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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.
Q1Tests: 2,4-DNPH, Fehling's, Tollens' · Core · Paper 1 style · 1 mark
(a)Which compound gives an orange precipitate with 2,4-dinitrophenylhydrazine but no change when warmed with Tollens' reagent?[1]
Show answer
(a) B propanone
Propanone is a ketone: it has C=O (positive 2,4-DNPH) but cannot be oxidised by Tollens'. Propanal (A) gives both a precipitate and a silver mirror.
Q2Tests: 2,4-DNPH, Fehling's, Tollens' · Core · Paper 2 style · 4 marks
A technician at a school in Rawalpindi finds two unlabelled bottles; one contains propanal and the other propanone.
(a)Describe how Tollens' reagent can be used to distinguish between propanal and propanone. Give the condition and the observation with each compound.[3]
(b)Write an equation, using [O], for the oxidation of propanal.[1]
Show answer
(a) Warm each with Tollens' reagent (in a water bath). Propanal: silver mirror (grey/black precipitate). Propanone: no change (stays colourless).
B1 warm. B1 propanal silver mirror. B1 propanone no change.
(b) CH₃CH₂CHO + [O] → CH₃CH₂COOH
B1
Q3Tests: 2,4-DNPH, Fehling's, Tollens' · Core · Paper 3 style · 6 marks
Four liquids W, X, Y and Z are ethanol, ethanal, propanone and ethanoic acid (in some order). A student carried out four tests:test | W | X | Y | Z2,4-DNPH | no change | orange ppt | orange ppt | no changeFehling's, warm | stays blue | red-brown ppt | stays blue | stays blueNa₂CO₃(aq) | no change | no change | no change | effervescenceacidified K₂Cr₂O₇, warm | orange to green | orange to green | stays orange | stays orange
(a)Identify W, X, Y and Z.[4]
(b)Explain why X, but not Y, turns acidified dichromate green.[1]
(c)Name the gas given off when Z reacts with sodium carbonate and give a test for it.[1]
Show answer
(a) W ethanol; X ethanal; Y propanone; Z ethanoic acid
1 mark each.
(b) X is an aldehyde and is oxidised (to ethanoic acid); Y is a ketone and is not oxidised.
B1
(c) Carbon dioxide: turns limewater milky.
B1 both needed.
Q4Tests: 2,4-DNPH, Fehling's, Tollens' · Standard · Paper 1 style · 1 mark
(a)[1]
Show answer
(a) B 2
Only aldehydes give a silver mirror: butanal, CH₃CH₂CH₂CHO, and 2-methylpropanal, (CH₃)₂CHCHO. Butanone is also C₄H₈O but is a ketone, so 3 (C) is wrong.
Q5Tests: 2,4-DNPH, Fehling's, Tollens' · Standard · Paper 2 style · 5 marks
Unknown carbonyl compound G gives an orange precipitate with 2,4-DNPH. After recrystallisation the precipitate melts at 124 °C. G also gives a red-brown precipitate with warm Fehling's solution.Melting points of 2,4-DNPH derivatives from a data book:compound | melting point / °Cpropanal | 156propanone | 126butanal | 123butanone | 117
(a)Name the type of reaction between a carbonyl compound and 2,4-DNPH.[1]
(b)Explain why the orange solid is recrystallised before its melting point is measured.[1]
(c)Identify G, giving your reasoning.[3]
Show answer
(a) Condensation (addition–elimination)
B1
(b) To purify it; impurities lower the melting point (and widen its range).
B1
(c) G is butanal. 124 °C is close to the values for both butanal (123 °C) and propanone (126 °C), but the positive Fehling's test shows that G is an aldehyde, not a ketone.
M1 melting point matches butanal or propanone. M1 Fehling's positive shows aldehyde. A1 butanal.
Q6Tests: 2,4-DNPH, Fehling's, Tollens' · Standard · Paper 2 style · 5 marks
Fehling's solution is a deep blue alkaline solution containing a complex of copper.
(a)Name the metal ion in Fehling's solution that acts as the oxidising agent, and give the formula of the red-brown precipitate.[2]
(b)Give the change in oxidation number of copper.[1]
(c)Glucose has an aldehyde group but sucrose has neither an aldehyde nor a ketone group. Predict what is seen when each is warmed with Fehling's solution.[2]
Show answer
(a) Copper(II), Cu²⁺; copper(I) oxide, Cu₂O
B1 each.
