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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.
Q1Phenylamine & azo dyes · Core · Paper 4 style · 5 marks
(a)Give the reagents and conditions for the two stages needed to convert nitrobenzene into phenylamine.[3]
(b)Phenylamine is shaken with bromine water at room temperature. Describe what is seen and name the organic product.[2]
Show answer
(a) Stage 1: tin and concentrated hydrochloric acid, heat under reflux. Stage 2: add aqueous sodium hydroxide (excess)
B1 Sn + conc HCl. B1 heat/reflux. B1 then NaOH(aq) to release the free amine from its salt.
(b) orange/brown bromine water decolourised and a white precipitate forms; 2,4,6-tribromophenylamine
B1 decolourises + white ppt. B1 name.
Q2Phenylamine & azo dyes · Core · Paper 4 style · 5 marks
(a)Phenylamine reacts with nitrous acid to form benzenediazonium chloride. State the reagents used to produce nitrous acid in the flask and the temperature needed.[2]
(b)Give the formula of the benzenediazonium ion.[1]
(c)Explain why the temperature must be kept low, and write an equation for what happens if the solution is warmed.[2]
Show answer
(a) sodium nitrite (NaNO₂) and dilute hydrochloric acid; below 10 °C (ice bath)
B1 NaNO₂ + HCl. B1 temperature below 10 °C (accept 0–10 °C).
(b) C₆H₅N₂⁺ (C₆H₅–N≡N⁺)
B1.
(c) Diazonium ion decomposes above about 10 °C; C₆H₅N₂⁺ + H₂O → C₆H₅OH + N₂ + H⁺
B1 decomposes / unstable at higher temperature. B1 equation (phenol and nitrogen formed).
Q3Phenylamine & azo dyes · Standard · Paper 4 style · 6 marks
(a)A solution of benzenediazonium chloride is added to a cold solution of phenol in aqueous sodium hydroxide. A yellow-orange precipitate forms. Give the structural formula of the product and name the type of reaction.[3]
(b)Explain why the product is coloured.[2]
(c)Suggest one use of compounds of this type.[1]
Show answer
(a) 4-hydroxyazobenzene: C₆H₅–N=N–C₆H₄–OH with the OH para (4-position) to the azo group; electrophilic substitution (coupling)
B1 –N=N– linking two benzene rings. B1 OH in 4-position. B1 electrophilic substitution / coupling.
(b) The –N=N– group links two benzene rings so there is an extended delocalised system; it absorbs light in the visible region
B1 extended delocalisation / conjugation. B1 absorbs some visible wavelengths; the colour seen is the light not absorbed.
(c) dyes (for fabric, e.g. cotton in Faisalabad mills) / food colouring / indicators
B1.
Q4Phenylamine & azo dyes · Standard · Paper 4 style · 4 marks
(a)Write an equation for the reduction of nitrobenzene to phenylamine using [H] to represent the reducing agent.[1]
(b)12.3 g of nitrobenzene gave 6.51 g of pure phenylamine. Calculate the percentage yield. Give your answer to 3 significant figures. (Mr: C₆H₅NO₂ 123.0, C₆H₅NH₂ 93.0)[3]
Q5Phenylamine & azo dyes · Standard · Paper 4 style · 6 marks
(a)Explain why phenylamine reacts with aqueous bromine at room temperature without a catalyst, whereas benzene does not react.[3]
(b)Write an equation for the reaction and calculate the maximum mass of organic product from 2.79 g of phenylamine. Give your answer to 3 significant figures. (Mr: C₆H₅NH₂ 93.0, product 329.7)[3]
Show answer
(a) The lone pair on N is delocalised into the ring; this increases the electron density of the ring (especially at 2, 4, 6 positions); the ring is more easily attacked by electrophiles / Br₂ is polarised
B1 lone pair on N overlaps with / delocalised into π system. B1 increased electron density in ring. B1 more susceptible to electrophilic attack, so no halogen carrier needed.
(b) C₆H₅NH₂ + 3Br₂ → C₆H₂Br₃NH₂ + 3HBr; 9.89 g
B1 equation. M1 n = 2.79 ÷ 93.0 = 0.0300 mol. A1 × 329.7 = 9.89 g.
