Give the IUPAC name of CH<sub>3</sub>CH(CH<sub>3</sub>)CH<sub>2</sub>CH<sub>2</sub>OH.
Organic chemistry
Notes and three levels of exam-style practice for each registered specification leaf in this section.
Open the printable packNomenclature
- Organic compounds may be represented by empirical, molecular, general, structural, displayed or skeletal formulae; each representation shows a different amount of structural detail.
- Members of a homologous series share a functional group and general formula, have similar chemical properties, and successive members differ by CH<sub>2</sub>.
- For an IUPAC name, choose the longest chain or ring containing the principal functional group, number it to give that group the lowest possible locant, then add and alphabetise substituents; AQA limits naming and drawing to chains and rings of up to six carbon atoms each.
- In a skeletal formula, every unlabelled line end and vertex is a carbon atom; forgetting an end carbon or writing carbon-bound hydrogen atoms explicitly is a common error.
Tier 1 · Easy
Tier 2 · Standard
Write a condensed structural formula for 3-ethyl-2-methylhexane.
Tier 3 · Hard
A compound has condensed formula CH<sub>3</sub>C(CH<sub>3</sub>)<sub>2</sub>CH<sub>2</sub>CH(OH)CH<sub>3</sub>. Give its IUPAC name and molecular formula.
Reaction mechanisms
- A mechanism accounts for bond making and bond breaking through a sequence of elementary steps and must show every reacting species, intermediate and product required by that sequence.
- An electron-pair curly arrow begins at a lone pair or covalent bond and ends at the atom or bond receiving that pair; an arrow must never begin at a positive charge.
- When a bond breaks heterolytically, the curly arrow starts in that bond and points to the atom that receives both bonding electrons.
- A radical is shown by a single dot. Free-radical steps use balanced equations and dots but no electron-pair curly arrows; mixing the two notations is a common error.
Tier 1 · Easy
In the first step of attack by :NH<sub>3</sub> on a positive carbon centre, state where the bond-forming curly arrow starts and ends.
Tier 2 · Standard
Write the radical propagation step in which a methyl radical reacts with chlorine, and explain the required dot-and-arrow notation.
Tier 3 · Hard
A student draws hydroxide attacking CH<sub>3</sub>CH<sub>2</sub>I 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.
Isomerism
- Structural isomers have the same molecular formula but different structural formulae; chain, position and functional-group isomerism describe different ways in which their connectivity can change.
- Stereoisomers have the same structural formula but a different arrangement of atoms in space.
- E–Z isomerism requires restricted rotation about C=C and two different groups attached to each carbon of that double bond.
- Apply CIP priorities separately at the two alkene carbons using atomic number: the higher-priority groups on the same side give Z, while opposite sides give E; comparing group size by eye is a common error.
Tier 1 · Easy
Butan-1-ol and butan-2-ol have the same molecular formula. State their type of isomerism.
Tier 2 · Standard
At the left carbon of a C=C bond the groups are H and CH<sub>3</sub>; at the right carbon they are CH<sub>3</sub> and CH<sub>2</sub>CH<sub>3</sub>. The left CH<sub>3</sub> and right CH<sub>2</sub>CH<sub>3</sub> groups are drawn on opposite sides. Assign E or Z and justify your choice using CIP priorities.
Tier 3 · Hard
Draw or give condensed structural formulae for every aldehyde and ketone with molecular formula C<sub>4</sub>H<sub>8</sub>O. Name each one and classify the relationship between the aldehydes and the ketone.
Fractional distillation of crude oil
- Crude oil is a mixture consisting mainly of alkane hydrocarbons, and alkanes are saturated because they contain only carbon–carbon single bonds.
- Fractional distillation separates hydrocarbons by their different boiling points: crude oil is vaporised and the vapours pass into a column that is hot at the bottom and cooler at the top.
- Longer alkane molecules have stronger London forces, higher boiling points and condense lower in the column; shorter molecules condense higher up.
- Fractional distillation breaks no covalent bonds and causes no chemical reaction; claiming that C–C bonds are broken confuses distillation with cracking.
Tier 1 · Easy
State the physical property used to separate crude oil into fractions and name the separation process.
Tier 2 · Standard
Explain why a C<sub>5</sub> alkane is collected nearer the top of a fractionating column than a C<sub>15</sub> alkane.
Tier 3 · Hard
A vapour mixture contains C<sub>7</sub>H<sub>16</sub>, C<sub>12</sub>H<sub>26</sub> and C<sub>18</sub>H<sub>38</sub>. Predict their order of condensation from highest to lowest in the column, and explain why separating them does not change any molecular formula.
Modification of alkanes by cracking
- Cracking converts larger alkane molecules into smaller, more useful molecules by breaking carbon–carbon bonds.
- 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, mainly producing motor fuels and aromatic hydrocarbons; its mechanism is not required.
- Cracking is economically valuable because demand for shorter-chain fuels and alkenes can exceed their supply from crude-oil fractions, while longer fractions may be less useful.
Tier 1 · Easy
Complete and balance this cracking equation: .
