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Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Arenes: electrophilic substitution · Core · Paper 4 style · 5 marks
(a)Give the reagents and conditions for making chlorobenzene from benzene.[2]
(b)Write equations to show how the catalyst forms the electrophile and how it is regenerated.[2]
(c)Write the overall equation.[1]
Show answer
(a) Chlorine with a halogen carrier such as AlCl₃ (or FeCl₃ / iron); room temperature, anhydrous, in the dark
Q2Arenes: electrophilic substitution · Core · Paper 4 style · 4 marks
(a)Cyclohexene decolourises bromine water quickly at room temperature, but benzene does not. Explain.[2]
(b)Explain why benzene undergoes substitution rather than addition reactions.[2]
Show answer
(a) In cyclohexene the π electrons are localised in the C=C bond, giving high electron density that polarises Br₂. In benzene the π electrons are delocalised over six carbons, so electron density is lower and the ring is not attacked without a catalyst.
B1 localised C=C high electron density; B1 delocalised π in benzene lower electron density.
(b) Addition would destroy the delocalised π system, which gives extra stability; substitution keeps the delocalised ring intact.
B1 delocalisation gives stability; B1 substitution retains it.
Q3Arenes: electrophilic substitution · Standard · Paper 4 style · 7 marks
(a)State the reagents and conditions for the nitration of benzene.[2]
(b)Write an equation for the formation of the electrophile.[1]
(c)Describe the mechanism of the reaction of benzene with this electrophile. Describe the curly arrows in words.[3]
(d)Name this type of mechanism.[1]
Show answer
(a) Concentrated nitric acid and concentrated sulfuric acid; temperature below about 55 °C
B1 both concentrated acids; B1 temperature (accept 25–55 °C).
(b) HNO₃ + 2H₂SO₄ → NO₂⁺ + 2HSO₄⁻ + H₃O⁺
B1 (accept HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O).
(c) Curly arrow from the delocalised π ring to N of NO₂⁺; intermediate cation with a broken (horseshoe) delocalised system, positive charge, and C bearing H and NO₂; curly arrow from the C–H bond back into the ring, H⁺ lost (taken by HSO₄⁻), regenerating the aromatic ring.
B1 arrow from π ring to NO₂⁺; B1 correct intermediate (partial delocalisation + charge); B1 C–H bond breaks, H⁺ lost, aromaticity restored.
(d) Electrophilic substitution
B1.
Q4Arenes: electrophilic substitution · Standard · Paper 4 style · 5 marks
(a)Give the reagent and catalyst needed to convert benzene into ethylbenzene, and write the equation.[2]
(b)Write an equation for the formation of the electrophile.[1]
(c)Write an equation to show how the catalyst is regenerated.[1]
(d)Ethylbenzene is made industrially as a step towards poly(phenylethene) packaging. Suggest why the AlCl₃ must be anhydrous.[1]
Q6Arenes: electrophilic substitution · Challenge · Paper 4 style · 7 marks
(a)Methylbenzene reacts with bromine in the presence of FeBr₃ in the dark. Name the two main organic products and the mechanism.[3]
(b)Methylbenzene reacts with bromine in ultraviolet light without a catalyst. Name the main organic product and the type of mechanism.[2]
(c)Explain why the conditions determine whether the ring or the side chain reacts.[2]
Show answer
(a) 1-bromo-2-methylbenzene and 1-bromo-4-methylbenzene; electrophilic substitution
B1 each product (ring substitution at 2 and 4 positions); B1 electrophilic substitution.
(b) (Bromomethyl)benzene, C₆H₅CH₂Br; free-radical substitution (in the side chain)
B1 product; B1 free-radical substitution.
(c) With a halogen carrier (FeBr₃), Br⁺ / polarised Br₂ is formed, an electrophile that attacks the electron-rich ring. In UV, Br₂ splits homolytically into Br• radicals, which abstract H from the CH₃ group.
B1 catalyst makes an electrophile for the ring; B1 UV makes radicals that attack the side chain.