Q6Phenylamine & azo dyes · Challenge · Paper 4 style · 7 marks
(a)Outline a four-step synthesis of the azo dye formed between benzene-derived diazonium ions and naphthalen-2-ol, starting from benzene. For each step give the reagents and conditions and the organic product.[6]
(b)Suggest why naphthalen-2-ol is dissolved in aqueous sodium hydroxide before coupling.[1]
Show answer
(a) 1 benzene → nitrobenzene: conc HNO₃ + conc H₂SO₄, below 55 °C. 2 nitrobenzene → phenylamine: Sn + conc HCl, reflux, then NaOH(aq). 3 phenylamine → benzenediazonium chloride: NaNO₂ + HCl, below 10 °C. 4 couple with naphthalen-2-ol in aqueous NaOH, below 10 °C → azo dye
B1 for each correct reagent set (4) and B1 for correct temperatures in steps 1 and 3; B1 all intermediates named correctly. Max 6.
(b) To form the naphthoxide (phenoxide-type) ion, which is more strongly activated / soluble in water
B1 the anion makes the ring more electron-rich and dissolves the reagent.
Q7Phenylamine & azo dyes · Challenge · Paper 4 style · 6 marks
(a)4-nitrophenylamine is diazotised and the product is coupled with phenol in alkaline solution. Give the structural formula and the molecular formula of the azo compound formed.[2]
(b)Calculate the Mr of the azo compound.[1]
(c)A dye manufacturer wants to make the compound HO–C₆H₄–N=N–C₆H₄–CH₃ (both substituents in 4-positions). Identify the amine and the coupling agent that should be used, and explain why the coupling agent must contain an OH or NH₂ group.[3]
Show answer
(a) O₂N–C₆H₄–N=N–C₆H₄–OH (both groups at 4-positions); C₁₂H₉N₃O₃
(c) amine: 4-methylphenylamine (diazotised); coupling agent: phenol. The OH group activates the ring so the weakly electrophilic diazonium ion can substitute it
B1 4-methylphenylamine. B1 phenol. B1 diazonium ion is a weak electrophile; it reacts only with activated rings.
Q8Amino acids, peptides & electrophoresis · Core · Paper 4 style · 3 marks
(a)Glycine is H₂NCH₂COOH. Give the structure of the zwitterion of glycine.[1]
(b)Glycine is a solid that melts (with decomposition) above 230 °C and dissolves well in water. Explain these properties.[2]
Show answer
(a) H₃N⁺CH₂COO⁻
B1.
(b) Zwitterions are held by strong ionic attractions in the lattice; ions are strongly hydrated by water molecules
B1 strong electrostatic attraction between zwitterions, so high melting point. B1 ion–dipole interactions with water / hydrogen bonding, so soluble.
Q9Amino acids, peptides & electrophoresis · Core · Paper 4 style · 4 marks
(a)Alanine is CH₃CH(NH₂)COOH. Give the structure of the main form of alanine present in (i) a solution at pH 1 and (ii) a solution at pH 13.[2]
(b)Define the term isoelectric point.[1]
(c)Explain why alanine is optically active but glycine is not.[1]
Show answer
(a) (i) CH₃CH(NH₃⁺)COOH; (ii) CH₃CH(NH₂)COO⁻
B1 each.
(b) The pH at which the amino acid exists as the zwitterion with no overall charge (does not move in an electric field)
B1.
(c) Alanine has a chiral carbon (four different groups: H, CH₃, NH₂, COOH); glycine has two H on its central carbon
B1.
Q10Amino acids, peptides & electrophoresis · Standard · Paper 4 style · 5 marks
(a)Glycine and alanine can combine to form two different dipeptides. Give the structural formula of each.[2]
(b)Name the linkage formed and the type of reaction.[2]
(c)Calculate the Mr of either dipeptide. (Mr: glycine 75.0, alanine 89.0, water 18.0)[1]
Show answer
(a) H₂NCH₂CONHCH(CH₃)COOH (Gly-Ala) and H₂NCH(CH₃)CONHCH₂COOH (Ala-Gly)
B1 each.