Tier 2 · Standard
A refinery cracks C<sub>15</sub>H<sub>32</sub> to one molecule of C<sub>8</sub>H<sub>18</sub>, one of C<sub>3</sub>H<sub>6</sub> and ethene only. Determine the number of ethene molecules formed, then explain the economic purpose of this conversion.
Tier 3 · Hard
Compare the conditions and main product emphasis of thermal cracking with catalytic cracking.
Combustion of alkanes
- Complete combustion in excess oxygen forms carbon dioxide and water; limited oxygen causes incomplete combustion and can form carbon monoxide and carbon.
- Internal-combustion engines emit NO<sub>x</sub>, CO, carbon particles and unburned hydrocarbons; catalytic converters change gaseous pollutants into less harmful products.
- Sulfur impurities in hydrocarbon fuels burn to sulfur dioxide, an acidic pollutant that can be removed from flue gases by basic calcium oxide or calcium carbonate.
- Balance combustion equations by balancing carbon, then hydrogen, then oxygen; changing a fuel subscript instead of using coefficients changes the compound and is incorrect.
Tier 1 · Easy
Write the balanced equation for the complete combustion of propane.
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.
Tier 3 · Hard
A flue gas contains 1.60 kg of SO<sub>2</sub>. Calculate the minimum mass of CaCO<sub>3</sub> needed for complete removal using . Use and .
Chlorination of alkanes
- Methane reacts with chlorine by free-radical substitution when ultraviolet radiation provides energy for homolytic fission of Cl–Cl.
- Initiation produces two chlorine radicals: .
- Propagation consumes a radical in one step and forms another, allowing a chain reaction: chlorine abstracts H, then the alkyl radical reacts with Cl<sub>2</sub>.
- Termination combines two radicals. Further substitutions can occur, so chlorination gives a mixture rather than only the desired monochloroalkane.
Tier 1 · Easy
Write the initiation step for methane chlorination and state the bond-fission type.
Tier 2 · Standard
Write both propagation equations that convert methane and chlorine into chloromethane during free-radical substitution.
Tier 3 · Hard
Give three different termination equations available in methane chlorination, and explain why prolonged irradiation lowers the purity of chloromethane.
Nucleophilic substitution
- The carbon–halogen bond is polar because the halogen is more electronegative, leaving the bonded carbon electron-deficient and open to nucleophilic attack.
- A nucleophile donates an electron pair. Required nucleophiles are OH<sup>−</sup>, CN<sup>−</sup> and NH<sub>3</sub>, forming an alcohol, a nitrile and an amine respectively.
- For the primary halogenoalkane mechanisms used here, OH<sup>−</sup> and CN<sup>−</sup> attack as C–X breaks in one concerted step. NH<sub>3</sub> first forms RNH<sub>3</sub><sup>+</sup>; a second NH<sub>3</sub> then removes H<sup>+</sup>, giving RNH<sub>2</sub> and NH<sub>4</sub><sup>+</sup>. CN<sup>−</sup> attacks through carbon.
- Hydrolysis is faster when the C–X bond enthalpy is lower, so comparable iodoalkanes react faster than bromoalkanes and chloroalkanes; arguing only from bond polarity gives the wrong trend.
Tier 1 · Easy
Name the organic product when 1-bromopropane undergoes nucleophilic substitution with CN<sup>−</sup>.
Tier 2 · Standard
Outline the one-step mechanism for CH<sub>3</sub>CH<sub>2</sub>CH<sub>2</sub>Br reacting with OH<sup>−</sup>, including both curly arrows and the products.
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.
Elimination
- A halogenoalkane can undergo substitution and elimination concurrently with potassium hydroxide; the reaction pathway depends on conditions.
- In substitution, OH<sup>−</sup> acts as a nucleophile by donating a lone pair to carbon. In elimination, it acts as a base by accepting a proton.
- The elimination mechanism uses three curly arrows: from an oxygen lone pair to a hydrogen on the carbon adjacent to C–X, from that C–H bond to form C=C, and from C–X to X.
- Hot ethanolic hydroxide favours elimination, whereas aqueous hydroxide favours substitution; omitting the solvent can make an otherwise correct reagent ambiguous.
Tier 1 · Easy
State the organic product of eliminating HBr from 2-bromopropane and name the role of OH<sup>−</sup>.
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<sup>−</sup>.
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.
Ozone depletion
- Stratospheric ozone is beneficial because it absorbs ultraviolet radiation that would otherwise reach Earth's surface.
- In the upper atmosphere, UV radiation breaks C–Cl bonds in CFC molecules and produces chlorine radicals.
- Chlorine radicals catalyse ozone decomposition through and .
- Because Cl is regenerated, one radical can destroy many ozone molecules. Evidence from different research groups supported legislation restricting CFCs, while chemists developed chlorine-free replacements.
Tier 1 · Easy
State why ozone in the upper atmosphere is beneficial.
Tier 2 · Standard
Combine the two chlorine-radical ozone steps to obtain the overall equation, and identify the catalyst and intermediate.
