References only, not question text. Open Official paper ↗ where Cambridge publishes the paper,
or see the official Cambridge past-papers page; for other papers, ask your
school. “Paper 42” means Paper 4, variant 2.
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.
Q1SN1/SN2 & reactivity · Core · Paper 2 style · 5 marks
(a)State which mechanism, SN1 or SN2, is mainly followed in the hydrolysis of (i) 1-bromobutane (ii) 2-bromo-2-methylpropane.[2]
(b)Explain why tertiary halogenoalkanes react mainly by the SN1 mechanism.[3]
Show answer
(a) (i) SN2 (ii) SN1
1 mark each.
(b) Tertiary carbocation is the most stable because three alkyl groups donate electrons (positive inductive effect) and stabilise the positive charge; the three bulky alkyl groups hinder (steric hindrance) attack by a nucleophile from the side opposite the halogen, so SN2 is difficult.
Q2SN1/SN2 & reactivity · Standard · Paper 2 style · 5 marks
1-bromobutane is warmed with aqueous sodium hydroxide.
(a)Describe the SN2 mechanism for the reaction of 1-bromobutane with hydroxide ions. Describe in words where each curly arrow starts and ends, and describe the transition state.[4]
(b)State the rate equation for this reaction.[1]
Show answer
(a) Arrow from a lone pair on O of OH⁻ to the δ+ carbon bonded to Br; arrow from the C–Br bond to Br; transition state [HO···C···Br]⁻ with partial bonds and a negative charge, OH and Br on opposite sides of carbon; products butan-1-ol and Br⁻.
B1 C–Br dipole (Cδ+, Brδ−); B1 arrow from OH⁻ lone pair to C; B1 arrow from C–Br bond to Br; B1 transition state with partial bonds and overall negative charge.
(b) rate = k[CH₃CH₂CH₂CH₂Br][OH⁻]
B1.
Q3SN1/SN2 & reactivity · Standard · Paper 4 style · 5 marks
Q4SN1/SN2 & reactivity · Standard · Paper 1 style · 1 mark
(a)Which halogenoalkane is most likely to undergo hydrolysis by an SN1 mechanism?[1]
Show answer
(a) C (CH₃)₃CCl
(CH₃)₃CCl is tertiary: it forms the most stable carbocation and is sterically hindered towards SN2. B is secondary and can react by either route, but less readily by SN1.
Q5SN1/SN2 & reactivity · Challenge · Paper 4 style · 5 marks
2-bromobutane, CH₃CHBrCH₂CH₃, can undergo nucleophilic substitution with hydroxide ions.
(a)Explain why 2-bromobutane is chiral.[1]
(b)A single enantiomer of 2-bromobutane reacts with OH⁻ by the SN2 mechanism. State and explain whether the butan-2-ol formed rotates plane-polarised light.[2]
(c)If the same reaction occurred by SN1, the product would not rotate plane-polarised light. Explain why.[2]
Show answer
(a) Carbon 2 is bonded to four different groups (H, Br, CH₃, C₂H₅).
B1.
(b) Yes – OH⁻ attacks from the side opposite Br, so the configuration is inverted and a single enantiomer of butan-2-ol forms, which is optically active.
B1 attack opposite to leaving group/inversion; B1 single enantiomer so optically active.
(c) The carbocation intermediate is planar, so the nucleophile can attack equally from either side; a racemic mixture (50 : 50 enantiomers) forms and the rotations cancel.
B1 planar carbocation, attack from both faces; B1 racemic mixture, no net rotation.
Q6SN1/SN2 & reactivity · Challenge · Paper 2 style · 6 marks
Three test-tubes each contain 1 cm³ of ethanol, 1 cm³ of aqueous silver nitrate and three drops of a halogenoalkane, warmed in a water bath at 50 °C: 1-chlorobutane, 1-bromobutane and 1-iodobutane.Bond energies / kJ mol⁻¹: C–Cl 340; C–Br 280; C–I 240
(a)Predict the order in which precipitates appear, fastest first, and state the colour of each precipitate.[2]
(b)Use the bond energies to explain the order of reactivity.[2]
(c)A student predicts that 1-chlorobutane should react fastest because the C–Cl bond is the most polar. Explain why this prediction is wrong.[1]
(d)Explain the purpose of the ethanol.[1]
Show answer
(a) 1-iodobutane (yellow AgI) first, then 1-bromobutane (cream AgBr), then 1-chlorobutane (white AgCl) last.