(b) +2 → +1
B1
(c) Glucose: blue solution gives a red-brown precipitate. Sucrose: remains a blue solution.
B1 each.
Q7Tests: 2,4-DNPH, Fehling's, Tollens' · Standard · Paper 3 style · 6 marks
(a)Describe how Tollens' reagent is prepared from aqueous silver nitrate.[2]
(b)Explain why the tubes are warmed in a beaker of hot water rather than over a Bunsen flame.[1]
(c)Give the formula of the silver complex ion in Tollens' reagent, and the change in oxidation number of silver in a positive test.[2]
(d)Suggest why the reagent is made fresh and any left over is washed away at once.[1]
Show answer
(a) Add a few drops of aqueous sodium hydroxide to give a brown precipitate (Ag₂O), then add dilute aqueous ammonia dropwise until the precipitate just dissolves.
B1 NaOH gives brown ppt. B1 aqueous ammonia until it dissolves.
(b) Carbonyl compounds are volatile and flammable; a water bath also gives gentle, even heating.
B1
(c) [Ag(NH₃)₂]⁺; +1 → 0
B1 each.
(d) On standing it can form explosive silver compounds.
B1
Q8Tests: 2,4-DNPH, Fehling's, Tollens' · Challenge · Paper 1 style · 1 mark
(a)Compound Y, C₅H₁₀O, gives an orange precipitate with 2,4-DNPH and does not react with Tollens' reagent. Reduction of Y with NaBH₄ gives an alcohol with no chiral centre. What is Y?[1]
Show answer
(a) B pentan-3-one
Y is a ketone (D is an aldehyde). Pentan-3-one gives pentan-3-ol, CH₃CH₂CH(OH)CH₂CH₃, whose C-3 carries two identical ethyl groups, so it is not chiral. A and C give pentan-2-ol and 3-methylbutan-2-ol, which are both chiral.
Q9Tests: 2,4-DNPH, Fehling's, Tollens' · Challenge · Paper 2 style · 6 marks
R and S are isomers with molecular formula C₄H₈O. Both give an orange precipitate with 2,4-DNPH. R gives a silver mirror with Tollens' reagent and has a branched carbon chain. S does not react with Tollens' reagent.
(a)Identify R and give its displayed or structural formula.[2]
(b)Identify S.[1]
(c)Give the structural formula of the organic product formed when R reacts with Tollens' reagent and the mixture is then acidified.[1]
(d)Give the structural formula of the product when S is reduced by NaBH₄, and state whether it is chiral.[2]
(d) CH₃CH(OH)CH₂CH₃, butan-2-ol; yes, C-2 is chiral
B1 product. B1 chiral.
Q10Tests: 2,4-DNPH, Fehling's, Tollens' · Challenge · Paper 2 style · 5 marks
(a)Each mole of aldehyde reduces two moles of Ag⁺. Calculate the maximum mass of silver formed from 0.580 g of propanal with excess Tollens' reagent. (Mr propanal = 58.0, Ar Ag = 107.9)[2]
(b)A 1.00 g mixture of propanal and propanone gives 1.62 g of silver with excess Tollens' reagent. Calculate the percentage by mass of propanal in the mixture.[3]
Show answer
(a) 2.16 g
M1 mol propanal = 0.0100; mol Ag = 0.0200. A1 × 107.9 = 2.16 g
(b) 43.5 %
M1 mol Ag = 1.62 ÷ 107.9 = 0.01501. M1 mol propanal = ÷ 2 = 0.00751; mass = 0.4354 g. A1 43.5 % (propanone does not react)
Q11Nucleophilic addition (HCN) · Core · Paper 1 style · 1 mark
(a)What type of reaction occurs when ethanal reacts with HCN in the presence of KCN?[1]
Show answer
(a) B nucleophilic addition
CN⁻ (a nucleophile) attacks the δ+ carbonyl carbon and the C=O π bond breaks; nothing is replaced. A is the typical reaction of alkenes, not carbonyls.
Q12Nucleophilic addition (HCN) · Core · Paper 2 style · 6 marks
(a)Give the reagents and conditions for the reaction of propanal with HCN.[1]
(b)Describe the mechanism of this reaction. In your answer, say where each curly arrow starts and ends, and give the intermediate.[4]
(c)Name the product.[1]
Show answer
(a) HCN with a small amount of KCN (or NaCN), at 10–20 °C
B1
(b) Step 1: a curly arrow from the lone pair on the carbon of CN⁻ to the δ+ carbonyl carbon; a second arrow from the C=O π bond to the oxygen, forming the anion CH₃CH₂CH(O⁻)CN. Step 2: an arrow from a lone pair on O⁻ to the H of HCN, and an arrow from the H–C bond of HCN to its C, giving CH₃CH₂CH(OH)CN and regenerating CN⁻.