Tier 3 · Hard
For the CFC CF<sub>2</sub>Cl<sub>2</sub>, write a UV photodissociation equation that releases a chlorine radical, then explain why a low concentration of this CFC can cause extensive ozone loss.
Structure, bonding and reactivity
- Alkenes are unsaturated hydrocarbons containing at least one carbon–carbon double covalent bond.
- The C=C bond is a double covalent bond and a centre of high electron density compared with a carbon–carbon single bond.
- This high electron density attracts electrophiles, which accept an electron pair from the double bond as a new covalent bond forms.
- During addition the C=C becomes C–C as two new bonds form to the attacking species; calling an alkene reactive without linking this to its high electron density is incomplete.
Tier 1 · Easy
Define an unsaturated hydrocarbon and state the feature that makes ethene unsaturated.
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.
Tier 3 · Hard
Explain, using electron density and bond changes, why an alkene reacts readily with an electrophile to form an addition product.
Addition reactions of alkenes
- Alkenes undergo electrophilic addition with HBr, H<sub>2</sub>SO<sub>4</sub> and Br<sub>2</sub>; bromine water changes from orange to colourless as Br<sub>2</sub> adds across C=C.
- With HBr or H<sub>2</sub>SO<sub>4</sub>, an arrow goes from C=C to H and another from H–Br or O–H to the remaining ion; Br<sup>−</sup> or HSO<sub>4</sub><sup>−</sup> then attacks the carbocation.
- The electron-rich C=C induces a dipole in Br<sub>2</sub>; arrows go from C=C to Br<sup>δ+</sup> and from Br–Br to Br<sup>δ−</sup>, then Br<sup>−</sup> attacks the positive intermediate.
- For an unsymmetrical alkene, the major product forms through the more stable carbocation: tertiary is more stable than secondary, then primary. A product rule without this comparison is not an explanation.
Tier 1 · Easy
State the observation when bromine water is shaken with cyclohexene and name the organic product.
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.
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.
Addition polymers
- Addition polymerisation joins many alkene or substituted-alkene molecules after their C=C bonds open; ordinary non-polar polyalkene chains attract one another through London forces caused by temporary and induced dipoles.
- To move from monomer to repeat unit, change C=C to C-C, retain every substituent on its original carbon, put the unit in brackets and extend a bond through each bracket.
- For example, propene gives poly(propene), whose repeat unit is [-CH<sub>2</sub>-CH(CH<sub>3</sub>)-]<sub>n</sub>; rigid PVC is used for drainpipes and window frames, while plasticised PVC is used for flexible cable insulation and flooring.
- A common error is to leave a C=C bond in an addition-polymer repeat unit. Polyalkenes are unreactive because their backbones contain strong non-polar C-C and C-H bonds; a plasticiser separates PVC chains and lets them move more easily.
Tier 1 · Easy
2-Methylpropene has the structure CH<sub>2</sub>=C(CH<sub>3</sub>)<sub>2</sub>. Give the displayed repeat unit of its addition polymer.
Tier 2 · Standard
A polymer segment is -CH<sub>2</sub>-CH(CN)-CH<sub>2</sub>-CH(CN)-. Deduce the monomer, name the addition polymer and state the strongest intermolecular force between its chains.
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.
Alcohol production
- Ethanol can be made by fermenting renewable glucose or by hydrating ethene with steam over an acid catalyst; a biofuel is a fuel made from recently living material.
- Fermentation uses yeast enzymes, an aqueous sugar solution, about 30-40 °C and anaerobic conditions; fractional distillation then separates ethanol from the dilute mixture.
- For hydration, protonate the alkene to form a carbocation, let water attack, then remove H<sup>+</sup> to regenerate the acid catalyst and form the alcohol.
- Calling fermented ethanol carbon neutral is an oversimplification: crop growth removes CO<sub>2</sub>, but farming, fertiliser manufacture, processing and transport can add emissions, while land and food use create ethical costs.
Tier 1 · Easy
Write the equation for fermentation of glucose to ethanol and state two conditions that keep the yeast working effectively.
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.
Tier 3 · Hard
A fermenter produces of ethanol. Complete combustion releases all of its carbon as CO<sub>2</sub>. Calculate the CO<sub>2</sub> mass using C<sub>2</sub>H<sub>5</sub>OH + 3O<sub>2</sub> → 2CO<sub>2</sub> + 3H<sub>2</sub>O, then assess whether describing the fuel as carbon neutral is justified.
Oxidation of alcohols
- Primary alcohols oxidise first to aldehydes and then to carboxylic acids; secondary alcohols form ketones, while tertiary alcohols are not easily oxidised.
- Warm with acidified potassium dichromate(VI): distil an aldehyde as it forms, but heat under reflux with excess oxidant to obtain a carboxylic acid.
- For propan-1-ol, CH<sub>3</sub>CH<sub>2</sub>CH<sub>2</sub>OH + [O] → CH<sub>3</sub>CH<sub>2</sub>CHO + H<sub>2</sub>O, followed by CH<sub>3</sub>CH<sub>2</sub>CHO + [O] → CH<sub>3</sub>CH<sub>2</sub>COOH.