B1 order I > Br > Cl; B1 all colours.
(b) C–I is the weakest bond (240 kJ mol⁻¹), so it breaks most easily; C–Cl is the strongest (340 kJ mol⁻¹), so the chloroalkane is hydrolysed most slowly.
B1 bond strength decreases C–Cl > C–Br > C–I; B1 weaker bond broken more easily so faster rate.
(c) Bond strength, not bond polarity, is the dominant factor controlling the rate of hydrolysis.
B1.
(d) It is a solvent in which both the halogenoalkane and aqueous silver nitrate dissolve (they mix).
B1.
Q7Elimination · Core · Paper 2 style · 4 marks
(a)Give the reagent and conditions needed to convert 2-bromopropane into propene.[2]
(b)Write an equation for the reaction.[1]
(c)State the role of the hydroxide ion in this reaction.[1]
Show answer
(a) NaOH (or KOH) dissolved in ethanol; heat (under reflux).
B1 ethanolic NaOH/KOH; B1 heat.
(b) CH₃CHBrCH₃ + NaOH → CH₃CH=CH₂ + NaBr + H₂O
B1.
(c) Base (removes H⁺ from the carbon next to the C–Br carbon).
B1.
Q8Elimination · Core · Paper 1 style · 1 mark
(a)Which reagent and conditions convert bromoethane into ethene?[1]
Show answer
(a) B NaOH in ethanol, heat
Hot ethanolic hydroxide removes HBr (elimination). A gives ethanol (substitution); C is used to dehydrate ethanol, not bromoethane.
Q9Elimination · Standard · Paper 2 style · 5 marks
2-bromobutane is heated with potassium hydroxide in ethanol. Three different alkenes are formed.
(a)Name all three alkenes formed.[3]
(b)Explain why there are two products with the formula CH₃CH=CHCH₃.[2]
(b) Restricted rotation about the C=C bond; and each carbon of the double bond carries two different groups (H and CH₃), so cis/trans (geometrical) isomers exist.
B1 restricted rotation; B1 two different groups on each C of C=C.
Q10Elimination · Standard · Paper 2 style · 6 marks
(a)Name the organic product and write an equation when 1-bromopropane is heated with (i) NaOH(aq) (ii) NaOH in ethanol.[4]
(b)Explain the different behaviour of the hydroxide ion in (i) and (ii).[2]
(b) 1-bromo-2-methylpropane and 2-bromo-2-methylpropane
B1 each.
(c) 2-bromobutane
B1 gives cis- and trans-but-2-ene.
(d) 1-bromobutane (gives but-1-ene only)
B1.
Q12Halogenoarenes vs halogenoalkanes · Core · Paper 4 style · 5 marks
(a)Explain why chlorobenzene does not undergo nucleophilic substitution with NaOH(aq) under normal laboratory conditions.[3]
(b)Predict what is seen when chlorobenzene and 1-chlorobutane are each warmed with ethanolic silver nitrate.[2]
Show answer
(a) A lone pair (p-orbital) on Cl overlaps with the delocalised π-electrons of the ring; so the C–Cl bond has partial double-bond character and is stronger/harder to break; the electron-rich ring also repels nucleophiles.
B1 p-orbital overlap with π system; B1 stronger C–Cl bond; B1 π electron cloud repels OH⁻.
(b) Chlorobenzene: no precipitate; 1-chlorobutane: white precipitate (slowly).
B1 each.
Q13Halogenoarenes vs halogenoalkanes · Standard · Paper 4 style · 5 marks
Consider three chlorine compounds: benzoyl chloride (C₆H₅COCl), 1-chlorobutane (CH₃CH₂CH₂CH₂Cl) and chlorobenzene (C₆H₅Cl).
(a)Put the three compounds in order of ease of hydrolysis, most easily hydrolysed first.[1]
(b)Describe what is observed when each compound is added to water containing AgNO₃(aq) at room temperature.[2]
(c)Explain why benzoyl chloride is hydrolysed most easily.[2]
(b) C₆H₅COCl: immediate white precipitate (and steamy fumes of HCl); 1-chlorobutane: no precipitate at room temperature (slow, faint on warming); C₆H₅Cl: no precipitate.