Q7Arenes: electrophilic substitution · Challenge · Paper 4 style · 6 marks
(a)Predict the main organic product(s) of mononitration of: (i) methylbenzene; (ii) nitrobenzene; (iii) benzoic acid.[3]
(b)Classify the substituents CH₃, NO₂ and COOH as 2,4-directing or 3-directing.[1]
(c)A chemist needs 3-nitrobenzoic acid. Suggest whether to nitrate methylbenzene and then oxidise the side chain, or oxidise first and then nitrate. Explain.[2]
Show answer
(a) (i) 1-methyl-2-nitrobenzene and 1-methyl-4-nitrobenzene; (ii) 1,3-dinitrobenzene; (iii) 3-nitrobenzoic acid
B1 each.
(b) CH₃ 2,4-directing; NO₂ and COOH 3-directing
B1 all.
(c) Oxidise first to benzoic acid, then nitrate: COOH is 3-directing so NO₂ goes to position 3. Nitrating methylbenzene first would put NO₂ at 2 or 4.
B1 oxidise first; B1 reason based on directing effects.
Q8Arenes: electrophilic substitution · Challenge · Paper 4 style · 5 marks
(a)State the number of different mononitro products that can be formed from each dimethylbenzene: (i) 1,2-dimethylbenzene; (ii) 1,3-dimethylbenzene; (iii) 1,4-dimethylbenzene. Ignore any directing effects.[3]
(b)A student mononitrates an unknown dimethylbenzene and obtains a single product. Identify it and give the name of the product.[2]
Show answer
(a) (i) 2; (ii) 3; (iii) 1
B1 each. 1,2: positions 3 (=6) and 4 (=5). 1,3: positions 2, 4 (=6) and 5. 1,4: all four ring H equivalent.
Q9Alkenes: electrophilic addition · Core · Paper 1 style · 1 mark
(a)What is the major product when propene reacts with hydrogen bromide?[1]
Show answer
(a) A 2-bromopropane
H adds to the CH₂ end to give the more stable secondary carbocation CH₃C⁺HCH₃, which Br⁻ attacks → 2-bromopropane. 1-bromopropane would form via the less stable primary carbocation, so it is the minor product.
Q10Alkenes: electrophilic addition · Core · Paper 1 style · 1 mark
(a)In the reaction of ethene with bromine, how does the bromine molecule behave?[1]
Show answer
(a) A as an electrophile after a dipole is induced in it
The high π electron density of C=C induces a dipole in Br₂; the δ+ Br accepts a pair of electrons, so Br₂ acts as an electrophile. A nucleophile would donate electrons, which is what the alkene does.
Q11Alkenes: electrophilic addition · Standard · Paper 2 style · 5 marks
(a)Describe the mechanism of the reaction between ethene and bromine. Describe the curly arrows and intermediate in words.[4]
(b)Name the mechanism.[1]
Show answer
(a) The C=C π electrons induce a dipole in Br₂ (Brδ+–Brδ−). Curly arrow from the C=C to Brδ+, and from the Br–Br bond to Brδ−. A carbocation CH₂Br–CH₂⁺ and Br⁻ form. Curly arrow from a lone pair on Br⁻ to the C⁺, giving 1,2-dibromoethane.
B1 induced dipole in Br₂; B1 arrows from C=C to Br and Br–Br bond to Br; B1 carbocation intermediate + Br⁻; B1 Br⁻ lone pair attacks C⁺.
(b) Electrophilic addition
B1.
Q12Alkenes: electrophilic addition · Standard · Paper 2 style · 5 marks
(a)State the conditions for the industrial hydration of ethene and write the equation.[2]
(b)But-1-ene is hydrated in the same way. Name the major and minor products.[2]
(c)Explain why one product of (b) is formed in greater amount.[1]
Show answer
(a) Steam, phosphoric acid catalyst, about 300 °C and 6–7 MPa; C₂H₄ + H₂O → C₂H₅OH
B1 catalyst + two of T/p; B1 equation.
(b) Major: butan-2-ol; minor: butan-1-ol
B1 each.
