Separate reaction from isolation. Reflux retains volatile material during the reversible esterification; distillation afterward uses volatility to collect the desired product.	5
03.1	Using excess alcohol shifts the equilibrium towards ester and water, increasing the equilibrium ester yield. Removing water also shifts the equilibrium to the product side to replace the removed water, increasing ester yield. More catalyst increases the rates of both forward and reverse reactions and makes equilibrium establish faster. A catalyst does not change the equilibrium constant or equilibrium composition, so it does not by itself increase the equilibrium yield.	Apply equilibrium reasoning separately from kinetics. Changing a reactant amount or removing a product changes the reaction quotient and moves the position of equilibrium; a catalyst supplies a lower-activation-energy route in both directions and leaves $K$ unchanged at fixed temperature.	4

Tier 3 · Hard

Mark scheme for 3.3.9.1 Tier 3 · Hard
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01.1	$2.40\,\text{g}$ of NaOH. Glycerol and the sodium salts of long-chain carboxylic acids (soap).	The triglyceride amount is $17.8/890=0.0200\,\text{mol}$ . Three ester links require three hydroxide ions, so $n(\mathrm{NaOH})=3(0.0200)=0.0600\,\text{mol}$ . Therefore $m=0.0600\times40.0=2.40\,\text{g}$ . Breaking all three ester links releases glycerol and three carboxylate salts per triglyceride molecule.	5
02.1	A is ethanol: it is a primary alcohol and controlled oxidation gives ethanal, which reduces Tollens' reagent. B is propanoic acid: its formula and carbonate reaction identify the carboxylic acid. The ester is ethyl propanoate. CH₃CH₂COOCH₂CH₃ + H₂O ⇌ CH₃CH₂COOH + CH₃CH₂OH, using dilute acid and heat.	Use the oxidation evidence to identify the two-carbon alcohol and the carbonate evidence to identify the three-carbon acid. Recombine the alcohol-derived ethyl group with the acid-derived propanoate group, keeping the ester name in alkyl alkanoate order.	7
03.1	$n(\text{triester})=44.2/884=0.0500\,\mathrm{mol}$ . The glycerol amount is $0.0500\,\mathrm{mol}$ , so its theoretical mass is $0.0500\times92.0=4.60\,\mathrm g$ . The methyl-ester amount is $3(0.0500)=0.150\,\mathrm{mol}$ . The theoretical methyl-ester mass is $0.150\times296=44.4\,\mathrm g$ . The recovered biodiesel mass is $44.4\times0.750=33.3\,\mathrm g$ to three significant figures. The biodiesel is the mixture of long-chain methyl esters, not the glycerol coproduct.	Convert the oil mass to triester moles, then use different stoichiometric multipliers for the two products: one glycerol but three methyl esters per triester. Apply the recovery percentage only to the biodiesel amount requested, retaining unrounded values until the final mass.	6
04.1	The alcohol is 3-methylbutan-1-ol, HOCH₂CH₂CH(CH₃)₂. The carboxylic acid is ethanoic acid, CH₃COOH. CH₃COOH + HOCH₂CH₂CH(CH₃)₂ ⇌ CH₃COOCH₂CH₂CH(CH₃)₂ + H₂O. Another common use of esters is as solvents; uses as plasticisers or perfumes are also acceptable. Ethanoic acid is a weak acid that donates H⁺ to carbonate ions, producing CO₂ gas and hence effervescence. The ester has no acidic hydrogen, so it does not liberate CO₂ from aqueous sodium carbonate.	Split the ester at its C(O)-O bond: the 3-methylbutyl group comes from 3-methylbutan-1-ol and the ethanoate group comes from ethanoic acid. Rejoin those reactants in the reversible condensation equation and include water. For the carbonate comparison, identify the acidic hydrogen in the carboxylic acid and note that the ester has no such hydrogen.	6

3.3.9.2 · Acylation (A-level only)

Tier 1 · Easy

Mark scheme for 3.3.9.2 Tier 1 · Easy
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01.1	Ethyl ethanoate and hydrogen chloride. Steamy or misty fumes are observed as HCl meets moist air.	Replace Cl in CH₃COCl by the ethoxy group from ethanol to form CH₃COOCH₂CH₃. The removed proton and chloride form HCl, which produces visible acidic mist in damp air.	3
02.1	The organic product is methyl propanoate, not propyl methanoate. Hydrogen chloride is the other product, not water. Steamy fumes of HCl are observed.	The alcohol supplies methyl and the acyl chloride supplies propanoate. Replacing Cl in CH₃CH₂COCl by OCH₃ forms the ester and releases HCl.	3

Tier 2 · Standard

Mark scheme for 3.3.9.2 Tier 2 · Standard
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01.1	N-Methylpropanamide and methylammonium chloride. The nitrogen lone pair attacks the carbonyl carbon and the C=O pi pair moves to oxygen, giving a tetrahedral intermediate with O^- and N⁺. The oxygen pair reforms C=O as Cl^- leaves; a second methylamine molecule removes H⁺, forming CH₃NH₃⁺Cl^-.	Use methylamine as both nucleophile and base. Its first molecule makes the new C-N bond through the charged tetrahedral intermediate: oxygen is negative after receiving the pi pair and nitrogen is positive after donating its lone pair. Collapse expels chloride, then the excess second amine accepts the proton to give the neutral amide and ammonium salt.	5
02.1	Choose propanoyl chloride. The ester is propyl propanoate and the other product is HCl. The carbonyl carbon in an acyl chloride is more susceptible to nucleophilic attack because chlorine withdraws electron density and Cl⁻ is a suitable leaving group. The acyl chloride therefore reacts rapidly without the acid catalyst and heating used for reversible esterification with a carboxylic acid.	Select the route from reactivity rather than only naming a reagent. Link the electron-deficient acyl carbon and leaving group to rapid nucleophilic addition-elimination.	5
03.1	The products are ethyl propanoate and propanoic acid. (CH₃CH₂CO)₂O + CH₃CH₂OH → CH₃CH₂COOCH₂CH₃ + CH₃CH₂COOH. The mechanism is nucleophilic addition-elimination. The leaving acyl fragment becomes propanoic acid; the reactant contains no chlorine from which HCl could form.	Treat the anhydride as two propanoyl groups joined through oxygen. Ethanol replaces one acyl-oxygen link to form the ester, and the other acyl fragment is protonated to the carboxylic acid.	4

Tier 3 · Hard

Mark scheme for 3.3.9.2 Tier 3 · Hard
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01.1	Ethanoic anhydride route: $180/(180+60.0)\times100=75.0\%$ . Ethanoyl chloride route: $180/(180+36.5)\times100=83.1\%$ . Despite its lower atom economy, ethanoic anhydride is safer and easier to handle and produces ethanoic acid rather than corrosive, harmful HCl fumes.	For each stated one-to-one route, divide the desired-product formula mass by the total formula mass of products. This gives $75.0\%$ for the anhydride and $83.1\%$ for the chloride. Atom economy is only one industrial criterion, so compare the hazards and usefulness of the by-products before deciding which reagent is preferable.	6
02.1	The first arrow starts at the ammonia nitrogen lone pair and ends at the carbonyl carbon. A second arrow moves the C=O pi pair onto oxygen, giving O⁻ in the tetrahedral intermediate. When C=O reforms, the C-Cl bond breaks and Cl⁻ leaves; chlorine is not retained in butanamide. Proton transfer gives butanamide, and a second ammonia molecule neutralises HCl to form ammonium chloride. CH₃CH₂CH₂COCl + 2NH₃ → CH₃CH₂CH₂CONH₂ + NH₄Cl.	Audit every arrow by its electron-pair source and every intermediate by charge. Addition temporarily places the pi pair on oxygen; elimination reforms C=O and expels chloride. The second ammonia accounts for the ammonium chloride product.	8
03.1	The products are N-methylethanamide and methylammonium ethanoate. (CH₃CO)₂O + 2CH₃NH₂ → CH₃CONHCH₃ + CH₃NH₃⁺CH₃COO⁻. A nitrogen lone pair attacks one carbonyl carbon and the C=O pi pair moves to oxygen, producing a tetrahedral intermediate with O⁻ and N⁺. An O⁻ lone pair reforms C=O while the bond from the acyl carbon to the anhydride oxygen breaks, expelling ethanoate. Proton transfer gives the neutral amide; a second methylamine accepts the proton and pairs with ethanoate as methylammonium ethanoate.	Use methylamine first as the electron-pair donor and then, because it is in excess, as a base. The leaving group is ethanoate rather than chloride, so the final salt contains methylammonium and ethanoate ions.	6
04.1	The trapped HCl amount is $0.800(25.0/1000)=0.0200\,\mathrm{mol}$ . The reacting acyl chloride amount is therefore $0.0200\,\mathrm{mol}$ . Its relative molecular mass is $2.13/0.0200=106.5$ . The acyl chloride is butanoyl chloride, CH₃CH₂CH₂COCl. CH₃CH₂CH₂COCl + CH₃CH₂OH → CH₃CH₂CH₂COOCH₂CH₃ + HCl; the ester is ethyl butanoate.	Use the NaOH titre as a one-to-one measurement of trapped HCl, then use the supplied acyl-chloride-to-HCl ratio. Dividing sample mass by amount gives $M_r=106.5$ , matching C₄H₇OCl. Replace chloride at the acyl carbon with the ethoxy group from ethanol to construct and name the ester.	5
05.1	(CH₃CO)₂O + H₂O → 2CH₃COOH. The anhydride reacts more slowly and less vigorously than ethanoyl chloride. The anhydride produces no steamy fumes because HCl is not formed. Ethanoyl chloride reacts vigorously: CH₃COCl + H₂O → CH₃COOH + HCl. Steamy fumes of hydrogen chloride are observed with ethanoyl chloride.	Hydrolysis replaces the acyl chloride's Cl with OH, so ethanoyl chloride gives ethanoic acid and HCl. In the anhydride, water splits the acyl-oxygen-acyl linkage to give two acid molecules; no chloride is present, accounting for the slower, less vigorous reaction and absence of steamy HCl fumes.	5

3.3.10.1 · Bonding (A-level only)

Tier 1 · Easy

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01.1	Benzene is planar and hexagonal; all six C-C bonds have equal length intermediate between a single and double bond; p orbitals overlap to give delocalised pi electrons above and below the ring.	State one structural point at each scale: the whole ring is planar, the six C-C links are equivalent, and the unhybridised p orbitals overlap continuously to delocalise the pi electrons.	3
02.1	Every carbon atom in benzene has a p orbital. The six parallel p orbitals overlap sideways around the ring. Six pi electrons are delocalised in electron-density regions above and below the plane of the ring.	A continuous delocalised system requires an unbroken ring of overlapping p orbitals. Omitting one p orbital would interrupt that overlap and cannot represent benzene.	3

Tier 2 · Standard

Mark scheme for 3.3.10.1 Tier 2 · Standard
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01.1	$156\,\text{kJ mol}^{-1}$	Three isolated double bonds would give $3(-121)=-363\,\text{kJ mol}^{-1}$ . Benzene releases only $207\,\text{kJ mol}^{-1}$ , so it begins lower in energy by $363-207=156\,\text{kJ mol}^{-1}$ . This is the delocalisation stability for the supplied data.	3
02.1	Cyclohexene has a localised C=C pi bond that polarises bromine and undergoes electrophilic addition, so bromine water is decolourised. Benzene's six pi electrons are delocalised around the ring. Addition to benzene would destroy this delocalisation and form a less stable non-aromatic product. Benzene therefore does not undergo this addition under the test conditions and the bromine colour remains.	Do not argue that benzene has no pi electrons. Contrast a localised alkene pi bond with a stabilised delocalised ring, then connect loss of delocalisation to the unfavourable addition pathway.	5
03.1	Methylbenzene has seven C-C sigma bonds and eight C-H sigma bonds. It has six delocalised pi electrons. Each of the six planar carbon atoms has a parallel p orbital containing one electron. The two electrons in the bond from the ring to the methyl carbon form a sigma bond; the methyl carbon has no p orbital in the ring's continuous overlap, so these electrons are not part of the delocalised pi system.	Count six C-C sigma bonds around the ring plus the bond to CH₃, and five ring C-H bonds plus three methyl C-H bonds. Only the six p electrons, one from each ring carbon, occupy the delocalised pi system; the ring-methyl bond is part of the sigma framework.	4