B1 immediate precipitate with acyl chloride; B1 none/very slow for the other two.
(c) The carbonyl carbon is attached to both O and Cl, two electronegative atoms, so it is strongly δ+; the planar C=O carbon is easily attacked by water (addition–elimination).
B1 two electronegative atoms make C very δ+; B1 attack by water easy (lone pair of H₂O).
Q14Halogenoarenes vs halogenoalkanes · Standard · Paper 4 style · 5 marks
Methylbenzene reacts with chlorine under two different sets of conditions.
(a)Give the conditions needed to make (i) 1-chloro-4-methylbenzene (and its 2-isomer) (ii) (chloromethyl)benzene, C₆H₅CH₂Cl, from methylbenzene and chlorine.[2]
(b)Name the type of mechanism in each case.[2]
(c)Write the equation for the formation of the electrophile in (i).[1]
Show answer
(a) (i) AlCl₃ (or FeCl₃) catalyst, room temperature, no UV (ii) UV light (or boiling)
1 mark each.
(b) (i) electrophilic substitution (ii) free-radical substitution
1 mark each.
(c) Cl₂ + AlCl₃ → Cl⁺ + AlCl₄⁻
B1.
Q15Halogenoarenes vs halogenoalkanes · Standard · Paper 4 style · 5 marks
Equal amounts of bromobenzene, 1-bromobutane and 2-bromo-2-methylpropane are added to separate tubes of warm aqueous ethanolic silver nitrate.
(a)Predict the order in which a cream precipitate appears in the three tubes, fastest first.[1]
(b)Explain your order for the two halogenoalkanes.[2]
(c)Explain why bromobenzene gives no precipitate.[2]
Show answer
(a) 2-bromo-2-methylpropane, then 1-bromobutane; bromobenzene gives no precipitate
B1.
(b) 2-bromo-2-methylpropane is tertiary and reacts by SN1 via a stable tertiary carbocation, faster than the primary 1-bromobutane (SN2).
B1 tertiary via SN1/carbocation; B1 faster than primary.
(c) The C–Br bond is strengthened by overlap of a Br lone pair with the ring π system, and the π electrons repel the nucleophile, so no Br⁻ is released.
B1 stronger C–Br bond (delocalisation); B1 nucleophile repelled.
Q16Halogenoarenes vs halogenoalkanes · Challenge · Paper 4 style · 6 marks
(a)Give the number of structural isomers of C₇H₇Cl that contain a benzene ring and name the one with Cl on the side chain.[2]
(b)Describe a chemical test to distinguish (chloromethyl)benzene from 1-chloro-4-methylbenzene. Give the results.[3]
(c)Explain why dilute nitric acid is added before the silver nitrate.[1]
Show answer
(a) 4; (chloromethyl)benzene
B1 4 (three ring isomers + one side-chain); B1 name.
(b) Warm each with NaOH(aq); acidify with dilute HNO₃; add AgNO₃(aq). (Chloromethyl)benzene gives a white precipitate; 1-chloro-4-methylbenzene gives no precipitate.
B1 warm NaOH(aq) then HNO₃ then AgNO₃; B1 white ppt with side-chain compound; B1 no ppt with ring-substituted.
(c) To neutralise excess OH⁻, which would otherwise form a brown precipitate of silver oxide.
B1.
Q17Reactions of halogenoalkanes (reagents & conditions) · Core · Paper 2 style · 6 marks
Bromoethane, C₂H₅Br, is a useful starting material in synthesis.
(a)Give the reagent and conditions to convert bromoethane into (i) ethanol (ii) propanenitrile (iii) ethylamine.[6]
Show answer
(a) (i) NaOH(aq), heat under reflux (ii) KCN in ethanol, heat under reflux (iii) excess concentrated NH₃ in ethanol, heat in a sealed tube (under pressure)
1 mark for each reagent and 1 mark for each set of conditions.
Q18Reactions of halogenoalkanes (reagents & conditions) · Standard · Paper 1 style · 1 mark
(a)1-bromopropane is heated under reflux with KCN in ethanol. What is the organic product?[1]
Show answer
(a) B butanenitrile
CN⁻ replaces Br, adding one carbon: CH₃CH₂CH₂CN is butanenitrile. A forgets that the nitrile carbon is counted in the name.