- To distinguish an aldehyde from a ketone, warm fresh portions with Tollens' reagent or Fehling's solution: an aldehyde gives a silver mirror or brick-red precipitate, while a ketone gives neither. A common error is to expect easy oxidation from a tertiary alcohol.
Tier 1 · Easy
Butan-2-ol is warmed with acidified potassium dichromate(VI). Name the organic product and state the colour change.
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.
Tier 3 · Hard
Three C<sub>4</sub>H<sub>10</sub>O 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.
Elimination
- Acid-catalysed elimination removes H<sub>2</sub>O from an alcohol and forms a C=C bond, so the reaction is also called dehydration.
- Protonate OH to make H<sub>2</sub>O a good leaving group, remove water, then show the adjacent C-H bond pair forming the pi bond as H<sup>+</sup> is lost; do not add a base-attack arrow.
- Heating butan-2-ol with concentrated acid can form but-1-ene and but-2-ene because a beta hydrogen is available on either side of the OH-bearing carbon.
- A common error is to remove a hydrogen from a non-adjacent carbon or to draw a curly arrow backwards; the electron pair must start at the bond that supplies it.
Tier 1 · Easy
Give the organic product and reaction type when ethanol vapour is heated with an acid catalyst and loses water.
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.
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.
Identification of functional groups by test-tube reactions
- Functional groups can be distinguished by their specified reactions: bromine water tests C=C, carbonate tests carboxylic acids, Tollens' or Fehling's tests aldehydes, and acidified dichromate tests oxidisable alcohols.
- Plan a sequence that gives an unambiguous observation, use a fresh portion for each reagent and record a colour or precipitate rather than the vague word 'clear'.
- For example, sodium carbonate gives CO<sub>2</sub> effervescence with ethanoic acid, while warmed Tollens' reagent gives a silver mirror with ethanal but not with a ketone.
- A common error is to name a reagent without the observation or to ignore cross-reactions; acidified dichromate turns green with both a primary alcohol and an aldehyde, so it cannot distinguish that pair on its own.
Tier 1 · Easy
A colourless liquid may contain a C=C bond. State a test-tube reagent and the positive observation.
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.
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.
Mass spectrometry
- A high-resolution molecular-ion mass can distinguish molecular formulas that share the same nominal relative molecular mass because isotopes have precise, non-integer masses.
- For each candidate formula, multiply every isotope's precise mass by its atom count, add the contributions and compare the total with the measured molecular mass.
- Using <sup>12</sup>C = 12.00000, <sup>1</sup>H = 1.00783 and <sup>16</sup>O = 15.99491, C<sub>3</sub>H<sub>6</sub>O<sub>2</sub> has precise mass .
- A common error is to use rounded relative atomic masses for an exact-mass deduction; candidates with the same nominal mass then appear indistinguishable.
Tier 1 · Easy
Calculate the precise molecular mass of C<sub>2</sub>H<sub>4</sub>O using <sup>12</sup>C = 12.00000, <sup>1</sup>H = 1.00783 and <sup>16</sup>O = 15.99491.
Tier 2 · Standard
A compound containing only C, H and O has a high-resolution molecular-ion mass of . Choose between C<sub>4</sub>H<sub>10</sub>O and C<sub>3</sub>H<sub>6</sub>O<sub>2</sub>. Use <sup>12</sup>C = 12.00000, <sup>1</sup>H = 1.00783 and <sup>16</sup>O = 15.99491.
Tier 3 · Hard
An unknown has measured molecular-ion mass . Candidate formulas are C<sub>4</sub>H<sub>8</sub>O<sub>2</sub> and C<sub>5</sub>H<sub>12</sub>O. Calculate both precise masses, identify the formula and calculate the absolute error of the chosen value. Use <sup>12</sup>C = 12.00000, <sup>1</sup>H = 1.00783 and <sup>16</sup>O = 15.99491.
Infrared spectroscopy
- Bonds absorb infrared radiation at characteristic wavenumbers when the radiation matches a vibrational energy change, so an IR spectrum reveals bonds and functional groups.
- Use the Data Booklet to match diagnostic absorptions, then compare the fingerprint region with a reference spectrum to identify a particular molecule.
- A strong C=O absorption near 1700 cm<sup>-1</sup> plus a very broad O-H absorption across roughly 2500-3300 cm<sup>-1</sup> supports a carboxylic acid.
- A common error is to declare a sample pure because the expected peaks are present; unexpected absorptions can reveal an impurity, while a missing absorption can rule out a proposed structure.
Tier 1 · Easy
An IR spectrum has a strong absorption at 1718 cm<sup>-1</sup>. Use the Data Booklet to identify the bond indicated by this absorption.
Tier 2 · Standard
A liquid's IR spectrum contains a strong peak at 1740 cm<sup>-1</sup>, 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.
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 3200-3550 cm<sup>-1</sup>. Suggest an impurity and explain how IR absorption by CO<sub>2</sub>, CH<sub>4</sub> and water vapour contributes to global warming.