Tier 3 · Hard

Mark scheme for 3.3.10.1 Tier 3 · Hard
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01.1	The claim is inconsistent with six equal $0.140\,\text{nm}$ bonds; this value lies between the single- and double-bond lengths instead of alternating between them. Continuous p-orbital overlap delocalises six pi electrons around the ring and lowers its energy. Addition would remove part of this delocalisation, whereas electrophilic substitution temporarily disrupts it but then restores the aromatic ring, so substitution is preferred.	Compare the one measured benzene length with both reference values and note that every ring bond has that same intermediate value. Account for this using delocalisation rather than bond switching. Finally compare the electronic result of the reaction types: addition leaves a less-delocalised product, while substitution replaces H and recovers the stable ring.	6
02.1	Three independent C=C bonds would have hydrogenation enthalpy $3(-119)=-357\,\text{kJ mol}^{-1}$ . Benzene releases only $208\,\text{kJ mol}^{-1}$ on reaching the same cyclohexane product, which is $149\,\text{kJ mol}^{-1}$ less exothermic. Benzene is therefore $149\,\text{kJ mol}^{-1}$ more stable, not less stable, than the hypothetical localised structure. The extra stability arises from delocalisation of the pi electrons.	Both reactants are compared with the same hydrogenated product. A less exothermic route starts from the lower-energy, more stable reactant; the negative sign in the student's subtraction does not mean lower stability.	5
03.1	$\Delta_{\mathrm r}H^\circ=-156-[49+3(0)]=-205\,\mathrm{kJ\,mol^{-1}}$ . Three localised C=C bonds would give $3(-117)=-351\,\mathrm{kJ\,mol^{-1}}$ . The actual hydrogenation is $351-205=146\,\mathrm{kJ\,mol^{-1}}$ less exothermic. Because both routes finish at cyclohexane, benzene starts $146\,\mathrm{kJ\,mol^{-1}}$ lower in energy and is that much more stable than the localised model. The extra stability arises from delocalisation of six pi electrons around the ring.	First apply formation enthalpies as products minus reactants for C₆H₆ + 3H₂ → C₆H₁₂. Then compare the actual route with three independent alkene hydrogenations leading to the same product; the smaller energy release means the real benzene reactant was already lower in energy.	5

3.3.10.2 · Electrophilic substitution (A-level only)

Tier 1 · Easy

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01.1	Concentrated nitric acid and concentrated sulfuric acid; sulfuric acid is the catalyst; the electrophile is NO₂⁺, the nitronium ion.	Use the nitrating mixture of concentrated HNO₃ and H₂SO₄. The stronger sulfuric acid protonates nitric acid, allowing loss of water and formation of NO₂⁺; sulfuric acid is regenerated.	3
02.1	The arrow starts at a benzene pi bond and ends at the nitrogen atom of NO₂⁺. The benzene pi system supplies the electron pair, while NO₂⁺ accepts it as the electrophile.	Curly arrows begin where the moving electron pair is located. An electrophile is electron-pair deficient, so an arrow cannot start at the nitronium ion in this bond-forming step.	3

Tier 2 · Standard

Mark scheme for 3.3.10.2 Tier 2 · Standard
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01.1	1-Phenylpropan-1-one. C₆H₆ + CH₃CH₂COCl → C₆H₅COCH₂CH₃ + HCl. The electrophile is CH₃CH₂CO⁺.	AlCl₃ accepts chloride from propanoyl chloride to generate the acylium ion. Replace one benzene H by CH₃CH₂CO-, then combine the removed H with Cl to balance the equation and name the resulting aromatic ketone.	4
02.1	(CH₃)₂CHCOCl + AlCl₃ → (CH₃)₂CHCO⁺ + AlCl₄⁻. The electrophile is the 2-methylpropanoyl ion, (CH₃)₂CHCO⁺. The product is 2-methyl-1-phenylpropan-1-one, C₆H₅COCH(CH₃)₂. The reaction is electrophilic substitution.	Use AlCl₃ to remove chloride and generate the positively charged acyl electrophile. Substitution attaches its carbonyl carbon to the benzene ring without reducing the carbonyl group.	5
03.1	C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O $n(\text{benzene})=7.80/78.0=0.100\,\mathrm{mol}$ , so the 1:1 reaction gives a theoretical nitrobenzene mass of $0.100\times123=12.3\,\mathrm g$ . Percentage yield $=(9.84/12.3)\times100=80.0\%$ . The reaction may be incomplete, or product is lost during transfer, separation or purification (any one reason).	Write the substitution equation with water as the co-product. Convert benzene mass to moles using $M_{\mathrm r}=78.0$ , apply the 1:1 stoichiometry and use $M_{\mathrm r}(\text{nitrobenzene})=123$ to obtain $12.3\,\mathrm g$ . Then divide the actual mass by the theoretical mass.	5

Tier 3 · Hard

Mark scheme for 3.3.10.2 Tier 3 · Hard
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01.1	CH₃COCl + AlCl₃ → CH₃CO⁺ + AlCl₄^-. A curly arrow starts inside the benzene ring and goes to the positive carbon of CH₃CO⁺, forming a C-C bond. In the sigma complex, H and COCH₃ remain bonded to the attacked carbon and the positive charge is delocalised around the other five ring carbons. AlCl₄^- removes H⁺; the C-H bond pair moves back into the ring, restoring delocalisation and forming phenylethanone, HCl and AlCl₃.	First use the Lewis acid to remove chloride and create the acylium electrophile. The benzene pi pair then forms the new bond; draw the attacked carbon bonded to both H and the acyl group, with delocalised positive charge on the remaining ring. Finally the C-H electron pair restores the ring as AlCl₄^- supplies chloride to H; this produces HCl and returns AlCl₃, confirming its catalytic role.	6
02.1	The attacking electrophile is NO₂⁺, generated from concentrated nitric and sulfuric acids, not an HNO₃ molecule. The first arrow starts at a benzene pi bond and ends at nitrogen in NO₂⁺. The intermediate has lost full delocalisation and carries a positive charge spread over the ring. The electron-pair arrow that restores the ring starts at the C-H bond and ends in the ring to remake the pi system as H⁺ is removed. HSO₄⁻ accepts the proton, regenerating H₂SO₄.	Check the mechanism in order: generate the electrophile, follow the ring electron pair to it, retain the positive arenium intermediate, then use the C-H bond electrons to restore aromatic delocalisation. Catalyst regeneration completes the substitution cycle.	8
03.1	Water hydrolyses propanoyl chloride to propanoic acid and HCl, consuming the acylating reagent and explaining the acid product. Water also hydrates or hydrolyses the Lewis-acid catalyst, so AlCl₃ is no longer available in the anhydrous form needed to accept chloride and generate the electrophile. Use dry glassware, exclude moisture and use anhydrous AlCl₃. CH₃CH₂COCl + AlCl₃ → CH₃CH₂CO⁺ + AlCl₄⁻. The electrophile is the propanoyl ion, CH₃CH₂CO⁺.	Use the observed propanoic acid to identify acyl-chloride hydrolysis, then separately account for failure to generate the acylium ion when the Lewis acid is wet. The corrected dry setup preserves both propanoyl chloride and anhydrous AlCl₃ for chloride abstraction.	5
04.1	Nitration introduces an NO₂ group that can be used as an intermediate in further organic synthesis. The nitro group can be reduced, for example using Sn/HCl then NaOH, to give phenylamine, C₆H₅NH₂. Monosubstitution of benzene means that only one hydrogen atom is replaced under these conditions. Nitrating benzene once produces nitrobenzene. Nitration is also used in the manufacture of explosives.	Treat nitration as a route to a functional-group intermediate rather than only as a benzene reaction. Reduce the nitro group with Sn/HCl, then use NaOH to obtain the free aromatic amine. Distinguish replacing one ring hydrogen, which gives nitrobenzene, from further substitution; separately recognise that highly nitrated compounds are important explosives.	5

3.3.11.1 · Preparation (A-level only)

Tier 1 · Easy

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01.1	Excess ammonia in ethanol; heat in a sealed vessel.	Use ethanolic ammonia and heat the sealed reaction mixture. Stating that ammonia is in excess is important because it makes collision with ammonia more likely than collision with the propylamine product, reducing formation of secondary and tertiary amines.	2
02.1	Route (i) forms propylamine (propan-1-amine); route (ii) forms butan-1-amine.	Direct substitution by ammonia replaces Br with NH₂, so the three-carbon chain is retained. In the cyanide route, the carbon in CN⁻ becomes the nitrile carbon and remains in the chain after reduction, so the product has four carbons.	2

Tier 2 · Standard

Mark scheme for 3.3.11.1 Tier 2 · Standard
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01.1	Heat bromoethane with potassium cyanide in ethanol under reflux to form propanenitrile; reduce propanenitrile using hydrogen and nickel, or lithium aluminium hydride in dry ether, to form propylamine.	First use ethanolic KCN under reflux. The cyanide carbon joins the two-carbon halogenoalkane, so the intermediate is the three-carbon nitrile propanenitrile. Reduction converts the nitrile group into a primary amine without removing that carbon, giving propylamine. Either H₂/Ni or LiAlH₄ in dry ether is an acceptable reduction route.	5
02.1	Route A can undergo further alkylation because its amine product is also nucleophilic. Route B forms one nitrile product, whose reduction gives the primary amine; cyanide substitution and nitrile reduction are two separate steps.	In Route A, butylamine can attack more 1-bromobutane and form secondary, tertiary and quaternary products, even though excess ammonia reduces this problem. Route B first substitutes CN for Br, adding one carbon, and then reduces –C≡N to –CH₂NH₂. It therefore avoids successive alkylation but needs both substitution and reduction.	4
03.1	C₆H₄(NO₂)₂ + 12[H] → C₆H₄(NH₂)₂ + 4H₂O; $0.300\,\mathrm{mol}$ of $[\mathrm{H}]$ is required.	Reducing one –NO₂ group to –NH₂ uses 6[H] and forms 2H₂O. The molecule has two nitro groups, so it uses 12[H] and forms 4H₂O. Therefore $n([\mathrm{H}])=12\times0.0250=0.300\,\mathrm{mol}$ .	4

