3.3 Organic chemistry — notes on one page

Every AQA A-level Chemistry 7405 specification point in 3.3, read straight through: 38 points with definitions, explanations, mark-scheme wording and worked examples. Practice questions are in the printable question pack.

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.

How this checking works

3.3.1.1 · Nomenclature

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.

3.3.1.2 · Reaction mechanisms

Definition. A reaction mechanism shows how bonds break and form. A curly arrow starts at a lone electron pair or at a covalent bond and points to where that pair of electrons goes; in free-radical mechanisms a dot represents the unpaired electron and curly arrows are not required.

The specification is precise about the convention, and so is the marking. A curly arrow that forms a covalent bond must start from a lone pair or from another covalent bond and end at the atom where the new bond forms;
an arrow that breaks a covalent bond must start from the bond itself and end on the atom that keeps the electrons. Every arrow is double-headed, because it represents a pair of electrons, and arrows are never drawn from a positive charge or from an atom with no lone pair.
Free-radical mechanisms are the exception: the unpaired electron is shown as a dot, and curly arrows are not required — you write balanced equations for initiation, propagation and termination instead.
In polar mechanisms, an intermediate must be drawn correctly and in full to score, and the mark scheme distinguishes primary, secondary and tertiary carbocations sharply, since the more substituted carbocation is more stable and is the one that forms.
Naming the mechanism is a separate mark from drawing it, and the full name is required: 'nucleophilic addition', 'electrophilic addition', 'nucleophilic substitution', 'electrophilic substitution' or 'nucleophilic addition-elimination'.

The two curly-arrow conventions: an arrow that forms a bond starts at a lone pair, and an arrow that breaks a bond starts at the bond itself.

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.

Mark-scheme wording to use

“curly arrow from a lone pair on the carbon of CN- to the carbonyl carbon; allow the negative charge anywhere on cyanide but the attacking lone pair must be on C, and penalise covalent KCN” — AQA 7405, June 2022 Paper 2, Q01.8
“M₄ correct tertiary carbocation, not a primary carbocation. M₅ arrow from a lone pair on the correct negatively charged oxygen of HSO₄- to the positive carbon” — AQA 7405, electrophilic addition mechanism, June 2022 Paper 2, Q03.1
“NB: The arrows are double-headed” — AQA 7405, June 2017 Paper 3, Q02.2

What the examiner wants. Arrows drawn from the lone pair or the bond, not from a charge, and the full two-word mechanism name. The list principle applies, so adding a second mechanism name cancels the mark.

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.

3.3.1.3 · Isomerism

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.

3.3.2.1 · Fractional distillation of crude oil

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.

3.3.2.2 · Modification of alkanes by cracking

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.

3.3.2.3 · Combustion of alkanes

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.

3.3.2.4 · Chlorination of alkanes

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.

3.3.3.1 · Nucleophilic substitution

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.

3.3.3.2 · Elimination

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.

3.3.3.3 · Ozone depletion

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.

3.3.4.1 · Structure, bonding and reactivity

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'.

3.3.4.2 · Addition reactions of alkenes

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.

3.3.4.3 · Addition polymers

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.

3.3.5.1 · Alcohol production

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.

3.3.5.2 · Oxidation of alcohols

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.

3.3.5.3 · Elimination

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.

3.3.6.1 · Identification of functional groups by test-tube reactions

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.

3.3.6.2 · Mass spectrometry

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.

3.3.6.3 · Infrared spectroscopy

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.

3.3.7 · Optical isomerism (A-level only)

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.

3.3.8 · Aldehydes and ketones (A-level only)

Definition. Aldehydes and ketones both contain the carbonyl group, which is attacked by nucleophiles. Aldehydes are readily oxidised to carboxylic acids; ketones are not. Both are reduced by NaBH₄ in aqueous solution — aldehydes to primary alcohols and ketones to secondary alcohols — by nucleophilic addition.

That oxidation difference is the basis of the two distinguishing tests. Tollens' reagent, ammoniacal silver nitrate, gives a silver mirror with an aldehyde and no change with a ketone; Fehling's solution gives a red precipitate with an aldehyde and stays blue with a ketone.
Acidified dichromate is not accepted for this test, because the alcohol product also reacts. In reduction reactions the reductant is written as [H] in the overall equation, and the nucleophile in the mechanism is shown as the hydride ion H⁻.
Carbonyl compounds also undergo nucleophilic addition with KCN followed by dilute acid to give hydroxynitriles, which extends the carbon chain by one. KCN is used rather than HCN because HCN is a weak acid and gives too low a cyanide concentration, and its hazard — highly toxic — must be stated when asked.
The stereochemical consequence is a regular exam target: the carbonyl group is planar, so the nucleophile can attack from either side with equal probability, and an aldehyde or an unsymmetrical ketone therefore gives an equal mixture of two enantiomers, a racemate that is optically inactive.
A symmetrical ketone such as pentan-3-one gives a product with no chiral centre.

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.

Mark-scheme wording to use

“M₁ Tollens'/ammoniacal silver nitrate or correctly described preparation, or Fehling's/Benedict's solution. M₂ corresponding silver mirror/solid or red precipitate. Do not accept dichromate because both aldehyde and product alcohol react” — AQA 7405, June 2019 Paper 2, Q13.2
“planar carbonyl group … Attack from either side … With equal probability OR produces equal amounts (of the two isomers/enantiomers) … Not planar molecule, not planar bond, not planar C=O” — AQA 7405, June 2017 Paper 2, Q04.3
“nucleophilic addition; both words are required and no additional mechanism names are allowed” — AQA 7405, June 2019 Paper 3, Q04.1

What the examiner wants. The word 'planar' attached to the CARBONYL GROUP, not to the molecule or the C=O bond, and 'equal amounts' of the two enantiomers rather than a vague 'a mixture'.