(b) peptide (amide) link –CONH–; condensation
B1 each.
(c) 146.0
B1 75.0 + 89.0 − 18.0 = 146.0.
Q11Amino acids, peptides & electrophoresis · Standard · Paper 4 style · 4 marks
(a)A mixture of lysine (isoelectric point 9.7), glycine (isoelectric point 6.0) and aspartic acid (isoelectric point 2.8) is placed in the centre of a strip of paper soaked in a buffer at pH 6.0. A voltage is applied. Predict and explain the movement of each amino acid.[3]
(b)Suggest how the spots can be made visible after electrophoresis.[1]
Show answer
(a) lysine is positively charged at pH 6.0 and moves to the cathode (negative electrode); glycine is at its isoelectric point – zwitterion – does not move; aspartic acid is negatively charged and moves to the anode
B1 each, with reason based on pH relative to isoelectric point.
(b) spray with ninhydrin (or a locating agent) and warm
B1 locating agent, e.g. ninhydrin.
Q12Amino acids, peptides & electrophoresis · Standard · Paper 4 style · 5 marks
(a)A tripeptide has the sequence Ala-Gly-Ser. State the reagents and conditions for hydrolysing it completely.[2]
(b)Serine is HOCH₂CH(NH₂)COOH. Give the structures of the three species formed when the hydrolysis is carried out in excess acid.[3]
Show answer
(a) 6 mol dm⁻³ hydrochloric acid, heat under reflux for several hours
B1 HCl(aq) (accept NaOH(aq)). B1 heat/reflux for a prolonged time.
Q13Amino acids, peptides & electrophoresis · Challenge · Paper 4 style · 5 marks
(a)Electrophoresis is carried out at pH 1.0 on a mixture of glycine, the dipeptide Gly-Gly and lysine (which has two NH₂ groups). State the charge carried by each species at pH 1.0.[2]
(b)Predict the order in which the three species are found from furthest to nearest from the starting point, towards the cathode. Explain your answer.[3]
Show answer
(a) glycine +1; Gly-Gly +1; lysine +2
B1 glycine and Gly-Gly +1 (amino group protonated, COOH un-ionised at pH 1.0). B1 lysine +2 (both amino groups protonated).
B1 order. B1 lysine has the largest charge (charge/mass ratio highest). B1 glycine and Gly-Gly have the same charge but glycine is smaller, so moves further.
Q14Amino acids, peptides & electrophoresis · Challenge · Paper 4 style · 6 marks
(a)A pentapeptide has the sequence Gly-Ala-Val-Ala-Gly. Calculate its Mr. (Mr: glycine 75.0, alanine 89.0, valine 117.0, water 18.0)[2]
(b)State the number of chiral centres in the pentapeptide and explain your answer.[2]
(c)Partial hydrolysis gives a mixture of dipeptides. List all the different dipeptides that could form.[2]
Show answer
(a) 373.0
M1 four peptide bonds so 4 H₂O lost. A1 2(75.0) + 2(89.0) + 117.0 − 72.0 = 373.0.
(b) 3; each alanine and the valine residue has a chiral carbon, glycine does not
B1 3. B1 glycine's α-carbon carries two H atoms.
(c) Gly-Ala, Ala-Val, Val-Ala, Ala-Gly
B1 for three correct. B1 all four with no extras.
Q15Amines: basicity & preparation · Core · Paper 4 style · 4 marks
(a)Place ammonia, ethylamine and phenylamine in order of increasing base strength.[1]
(b)Explain your order.[3]
Show answer
(a) phenylamine < ammonia < ethylamine
B1.
(b) Ethyl group is electron-donating so the lone pair on N is more available to accept H⁺; in phenylamine the lone pair is delocalised into the benzene ring, so it is less available
B1 base strength depends on availability of N lone pair to accept a proton. B1 alkyl group pushes electron density onto N. B1 lone pair delocalised into ring in phenylamine.
Q16Amines: basicity & preparation · Core · Paper 2 style · 4 marks
(a)Give the reagent and conditions for making ethylamine from bromoethane.[2]
(b)Write an equation for the reaction.[1]
(c)Explain why ammonia is used in excess.[1]
Show answer
(a) excess concentrated ammonia in ethanol; heat in a sealed tube (under pressure)
B1 NH₃ in ethanol. B1 heat in sealed tube / under pressure.