Tier 3 · Hard

Mark scheme for 3.3.11.1 Tier 3 · Hard
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01.1	Heat nitrobenzene under reflux with tin and concentrated hydrochloric acid, then add excess sodium hydroxide to release phenylamine from the phenylammonium salt and separate the organic product.	Sn/HCl reduces the nitro group. In the acidic mixture the amine is protonated, so the immediate product is a phenylammonium salt rather than free phenylamine. Adding excess NaOH removes the proton and liberates phenylamine. The organic product can then be separated from the aqueous mixture and purified, for example by distillation.	5
02.1	Overall yield = 67.2%; amount of butan-1-amine = 0.202 mol. The nitrile route gives 0.0516 mol more product (allow 0.052 from a correctly rounded 0.202) and avoids successive alkylation of an amine product.	Multiply step yields as decimals: 0.800 × 0.840 = 0.6720, so the overall yield is 67.20%, or 67.2% to three significant figures. The substitution and reduction are each 1:1, so 0.300 × 0.6720 = 0.20160 mol, or 0.202 mol to three significant figures. Compared with 0.150 mol, the increase is 0.05160 mol, or 0.0516 mol to three significant figures. Reduction of a single nitrile does not create the successive-substitution mixture possible when a halogenoalkane reacts with ammonia.	5
03.1	The intermediate is propanenitrile and the product is propan-1-amine. The ¹³C becomes the carbon of the terminal –¹³CH₂NH₂ group, and the molecular ion is at $m/z=60$ .	The cyanide carbon bonds to the carbon bearing bromine, so CH₃CH₂Br becomes CH₃CH₂¹³CN. Reduction changes –¹³C≡N into –¹³CH₂NH₂ without moving that carbon. Unlabelled propan-1-amine has $M_r=59$ , so replacing ¹²C by ¹³C raises the molecular-ion value by one to $m/z=60$ .	4
04.1	CH₃CH₂CH₂CN + 2H₂ → CH₃CH₂CH₂CH₂NH₂. $n(\mathrm{H_2})=0.200/2.00=0.100\,\mathrm{mol}$ . Reducing all the nitrile would require $2\times0.0750=0.150\,\mathrm{mol}$ of H₂, so hydrogen is limiting. The maximum amine amount is $0.100/2=0.0500\,\mathrm{mol}$ . The maximum amine mass is $0.0500\times73.0=3.65\,\mathrm{g}$ .	Nitrile hydrogenation uses two moles of H₂ per mole of nitrile and converts –C≡N into –CH₂NH₂. The available hydrogen is $0.200/2.00=0.100\,\mathrm{mol}$ H₂, less than the $0.150\,\mathrm{mol}$ needed for all the butanenitrile. Hydrogen therefore limits production to $0.0500\,\mathrm{mol}$ of butan-1-amine, whose mass is $3.65\,\mathrm{g}$ .	5
05.1	The mass increase per mole of reduced nitrile is $4.00\,\mathrm{g\,mol^{-1}}$ . The initial butanenitrile amount is $0.360/4.00=0.0900\,\mathrm{mol}$ . The initial butan-1-amine amount is $0.150-0.0900=0.0600\,\mathrm{mol}$ . Reduction converts the nitrile carbon into the terminal –CH₂NH₂ carbon without cleaving a C–C bond.	Only the nitrile reacts. The equation $\mathrm{RCN+4[H]\rightarrow RCH_2NH_2}$ shows a gain of four hydrogen atoms, or $4.00\,\mathrm{g}$ per mole. Therefore $n(\text{nitrile})=0.360/4.00=0.0900\,\mathrm{mol}$ . Subtract this from the stated total to obtain $0.0600\,\mathrm{mol}$ of amine initially. The nitrile carbon remains bonded to the same alkyl group throughout reduction, so no carbon is added, removed or rearranged.	4

3.3.11.2 · Base properties (A-level only)

Tier 1 · Easy

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01.1	CH₃CH₂NH₂ + H₂O ⇌ CH₃CH₂NH₃⁺ + OH⁻	The nitrogen lone pair accepts H⁺ from water. Ethylamine therefore becomes ethylammonium, while the water molecule that loses H⁺ becomes OH⁻. Use an equilibrium arrow because ethylamine is a weak base.	2
02.1	A base is a proton acceptor; C₆H₅NH₂ + HCl → C₆H₅NH₃⁺Cl⁻.	The nitrogen lone pair accepts H⁺. Phenylamine is therefore protonated to the phenylammonium ion, while chloride is the counter-ion.	2

Tier 2 · Standard

Mark scheme for 3.3.11.2 Tier 2 · Standard
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01.1	Ethylamine > ammonia > phenylamine; the ethyl group increases electron density on nitrogen, whereas delocalisation into the benzene ring makes the phenylamine lone pair less available.	Start by comparing lone-pair availability. The electron-releasing ethyl group has a positive inductive effect, so ethylamine accepts H⁺ more readily than ammonia. In phenylamine, the lone pair overlaps with the aromatic π system and is delocalised, so it is less available than the localised lone pair in ammonia.	4
02.1	Methylamine: 11.8; ammonia: 11.1; phenylamine: 8.7. The methyl group releases electron density towards nitrogen, while the phenylamine lone pair is delocalised into the benzene ring.	At the same concentration, the stronger weak base produces the larger hydroxide-ion concentration and hence the higher pH. The positive inductive effect of CH₃ makes the methylamine lone pair more available than the ammonia lone pair. Delocalisation makes the phenylamine lone pair least available, giving methylamine > ammonia > phenylamine.	4
03.1	The hydroxide-ion concentration ratio is $10^{12.05-11.35}=5.0$ . Propylamine is the stronger base because the propyl group releases electron density towards nitrogen, making its lone pair more available to accept a proton.	At one temperature, subtracting pH values gives the opposite difference in pOH values, so the hydroxide concentration ratio is $10^{0.70}=5.0119$ . Report 5.0 to two significant figures; allow 5.01. The positive inductive effect of the alkyl group increases nitrogen lone-pair availability, so propylamine establishes the larger hydroxide concentration.	4

Tier 3 · Hard

Mark scheme for 3.3.11.2 Tier 3 · Hard
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01.1	The claim is incorrect. The phenylamine lone pair is delocalised into the ring, whereas the propyl group releases electron density towards nitrogen; propylamine therefore forms the greater hydroxide-ion concentration.	A ring being electron-rich does not by itself make the nitrogen lone pair available. Conjugation lets the phenylamine lone pair spread into the benzene π system, stabilising the unprotonated molecule and reducing its tendency to accept H⁺. The propyl group instead increases electron density at nitrogen through the positive inductive effect. Propylamine is the stronger weak base and shifts its reaction with water further towards alkylammonium and OH⁻ ions.	5
02.1	Decreasing base strength: phenylmethylamine > ammonia > phenylamine. The CH₂ group prevents the nitrogen lone pair in phenylmethylamine from delocalising into the benzene ring, whereas the lone pair in phenylamine is delocalised into the ring.	The CH₂ spacer breaks conjugation between nitrogen and the benzene ring, so the phenylmethylamine lone pair remains localised and available to accept H⁺, giving aliphatic-amine-like basicity. It is therefore stronger than ammonia. In phenylamine, the nitrogen lone pair overlaps with the benzene π system and is delocalised, making it less available to accept H⁺. Hence phenylmethylamine > ammonia > phenylamine.	4
03.1	The equilibrium favours the products. Ethylamine is a stronger base than phenylamine, so it accepts a proton from phenylammonium; the products are the weaker acid, C₂H₅NH₃⁺, and the weaker base, C₆H₅NH₂, so the position of equilibrium lies to the right.	The ethyl group releases electron density towards nitrogen, so the ethylamine lone pair is relatively available. In phenylamine the lone pair is delocalised into the benzene ring and is less available. Proton transfer therefore proceeds mainly from phenylammonium to ethylamine, forming ethylammonium and phenylamine; equilibrium favours the weaker acid–base pair.	4
04.1	Phenylamine is the stronger base. In both amines, the nitrogen lone pair can be delocalised into the benzene ring. The 4-nitro group withdraws electron density through the conjugated ring and increases the withdrawal of the amine lone pair from nitrogen. The lone pair in 4-nitrophenylamine is therefore less available to accept H⁺.	Compare the same basic site in the two primary aromatic amines. Phenylamine already has a less available nitrogen lone pair than an aliphatic amine because that pair overlaps with the ring π system. A nitro group at the 4-position withdraws electron density through the conjugated ring, stabilising further delocalisation away from the amine nitrogen. Proton acceptance is therefore less favourable for 4-nitrophenylamine, so phenylamine is the stronger base.	4

3.3.11.3 · Nucleophilic properties (A-level only)

Tier 1 · Easy

Mark scheme for 3.3.11.3 Tier 1 · Easy
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01.1	A lone pair of electrons on the nitrogen atom.	A nucleophile donates an electron pair. In an amine, the available electron pair is the lone pair on nitrogen.	1
02.1	CH₃CONHCH₂CH₂CH₂CH₃	A primary amine undergoes nucleophilic addition–elimination with an acid anhydride. An ethanoyl group bonds to the nitrogen of butylamine, giving N-butylethanamide, CH₃CONHCH₂CH₂CH₂CH₃.	1

Tier 2 · Standard

Mark scheme for 3.3.11.3 Tier 2 · Standard
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01.1	N-ethylethanamide; nucleophilic addition–elimination; an arrow from the nitrogen lone pair to the carbonyl carbon and an arrow from the C=O π bond to oxygen.	The nitrogen lone pair attacks the δ⁺ carbonyl carbon, so the first curly arrow begins at that lone pair and ends at the carbonyl carbon. To avoid giving carbon five bonds, the C=O π pair moves onto oxygen. Elimination then reforms C=O and removes Cl⁻; deprotonation gives N-ethylethanamide.	4
02.1	CH₃COCl + 2CH₃NH₂ → CH₃CONHCH₃ + CH₃NH₃⁺Cl⁻; the second methylamine accepts a proton and neutralises the HCl formed.	One methylamine molecule attacks the acyl carbon and becomes part of the amide. Deprotonation is required to give the neutral N-methylethanamide. A second methylamine molecule accepts that proton; chloride is its counter-ion, so methylammonium chloride is the other product.	4
03.1	CH₃COCl + 2C₂H₅NH₂ → CH₃CONHC₂H₅ + C₂H₅NH₃⁺Cl⁻. (CH₃CO)₂O + C₂H₅NH₂ → CH₃CONHC₂H₅ + CH₃COOH. The second ethylamine accepts the proton removed from the attacking amine, neutralising the HCl that would otherwise form.	In both reactions one ethylamine nitrogen attacks an acyl carbon and becomes part of the amide. Ethanoyl chloride supplies Cl⁻, so another ethylamine accepts the proton removed from the attacking molecule and forms ethylammonium chloride. With ethanoic anhydride, the expelled ethanoate group accepts the proton and becomes ethanoic acid in the overall equation.	4

Tier 3 · Hard

Mark scheme for 3.3.11.3 Tier 3 · Hard
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01.1	A tetrabutylammonium salt forms; each amine product retains a nitrogen lone pair until the quaternary ion is reached, and the ion combines a positive hydrophilic head with hydrophobic butyl groups.	Butylamine first attacks another 1-bromobutane molecule. After deprotonation the secondary amine still has a lone pair, so it can attack again; the tertiary amine can also attack, forming the quaternary ion (C₄H₉)₄N⁺ with Br⁻. The permanent positive charge interacts with water, while the hydrocarbon groups interact with grease or non-polar material, giving surfactant behaviour.	5
02.1	The first arrow must start at the nitrogen lone pair and end at the carbonyl carbon; a second arrow must move the C=O π electrons to oxygen; the product is N-ethylpropanamide.	Ethylamine is the electron-pair donor, so its nitrogen lone pair is the origin of the attack arrow. Moving the C=O π pair onto oxygen prevents the carbonyl carbon from having five bonds and creates the tetrahedral intermediate. The ethyl substituent is attached to the amide nitrogen, so the correct name needs the N- locant: N-ethylpropanamide.	5
03.1	$0.0400\,\mathrm{mol}$ of N-ethylpropanamide forms and $0.0100\,\mathrm{mol}$ of ethylamine remains. The nitrogen-containing products are N-ethylpropanamide and ethylammonium chloride.	Each mole of amide formation uses one ethylamine as the nucleophile and a second as the proton acceptor. Reacting $0.0400\,\mathrm{mol}$ of propanoyl chloride therefore uses $2\times0.0400=0.0800\,\mathrm{mol}$ of ethylamine, so the acid chloride is limiting. The ethylamine remaining is $0.0900-0.0800=0.0100\,\mathrm{mol}$ . The organic acylation product is N-ethylpropanamide and the proton-accepting amine forms ethylammonium chloride.	5
04.1	A curly arrow starts at the methylamine nitrogen lone pair and ends at the carbon bonded to bromine. A simultaneous curly arrow starts at the C–Br bond and ends at Br, forming Br⁻. The first organic product is [CH₃NH₂CH₂CH₃]⁺. A second methylamine lone pair removes a proton from the positively charged nitrogen. The N–H bond pair returns to nitrogen, giving CH₃NHCH₂CH₃ and CH₃NH₃⁺Br⁻.	Methylamine is the electron-pair donor, so attack must begin at its nitrogen lone pair. The attacked carbon loses bromide in the same substitution step and the new nitrogen has four bonds, hence a positive charge. Another methylamine molecule acts as a base: its lone pair attacks an N–H proton while the corresponding N–H bond electrons return to the substituted nitrogen. This produces neutral ethylmethylamine and methylammonium bromide.	5
05.1	$n(\text{anhydride})=6.12/102.0=0.0600\,\mathrm{mol}$ and $n(\text{propylamine})=2.95/59.0=0.0500\,\mathrm{mol}$ , so propylamine is limiting for the 1:1 reaction. $0.0500\,\mathrm{mol}$ of amide forms, with maximum mass $0.0500\times101.0=5.05\,\mathrm{g}$ . The organic products are N-propylethanamide and ethanoic acid. Propylamine is primary, RNH₂; acylation replaces only one N–H hydrogen with an ethanoyl group, so N-propylethanamide, CH₃CONHCH₂CH₂CH₃, retains one N–H bond.	Ethanoic anhydride and propylamine react 1:1. The calculated amounts are $0.0600\,\mathrm{mol}$ and $0.0500\,\mathrm{mol}$ respectively, so propylamine limits the amide amount to $0.0500\,\mathrm{mol}$ and gives $5.05\,\mathrm{g}$ at 100% yield. The expelled ethanoate group accepts a proton and forms ethanoic acid. Because propylamine begins with two N–H bonds, loss of one proton during addition–elimination leaves the product as a secondary amide with one N–H bond.	4