(c) It forms via the more stable secondary carbocation (Markovnikov addition).
B1.
Q13Alkenes: electrophilic addition · Challenge · Paper 2 style · 5 marks
(a)2-methylpropene, (CH₃)₂C=CH₂, reacts with HCl. Draw (describe) the two possible carbocation intermediates and classify each as primary, secondary or tertiary.[2]
(b)Name the major product and explain why it is formed in greater amount.[3]
Show answer
(a) (CH₃)₃C⁺ tertiary; (CH₃)₂CHCH₂⁺ primary
B1 each with classification.
(b) 2-chloro-2-methylpropane. The tertiary carbocation is more stable because three alkyl groups donate electrons (positive inductive effect) and spread the positive charge, so it forms faster/more readily.
B1 product; B1 tertiary carbocation more stable; B1 alkyl groups electron-donating / inductive effect.
Q14Alkenes: electrophilic addition · Challenge · Paper 2 style · 4 marks
(a)Explain why but-2-ene gives only one structural isomer with HBr, while but-1-ene gives two.[2]
(b)2-bromobutane is formed as a racemic mixture. Explain why.[2]
Show answer
(a) But-2-ene is symmetrical, so H adding to either carbon gives the same secondary carbocation and 2-bromobutane only. But-1-ene is unsymmetrical: primary or secondary carbocation → 1-bromobutane or 2-bromobutane.
B1 symmetry of but-2-ene; B1 two different carbocations for but-1-ene.
(b) C2 is a chiral centre; the carbocation is planar, so Br⁻ attacks from either side with equal probability, giving equal amounts of the two enantiomers.
B1 planar carbocation attacked from both sides; B1 equal amounts of two optical isomers (optically inactive mixture).
Q15Alkanes: free-radical substitution · Core · Paper 1 style · 1 mark
(a)Which equation shows a propagation step in the reaction of methane with chlorine in ultraviolet light?[1]
Show answer
(a) A CH₄ + Cl• → CH₃• + HCl
A propagation step uses one radical and produces another radical. B is initiation; C and D are termination steps (two radicals combine).
Q16Alkanes: free-radical substitution · Core · Paper 2 style · 5 marks
(a)Methane is the main component of natural gas from the Sui field. Explain why alkanes are generally unreactive.[2]
(b)Methane reacts with bromine in bright sunlight. Write the overall equation for the formation of bromomethane and name the type of reaction.[2]
(c)Suggest how ethane can be formed as a by-product.[1]
Show answer
(a) C–C and C–H bonds are strong and almost non-polar, so they are not attacked by electrophiles or nucleophiles.
B1 strong bonds; B1 non-polar / no partial charges.
(c) Two methyl radicals combine in a termination step: 2CH₃• → C₂H₆.
B1.
Q17Alkanes: free-radical substitution · Standard · Paper 2 style · 6 marks
(a)Ethane reacts with chlorine in ultraviolet light. Write equations for the initiation step, two propagation steps and two termination steps that lead to chloroethane or other products.[5]
H environments in (CH₃)₂CHCH₂CH₃: 6H (two equivalent CH₃), 1H (CH), 2H (CH₂), 3H (end CH₃) → 4 monochloro isomers. B1 four; B2 names (B1 for two or three correct).
(b) Cl• can substitute any hydrogen, giving several isomers; further substitution gives di- and polychloro compounds; termination steps give other products.
B1 mixture of positional isomers; B1 multiple substitution / other products.
Q19Alkenes: oxidation with KMnO4 · Core · Paper 2 style · 4 marks
(a)Propene is shaken with cold, dilute, acidified potassium manganate(VII). State the colour change and name the organic product.[2]
(b)Write an equation for this reaction using [O] for the oxidising agent.[1]
(c)Give one use of this colour change.[1]
Show answer
(a) Purple to colourless; propane-1,2-diol
B1; B1.
(b) CH₃CH=CH₂ + [O] + H₂O → CH₃CH(OH)CH₂OH
B1.
(c) Test for C=C (unsaturation) in a compound.
B1.