Optical isomerism (A-level only)
- Optical isomerism is stereoisomerism caused here by a single chiral carbon attached to four different groups; its enantiomers are non-superimposable mirror images.
- To find a chiral centre, inspect each tetrahedral carbon and list its four attached groups, tracing along chains far enough to detect a difference.
- The two enantiomers rotate plane-polarised light by equal amounts in opposite directions, whereas a 1:1 racemic mixture is optically inactive because the rotations cancel.
- A common error is to call any carbon with four bonds chiral; repeated substituents make it achiral, and rotating one drawing in space does not create a new enantiomer.
Tier 1 · Easy
Identify the chiral carbon in CH<sub>3</sub>CH(OH)CH<sub>2</sub>CH<sub>3</sub> and name the compound.
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.
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.
Aldehydes and ketones (A-level only)
- Aldehydes oxidise readily to carboxylic acids and give positive Tollens' and Fehling's tests; ketones do not under those test conditions.
- For NaBH<sub>4</sub> reduction, draw H<sup>-</sup> attacking the delta-positive carbonyl carbon and move the C=O pi pair to oxygen before protonating the alkoxide.
- Aldehydes reduce to primary alcohols and ketones to secondary alcohols; KCN followed by dilute acid instead adds HCN and extends the carbon skeleton by one carbon to a hydroxynitrile. KCN is highly toxic and must be handled with strict safety controls.
- A common error is to describe carbonyl addition as addition-elimination: no group leaves. A racemate forms only when the planar carbonyl gives a product with four different groups at the new tetrahedral carbon.
Tier 1 · Easy
Butanal is treated with aqueous sodium tetrahydridoborate. Name the organic product and write an equation using [H].
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.
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.
Carboxylic acids and esters (A-level only)
- Carboxylic acids are weak acids but react with carbonates to give a carboxylate salt, water and CO<sub>2</sub>; alcohols form useful solvent, perfume, flavouring and plasticiser esters with them in a reversible acid-catalysed condensation.
- For ester hydrolysis, identify the acyl and alkoxy parts: acid hydrolysis gives a carboxylic acid and alcohol, whereas alkaline hydrolysis gives a carboxylate salt and alcohol.
- A triglyceride is a triester of propane-1,2,3-triol; alkaline hydrolysis gives glycerol and soap, while reaction of vegetable oil with methanol and a catalyst gives biodiesel, a mixture of long-chain methyl esters.
- A common error is to reverse the ester link or name it from the wrong side: the alkyl part comes from the alcohol and the alkanoate part comes from the acid.
Tier 1 · Easy
Name the ester made from propan-1-ol and ethanoic acid, state the catalyst and write the equation using condensed structures.
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.
Tier 3 · Hard
A triglyceride has . Calculate the minimum mass of NaOH needed to hydrolyse completely, then state the two types of organic product. Use .
Acylation (A-level only)
- Acyl chlorides, RCOCl, and acid anhydrides, RCOOCOR', acylate water, alcohols, ammonia and primary amines by nucleophilic addition-elimination; possible products include acids, esters and amides.
- Draw the nucleophile attacking the delta-positive acyl carbon, move the C=O pi pair to oxygen, restore C=O as the leaving group departs, then complete any proton transfer.
- Ethanoyl chloride plus ethanol gives ethyl ethanoate and HCl, while reaction with excess ethylamine gives N-ethylethanamide and ethylammonium chloride.
- A common error is to call acylation simple nucleophilic substitution and omit the tetrahedral intermediate; in aspirin manufacture ethanoic anhydride is preferred because it is safer to handle and produces ethanoic acid rather than corrosive HCl.
Tier 1 · Easy
Ethanoyl chloride is added to ethanol. Name both products and state one visible observation.
Tier 2 · Standard
Propanoyl chloride reacts with excess methylamine. Name the amide and salt formed, and outline the addition-elimination steps.
Tier 3 · Hard
Aspirin manufacture can use ethanoic anhydride or ethanoyl chloride as the acylating agent. The desired aspirin has ; the anhydride route also makes ethanoic acid, , while the chloride route also makes HCl, . Calculate each route's atom economy and explain why industry can still prefer the anhydride route.
Bonding (A-level only)
- Benzene is a planar hexagonal ring in which all six C-C bonds have the same length, intermediate between ordinary single and double bonds.
- Each carbon contributes one electron in a p orbital; sideways overlap around the ring produces delocalised pi electron density above and below the carbon plane.
- Three isolated C=C bonds would hydrogenate three times as exothermically as cyclohexene, but benzene releases less energy, showing that delocalisation gives extra stability.
- A common error is to describe rapidly alternating single and double bonds. Benzene's equal bonds and extra stability are permanent features, and substitution is favoured because it restores the delocalised ring whereas addition would disrupt it.
Tier 1 · Easy
Describe the shape, carbon-carbon bond lengths and pi bonding in a benzene molecule.