3.3.12.1 · Condensation polymers (A-level only)

Tier 1 · Easy

Mark scheme for 3.3.12.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	An amide linkage; water.	A carboxyl group and an amino group condense to form –CONH–. The –OH from the acid and an H from the amine form H₂O.	2
02.1	A polyester; an ester linkage; water.	Each molecule contains both an alcohol group and a carboxylic acid group. The two different functional groups on neighbouring molecules condense to form –COO– links, eliminating H₂O.	3

Tier 2 · Standard

Mark scheme for 3.3.12.1 Tier 2 · Standard
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01.1	[–NH–(CH₂)₆–NH–CO–(CH₂)₄–CO–]_n; hydrogen bonding.	Remove H from each terminal –NH₂ and OH from each –COOH as the two monomers join repeatedly. This gives the chain fragment –NH–(CH₂)₆–NH–CO–(CH₂)₄–CO– inside repeat brackets. N–H groups donate and carbonyl oxygen atoms accept hydrogen bonds between chains.	4
02.1	Butane-1,4-diamine, H₂N(CH₂)₄NH₂, and propanedioic acid, HOOCCH₂COOH.	Split both amide links at C–N. Restore H at each nitrogen end to recover the diamine and restore OH at each carbonyl carbon to recover the dicarboxylic acid.	4
03.1	Six ester links and six water molecules form; an unreacted hydroxyl group remains at each end of the chain.	Seven monomer molecules joined into one linear chain require six joins. Every join is between an –OH group and a –COOH group, so each creates one ester link and eliminates one H₂O molecule. Because the chain begins and ends with diol molecules, their outer –OH groups are not used and remain as the two end groups.	4

Tier 3 · Hard

Mark scheme for 3.3.12.1 Tier 3 · Hard
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01.1	Ethane-1,2-diol and benzene-1,4-dicarboxylic acid; a polyester; polar C=O and C–O bonds give permanent dipole–dipole attractions, with London forces also present.	Split each –COO– linkage between the acyl carbon and oxygen. Restore H to each chain oxygen to obtain HO–CH₂CH₂–OH, and restore OH to each acyl carbon to obtain HOOC–C₆H₄–COOH in the para arrangement shown. Ester groups identify a polyester. Its polar carbonyl and C–O bonds produce permanent dipoles, so neighbouring chains attract by permanent dipole–dipole forces as well as London forces.	6
02.1	Hexanedioic acid route: 82.7%; hexanedioyl dichloride route: 70.2%. The acid route has better atom economy, but atom economy alone does not determine reaction rate, equilibrium conversion, isolated yield, energy use or ease of purification.	M_r(repeat) = (8 × 12.0) + (12 × 1.0) + (4 × 16.0) = 172.0. For the acid route, total reactant M_r = 146.0 + 62.0 = 208.0, so atom economy = 172.0/208.0 × 100 = 82.7%. For the acyl chloride route, total reactant M_r = 183.0 + 62.0 = 245.0, so atom economy = 172.0/245.0 × 100 = 70.2%. The comparison measures formula mass incorporated into the repeat, not the practical yield or whole-process environmental impact.	6
03.1	Polymer A should resist stretching and thermal softening more strongly. Its amide groups form interchain hydrogen bonds and its aromatic rings make the chains relatively rigid; Polymer B has permanent dipole attractions but more flexible chains and no N–H hydrogen-bond donors.	The N–H groups in A donate hydrogen bonds to carbonyl oxygen atoms on neighbouring chains, producing strong repeated interchain attractions. Benzene rings restrict rotation and make the backbone harder to deform. Ester groups in B are polar and give permanent dipole–dipole attractions, but they cannot provide the same N–H···O hydrogen-bond network, while the aliphatic sections allow more chain movement.	5
04.1	The –NH₂ group of propylamine can condense with a terminal –COOH group, forming an amide link and water. After that reaction, the propyl group supplies no second reactive functional group. The capped chain end therefore cannot form another condensation link and the average chain length is reduced. Hexane-1,6-diamine has two –NH₂ groups, so after one reacts the other can still extend the chain.	Long-chain condensation requires a reacting molecule to retain another functional group after making one link. Propylamine can use its only amino group to form –CONHCH₂CH₂CH₃ at a carboxyl-terminated chain end, but its hydrocarbon end cannot react further. It is therefore a monofunctional chain stopper. A diamine is bifunctional: one amino group can join the existing chain while the second remains available for the next diacid molecule.	4
05.1	One repeat-unit representation is [–NH–(CH₂)₃–NH–CO–(CH₂)₅–CO–]_n. An equivalent representation is [–CO–(CH₂)₅–CO–NH–(CH₂)₃–NH–]_n. It is a polyamide because the continuing chain contains –CONH– linkages. In each representation, both bracket bonds pass through bonds that continue into neighbouring repeats. The second representation starts at a different point in the same repeating atom sequence; translating the brackets along the chain does not change the polymer.	Find the shortest atom sequence that reproduces the given chain: one diamine residue followed by one diacid residue. Brackets may start immediately before the first nitrogen or immediately before the first carbonyl carbon, provided the complete repeating sequence is retained and the continuation bonds connect correctly. The two displayed repeats are cyclic shifts of the same sequence, so repeating either gives the supplied chain.	5

3.3.12.2 · Biodegradability and disposal of polymers (A-level only)

Tier 1 · Easy

Mark scheme for 3.3.12.2 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	A polyalkene has an inert C–C backbone with no hydrolysable polar linkage, whereas a polyester contains ester bonds that can be hydrolysed.	Identify the bond that water, acid or alkali can attack. The polyester has ester links containing an electron-deficient carbonyl carbon; the polyalkene backbone contains only strong C–C bonds and no comparable functional group.	2
02.1	The main chain contains only C–C bonds; the ester groups are side groups, so hydrolysing them does not break the carbon backbone.	Locate the bonds that continue through the repeat brackets. They are C–C bonds made by addition polymerisation. Although each repeat has an ester group, that group is not part of the backbone, so its hydrolysis changes a side group rather than shortening the chain.	2

Tier 2 · Standard

Mark scheme for 3.3.12.2 Tier 2 · Standard
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01.1	The polyamide is more biodegradable because its polar amide links can be hydrolysed, breaking the chain; poly(propene) has a chemically inert carbon–carbon backbone and is non-biodegradable.	Relate biodegradation to bond cleavage. Hydrolysis of –CONH– links divides a polyamide into smaller molecules, so biological processes can eventually break it down. Poly(propene) lacks hydrolysable links: its backbone is made from C–C bonds with C–H and methyl substituents, so it persists much longer.	4
02.1	The product is 2-hydroxypropanoic acid, HOCH(CH₃)COOH; water cleaves the ester links in the backbone, producing smaller molecules and ultimately the monomer.	Split each backbone –COO– link and restore OH to the acyl carbon and H to the single-bonded oxygen. Both restored ends belong to the same bifunctional monomer, 2-hydroxypropanoic acid. Because the ester link lies in the backbone, every cleavage reduces chain length.	3
03.1	Propane-1,3-diamine, H₂N(CH₂)₃NH₂, and disodium butanedioate, NaOOC(CH₂)₂COONa; hydrolysis cleaves the amide links in the backbone.	Split each –CO–NH– bond. Restore H at nitrogen to recover the diamine. Under alkaline conditions each carboxyl product remains deprotonated, so restore –COO⁻Na⁺ at both ends of the four-carbon diacid fragment. Because these amide bonds lie on the path through the repeat brackets, cleaving them shortens and ultimately dismantles the chain.	4

Tier 3 · Hard

Mark scheme for 3.3.12.2 Tier 3 · Hard
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01.1	Recycling conserves raw materials but needs clean separated streams; landfill uses land and leaves persistent waste; incineration recovers energy and reduces volume but emits carbon dioxide and may form harmful gases. Sorting makes useful recycling streams possible and separates polymers needing different emission controls.	Landfill requires little processing but occupies land and retains non-biodegradable material. Mechanical recycling reduces demand for new petrochemical feedstock, but mixed or contaminated polymers give poor products and separation costs energy. Incineration handles mixed waste, releases useful energy and greatly reduces volume, but adds CO₂ and can produce acidic or toxic emissions. Sorting raises the quality of recycled material and allows unsuitable fractions to be treated separately, so the justified strategy is usually a combination rather than one universal method.	6
02.1	The claim is too broad. Ester links can hydrolyse under suitable warm, moist biological conditions, so industrial composting may be effective. Degradation may be slow in a cool, dry or oxygen-poor landfill. A clean polyalkene stream can be mechanically recycled and conserve feedstock, whereas mixing a compostable polymer into that stream can contaminate the recycled product.	Link each judgement to the disposal conditions. Chemical structure makes hydrolysis possible but does not guarantee a useful rate in every environment. Landfill can suppress the conditions needed for biodegradation. Mechanical recycling needs sorted, compatible material and avoids making new polymer from fresh feedstock. A valid decision therefore depends on collection, sorting, actual degradation conditions, emissions and the number of useful reuse cycles.	6
03.1	$0.0390\,\mathrm{mol}$ of each monomer is recovered; the mass of benzene-1,4-dicarboxylic acid is $6.47\,\mathrm{g}$ . Yield alone does not include energy use, reagent use, emissions or the fate of unrecovered material.	The number of repeat sections represented is $9.60/192=0.0500\,\mathrm{mol}$ . Each section gives one molecule of each monomer, so the recovered amount of each is $0.0500\times0.780=0.0390\,\mathrm{mol}$ . The acid mass is $0.0390\times166=6.474\,\mathrm{g}$ , reported as $6.47\,\mathrm{g}$ to three significant figures. A life-cycle judgement also needs the process inputs and wastes, not only recovered yield.	6
04.1	C₂H₄ + 3O₂ → 2CO₂ + 2H₂O. C₃H₄O₂ + 3O₂ → 3CO₂ + 2H₂O. Poly(ethene) forms $(1000/28.0)\times2\times44.0=3.14\,\mathrm{kg}$ of CO₂. The polyester forms $(1000/72.0)\times3\times44.0=1.83\,\mathrm{kg}$ of CO₂. The claim is not supported: the polyester's lower figure reflects carbon already partly oxidised, and neither figure credits energy recovered from incineration or the option of recycling instead.	Balance complete combustion per repeat formula. One mole of C₂H₄ repeat mass produces two moles of CO₂, while one mole of C₃H₄O₂ produces three. Their repeat masses are 28.0 and 72.0, so $1.00\,\mathrm{kg}$ gives $3.142857\ldots\,\mathrm{kg}$ and $1.83333\ldots\,\mathrm{kg}$ CO₂. The polyester already contains oxygen, so its carbon is partly oxidised and less CO₂ is formed per kilogram; neither calculated figure includes any benefit from recovering energy during incineration.	5
05.1	Chemical recycling recovers $17.6/24.0\times100=73.3\%$ . Mechanical recycling recovers $20.4/24.0\times100=85.0\%$ . Mechanical recycling gives the larger usable mass. Mechanical recycling also uses $26.5\,\mathrm{kWh}$ less energy for this batch. Chemical recycling can supply monomers suitable for making higher-quality material rather than a degraded polymer. The preferred process depends on required product quality, energy source, further processing, emissions and the fate of unrecovered material.	Divide each recovered mass by the common $24.0\,\mathrm{kg}$ feed: chemical recovery is 73.333...%, reported as 73.3%, and mechanical recovery is exactly 85.0%. The energy difference is $38.0-11.5=26.5\,\mathrm{kWh}$ . Those two measures favour mechanical recycling, but recovered monomers may be more valuable because they can be repolymerised without carrying forward the same loss of material properties. A justified decision must therefore state the intended use and include wider process impacts.	6