Q20Alkenes: oxidation with KMnO4 · Standard · Paper 1 style · 1 mark
(a)2-methylbut-2-ene, (CH₃)₂C=CHCH₃, is heated with hot, concentrated, acidified KMnO₄. What are the organic products?[1]
Show answer
(a) A propanone and ethanoic acid
The C=C is broken. (CH₃)₂C= becomes (CH₃)₂C=O (propanone); =CHCH₃ becomes CH₃CHO, which is further oxidised to CH₃COOH under these conditions. Ethanal (B) is not the final product; the diol (D) forms only with cold dilute KMnO₄.
Q21Alkenes: oxidation with KMnO4 · Challenge · Paper 2 style · 6 marks
(a)Alkene F, C₆H₁₂, is heated with hot concentrated acidified KMnO₄. The products are propanone and propanoic acid. Identify F.[2]
(b)Alkene G, C₅H₁₀, gives carbon dioxide and butanone under the same conditions. Identify G and explain the formation of CO₂.[3]
(c)Predict the products from hex-3-ene under the same conditions.[1]
Show answer
(a) 2-methylpent-2-ene, (CH₃)₂C=CHCH₂CH₃
B2 (B1 for a structure with C=C between the correct carbons but wrong name).
(b) 2-methylbut-1-ene, CH₂=C(CH₃)CH₂CH₃. The =CH₂ end is oxidised to methanoic acid then to CO₂ and H₂O.
Q22Alkenes: addition reactions · Core · Paper 2 style · 5 marks
(a)Name the organic product when cyclohexene reacts with: (i) H₂ / Ni; (ii) Br₂; (iii) HBr; (iv) steam with H₃PO₄ catalyst.[4]
(b)State the observation when bromine water is added to cyclohexene.[1]
Show answer
(a) (i) cyclohexane; (ii) 1,2-dibromocyclohexane; (iii) bromocyclohexane; (iv) cyclohexanol
B1 each.
(b) Orange/brown to colourless
B1.
Q23Alkenes: addition reactions · Standard · Paper 2 style · 6 marks
(a)Vanaspati ghee is made by partially hydrogenating vegetable oils. State the catalyst and conditions used for hydrogenation.[2]
(b)Linoleic acid, C₁₈H₃₂O₂, has two C=C bonds per molecule. Calculate the volume of hydrogen, at room conditions, needed to saturate 14.0 g of linoleic acid completely. (Ar: C 12.0, H 1.0, O 16.0; molar gas volume 24.0 dm³)[3]
(c)Explain why hydrogenation raises the melting point of the oil.[1]
Show answer
(a) Nickel catalyst, about 150 °C (and moderate pressure)
(c) Saturated chains are straighter and pack more closely, so stronger id–id forces.
B1.
Q24Alkenes: addition reactions · Challenge · Paper 2 style · 5 marks
0.410 g of a hydrocarbon C₆H₁₀, which contains no triple bonds, reacts completely with 1.60 g of bromine, Br₂. Ar: C 12.0, H 1.0; Mr Br₂ = 159.8.
(a)Calculate the amounts, in mol, of hydrocarbon and of bromine.[2]
(b)Deduce the number of C=C bonds in one molecule of the hydrocarbon and suggest a possible structure.[2]
(c)A second hydrocarbon, also C₆H₁₀, reacts with only one mole of Br₂ per mole. Suggest a structure.[1]
Show answer
(a) hydrocarbon 0.00500 mol; Br₂ 0.0100 mol
B1 0.410/82.0 = 0.00500; B1 1.60/159.8 = 0.0100.
(b) 2 C=C; e.g. hexa-1,5-diene, CH₂=CHCH₂CH₂CH=CH₂ (or hexa-1,3-diene etc.)
M1 ratio Br₂ : hydrocarbon = 2.00 ≈ 2; A1 any acyclic C₆H₁₀ diene.
(c) Cyclohexene
B1 (any cyclic C₆H₁₀ with one C=C).