Tier 2 · Standard
Hydrogenation of cyclohexene has , while hydrogenation of benzene to cyclohexane has . Calculate the extra stability of benzene relative to a ring with three isolated C=C bonds.
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 in benzene, for a C-C bond and for a C=C bond, then use delocalisation to explain why substitution is preferred to addition.
Electrophilic substitution (A-level only)
- Benzene undergoes electrophilic monosubstitution: its pi electrons attack an electrophile, a sigma complex forms, and loss of H<sup>+</sup> restores delocalisation. Nitration is used in explosive manufacture and as a step towards aromatic amines.
- For nitration, use concentrated nitric acid with concentrated sulfuric acid and generate the nitronium ion: HNO<sub>3</sub> + H<sub>2</sub>SO<sub>4</sub> → NO<sub>2</sub><sup>+</sup> + HSO<sub>4</sub><sup>-</sup> + H<sub>2</sub>O.
- For Friedel-Crafts acylation, an acyl chloride and anhydrous AlCl<sub>3</sub> generate an acylium ion, RCO<sup>+</sup>, and the aromatic product is a ketone.
- A common error is to start a curly arrow at the electrophile or to omit the positive sigma complex; arrows start at electron pairs, and the final C-H bond arrow must return electrons to the ring.
Tier 1 · Easy
State the two reagents for nitrating benzene, name the catalyst and identify the electrophile.
Tier 2 · Standard
Benzene reacts with propanoyl chloride in the presence of anhydrous AlCl<sub>3</sub>. Name the organic product, write the overall equation and identify the attacking electrophile.
Tier 3 · Hard
Outline the complete electrophilic-substitution mechanism for reacting benzene with ethanoyl chloride and anhydrous AlCl<sub>3</sub>. Include electrophile generation, both ring electron movements and catalyst regeneration.
Preparation (A-level only)
- Primary aliphatic amines are made by reacting a halogenoalkane with excess ammonia or by reducing a nitrile. Reducing a nitrile retains the nitrile carbon, so it gives an amine with one more carbon than the starting halogenoalkane used to make that nitrile.
- For the halogenoalkane route, heat with excess ethanolic ammonia in a sealed vessel. Excess ammonia favours the primary amine because the amine product is also nucleophilic and can otherwise undergo further substitution.
- A nitrile can be reduced using hydrogen with a nickel catalyst or lithium aluminium hydride in dry ether. A nitro compound is reduced, commonly with tin and hydrochloric acid followed by sodium hydroxide, to prepare an aromatic amine such as phenylamine; aromatic amines are used in dye manufacture.
- A common error is to lose track of the carbon skeleton: halogenoalkane plus ammonia leaves the carbon count unchanged, whereas halogenoalkane to nitrile to amine lengthens the chain by one carbon.
Tier 1 · Easy
Name a reagent and a condition used to convert 1-bromopropane into propylamine while limiting further substitution.
Tier 2 · Standard
A two-stage route changes bromoethane into propylamine. Give the reagent and condition for each stage and name the intermediate.
Tier 3 · Hard
Describe how nitrobenzene can be converted into a pure sample of phenylamine. Include the purpose of the final alkaline treatment.
Base properties (A-level only)
- Amines are weak Brønsted–Lowry bases because the nitrogen lone pair can accept a proton, forming an alkylammonium or arylammonium ion.
- Alkyl groups push electron density towards nitrogen by the positive inductive effect, making the lone pair in a primary aliphatic amine more available than the lone pair in ammonia.
- In phenylamine, the nitrogen lone pair is delocalised into the benzene ring. It is therefore less available to bond to a proton, so phenylamine is a weaker base than ammonia.
- A common error is to explain base strength using N–H bond strength. The required comparison concerns the availability of the lone pair on nitrogen, not how easily an N–H bond breaks.
Tier 1 · Easy
Write an equation showing ethylamine acting as a base in water.
Tier 2 · Standard
Place ethylamine, ammonia and phenylamine in decreasing order of base strength. Explain both differences in the order.
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.
Nucleophilic properties (A-level only)
- Amines act as nucleophiles because the nitrogen lone pair can form a bond to an electron-deficient carbon. With halogenoalkanes they undergo nucleophilic substitution.
- Successive alkylations can form primary, secondary and tertiary amines and finally a quaternary ammonium salt. Excess ammonia favours a primary amine; excess halogenoalkane favours more highly substituted products.
- Ammonia and primary amines react with acyl chlorides and acid anhydrides by nucleophilic addition–elimination. For an acyl chloride, the first product loses H<sup>+</sup> and a second molecule of ammonia or amine neutralises HCl; students must be able to outline this mechanism.
- Quaternary ammonium ions have a charged hydrophilic end and long hydrophobic hydrocarbon groups in cationic surfactants. A common mechanism error is to draw a curly arrow from the electrophile rather than from the nitrogen lone pair.
Tier 1 · Easy
State the feature of an amine molecule that allows it to act as a nucleophile.
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.
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.