3.3.13.1 · Amino acids (A-level only)

Tier 1 · Easy

Mark scheme for 3.3.13.1 Tier 1 · Easy
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01.1	A zwitterion contains both a positive and a negative charge in the same species and has zero overall charge.	For an amino acid, the positive site is normally –NH₃⁺ and the negative site is –COO⁻. The charges cancel, but both must be shown.	2
02.1	It contains an amino group and a carboxylic acid group. The amino group can accept H⁺ and the carboxylic acid group can donate H⁺.	Amphoteric substances can react as both acids and bases. The nitrogen lone pair makes –NH₂ basic, while the –COOH group can lose a proton and act as an acid.	3

Tier 2 · Standard

Mark scheme for 3.3.13.1 Tier 2 · Standard
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01.1	Acid: CH₃CH(NH₃⁺)COOH; alkali: CH₃CH(NH₂)COO⁻.	In acid, the amino group accepts H⁺ while the carboxyl group remains protonated, giving an overall $+1$ ion. In alkali, OH⁻ removes the carboxyl proton while the amino group is unprotonated, giving an overall $-1$ ion.	4
02.1	CH₃CH₂CH(NH₂)COOH; 2-aminobutanoic acid; CH₃CH₂CH(NH₃⁺)COO⁻.	An α-amino acid has NH₂ on the carbon next to COOH. With an unbranched four-carbon chain, the structure is HOOC–CH(NH₂)–CH₂–CH₃, whose α-carbon has four different groups. Internal proton transfer from COOH to NH₂ gives NH₃⁺ and COO⁻.	4
03.1	$0.0350\,\mathrm{mol}$ HCl and $0.0350\,\mathrm{mol}$ NaOH are needed. H⁺ protonates –COO⁻; OH⁻ removes a proton from –NH₃⁺.	Each zwitterion has one carboxylate site that can accept one H⁺, so protonation is 1:1 with HCl. It also has one ammonium site that can lose one H⁺ to OH⁻, forming water, so deprotonation is 1:1 with NaOH. Each separate $0.0350\,\mathrm{mol}$ sample therefore needs $0.0350\,\mathrm{mol}$ of the stated reagent.	3

Tier 3 · Hard

Mark scheme for 3.3.13.1 Tier 3 · Hard
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01.1	CH₃CH₂CH(NH₃⁺)COO⁻ + H⁺ → CH₃CH₂CH(NH₃⁺)COOH; it acts as a base. CH₃CH₂CH(NH₃⁺)COO⁻ + OH⁻ → CH₃CH₂CH(NH₂)COO⁻ + H₂O; it acts as an acid.	Represent the zwitterion as CH₃CH₂CH(NH₃⁺)COO⁻. HCl protonates –COO⁻ to –COOH, so the zwitterion accepts a proton and behaves as a base. OH⁻ removes a proton from –NH₃⁺, producing –NH₂, H₂O and the carboxylate ion; the zwitterion behaves as an acid.	6
02.1	The amino acid is present as a zwitterion. Strong electrostatic attractions between oppositely charged ions require substantial energy to overcome, and the charged groups interact strongly with polar water but not with non-polar hexane.	In the solid, proton transfer gives –NH₃⁺ and –COO⁻ within each amino-acid species. The crystal is held by strong attractions between charges, so melting needs more energy than overcoming ordinary intermolecular forces in neutral molecules. Ion–dipole interactions with water stabilise dissolved zwitterions, whereas hexane cannot stabilise the charged groups effectively.	5
03.1	$0.0375\,\mathrm{mol}$ OH⁻; the final ion is ⁻OOCCH₂CH(NH₂)COO⁻ and has overall charge $2-$ .	There are three removable protons in the stated species: one on each of the two –COOH groups and one extra proton on –NH₃⁺. Complete conversion therefore uses three moles of OH⁻ per mole of amino acid. Thus $n(\mathrm{OH^-})=3\times0.0125=0.0375\,\mathrm{mol}$ . Both carboxyl groups become –COO⁻ and nitrogen becomes neutral –NH₂, giving charge $2-$ .	5
04.1	Each glycine zwitterion accepts one H⁺ at –COO⁻. Each diamino-acid zwitterion accepts two H⁺: one at –COO⁻ and one at the neutral –NH₂ group. If $x$ is the diamino-acid amount, $(0.0500-x)+2x=0.0800$ . $x=0.0300\,\mathrm{mol}$ diamino acid and $0.0200\,\mathrm{mol}$ glycine.	Excess acid protonates every carboxylate and every unprotonated amino group. Glycine therefore uses one proton per zwitterion, while H₃N⁺(CH₂)₂CH(NH₂)COO⁻ uses two. With $x$ moles of diamino acid and $0.0500-x$ moles of glycine, proton balance gives $0.0500-x+2x=0.0800$ , so $x=0.0300\,\mathrm{mol}$ and the glycine amount is $0.0200\,\mathrm{mol}$ .	4
05.1	A zwitterionic form is H₃N⁺(CH₂)₄CH(NH₂)COO⁻. The zwitterion has overall charge zero. In excess acid the ion is H₃N⁺(CH₂)₄CH(NH₃⁺)COOH and has charge $2+$ . In excess alkali the ion is H₂N(CH₂)₄CH(NH₂)COO⁻. The alkaline ion has overall charge $1-$ .	Internal proton transfer can protonate one amino group while the carboxyl group becomes –COO⁻, giving cancelling charges. In excess acid, both amino groups are protonated and the carboxyl group is neutral –COOH, so the two ammonium groups give charge $2+$ . In excess alkali, both amino groups are neutral –NH₂ and the single carboxyl group is –COO⁻, giving charge $1-$ .	5

3.3.13.2 · Proteins (A-level only)

Tier 1 · Easy

Mark scheme for 3.3.13.2 Tier 1 · Easy
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01.1	The sequence or order of amino acids in the polypeptide chain.	Primary structure refers only to which amino acids occur and their order along the chain, not to the helix, sheet or overall folding.	1
02.1	Hydrogen bonding between a peptide-link C=O group and a peptide-link N–H group.	The carbonyl oxygen is a hydrogen-bond acceptor and the δ⁺ hydrogen bonded to nitrogen is a donor. Repeated C=O···H–N attractions help hold the chain in an α-helix.	2

Tier 2 · Standard

Mark scheme for 3.3.13.2 Tier 2 · Standard
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01.1	H₂NCH₂CONHCH(CH₃)COOH; peptide or amide link; water.	Remove OH from glycine's –COOH and H from alanine's –NH₂. Join the glycine carbonyl carbon to the alanine nitrogen, giving –CO–NH– and H₂O. Keep the stated order: glycine is on the N-terminal side and alanine on the C-terminal side.	4
02.1	3 mol H₂O; 2 mol glycine, 1 mol alanine and 1 mol cysteine.	A chain of four residues contains three peptide links. Complete hydrolysis consumes one water molecule per peptide link, so it uses 3 mol H₂O per mole of tetrapeptide. Reading the sequence gives two Gly residues and one each of Ala and Cys.	4
03.1	Gly–Ala–Ser; H₂NCH₂CONHCH(CH₃)CONHCH(CH₂OH)COOH.	The shared alanine residue must follow glycine in one fragment and precede serine in the other, so the fragments overlap as Gly–Ala–Ser. Join glycine's carboxyl group to alanine's amino group and alanine's carboxyl group to serine's amino group, producing two –CONH– links while retaining the CH₃ and CH₂OH side chains.	4

Tier 3 · Hard

Mark scheme for 3.3.13.2 Tier 3 · Hard
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01.1	Both peptide links are hydrolysed; hydrogen bonds and sulfur–sulfur bonds can maintain tertiary structure; $R_f=0.59$ .	Complete hydrolysis adds the elements of water across each –CONH– link and releases the three amino acids. In proteins, two cysteine residues can form an S–S bond, and polar groups can form hydrogen bonds during folding. For the chromatogram, $R_f=38/64=0.59375$ , so $R_f=0.59$ to two decimal places.	6
02.1	The products are alanine, glycine and cysteine; two peptide links are broken. Two cysteine side-chain –SH groups can be oxidised to form a covalent S–S disulfide bridge between parts of the protein chain.	Split at each –CO–NH– link and restore OH to each carbonyl carbon and H to each nitrogen. The side groups are CH₃, H and CH₂SH, identifying alanine, glycine and cysteine. Three residues are joined by two peptide links. Oxidation of two thiol groups removes hydrogen and creates an S–S bond, providing a covalent link that helps fix the folded shape.	6
03.1	Heating disrupts secondary and may disrupt tertiary structure; reducing S–S bonds disrupts tertiary structure; complete peptide hydrolysis destroys primary structure and consequently the higher structures. The first two treatments leave the amino-acid sequence joined by peptide bonds, whereas hydrolysis cleaves those bonds.	Hydrogen bonds between peptide C=O and N–H groups stabilise α-helices and also contribute to tertiary folding, so disrupting them changes those higher levels. Disulfide bridges are covalent links between cysteine side chains at different positions in the fold, so reducing them changes tertiary structure. Primary structure is the covalently joined residue sequence: it remains while peptide links are intact, but complete hydrolysis splits those links and releases amino acids.	6
04.1	The sample $R_f$ values are 0.300, 0.531 and 0.750. The spots are consistent with glycine, alanine and leucine respectively. Matching $R_f$ values support but do not uniquely prove identity because another amino acid could coincide under these conditions. A spot with $R_f=0.42$ would travel $0.42\times7.20=3.02\,\mathrm{cm}$ .	Use the distance of each spot centre divided by the common solvent-front distance. The values are 0.300, 0.53056 and 0.750; the middle value is 0.531 to three significant figures and agrees with 0.53 to the standard's precision. Match each only to the standard developed alongside it. TLC retention is not a unique structural measurement, so another analytical technique would strengthen the assignments. For the final distance, rearrange the definition to give $0.42\times7.20=3.024\,\mathrm{cm}$ , reported as $3.02\,\mathrm{cm}$ .	4
05.1	Each mole of S–S bridges corresponds to a mass loss of $2.00\,\mathrm{g}$ . $n(\mathrm{S-S})=3.60\times10^{-3}/2.00=1.80\times10^{-3}\,\mathrm{mol}$ . The new covalent links can stabilise or alter tertiary structure. Primary structure is unchanged because the amino-acid sequence and peptide links remain intact. Secondary structure is not defined by disulfide bridges; it is maintained principally by peptide-group hydrogen bonding.	Two –SH groups form one –S–S– link by losing two H atoms, so one mole of bridges removes $2.00\,\mathrm{g}$ . Convert $3.60\,\mathrm{mg}$ to $3.60\times10^{-3}\,\mathrm{g}$ and divide to obtain $1.80\times10^{-3}\,\mathrm{mol}$ . The bridge connects side chains at different positions in a folded protein, affecting tertiary structure without changing residue order or breaking peptide bonds.	5