Q25Arenes: side-chain oxidation & other reactions · Core · Paper 4 style · 4 marks
(a)State the reagents and conditions to convert ethylbenzene into benzoic acid.[2]
(b)Name the organic product formed from propylbenzene under the same conditions.[1]
(c)Benzene itself is not oxidised under these conditions. Suggest why.[1]
Show answer
(a) Hot alkaline KMnO₄ (heat under reflux), followed by dilute acid (e.g. H₂SO₄)
B1 hot alkaline KMnO₄; B1 then acidify.
(b) Benzoic acid
B1.
(c) The delocalised ring is very stable; only the alkyl side chain (C attached to the ring bearing H) is oxidised.
B1.
Q26Arenes: side-chain oxidation & other reactions · Challenge · Paper 4 style · 5 marks
(a)Compound H, C₉H₁₂, is a benzene derivative. On oxidation with hot alkaline KMnO₄ followed by acid it gives benzene-1,4-dicarboxylic acid. Suggest a structure for H.[2]
(b)Compound J is an isomer of H that gives benzoic acid. Suggest two possible structures for J.[2]
(c)Benzene-1,4-dicarboxylic acid is used to make polyester fibres. Name the other type of monomer needed.[1]
Show answer
(a) 1-ethyl-4-methylbenzene, CH₃C₆H₄C₂H₅ (para)
B1 two side chains totalling C₃; B1 para positions.
Q27Alkanes: properties, combustion & cracking · Core · Paper 2 style · 4 marks
(a)Decane, C₁₀H₂₂, is cracked to give octane and one other product. Write the equation and give typical conditions.[2]
(b)Write an equation for the complete combustion of octane.[1]
(c)Give one reason why refineries crack long-chain alkanes.[1]
Show answer
(a) C₁₀H₂₂ → C₈H₁₈ + C₂H₄; about 500 °C with a catalyst (Al₂O₃ / zeolite), or high temperature alone
B1 equation; B1 conditions.
(b) 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O
B1 (accept C₈H₁₈ + 12½O₂ → 8CO₂ + 9H₂O).
(c) Demand for shorter chains (petrol) exceeds supply / alkenes are needed for polymers.
B1.
Q28Alkanes: properties, combustion & cracking · Standard · Paper 2 style · 5 marks
(a)Calculate the mass of carbon dioxide, in kg, produced when 1.00 kg of octane burns completely. Give your answer to 3 significant figures. (Mr: octane 114.0; CO₂ 44.0)[2]
(b)In heavy traffic in Karachi, some octane burns incompletely. Write an equation for the formation of carbon monoxide.[1]
(c)Catalytic converters remove CO and NO. Write an equation for this reaction and explain why it reduces harm.[2]
Show answer
(a) 3.09 kg
M1 n = 1000/114.0 = 8.772 mol → CO₂ = 8 × = 70.2 mol; A1 × 44.0 = 3088 g = 3.09 kg.
(b) 2C₈H₁₈ + 17O₂ → 16CO + 18H₂O
B1.
(c) 2CO + 2NO → 2CO₂ + N₂; CO is toxic and NO causes acid rain/smog, products are less harmful.
B1 equation; B1 reason.
Q29Arenes: structure & bonding · Core · Paper 4 style · 4 marks
(a)Describe the bonding in a benzene molecule, including hybridisation and the arrangement of the π electrons.[3]
(b)All C–C bonds in benzene are 0.139 nm long. Explain how this is evidence against a structure with alternating single and double bonds.[1]
Show answer
(a) Each C is sp² hybridised, forming σ bonds to two C and one H (planar, 120°). Each C has one unhybridised p orbital; these overlap sideways all round the ring to form a delocalised π system above and below the plane, containing 6 electrons.
B1 sp², σ bonds, planar 120°; B1 p orbitals overlap sideways; B1 delocalised ring of 6 π electrons above and below plane.
(b) Alternating bonds would have two different lengths (C–C ≈ 0.154 nm and C=C ≈ 0.134 nm); equal lengths show delocalisation.
B1.
Q30Arenes: structure & bonding · Challenge · Paper 4 style · 5 marks