Q19Reactions of halogenoalkanes (reagents & conditions) · Standard · Paper 2 style · 5 marks
(a)Give reagents and conditions for each step of a two-step synthesis of butanoic acid from 1-bromopropane, naming the intermediate.[4]
(b)Explain why this route is useful in synthesis.[1]
Show answer
(a) Step 1: KCN in ethanol, heat under reflux → butanenitrile; Step 2: dilute HCl(aq) (or H₂SO₄), heat under reflux → butanoic acid.
(b) It increases the length of the carbon chain by one carbon atom.
B1.
Q20Reactions of halogenoalkanes (reagents & conditions) · Challenge · Paper 2 style · 5 marks
(a)Write the equation for the reaction of 1-chloropropane with excess ammonia to form propylamine.[1]
(b)Explain why ammonia is used in large excess.[2]
(c)Explain why the reaction is carried out in a sealed tube.[1]
(d)Give the structural formula of the secondary amine that could form as a by-product.[1]
Show answer
(a) CH₃CH₂CH₂Cl + 2NH₃ → CH₃CH₂CH₂NH₂ + NH₄Cl
B1.
(b) Propylamine has a lone pair and is itself a nucleophile, so it can react with more 1-chloropropane to form secondary and tertiary amines (and quaternary salts); excess NH₃ makes collision of the halogenoalkane with NH₃ more likely.
(c) Ammonia is volatile; the sealed tube keeps it in (under pressure) when heated.
B1.
(d) (CH₃CH₂CH₂)₂NH
B1.
Q21Nucleophilic substitution mechanism · Core · Paper 2 style · 4 marks
(a)Define the term nucleophile.[1]
(b)From the list, identify all the nucleophiles: OH⁻, H⁺, CN⁻, Br₂, NH₃, NO₂⁺, H₂O, AlCl₃[2]
(c)Explain why the carbon atom bonded to the halogen in a halogenoalkane is attacked by nucleophiles.[1]
Show answer
(a) A species that donates a lone pair of electrons (to an electron-deficient carbon) to form a new covalent bond.
B1.
(b) OH⁻, CN⁻, NH₃, H₂O
B2 all four and no extras (B1 three correct).
(c) The halogen is more electronegative than carbon, so the C–X bond is polar and the carbon is δ+.
B1.
Q22Nucleophilic substitution mechanism · Standard · Paper 2 style · 5 marks
(a)Describe, with curly arrows described in words, the SN1 mechanism for the hydrolysis of 2-chloro-2-methylpropane by water.[4]
(b)Identify the rate-determining step.[1]
Show answer
(a) Step 1: C–Cl bond breaks heterolytically (arrow from C–Cl bond to Cl) to give (CH₃)₃C⁺ and Cl⁻; Step 2: a lone pair on the O of H₂O attacks the carbocation (arrow from O lone pair to C⁺); Step 3: loss of H⁺ from the –OH₂⁺ group gives (CH₃)₃COH.
B1 heterolytic fission arrow; B1 carbocation and Cl⁻; B1 arrow from O lone pair to C⁺; B1 loss of H⁺ to give 2-methylpropan-2-ol.
(b) Step 1 – formation of the carbocation.
B1.
Q23Nucleophilic substitution mechanism · Standard · Paper 1 style · 1 mark
(a)Bromomethane reacts with hydroxide ions by the SN2 mechanism. Which statement describes the transition state?[1]
Show answer
(a) B It has a partial C–O bond and a partial C–Br bond and carries a negative charge.
In SN2, bond-making and bond-breaking happen together: [HO···CH₃···Br]⁻ with OH and Br opposite each other. A describes the SN1 intermediate, not a transition state.
Q24Nucleophilic substitution mechanism · Challenge · Paper 2 style · 5 marks
(a)Describe the mechanism of the reaction between bromoethane and ammonia to form ethylamine. Describe each curly arrow in words and show the intermediate ion.[4]
(b)Name the type of mechanism.[1]
Show answer
(a) Lone pair on N of NH₃ attacks δ+ C of C–Br (arrow N lone pair → C); C–Br bond breaks (arrow C–Br → Br), giving CH₃CH₂NH₃⁺ and Br⁻; a second NH₃ uses its lone pair to remove H⁺ from CH₃CH₂NH₃⁺ (arrow from N lone pair to H, arrow from N–H bond to N) giving CH₃CH₂NH₂ and NH₄⁺.