3.3.13.3 · Enzymes (A-level only)

Tier 1 · Easy

Mark scheme for 3.3.13.3 Tier 1 · Easy
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01.1	It increases reaction rate by providing a route with lower activation energy and is regenerated or not consumed overall.	Give both catalytic ideas: the enzyme lowers the activation-energy barrier and is available again after products leave the active site.	2
02.1	The active site has a three-dimensional shape and arrangement of groups that allow the substrate to fit and form suitable interactions.	Complementary does not mean merely the same outline. The substrate must fit the pocket and position its polar or charged groups so that the required intermolecular attractions can form.	2

Tier 2 · Standard

Mark scheme for 3.3.13.3 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	The active site is three-dimensional and stereospecific; only one enantiomer places its groups in the correct positions to form the required intermolecular interactions at the same time.	Enantiomers have mirror-image arrangements. The active site is itself chiral because it is built from a folded protein. One enantiomer can align complementary charged, polar or non-polar groups with binding groups in the site, whereas the mirror image cannot make the same set of contacts simultaneously.	4
02.1	The inhibitor binds reversibly at the active site and competes with the substrate, so fewer enzyme–substrate complexes form. It changes the rate of reaching equilibrium but not the equilibrium position.	Shape and group complementarity allow the inhibitor to occupy the active site. While it is bound, substrate cannot bind there, so productive collisions are less frequent. Because the binding is reversible and the enzyme is a catalyst, it changes activation pathways and rates rather than the thermodynamic equilibrium composition.	4
03.1	The inhibitor lowers the rate strongly when substrate is scarce, but its effect is much smaller when substrate is abundant. Extra substrate can outcompete a reversible inhibitor for the same active sites, supporting active-site competition.	At low substrate concentration the rate falls by $8.0\,\mathrm{mmol\,min^{-1}}$ , to one third of the uninhibited value. At high substrate concentration it falls by only $6.0\,\mathrm{mmol\,min^{-1}}$ , remaining at 87.5% of the uninhibited value. A high frequency of substrate–active-site encounters therefore reduces inhibitor occupancy, as expected when both molecules seek the same site.	4

Tier 3 · Hard

Mark scheme for 3.3.13.3 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	The drug can bind strongly in the active site and block substrate entry; only a suitably oriented stereoisomer makes the required interactions; modelling can screen fit and interactions before laboratory synthesis.	A transition-state-like arrangement may be complementary to the binding groups in the active site, so the drug occupies the site and prevents the normal substrate forming an enzyme–substrate complex. Because the site is stereospecific, the wrong stereoisomer may not align its groups and may bind weakly. A computer model can dock many candidate structures, estimate shape complementarity and interactions, and prioritise the most promising molecules for synthesis and testing.	6
02.1	The model accounts for the three-dimensional fit and predicted interactions, so it can prioritise the matching enantiomer and reduce unnecessary syntheses. Its result depends on the model and does not itself show inhibition in a real system. Measure enzyme reaction rate with and without the candidate, and compare the two enantiomers under identical conditions.	An asymmetric active site can distinguish mirror-image molecules, making an enantiomer-specific prediction chemically reasonable. A predicted binding pose is still evidence from a model, not a measured rate change. A controlled rate experiment tests inhibition directly; testing both pure enantiomers checks the predicted stereospecificity. Replicates and a range of candidate concentrations would strengthen both comparisons.	6
03.1	The changed sequence can alter local interactions, tertiary folding, active-site shape and the binding interaction with the substrate or transition state. Fewer correctly bound complexes or poorer transition-state stabilisation raises the effective activation barrier and lowers rate, but the equilibrium position is unchanged.	Replacing one residue changes the primary structure. Loss of a charged side chain can remove an ionic or hydrogen-bonding interaction and can also alter how the protein folds, so the substrate may no longer have the required three-dimensional and chemical complementarity. The enzyme then provides a less effective lower-activation-energy pathway. Catalysts affect forward and reverse rates, not the thermodynamic equilibrium composition.	5
04.1	P is predicted to bind more strongly because it can form both an ionic attraction and a hydrogen bond. Only a stereoisomer that presents both groups in the required three-dimensional arrangement can make the two contacts simultaneously. A static model may not represent protein movement, solvent effects or the range of conformations in the real active site. Predicted binding does not demonstrate inhibition, toxicity, uptake or stability and therefore needs experimental testing.	Count simultaneous complementary contacts rather than functional groups in isolation. P has two favourable contacts in the stated pose, whereas Q has one. Mirror-image arrangements can place one group correctly while directing the other away from its partner, so enantiomers need separate consideration. Computer modelling prioritises candidates, but enzyme-rate measurements and biological tests are required before concluding that binding produces a useful drug effect.	4
05.1	$n(\text{racemate})=0.612/255.0=2.40\times10^{-3}\,\mathrm{mol}=2.40\,\mathrm{mmol}$ . The racemate contains $2.40/2=1.20\,\mathrm{mmol}$ of A. The $1.20\,\mathrm{mmol}$ of A in the racemate and the $1.20\,\mathrm{mmol}$ pure-A dose give equal initial inhibition because only A binds effectively to the stereospecific active site. Pure A is preferable because it avoids administering B, which could cause a different biological effect or add to the dose burden.	Divide mass by relative molecular mass: $0.612/255.0=0.00240\,\mathrm{mol}=2.40\,\mathrm{mmol}$ of racemate. A racemate is a 1:1 mixture, so half is the active A form: $1.20\,\mathrm{mmol}$ . This equals the stated pure-A amount and explains the equal initial inhibition. The mirror-image B molecule cannot reproduce the required three-dimensional contacts, but lack of inhibition at this enzyme does not prove that B is harmless elsewhere.	4

3.3.13.4 · DNA (A-level only)

Tier 1 · Easy

Mark scheme for 3.3.13.4 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	A phosphate group, 2-deoxyribose and one nitrogen-containing base.	List one component from each part of the nucleotide: phosphate, the pentose sugar 2-deoxyribose, and one of A, C, G or T.	3
02.1	Adenine is only a nitrogen-containing base. A nucleotide contains a base bonded to 2-deoxyribose, which is bonded to a phosphate group.	Do not use base and nucleotide as synonyms. Adenine supplies the base part of a nucleotide; phosphate and 2-deoxyribose must also be present.	2

Tier 2 · Standard

Mark scheme for 3.3.13.4 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	T–G–C–A–A–T; covalent bonds form the sugar–phosphate backbone, while hydrogen bonds join complementary bases across the strands.	Apply A–T and C–G pairing one position at a time to obtain T–G–C–A–A–T. The continuous backbone within either strand is covalently bonded. The two separate strands associate through hydrogen bonds between paired bases.	4
02.1	Thymine = 22%; guanine = 28%; cytosine = 28%.	Every adenine is paired with thymine, so their percentages are equal: A = T = 22%. This accounts for 44%, leaving 56% for G and C. Since guanine pairs with cytosine, each contributes 56/2 = 28%.	3
03.1	There are 36 nucleotides, 36 2-deoxyribose units and 36 phosphate groups. Covalent bonds join alternating sugar and phosphate units along each backbone; hydrogen bonds act between paired bases across the strands.	Each base pair contains one nucleotide on each of two strands, so $18\times2=36$ nucleotides and therefore 36 sugar units. Each strand has 17 internucleotide phosphates plus its stated 5′ terminal phosphate, giving $2\times18=36$ phosphate groups. The backbone is covalent, whereas complementary bases across the strands are hydrogen-bonded.	4

Tier 3 · Hard

Mark scheme for 3.3.13.4 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$49$ hydrogen bonds; C–G pairs form three hydrogen bonds but A–T pairs form two.	The A–T contribution is $8\times2=16$ hydrogen bonds and the C–G contribution is $11\times3=33$ . The total is $16+33=49$ . Replacing A–T pairs with C–G pairs increases the number of hydrogen bonds between the strands, so more energy is required to overcome the intermolecular attractions.	5
02.1	The statement is incorrect. Heating disrupts hydrogen bonds between complementary bases rather than hydrolysing the covalent sugar–phosphate backbone. On cooling, complementary bases can realign and reform hydrogen bonds if the strands remain intact.	Distinguish bonding locations. Strong covalent bonds form the sugar–phosphate backbone along each strand; hydrogen bonds join the paired bases across the two strands. Heating supplies enough energy to disrupt many intermolecular hydrogen bonds without necessarily breaking the covalent backbone. The unchanged base sequence then guides complementary re-pairing on cooling.	5
03.1	³²P remains in the covalent sugar–phosphate backbone of the separated strands, and ¹⁵N remains in thymine bases covalently attached to their sugars. Heating breaks hydrogen bonds between complementary bases but leaves the covalent backbone and nucleotide components intact.	The phosphate component is part of the repeating backbone, so its ³²P label travels with whichever strand contains that nucleotide. Thymine is the labelled base, so ¹⁵N stays within thymine and remains bonded through the nucleotide to its strand. Strand separation requires disruption of the weaker interstrand A–T and C–G hydrogen bonds, not cleavage of sugar–phosphate or sugar–base covalent bonds.	5
04.1	Let $x$ be A–T pairs and $y$ be C–G pairs, so $x+y=80$ . Hydrogen-bond counting gives $2x+3y=196$ . Solving the equations gives 44 A–T pairs. There are 36 C–G pairs.	Every base pair is one of the two complementary types. Substitute $x=80-y$ into $2x+3y=196$ : $160-2y+3y=196$ , so $y=36$ and $x=44$ .	4
05.1	Each fragment has $30\times2+20\times3=120$ hydrogen bonds in total. They need not have the same base sequence because identical pair counts can be arranged in different orders. The equal bond counts suggest similar total energy to separate all pairs if other factors are comparable, but bond count alone does not describe sequence-dependent local stability or conditions. Each fragment contains 30 adenine bases.	Total the interstrand hydrogen bonds using two per A–T pair and three per C–G pair. Composition fixes this total but not the order of the pairs, so many sequences share the same value. A total bond count supports a broad comparison of complete separation energy, but it cannot prove identical behaviour at every region of the fragments.	4

3.3.13.5 · Action of anticancer drugs (A-level only)

Tier 1 · Easy

Mark scheme for 3.3.13.5 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	Guanine; a nitrogen atom.	The specification requires the Pt–N interaction to be linked specifically to guanine. The nitrogen donates a lone pair to platinum.	2
02.1	A chloride ligand is replaced; a co-ordinate covalent bond forms between platinum and a nitrogen atom on guanine.	A guanine nitrogen atom replaces a chloride ligand and donates a lone pair to Pt(II). The resulting Pt–N link is a co-ordinate covalent bond, not a hydrogen bond.	2

Tier 2 · Standard

Mark scheme for 3.3.13.5 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Chloride ligands are replaced and Pt forms coordinate bonds to guanine nitrogen atoms; the resulting links distort or hold the DNA so the strands cannot separate or be copied normally.	A guanine nitrogen lone pair replaces a ligand at Pt(II), producing a Pt–N coordinate bond. Formation of more than one such bond links sites in the DNA and changes its shape. The damaged double helix cannot open and act as a normal template, so DNA replication and cell division are inhibited.	4
02.1	The chloride ligands are adjacent in cisplatin and opposite in transplatin. Ligand replacement in cisplatin allows platinum to form coordinate bonds at two suitably positioned sites in DNA, creating a cross-link that distorts the DNA and hinders replication.	In a square-planar complex, cis means adjacent and trans means opposite. Replacement of the labile chloride ligands creates sites through which platinum can bond to nitrogen donor atoms in DNA bases. The adjacent positions allow the characteristic cross-link and bend in DNA that prevents normal copying.	4
03.1	cis-[Pt(NH₃)₂Cl₂] + G → [Pt(NH₃)₂Cl(G)]⁺ + Cl⁻ A co-ordinate (dative covalent) bond forms from the nitrogen lone pair to platinum Oxidation state +2, co-ordination number 4.	Guanine is a neutral ligand and replaces one Cl⁻ ligand. The neutral cisplatin complex therefore forms a complex ion with charge +1 and releases Cl⁻. Nitrogen donates a lone pair to platinum to form a co-ordinate bond. Ligand replacement is not a redox reaction, so platinum remains at +2; replacing one ligand with one donor atom also keeps the co-ordination number at 4.	4