Condensation polymers (A-level only)
- Condensation polymerisation joins bifunctional monomers and eliminates a small molecule. A dicarboxylic acid plus a diol forms a polyester; a dicarboxylic acid plus a diamine, or amino acids reacting with one another, forms a polyamide.
- To draw a repeat unit, join the functional groups at both ends, retain the ester linkage –COO– or peptide linkage –CONH–, place brackets through bonds in the chain and add the subscript .
- To recover monomers from a polymer segment, split each ester or amide linkage at the acyl C–O or C–N bond, then restore –OH to the acyl carbon and H to the O or N atom.
- Polyamides form hydrogen bonds between N–H and C=O groups, while polyesters have permanent dipole–dipole attractions between ester groups; these support uses such as strong polyamide fibres and polyester fibres or bottles. A common error is to call the attractions covalent cross-links.
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.
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.
Tier 3 · Hard
A polymer contains the repeating segment [–O–CH<sub>2</sub>CH<sub>2</sub>–O–CO–C<sub>6</sub>H<sub>4</sub>–CO–]<sub>n</sub>, 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.
Biodegradability and disposal of polymers (A-level only)
- Polyalkenes have non-polar C–C and C–H backbones and are chemically inert, so microorganisms do not readily break them down and they are non-biodegradable.
- Polyesters and polyamides contain polar ester or amide links that can be hydrolysed. Cleaving these links shortens the chains, allowing biological degradation under suitable conditions.
- Mechanical recycling conserves feedstock and reduces landfill but requires collection, sorting and cleaning, and repeated processing can lower polymer quality. Feedstock recycling can recover useful chemicals but needs energy.
- Incineration reduces waste volume and can recover energy, but it releases carbon dioxide and may release toxic gases unless emissions are controlled. A balanced answer must compare impacts rather than claim that one method has no disadvantages.
Tier 1 · Easy
Explain why a polyalkene chain is much less susceptible to hydrolysis than a polyester chain.
Tier 2 · Standard
A disposable article can be made from a polyamide or from poly(propene). Compare their likely biodegradability using their chain structures.
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.
Amino acids (A-level only)
- An α-amino acid contains an amino group and a carboxylic acid group on the same carbon atom. It has basic and acidic properties because one group accepts H<sup>+</sup> while the other can donate H<sup>+</sup>.
- In the solid state and near neutral conditions, proton transfer gives a zwitterion containing –NH<sub>3</sub><sup>+</sup> and –COO<sup>−</sup> groups but no overall charge.
- In acid solution the amino acid is protonated overall, so the dominant form has –NH<sub>3</sub><sup>+</sup> and –COOH. In alkaline solution it has –NH<sub>2</sub> and –COO<sup>−</sup>.
- A common error is to draw the neutral –NH<sub>2</sub>/–COOH molecule when a zwitterion is requested, or to change both groups in acid or alkali without checking the ion's overall charge.
Tier 1 · Easy
Define the term zwitterion and state its overall charge.
Tier 2 · Standard
Give the displayed ionic forms of alanine, CH<sub>3</sub>CH(NH<sub>2</sub>)COOH, in strongly acidic solution and in strongly alkaline solution.
Tier 3 · Hard
A solution contains the zwitterion of 2-aminobutanoic acid. Write net ionic equations for its separate reactions with H<sup>+</sup> and OH<sup>−</sup>, and identify the role of the zwitterion in each reaction.
Proteins (A-level only)
- A peptide link is the amide group –CONH– formed when the carboxyl group of one amino acid condenses with the amino group of another. A chain has direction, so reversing the amino-acid order gives a different displayed structure.
- Primary structure is the amino-acid sequence; secondary structure includes α-helices and β-pleated sheets held by hydrogen bonds; tertiary structure is the overall three-dimensional folding maintained by interactions including hydrogen bonds and sulfur–sulfur bonds.
- Hydrolysis breaks peptide links and restores the constituent amino acids. Acidic hydrolysis gives protonated amino groups, whereas alkaline hydrolysis gives carboxylate groups.
- Amino acids can be separated by TLC and located with ninhydrin or ultraviolet light. Calculate using distance moved by the centre of the spot divided by distance moved by the solvent front; both distances must use the same origin.
Tier 1 · Easy
State what determines the primary structure of a protein.
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.
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 of an amino-acid spot that moves when the solvent front moves .
Enzymes (A-level only)
- Enzymes are protein catalysts. Their folded tertiary structure creates an active site whose shape and arrangement of functional groups are complementary to a particular substrate.
- An active site is stereospecific: only an enantiomer with the correct three-dimensional arrangement can make the required set of interactions and bind effectively.
- A drug can inhibit an enzyme by occupying its active site, preventing substrate binding and lowering the rate of the enzyme-catalysed reaction. Binding must be strong enough to inhibit but selective enough to limit effects on other proteins.
- Computer modelling can compare candidate shapes and interactions with an active site before synthesis. A common error is to say the enzyme changes the equilibrium position; a catalyst changes rate by providing a lower-activation-energy route.
Tier 1 · Easy
State why an enzyme is described as a catalyst.
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.
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.