(b) CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br
B1 (accept CH₃CH₂Br + NH₃ → CH₃CH₂NH₂ + HBr).
(c) To reduce further substitution – ethylamine can itself react with bromoethane to form secondary and tertiary amines
B1.
Q17Amines: basicity & preparation · Standard · Paper 4 style · 4 marks
(a)Give the reagents and conditions for reducing propanenitrile, CH₃CH₂CN, to an amine and give the structure of the amine.[2]
(b)N-methylethanamide, CH₃CONHCH₃, is reduced by LiAlH₄. Give the structure of the product and state whether it is a primary, secondary or tertiary amine.[2]
Show answer
(a) LiAlH₄ in dry ether (or H₂ with Ni catalyst / heat); CH₃CH₂CH₂NH₂ (propylamine)
B1 reagent and conditions. B1 product.
(b) CH₃CH₂NHCH₃; secondary amine
B1 structure. B1 secondary.
Q18Amines: basicity & preparation · Standard · Paper 4 style · 5 marks
(a)Write an equation to show why an aqueous solution of ethylamine is alkaline.[1]
(b)Ethylamine reacts with hydrochloric acid. Write an equation and name the product.[2]
(c)Explain why the salt formed in (b) has no fishy smell, while ethylamine does. Then explain how ethylamine could be recovered from the salt.[2]
(b) Nitrile reduction gives only the primary amine; with ammonia, the amine product is a nucleophile and reacts further with the halogenoalkane to give secondary/tertiary amines and quaternary salts
B1 no further substitution in the nitrile route. B1 mixture of amines in the ammonia route.
Q20Amines: basicity & preparation · Challenge · Paper 4 style · 6 marks
(a)When 1-bromopropane is heated with ammonia, a mixture of products forms. Give the structures of the primary, secondary and tertiary amines and the quaternary ammonium salt that can form.[4]
(b)Calculate the Mr of the quaternary ammonium salt.[1]
(c)Explain why the quaternary ammonium ion cannot act as a base.[1]
(a)Explain why ethanamide, CH₃CONH₂, is a much weaker base than ethylamine, CH₃CH₂NH₂.[3]
(b)Explain why ethanamide is a solid at room temperature, whereas ethanoyl chloride is a liquid.[2]
Show answer
(a) In ethanamide the lone pair on N is delocalised over the C=O group (the electronegative O withdraws electron density); the lone pair is much less available to accept a proton; in ethylamine the ethyl group donates electrons
B1 lone pair delocalised with the C=O π system. B1 so less available to bond to H⁺. B1 ethyl group electron-donating in ethylamine.
(b) Ethanamide molecules form hydrogen bonds (N–H…O=C); ethanoyl chloride molecules only have permanent dipole and London forces
B1 hydrogen bonding in amide. B1 weaker forces in acyl chloride.
Q24Amides · Standard · Paper 4 style · 4 marks
(a)N-methylpropanamide, CH₃CH₂CONHCH₃, is reduced by LiAlH₄ in dry ether. Give the structure of the product.[1]
(b)4.35 g of N-methylpropanamide (Mr 87.0) gave 2.92 g of the amine (Mr 73.0). Calculate the percentage yield. Give your answer to 3 significant figures.[2]
(c)Suggest how the amide could be made from an acyl chloride.[1]
Show answer
(a) CH₃CH₂CH₂NHCH₃
B1.
(b) 80.0 %
M1 n = 4.35 ÷ 87.0 = 0.0500 mol; theoretical = 3.65 g. A1 80.0 %.
(c) propanoyl chloride + methylamine
B1 CH₃CH₂COCl + CH₃NH₂.