Tier 3 · Hard

Mark scheme for 3.3.13.5 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Cisplatin blocks DNA replication in rapidly dividing tumour cells, but it can also block replication in healthy rapidly dividing cells such as bone-marrow and gut-lining cells; the therapeutic benefit must therefore be weighed against toxicity.	Cancer cells divide rapidly and need repeated DNA replication, so Pt–guanine links can restrict tumour growth. The drug is not perfectly selective: bone-marrow cells and cells replacing the digestive lining also divide frequently and can suffer the same DNA damage. A suitable dose and treatment plan must provide enough anticancer effect while keeping harm to healthy tissue acceptable.	5
02.1	The data support stronger inhibition of DNA synthesis by the cis isomer under these conditions, consistent with cisplatin cross-linking DNA more effectively. They do not prove that geometry is the sole cause of cell death: the experiment reports DNA synthesis rather than cell survival, gives no uncertainty or replication, and other differences in uptake, ligand replacement or side reactions could contribute.	Compare the controlled outcome first: 24% is much lower than 81%, so the cis isomer has the larger measured effect. Then limit the inference. One end-point cannot establish the whole causal chain from geometry to cell death, and no conclusion about reliability can be made without repeats or uncertainties. Measurements of cell viability, platinum–DNA binding and repeated dose-response data would test the proposed explanation more directly.	6
03.1	The drug can attach at one guanine site but cannot use a second replacement site to cross-link two DNA positions, so it should distort or lock the helix less effectively and inhibit replication less. Its Pt–N binding can still damage DNA in healthy dividing cells, so adverse effects remain possible.	One guanine bond anchors the complex, but a cross-link needs two suitably placed Pt–N bonds. Cisplatin's adjacent replaceable sites allow this second attachment and the resulting distortion blocks normal strand copying. Removing that capacity weakens the causal chain to replication failure, but does not make binding selective for tumour DNA; healthy cells that copy DNA can still be harmed.	5
04.1	Tumour-shrinkage percentages are $64/160\times100=40.0\%$ for the standard treatment and $104/160\times100=65.0\%$ with cisplatin. Severe-adverse-effect percentages are $12/160\times100=7.50\%$ and $36/160\times100=22.5\%$ respectively. The data show a larger measured benefit but also substantially more severe harm, so use should depend on the patient's likely benefit, alternatives and tolerance of risk; the figures alone do not justify automatic treatment for everyone. Healthy dividing cells contain the same guanine in replicating DNA. Cisplatin cannot distinguish tumour DNA from healthy-cell DNA.	Divide each count by the common group size of 160. The measured response is 40.0% with standard treatment and 65.0% with cisplatin, while severe adverse effects occur in 7.50% and 22.5% respectively. A defensible treatment decision must balance the extra chance of tumour response against the extra risk of serious harm. Guanine occurs in the DNA of both tumour cells and healthy dividing cells, and cisplatin cannot distinguish between them, so Pt–N bonding can also inhibit replication in healthy cells.	5
05.1	$2.80\,\mathrm{nmol}$ of cisplatin requires $5.60\,\mathrm{nmol}$ of guanine sites. Cisplatin is limiting because $7.20\,\mathrm{nmol}$ of sites are available. The maximum cross-link amount is $2.80\,\mathrm{nmol}$ . $7.20-5.60=1.60\,\mathrm{nmol}$ of guanine sites remain unlinked. The Pt–N cross-links distort or constrain the double helix, preventing normal strand separation and copying.	Two guanine donor sites are consumed per cisplatin cross-link. The arriving drug therefore needs $2\times2.80=5.60\,\mathrm{nmol}$ of sites, fewer than the $7.20\,\mathrm{nmol}$ available, so all drug molecules can cross-link. Subtraction leaves $1.60\,\mathrm{nmol}$ of sites. Linking two DNA positions changes the helix geometry and blocks the molecular movements required for replication.	5

3.3.14 · Organic synthesis (A-level only)

Tier 1 · Easy

Mark scheme for 3.3.14 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	It produces less waste and uses a greater fraction of reactant atoms in the desired product; fewer steps can also reduce energy, solvent use or hazardous operations.	Link high atom economy to reduced by-product waste, then link fewer steps to a second environmental or safety benefit such as lower energy demand, less solvent, fewer reagents or fewer purification stages.	2
02.1	Route A maximises isolated product; Route B has the better atom economy and incorporates the greater fraction of reactant atoms.	Percentage yield compares actual product with the theoretical amount, so the larger 68% yield selects Route A for isolated quantity. Atom economy measures the fraction of reactant formula mass in the desired product, so the larger 82% selects Route B for incorporation of atoms.	2

Tier 2 · Standard

Mark scheme for 3.3.14 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Steam/H₃PO₄ catalyst gives ethanol; acidified potassium dichromate(VI), warm and distil gives ethanal; acidified potassium dichromate(VI), heat under reflux gives ethanoic acid.	Hydrate ethene with steam using an H₃PO₄ catalyst to make ethanol. Partially oxidise the primary alcohol with acidified K₂Cr₂O₇ and distil the ethanal as it forms. Return ethanal to excess oxidising mixture and heat under reflux to complete oxidation to ethanoic acid.	6
02.1	Aqueous sodium hydroxide can hydrolyse the ester: wash with aqueous sodium carbonate or aqueous sodium hydrogencarbonate, venting the separating funnel as carbon dioxide forms. Aqueous magnesium sulfate cannot dry the ester: use anhydrous magnesium sulfate or another suitable anhydrous drying agent.	The esterification, cooling, separating-funnel transfer and final distillation are valid. The washing reagent should neutralise acidic impurities without using strongly alkaline NaOH: use Na₂CO₃(aq) or NaHCO₃(aq), and release the CO₂ pressure while shaking. Drying requires an anhydrous solid such as MgSO₄; an aqueous solution would add water rather than remove it.	4
03.1	Route A overall yield = 58.0%. Route B: higher isolated yield (68.0% against 58.0%), one step rather than three, and no solvent required. Route A: higher atom economy (88% against 63%), so a greater fraction of the reactant mass ends up in the desired product.	Multiply the unrounded decimal step yields: $0.920\times0.840\times0.750=0.5796$ , so Route A gives 57.96%, or 58.0% to three significant figures. Both sustainability measures are legitimate and answer different questions: isolated yield, step count and solvent use favour Route B, while atom economy favours Route A.	4

Tier 3 · Hard

Mark scheme for 3.3.14 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Bromoethane → propanenitrile using ethanolic KCN under reflux → propylamine by reduction with H₂/Ni or LiAlH₄/dry ether → N-propylethanamide using ethanoyl chloride; cyanide contributes its carbon atom.	Use nucleophilic substitution with ethanolic KCN under reflux: CH₃CH₂Br becomes CH₃CH₂CN. The carbon in CN⁻ joins the chain, so the product has three carbons. Reduce propanenitrile with H₂/Ni or LiAlH₄ in dry ether to obtain CH₃CH₂CH₂NH₂. Finally react propylamine with CH₃COCl; nucleophilic addition–elimination gives CH₃CONHCH₂CH₂CH₃.	8
02.1	Step 1: heat propan-2-ol with concentrated H₂SO₄ at about 170–180 °C to form propene. Step 2: react propene with HBr at room temperature to form 2-bromopropane as the major product. Step 3: heat 2-bromopropane with excess ethanolic ammonia in a sealed tube to form propan-2-amine; ammonia acts as a nucleophile rather than a base here, so substitution dominates, and the excess limits further substitution.	Concentrated sulfuric acid and heat at about 170–180 °C eliminate water from propan-2-ol to give propene; passing the vapour over heated Al₂O₃ is an alternative. At room temperature, electrophilic addition of HBr across the C=C bond gives 2-bromopropane as the Markovnikov major product. Heating this halogenoalkane with excess ethanolic ammonia in a sealed tube substitutes NH₂ for Br. Excess ammonia favours reaction with ammonia rather than further substitution of the amine product.	7
03.1	Electrophilic addition of HBr gives mainly 2-bromopropane, CH₃CHBrCH₃. Heat the bromoalkane with ethanolic KCN under reflux. The nitrile is 2-methylpropanenitrile, (CH₃)₂CHCN. The cyanide carbon increases the carbon count from three to four without changing the existing branched connectivity. The claim is incorrect because 2-methylpropanenitrile is a branched isomer of straight-chain butanenitrile.	Electrophilic addition of HBr to propene proceeds mainly through the secondary carbocation, giving CH₃CHBrCH₃. Under ethanolic reflux, CN⁻ replaces Br and contributes its own carbon, producing (CH₃)₂CHCN. Carbon count alone is insufficient: the branch already present around the substituted carbon remains, so the product is 2-methylpropanenitrile rather than butanenitrile.	5
04.1	Warm propan-1-ol with acidified potassium dichromate(VI) and distil propanal as it forms. React propanal with KCN, followed by dilute acid, to form 2-hydroxybutanenitrile, CH₃CH₂CH(OH)CN. The cyanide carbon becomes the nitrile carbon and increases the chain length from three to four. Reduce the hydroxynitrile using $\mathrm{H_2/Ni}$ to convert –C≡N into –CH₂NH₂. The product is CH₃CH₂CH(OH)CH₂NH₂, 1-aminobutan-2-ol.	Work backwards from the amino alcohol: nitrile reduction supplies the terminal –CH₂NH₂ group, and nucleophilic addition using KCN followed by dilute acid converts propanal into the required hydroxynitrile. Propanal is obtained by controlled oxidation of propan-1-ol, with distillation preventing further oxidation. The cyanide carbon becomes the carbon of the terminal –CH₂NH₂ group, so the four-carbon skeleton is retained during reduction.	5
05.1	Heat 2-bromobutane under reflux with aqueous NaOH to form butan-2-ol. Heat butan-2-ol under reflux with acidified potassium dichromate(VI) to form butanone. React butanone with KCN, followed by dilute acid, to form 2-hydroxy-2-methylbutanenitrile. The nucleophilic-addition step increases the carbon count from four to five because cyanide supplies one carbon. The carbonyl group in butanone is planar, so CN⁻ can attack its carbonyl carbon from either face. The two attacks form equal amounts of the two enantiomers because the new chiral centre is bonded to OH, CN, CH₃ and CH₂CH₃; the product is therefore a racemate.	Aqueous OH⁻ replaces Br in 2-bromobutane during heating under reflux, giving butan-2-ol. Oxidation of this secondary alcohol gives butanone without changing the four-carbon skeleton. With KCN, followed by dilute acid, CN⁻ attacks either face of the planar carbonyl group and contributes its carbon to the molecule, producing a five-carbon hydroxynitrile. The two equally likely attack directions give mirror-image configurations at the new chiral centre, so an achiral reaction mixture produces a racemate.	6

3.3.15 · Nuclear magnetic resonance spectroscopy (A-level only)