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.

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

Definition. Carboxylic acids contain the -COOH group and are weak acids, but they are strong enough to liberate carbon dioxide from carbonates. Esters contain the -COO- link and are formed when a carboxylic acid and an alcohol react in the presence of an acid catalyst.

The carbonate test is the standard way to identify a carboxylic acid: adding sodium carbonate or sodium hydrogencarbonate produces effervescence with a carboxylic acid and no visible change with an alcohol. Esterification is reversible and slow, so concentrated sulfuric acid is used as catalyst and the ester is separated afterwards, typically with a separating funnel and distillation.
Esters are named alcohol-part first, so ethanol and ethanoic acid give ethyl ethanoate, and they are used as solvents, plasticisers, perfumes and food flavourings. Vegetable oils and animal fats are esters of propane-1,2,3-triol, glycerol, with three long-chain carboxylic acids.
Esters can be hydrolysed in either acid or alkaline conditions: acid hydrolysis is reversible and gives the carboxylic acid and the alcohol, while alkaline hydrolysis goes to completion and gives the salt of the carboxylic acid plus the alcohol.
Applying alkaline hydrolysis to a fat gives glycerol and the salts of long-chain carboxylic acids, which are soap — and because it goes to completion, the alkaline route is preferred industrially.
Biodiesel is a mixture of methyl esters of long-chain carboxylic acids, made by reacting vegetable oils with methanol over a catalyst.

The ester link drawn out, and the same link repeated three times on a glycerol backbone to give a fat or vegetable oil.

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

Mark-scheme wording to use

“M₁ add sodium carbonate/hydrogencarbonate. M₂ propanoic acid effervesces/bubbles; merely saying CO₂ gas is given off is insufficient. M₃ propan-1-ol shows no visible change/reaction; do not allow only nothing/no observation” — AQA 7405, November 2021 Paper 3, Q06.6
“M₁ correct glycerol and three ester-link framework; allow -O₂C-, -OOC- or -OCO-, not reversed -CO₂-/-COO-” — AQA 7405, June 2019 Paper 2, Q11.2
“M₁ CH₂OHCH(OH)CH₂OH. M₂ potassium carboxylate salt … M₃ soap; allow detergent or surfactant” — AQA 7405, November 2021 Paper 2, Q01.1

What the examiner wants. An OBSERVATION, not an inference. 'Carbon dioxide is given off' is explicitly insufficient where 'effervescence' is required, and 'no reaction' is rejected where 'no visible change' is wanted.

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.

3.3.9.2 · Acylation (A-level only)

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.

3.3.10.1 · Bonding (A-level only)

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.

3.3.10.2 · Electrophilic substitution (A-level only)

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.

3.3.11.1 · Preparation (A-level only)

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.

3.3.11.2 · Base properties (A-level only)

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.

3.3.11.3 · Nucleophilic properties (A-level only)

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.

3.3.12.1 · Condensation polymers (A-level only)

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.

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

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.

3.3.13.1 · Amino acids (A-level only)

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.

3.3.13.2 · Proteins (A-level only)

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.

3.3.13.3 · Enzymes (A-level only)

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.

3.3.13.4 · DNA (A-level only)

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.

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

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.

3.3.14 · Organic synthesis (A-level only)

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.

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

Definition. Nuclear magnetic resonance spectroscopy gives information about the position of 13C or 1H atoms in a molecule. Chemical shift, recorded on the delta scale, depends on the molecular environment, and integrated 1H spectra indicate the relative numbers of protons in each environment.

Read a spectrum in a fixed order. Count the peaks to count the environments — 13C NMR gives simpler spectra than 1H NMR precisely because there is no splitting to interpret. Use the chemical shift, with the Chemistry Data Booklet, to identify what each environment is attached to:
a proton on a carbon next to a carbonyl sits at δ 2.1–2.6, a proton on a carbon bonded to oxygen or a halogen is further downfield at 3.1–4.2, and an aldehyde proton appears at 9–10. Use the integration to get the ratio of protons in each environment, then scale it to the molecular formula.
Finally, apply the n+1 rule: a peak split into n+1 lines has n non-equivalent protons on the adjacent carbon, so a triplet means two adjacent protons and a quartet means three, and the classic quartet-plus-triplet pair identifies an ethyl group.
Splitting is limited to doublets, triplets and quartets in aliphatic compounds.
Samples are dissolved in deuterated solvents or CCl₄ so that the solvent contributes no signal, and tetramethylsilane is the standard: it is chemically inert, non-toxic, volatile so it is easily removed, and all twelve of its protons are equivalent, giving a single sharp peak defined as 0 ppm and upfield of almost every other signal.

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.

Mark-scheme wording to use

“M₂ triplet. M₃ two H on adjacent carbon, dependent on M₂; generic n+1 alone insufficient” — AQA 7405, June 2018 Paper 2, Q11.1
“M₁ the quartet is caused by a neighbouring carbon with 3 H … M₅ the fragment -C(=O)-O-CH₂CH₃” — AQA 7405, structure-deduction five-marker, June 2023 Paper 2, Q05.4
“Correct structure without explanation caps at 4; no structure or incoherent identification of two key groups caps at 6” — AQA 7405, Specimen Paper 2, Q10

What the examiner wants. The number of adjacent protons stated explicitly for each multiplet. Quoting the n+1 rule generically is not enough, and a correct structure with no supporting explanation is capped below full marks.

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.

3.3.16 · Chromatography (A-level only)

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.