B1 arrow from CN⁻ lone pair (on C) to carbonyl C. B1 arrow from C=O bond to O (dipole on C=O shown). B1 correct intermediate with negative charge on O. B1 O⁻ takes H⁺ from HCN (or H₂O), regenerating CN⁻.
(c) 2-hydroxybutanenitrile
B1
Q13Nucleophilic addition (HCN) · Core · Paper 2 style · 5 marks
(a)Write the equation for the reaction of ethanal with HCN.[1]
(b)Calculate the Mr of the product and the percentage by mass of nitrogen in it.[2]
(c)Name the functional groups present in the product.[1]
Q14Nucleophilic addition (HCN) · Standard · Paper 1 style · 1 mark
(a)What is the name of the organic product when propanone reacts with HCN?[1]
Show answer
(a) B 2-hydroxy-2-methylpropanenitrile
(CH₃)₂C(OH)CN: the CN joins the former carbonyl carbon, giving a 4-carbon nitrile with OH and CH₃ on C-2. A is the product from propanal.
Q15Nucleophilic addition (HCN) · Standard · Paper 2 style · 5 marks
Lactic acid is used in the food industry. One laboratory route to it starts from ethanal.
(a)Give the structural formula of the product when ethanal reacts with HCN.[1]
(b)Give the reagent and condition to convert this product into lactic acid, CH₃CH(OH)COOH.[1]
(c)Lactic acid made in this way is optically inactive, although it has a chiral centre. Explain why.[3]
Show answer
(a) CH₃CH(OH)CN
B1
(b) Heat under reflux with dilute hydrochloric (or sulfuric) acid
B1
(c) The carbonyl group (and the C atom attacked) is planar; CN⁻ can attack from either side with equal probability; so equal amounts of both enantiomers form (a racemic mixture), whose effects on plane-polarised light cancel.
B1 planar C=O. B1 attack equally from above and below. B1 racemic mixture / 50:50 enantiomers cancel.
Q16Nucleophilic addition (HCN) · Standard · Paper 2 style · 5 marks
(a)HCN is a weak acid. Explain why a little KCN is added to the reaction mixture.[2]
(b)Explain why the carbon atom of the carbonyl group is attacked by nucleophiles.[1]
(c)Explain why alkenes do not undergo nucleophilic addition.[1]
(d)Give one safety precaution needed for this reaction.[1]
Show answer
(a) HCN is only slightly dissociated, so there are very few CN⁻ ions; KCN provides CN⁻ ions, which are the nucleophile needed to start the reaction.
B1 HCN gives few CN⁻. B1 CN⁻ is the nucleophile.
(b) Oxygen is more electronegative than carbon, so the C=O bond is polar and the carbon is δ+.
B1
(c) The C=C bond is non-polar and electron-rich, so it repels nucleophiles and attracts electrophiles instead.
B1
(d) Work in a fume cupboard because HCN is extremely toxic (or avoid acidifying KCN in the open laboratory).
B1
Q17Nucleophilic addition (HCN) · Challenge · Paper 1 style · 1 mark
(a)Which carbonyl compound reacts with HCN to give a product with no chiral centre?[1]
Show answer
(a) B propanone
(CH₃)₂C(OH)CN has two identical CH₃ groups on the central carbon. The products from A, C and D each have four different groups on that carbon (they form racemic mixtures).
Q18Nucleophilic addition (HCN) · Challenge · Paper 2 style · 6 marks
2-Hydroxy-2-methylpropanoic acid, (CH₃)₂C(OH)COOH, can be made from propan-2-ol in three steps:propan-2-ol → (step 1) → ketone → (step 2) → hydroxynitrile → (step 3) → product
(a)Give the reagents and conditions for each of the three steps and the structural formula of each intermediate.[4]
(b)The yields of the three steps are 80 %, 70 % and 90 %. Calculate the mass of 2-hydroxy-2-methylpropanoic acid (Mr 104.0) made from 6.00 g of propan-2-ol (Mr 60.0).[2]
Show answer
(a) Step 1: K₂Cr₂O₇ / dilute H₂SO₄, heat (reflux) → propanone, CH₃COCH₃. Step 2: HCN with KCN, 10–20 °C → (CH₃)₂C(OH)CN. Step 3: heat under reflux with dilute HCl(aq) → (CH₃)₂C(OH)COOH.