B1 dipole and attack by N lone pair; B1 C–Br breaks, Br⁻; B1 intermediate CH₃CH₂NH₃⁺; B1 deprotonation by second NH₃.
(b) Nucleophilic substitution (SN2)
B1.
Q25Nucleophilic substitution (reactions & products) · Core · Paper 2 style · 4 marks
(a)Name the organic product formed when 1-iodopropane is heated with each reagent: (i) NaOH(aq) (ii) KCN in ethanol (iii) excess NH₃ in ethanol in a sealed tube.[3]
(b)Give the formula of the inorganic product in (i).[1]
Show answer
(a) (i) propan-1-ol (ii) butanenitrile (iii) propylamine (propan-1-amine)
1 mark each.
(b) NaI
B1.
Q26Nucleophilic substitution (reactions & products) · Core · Paper 1 style · 1 mark
(a)What is the organic product when 2-bromobutane is warmed with dilute aqueous sodium hydroxide?[1]
Show answer
(a) B butan-2-ol
OH⁻ replaces Br on the same carbon (C2), giving butan-2-ol. C would need ethanolic NaOH (elimination).
Q27Nucleophilic substitution (reactions & products) · Challenge · Paper 2 style · 5 marks
5.48 g of 1-bromobutane is heated under reflux with excess aqueous sodium hydroxide. 2.22 g of pure butan-1-ol is obtained.
(a)Calculate the percentage yield of butan-1-ol. [Mr: C₄H₉Br 136.9; C₄H₉OH 74.0][2]
(b)A different isomer, P, of C₄H₉Br is hydrolysed to an alcohol Q. Q does not change the colour of hot acidified potassium dichromate(VI). Identify P and Q.[2]
(c)State the colour change seen if butan-1-ol is warmed with acidified potassium dichromate(VI).[1]
M1 Q is a tertiary alcohol (not oxidised); A1 both names/structures.
(c) Orange to green
B1.
Q28Preparation of halogenoalkanes · Core · Paper 2 style · 6 marks
(a)Give a reagent for converting propan-2-ol into 2-chloropropane, and write the equation.[2]
(b)Give the reagent and conditions for making bromoethane from (i) ethene (ii) ethane.[3]
(c)Suggest why method (ii) is not a good way to make pure bromoethane.[1]
Show answer
(a) PCl₅ (or SOCl₂, or PCl₃ with heat, or conc HCl with ZnCl₂): CH₃CH(OH)CH₃ + PCl₅ → CH₃CHClCH₃ + POCl₃ + HCl
B1 reagent; B1 matching balanced equation.
(b) (i) HBr(g), room temperature (ii) Br₂, ultraviolet light
B1 HBr; B1 Br₂; B1 UV light.
(c) Free-radical substitution gives a mixture of multi-substituted products (e.g. dibromoethanes).
B1.
Q29Preparation of halogenoalkanes · Standard · Paper 5 style · 6 marks
2-chloro-2-methylpropane is prepared by shaking 7.40 g of 2-methylpropan-2-ol with concentrated hydrochloric acid in a separating funnel at room temperature. The upper organic layer is washed with sodium hydrogencarbonate solution, then with water, dried with anhydrous sodium sulfate and distilled. 6.94 g of product is collected.
(a)Explain the purpose of shaking with sodium hydrogencarbonate solution and how the pressure in the funnel is controlled.[2]
(b)Explain why anhydrous sodium sulfate is added.[1]
(c)Calculate the percentage yield. [Mr: (CH₃)₃COH 74.0; (CH₃)₃CCl 92.5][2]
(d)Suggest one safety precaution specific to this experiment.[1]
Show answer
(a) Removes/neutralises the remaining HCl acid; CO₂ is produced so the tap must be opened regularly (with the funnel inverted) to release pressure.
B1 neutralise acid; B1 release CO₂ pressure.
(b) It is a drying agent that removes traces of water from the organic layer.