DNA (A-level only)
- A nucleotide contains a phosphate group bonded to 2-deoxyribose, which is bonded to one base: adenine, cytosine, guanine or thymine.
- A DNA strand is a condensation polymer with covalent phosphodiester links between the phosphate group of one nucleotide and the sugar of another, producing a sugar–phosphate backbone with bases attached to the sugars.
- Two strands form a double helix through complementary hydrogen bonding: adenine pairs with thymine and cytosine pairs with guanine. Complementarity lets each strand act as a template.
- A common error is to describe hydrogen bonds as holding each sugar–phosphate backbone together. The backbone is covalent; hydrogen bonds act between complementary bases on different strands.
Tier 1 · Easy
Name the three types of component present in a DNA nucleotide.
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.
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.
Action of anticancer drugs (A-level only)
- Cisplatin is a square-planar Pt(II) complex used as an anticancer drug. Its cis arrangement places the two replaceable chloride ligands next to one another.
- In cells, ligand replacement allows platinum to form coordinate bonds to nitrogen atoms on guanine bases, often linking nearby sites on DNA.
- The platinum–DNA links distort the double helix and prevent the strands functioning normally during DNA replication, so rapidly dividing cancer cells cannot reproduce successfully.
- Cisplatin can also damage healthy dividing cells, causing adverse effects. A common error is to say that platinum forms hydrogen bonds to guanine; the new Pt–N bonds are coordinate covalent bonds formed by ligand replacement.
Tier 1 · Easy
Name the DNA base to which cisplatin bonds and identify the donor atom in that base.
Tier 2 · Standard
Explain, using ligand replacement and DNA structure, how cisplatin can stop a cancer cell from replicating.
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.
Organic synthesis (A-level only)
- A synthesis route may use any reactions in the specification and can contain up to four steps. Work backwards from the target functional group, then check that every forward step has an appropriate reagent and condition.
- When planning a route, track the carbon skeleton as well as the functional group. Cyanide substitution adds one carbon; oxidation, reduction, addition, elimination and acylation normally preserve the carbon framework unless another reactant contributes carbon.
- A strong answer names intermediates or shows their structures, specifies selective conditions such as distillation versus reflux, and includes purification or separation only when it is chemically relevant.
- Processes with fewer steps and high percentage atom economy usually consume less material and produce less waste. Avoiding solvents and choosing non-hazardous starting materials can further reduce environmental and safety impacts.
Tier 1 · Easy
Give two reasons, other than cost, why a chemist may prefer a high-atom-economy synthesis with fewer reaction steps.
Tier 2 · Standard
Devise 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.
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.
Nuclear magnetic resonance spectroscopy (A-level only)
- Each chemically distinct carbon environment normally gives one <sup>13</sup>C NMR signal. <sup>13</sup>C spectra are simpler than <sup>1</sup>H spectra; each distinct proton environment gives a <sup>1</sup>H signal, whose chemical shift is compared with the Chemistry Data Booklet.
- The area under a <sup>1</sup>H NMR signal is proportional to the number of equivalent protons in that environment. Convert integration values to the simplest whole-number ratio before matching them to a formula.
- For adjacent non-equivalent protons in simple aliphatic compounds, equivalent neighbouring protons split a signal into peaks: one neighbour gives a doublet, two a triplet and three a quartet.
- TMS gives one sharp signal, is inert and is volatile, and its protons are strongly shielded so its peak is set at . Samples use a deuterated solvent or CCl<sub>4</sub> to avoid a large solvent <sup>1</sup>H signal; a common error is to count exchangeable O–H protons as reliably splitting adjacent signals.
Tier 1 · Easy
Give two properties that make tetramethylsilane suitable as the standard for chemical shift in NMR spectroscopy.
Tier 2 · Standard
A compound gives three <sup>1</sup>H 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 C<sub>4</sub>H<sub>8</sub>O<sub>2</sub>.
Tier 3 · Hard
An ester has molecular formula C<sub>5</sub>H<sub>10</sub>O<sub>2</sub> and five <sup>13</sup>C NMR signals. Its <sup>1</sup>H 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.
Chromatography (A-level only)
- Chromatography separates mixture components by their different balances between the mobile phase and the stationary phase. Greater solubility in the mobile phase increases movement; stronger retention by the stationary phase decreases it.
- In TLC a solvent rises over a solid-coated plate; in column chromatography solvent moves through a packed solid; in GC a carrier gas moves compounds through a heated column containing a solid or a liquid-coated solid stationary phase.
- Calculate as distance moved by the component divided by distance moved by the solvent front. Compare values or GC retention times only with standards obtained under the same conditions.
- GC can be coupled to mass spectrometry: retention time separates components and the mass spectrum supports identification. A common error is to assume that matching one retention time alone proves identity.
Tier 1 · Easy
On a TLC plate, a spot travels from the start line while the solvent front travels . Calculate the value.
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 and explain your answer.
Tier 3 · Hard
A GC trace of a flavour mixture has peaks at , and . Under identical conditions, reference compounds X, Y and Z have retention times , and 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.