Q19Nucleophilic addition (HCN) · Challenge · Paper 2 style · 6 marks
(a)For each compound, give the structural formula of the product with HCN and state whether that product has a chiral centre: methanal, ethanal, pentan-3-one, butanone.[4]
(b)State the number of carbon atoms in the product from pentan-3-one and explain why this reaction is useful in synthesis.[2]
Show answer
(a) methanal → HOCH₂CN, not chiral; ethanal → CH₃CH(OH)CN, chiral; pentan-3-one → (CH₃CH₂)₂C(OH)CN, not chiral; butanone → CH₃CH₂C(CH₃)(OH)CN, chiral.
1 mark for each correct product with correct chirality statement.
(b) 6 carbon atoms; the reaction lengthens the carbon chain by one carbon atom.
B1 6. B1 increases chain length by one C.
Q20Tri-iodomethane test · Core · Paper 1 style · 1 mark
(a)Which compound gives a pale yellow precipitate when warmed with alkaline aqueous iodine?[1]
Show answer
(a) A propan-2-ol
Propan-2-ol has the CH₃CH(OH)– group, which is oxidised to CH₃CO– and then forms CHI₃. Pentan-3-one (C) is a ketone but has no CH₃CO– group.
Q21Tri-iodomethane test · Core · Paper 2 style · 6 marks
(a)Give the reagents and conditions for the tri-iodomethane (iodoform) test and the observation for a positive result.[2]
(b)State the two structural features that give a positive result.[1]
(c)State which of these give a positive result: ethanol, propanal, butanone, pentan-3-one, ethanal, methanol.[3]
Show answer
(a) Iodine and aqueous sodium hydroxide (alkaline I₂), warm; pale yellow precipitate (of CHI₃) with an antiseptic smell.
B1 reagents and warm. B1 pale yellow precipitate.
(b) CH₃CO– (methyl carbonyl) or CH₃CH(OH)– (secondary alcohol with an adjacent methyl group)
3 marks for all six correctly classified; 2 marks for four or five; 1 mark for two or three.
Q22Tri-iodomethane test · Core · Paper 3 style · 6 marks
A student tested four liquids P, Q, R and S, which are ethanal, propanone, propanal and propan-1-ol (in some order). For the iodine test, aqueous iodine was added to each sample, then aqueous NaOH dropwise until the brown colour just disappeared, and the tube was warmed in a 60 °C water bath.test | P | Q | R | S2,4-DNPH | orange ppt | orange ppt | orange ppt | no changeFehling's, warm | red-brown ppt | stays blue | red-brown ppt | stays bluealkaline iodine, warm | yellow ppt | yellow ppt | no change | no change
(a)Identify P, Q, R and S.[4]
(b)State the structural feature shown by a positive result in the alkaline iodine test.[1]
(c)State which single test in the table distinguishes P from Q.[1]
Show answer
(a) P ethanal; Q propanone; R propanal; S propan-1-ol
1 mark each.
(b) A CH₃CO– group (or a CH₃CH(OH)– group, which is oxidised to CH₃CO–)
B1
(c) Fehling's solution (P, an aldehyde, gives a red-brown ppt; Q does not)
B1
Q23Tri-iodomethane test · Standard · Paper 2 style · 5 marks
(a)Complete and balance the equation: CH₃COCH₃ + 3I₂ + 4NaOH → CHI₃ + ____ + 3NaI + 3H₂O[1]
(b)Calculate the maximum mass of CHI₃ formed from 0.290 g of propanone. (Mr propanone 58.0, Mr CHI₃ 393.7)[2]
(c)The student obtains 75.0 % of this mass after filtering and drying. Calculate the mass obtained and suggest one reason for the loss.[2]
Show answer
(a) CH₃COONa (sodium ethanoate)
B1
(b) 1.97 g
M1 mol = 0.00500. A1 × 393.7 = 1.97 g
(c) 1.48 g; e.g. some solid left on the filter paper/in the tube, or the reaction was incomplete.
B1 1.9685 × 0.750 = 1.48 g. B1 valid reason.