Q25Amides · Challenge · Paper 4 style · 6 marks
(a)Paracetamol, HO–C₆H₄–NHCOCH₃ (substituents 1,4), is manufactured from 4-aminophenol. Name the two functional groups in paracetamol.[2]
(b)Industry uses ethanoic anhydride: HO–C₆H₄–NH₂ + (CH₃CO)₂O → HO–C₆H₄–NHCOCH₃ + CH₃COOH. Calculate the atom economy of this reaction. (Mr: 4-aminophenol 109.0, ethanoic anhydride 102.0, paracetamol 151.0)[2]
(c)Paracetamol is heated with excess aqueous sodium hydroxide. Give the structures of the organic products.[2]
Show answer
(a) phenol (OH on ring) and amide (secondary amide –NHCO–)
(c) ⁻O–C₆H₄–NH₂ (sodium 4-aminophenoxide) and CH₃COO⁻Na⁺
B1 amide hydrolysed to ethanoate. B1 4-aminophenoxide (phenol deprotonated by excess NaOH).
Q26Amides · Challenge · Paper 4 style · 6 marks
(a)Three isomers, J, K and L, have the formula C₃H₇NO and are all amides. Each is heated with aqueous NaOH.J gives sodium propanoate and ammonia.K gives sodium ethanoate and methylamine.L gives sodium methanoate and dimethylamine, (CH₃)₂NH.Give the structures of J, K and L.[3]
(b)Classify J, K and L as primary, secondary or tertiary amides.[1]
(c)Only one of the three can be made from an acyl chloride and ammonia. Identify it and write the equation.[2]
Show answer
(a) J CH₃CH₂CONH₂; K CH₃CONHCH₃; L HCON(CH₃)₂
B1 each.
(b) J primary; K secondary; L tertiary
B1 all three.
(c) J: CH₃CH₂COCl + NH₃ → CH₃CH₂CONH₂ + HCl
B1 J. B1 equation.
Q27Amines & nitriles · Core · Paper 1 style · 1 mark
(a)Which reagent and conditions convert bromoethane into propanenitrile?[1]
Show answer
(a) A KCN in ethanol, heat under reflux
Cyanide ions substitute the bromine when KCN in ethanol is heated under reflux. D gives ethylamine, not a nitrile; B (HCN) is used for addition to carbonyl compounds.
Q28Amines & nitriles · Core · Paper 2 style · 4 marks
(a)Propanal reacts with HCN in the presence of a small amount of KCN. Give the structure and name of the product.[2]
(b)Name the mechanism of this reaction.[1]
(c)The product of (a) is heated under reflux with dilute hydrochloric acid. Give the structure of the organic product.[1]
Show answer
(a) CH₃CH₂CH(OH)CN; 2-hydroxybutanenitrile
B1 structure. B1 name.
(b) nucleophilic addition
B1.
(c) CH₃CH₂CH(OH)COOH (2-hydroxybutanoic acid)
B1.
Q29Amines & nitriles · Standard · Paper 1 style · 1 mark
(a)2-bromopropane is heated under reflux with KCN in ethanol. The organic product is then heated under reflux with dilute sulfuric acid. What is the final organic product?[1]
Show answer
(a) B 2-methylpropanoic acid
CH₃CHBrCH₃ → (CH₃)₂CHCN (2-methylpropanenitrile) → (CH₃)₂CHCOOH, 2-methylpropanoic acid. A assumes a straight chain, but the CN group attaches to the middle carbon.
Q30Amines & nitriles · Challenge · Paper 2 style · 6 marks
(a)Lactic acid, CH₃CH(OH)COOH, can be made from ethanal in two steps. Give the reagents and conditions for each step and the intermediate.[3]
(b)Lactic acid made this way is not optically active, although its molecules are chiral. Explain why.[3]
Show answer
(a) 1 HCN with KCN catalyst (about 10–20 °C) → CH₃CH(OH)CN (2-hydroxypropanenitrile); 2 dilute HCl or H₂SO₄, heat under reflux → lactic acid
B1 step 1. B1 intermediate. B1 step 2.
(b) The C=O group in ethanal is planar; CN⁻ attacks equally from either side; a racemic mixture (50:50) of the two enantiomers forms, so the rotations cancel
B1 planar carbonyl group. B1 equal chance of attack from above and below. B1 racemic mixture – no net rotation of plane-polarised light.
Similar questions
Questions that test the same subtopic in different years, grouped so you can work on one skill at a time.
References only: find each question in its paper (official link where available, otherwise through your school).