Tier 1 · Easy

Mark scheme for 3.3.15 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	Any two of: it gives one sharp signal, is chemically inert, is volatile, and its signal lies away from most organic signals at $\delta=0$ .	All twelve TMS protons are equivalent, so TMS produces a single intense reference peak. It does not normally react with the sample and can be removed easily because it is volatile; its highly shielded signal is assigned $\delta=0$ .	2
02.1	Propanone, CH₃COCH₃; the two methyl carbon atoms are equivalent and the carbonyl carbon is the second environment.	A signal near 205 ppm is in the aldehyde/ketone carbonyl region. The formula and a single alkyl-carbon environment fit propanone. Its plane of symmetry makes the two CH₃ groups chemically equivalent, so both methyl carbons produce one signal at about 30 ppm.	3

Tier 2 · Standard

Mark scheme for 3.3.15 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Ethyl ethanoate, CH₃COOCH₂CH₃; the terminal CH₃ is split by two neighbouring protons into a triplet, CH₂ is split by three into a quartet, and the acyl CH₃ has no adjacent proton and is a singlet.	The 3H triplet and 2H quartet are the paired pattern of an ethyl group: $n=2$ neighbours split CH₃ into three peaks, while $n=3$ neighbours split CH₂ into four. The remaining 3H singlet must have no neighbouring proton. With two oxygen atoms, CH₃COOCH₂CH₃ fits the formula and all three integrations.	5
02.1	Propan-2-ol, CH₃CH(OH)CH₃.	The two equivalent CH₃ groups contain six protons and each is next to one methine proton, so they give a 6H doublet. The one CH bonded to oxygen accounts for the downfield 1H signal at 3.9 ppm. The exchangeable O–H proton gives the broad 1H signal. Equivalent methyl carbons plus the central C–O carbon account for two ¹³C environments.	5
03.1	Two CH₂ environments remain with integration ratio 2:2 (or 1:1). The O–H and N–H protons exchange with deuterium, so their ¹H signals disappear.	The two carbon environments are different because one CH₂ is bonded to O and the other to N, so they give separate 2H signals. Labile protons on oxygen and nitrogen exchange for deuterium when shaken with D₂O. Deuterium is not detected in an ordinary ¹H NMR spectrum, so the original 1H O–H and 2H N–H signals are removed. A new HDO peak may appear after the shake.	4

Tier 3 · Hard

Mark scheme for 3.3.15 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Ethyl propanoate, CH₃CH₂COOCH₂CH₃.	Two separate 3H-triplet/2H-quartet pairs show two non-equivalent ethyl groups. The formula requires an ester, so placing one ethyl group on the acyl side and one on oxygen gives CH₃CH₂COOCH₂CH₃. Each CH₃ has two adjacent protons and is a triplet; each CH₂ has three adjacent protons and is a quartet. O–CH₂ is further downfield because electronegative oxygen deshields it. The carbonyl carbon and four different alkyl carbons account for five ¹³C signals.	7
02.1	2-Methylpropanoic acid, (CH₃)₂CHCOOH.	The molecular ion matches M_r = 88 for C₄H₈O₂. The broad 2500–3000 cm⁻¹ absorption and 1710 cm⁻¹ carbonyl absorption identify a carboxylic acid. The 180 ppm carbon is COOH, the 34 ppm carbon is the adjacent CH and the two equivalent methyl carbons give one 19 ppm signal. Six equivalent methyl protons split into a doublet by the one neighbouring CH proton; the adjacent CH accounts for the 1H signal at 2.6 ppm. The 11.5 ppm broad singlet is the acidic O–H proton.	8
03.1	Pentan-2-one has five ¹³C environments; pentan-3-one has three. Pentan-3-one gives only a 6H triplet and a 4H quartet for its two equivalent ethyl groups, whereas pentan-2-one has more proton environments, including a 3H acyl-methyl singlet.	Pentan-3-one has a plane of symmetry through the carbonyl group: its two terminal CH₃ carbons are equivalent and its two CH₂ carbons are equivalent, with the carbonyl as the third environment. The six methyl protons each have two neighbouring CH₂ protons and form a triplet; the four methylene protons each have three neighbouring methyl protons and form a quartet. Pentan-2-one has no such symmetry, so all five carbons differ and its COCH₃ group gives a separate singlet.	5
04.1	The ethyl ethanoate amount is proportional to $3.20/2=1.60$ because its quartet represents two protons per molecule. The propanone amount is proportional to $7.20/6=1.20$ because its singlet represents six protons per molecule. The mole ratio ethyl ethanoate : propanone is $1.60:1.20=4:3$ . The mole percentages are 57.1% ethyl ethanoate and 42.9% propanone. Integrated area is proportional to the total number of contributing protons, so each area must be divided by that signal's protons per molecule.	Convert each diagnostic area to a relative molecular amount. For ethyl ethanoate, $3.20/2=1.60$ ; for propanone, $7.20/6=1.20$ . Division by 0.40 gives the integer ratio 4:3. The total ratio units are seven, so the percentages are $4/7\times100=57.142...\%$ and $3/7\times100=42.857...\%$ , reported to 57.1% and 42.9%.	5
05.1	The compound is ethyl methanoate, HCOOCH₂CH₃. The three different carbons are the ester carbonyl carbon, OCH₂ carbon and CH₃ carbon. The 1H downfield singlet is the methanoate H bonded to the carbonyl carbon. That proton is a singlet because there is no proton on an adjacent carbon. The OCH₂ group integrates to 2 and is split by three neighbouring methyl protons into a quartet. The terminal CH₃ group integrates to 3 and is split by two neighbouring methylene protons into a triplet.	No effervescence with aqueous sodium carbonate excludes a carboxylic acid. An ester carbonyl plus only three carbons therefore suggests arranging the remaining atoms as HCOO– and an ethyl group. HCOOCH₂CH₃ has exactly three carbon environments. The formyl-type ester proton has no adjacent carbon bearing hydrogen, so it is a 1H singlet. The ethyl group produces the paired 2H quartet and 3H triplet by the $n+1$ rule, completing the formula and signal assignment.	6

3.3.16 · Chromatography (A-level only)

Tier 1 · Easy

Mark scheme for 3.3.16 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$R_f=0.60$	Use $R_f=\text{distance moved by spot}/\text{distance moved by solvent}=4.5/7.5=0.60$ . It has no unit because it is a ratio of two distances.	2
02.1	Draw the start line in pencil; keep the solvent level below the start line; cover the chamber with a lid.	Graphite does not dissolve and travel like ink may. Keeping the solvent below the spots prevents samples dissolving directly into the solvent reservoir. A lid reduces solvent evaporation and helps maintain a solvent-saturated atmosphere for more reproducible development.	3

Tier 2 · Standard

Mark scheme for 3.3.16 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Dye P has the larger $R_f$ because it spends a greater proportion of time in the moving solvent and is retained less strongly by the stationary phase.	Movement results from repeated transfer between phases. P is carried further whenever it is in the mobile phase and is not held strongly on the solid, so its average movement is faster. Q spends more time attached to the stationary phase and remains closer to the start line.	4
02.1	The result is consistent with the unknown containing the reference compound, but it is not conclusive because different compounds can have the same $R_f$ under one set of conditions. Repeat using a different solvent or solvent mixture and compare again, or co-spot the unknown with the reference.	An $R_f$ value is conditional on the stationary phase, solvent, temperature and procedure. A match supplies supporting evidence, not a unique molecular identity. Changing the mobile phase changes the balance of attractions and may separate compounds that coincided in the first solvent; a co-spot can reveal whether the combined sample still produces a single spot.	4
03.1	Components move down the column at different rates because they differ in mobile-phase solubility and attraction to the stationary phase. Spot each fraction on one TLC plate; combine fractions that each give a single spot at the same $R_f$ as the pure-product standard under identical conditions.	A component that is more soluble in the eluting solvent and less strongly retained by the packed solid travels faster and appears in earlier fractions. Collect small fractions so differently moving bands are less likely to be mixed. Apply samples of the fractions and a product standard to the same TLC plate, develop them together, and compare spot number and $R_f$ . A single matching spot supports purity and identity, so matching fractions can be pooled.	4

Tier 3 · Hard

Mark scheme for 3.3.16 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	The first two peaks are consistent with X and Y; the $7.1\,\text{min}$ peak cannot safely be assigned as Z from the near match alone; a mass spectrum supplies fragment and molecular-ion data for comparison with a reference spectrum.	Because conditions are identical, exact retention-time matches make X and Y plausible identities, although another compound could in principle co-elute. The final peak differs from Z's reference time, so it is not justified to label it Z without further evidence. GC first separates the components; the mass spectrometer then records a spectrum for each eluting peak. Matching molecular-ion and fragmentation patterns to standards provides independent structural evidence and makes the assignment stronger.	6
02.1	Solvent A: P = 0.80, Q = 0.83, R = 0.86. Solvent B: P = 0.20, Q = 0.50, R = 0.80. Solvent B gives better separation. The unknown result is consistent with Q but does not prove identity because another compound could share that $R_f$ under these conditions.	Use $R_f$ = distance travelled by spot / distance travelled by solvent front. For A: 6.40/8.00 = 0.80, 6.64/8.00 = 0.83 and 6.88/8.00 = 0.86. For B: 1.60/8.00 = 0.20, 4.00/8.00 = 0.50 and 6.40/8.00 = 0.80. The much wider spacing in B makes resolution more reliable. Matching Q at 0.50 is evidence only; repeat in another solvent or use an independent technique such as mass spectrometry for stronger identification.	7
03.1	The peak percentages are 15.0%, 35.0% and 50.0%. The middle peak represents 35.0% combined, but its two compounds cannot be assigned separate percentages from the single GC area alone. More than one molecular-ion or incompatible sets of fragment peaks across that elution reveal overlapping mass spectra.	The total area is $18+42+60=120$ . The three fractions are $18/120=0.150$ , $42/120=0.350$ and $60/120=0.500$ . Equal response makes these 15.0%, 35.0% and 50.0% mole proportions. Co-eluting substances reach the detector together, so their responses are added into the 42-unit area. The mass spectrometer can record ions characteristic of two structures, but separate amounts require deconvolution or another separation rather than assigning the whole 35.0% to either compound.	6
04.1	For F₂–F₄, target mass is $0.91\,\mathrm{g}$ and total residue mass is $0.95\,\mathrm{g}$ , so purity is $0.91/0.95\times100=95.8\%$ . Target recovery for F₂–F₄ is $0.91/1.00\times100=91.0\%$ . For F₃–F₄, all $0.63\,\mathrm{g}$ of residue is target, so purity is 100%. Target recovery for F₃–F₄ is $0.63/1.00\times100=63.0\%$ . Choose F₃–F₄ because 95.8% from the larger pool fails the 98.0% purity requirement. Collect smaller fractions around the boundary represented by F₂ so that its target-rich portion can be separated from the overlapping impurity and more pure target can be pooled.	Purity uses target mass divided by total pooled residue mass; recovery uses target mass divided by the $1.00\,\mathrm{g}$ loaded. Including F₂ raises recovery from 63.0% to 91.0% but introduces $0.04\,\mathrm{g}$ of impurity, giving only $95.789\ldots\%$ purity. The stricter pool therefore sacrifices recovery to meet the specification. Smaller collection intervals across an overlapping band reduce the amount of pure target discarded with a mixed boundary fraction.	6
05.1	Use TLC with the dye mixture and known standard on the same plate for task (i). A matching $R_f$ under identical conditions supports the known dye's presence, while the spot count gives a rapid mixture check. Use column chromatography for task (ii). A packed stationary phase separates bands that can be collected as preparative fractions, unlike analytical TLC spots. Use gas chromatography coupled to mass spectrometry for task (iii) because the fragrance components are volatile enough to pass through a heated column. Retention time supplies chromatographic comparison and the mass spectrum supplies molecular-ion and fragmentation evidence for each separated component.	Match method scale and sample properties to the task. TLC is fast and needs little sample, so it is suitable for comparison with a standard but not for collecting gram quantities. A column applies the same differential retention principle at preparative scale and permits fraction collection. GC separates substances transported in a carrier gas, so it suits volatile fragrance components; coupling to MS strengthens identification beyond retention time alone.	6