Q24Tri-iodomethane test · Standard · Paper 2 style · 5 marks
(a)Describe a chemical test that distinguishes pentan-2-one from pentan-3-one. Give the result with each.[2]
(b)Give the formulae of the two organic products formed in the positive test.[2]
(c)Explain why Tollens' reagent cannot be used to distinguish these two compounds.[1]
Show answer
(a) Warm with alkaline aqueous iodine: pentan-2-one gives a pale yellow precipitate; pentan-3-one gives no precipitate.
B1 reagent/condition. B1 both results.
(b) CHI₃ and CH₃CH₂CH₂COONa (sodium butanoate)
B1 each.
(c) Both are ketones, so neither is oxidised; both give no change.
B1
Q25Tri-iodomethane test · Standard · Paper 2 style · 4 marks
(a)There are four structural isomers of C₄H₁₀O that are alcohols. Identify the only one that gives a positive tri-iodomethane test and explain why.[2]
(b)Give the structural formula of the organic intermediate and the two final organic products.[2]
Show answer
(a) Butan-2-ol, CH₃CH(OH)CH₂CH₃: it has the CH₃CH(OH)– group, which alkaline iodine oxidises to CH₃CO–.
B1 butan-2-ol. B1 explanation.
(b) Intermediate butanone, CH₃COCH₂CH₃; products CHI₃ and CH₃CH₂COONa (sodium propanoate)
B1 butanone. B1 both products.
Q26Tri-iodomethane test · Standard · Paper 2 style · 4 marks
A 3.00 g sample of a solvent is a mixture of propan-1-ol and propan-2-ol. It is warmed with excess alkaline iodine. The dry tri-iodomethane formed has a mass of 5.91 g. Assume 1 mol of CHI₃ forms from each mole of the reacting alcohol.
(a)Explain why only one of the two alcohols reacts.[1]
(b)Calculate the percentage by mass of propan-2-ol in the mixture. (Mr propan-2-ol 60.0; Mr CHI₃ 393.7)[3]
Show answer
(a) Only propan-2-ol has the CH₃CH(OH)– group; propan-1-ol does not.
Q27Tri-iodomethane test · Challenge · Paper 2 style · 6 marks
T is an unbranched alcohol, C₅H₁₂O, with a chiral centre. T gives a pale yellow precipitate with alkaline aqueous iodine. Oxidation of T with acidified dichromate gives U, which gives an orange precipitate with 2,4-DNPH but no change with Fehling's solution. The alkaline iodine reaction of T gives CHI₃ and the sodium salt V of a carboxylic acid.
(a)Identify T, U and V, giving their structural formulae.[3]
(b)Explain how the information shows that U is a ketone and that T is a secondary alcohol.[2]
(c)Pentan-3-ol is also an unbranched secondary alcohol. Give two pieces of evidence from the data that rule it out.[1]
Show answer
(a) T pentan-2-ol, CH₃CH(OH)CH₂CH₂CH₃; U pentan-2-one, CH₃COCH₂CH₂CH₃; V CH₃CH₂CH₂COONa (sodium butanoate)
B1 each. T must be unbranched, chiral and have CH₃CH(OH)–.
(b) U gives an orange ppt with 2,4-DNPH (carbonyl) but does not reduce Fehling's (not an aldehyde); oxidation of an alcohol to a ketone shows T is secondary.
B1 U ketone reasoning. B1 T secondary.
(c) Pentan-3-ol has no chiral centre, and it does not give a positive tri-iodomethane test.
B1 both.
Q28Tri-iodomethane test · Challenge · Paper 2 style · 5 marks
The tri-iodomethane reaction can be used in synthesis to convert a methyl ketone into a carboxylic acid with one fewer carbon atom. Phenylethanone, C₆H₅COCH₃, is warmed with alkaline iodine, the CHI₃ is filtered off and the filtrate is acidified to give benzoic acid.
(a)Write the formula of the organic product formed in the alkaline iodine reaction before acidification.[1]
(b)Calculate the theoretical mass of benzoic acid, C₆H₅COOH (Mr 122.0), from 3.00 g of phenylethanone (Mr 120.0), and the percentage yield if 2.20 g is obtained.[3]
(c)Suggest how the benzoic acid can be purified.[1]
(c) CH₂=CHCH(OH)CH₃ (but-3-en-2-ol). NaBH₄ provides H⁻, a nucleophile, which attacks the δ+ carbon of the polar C=O; the non-polar C=C does not attract nucleophiles.
B1 product. B1 H⁻ nucleophile / C=C non-polar.
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).
