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Question 1 -- Four Unlabelled Bottles
Total: 12 marks
During a stock-take at a Delhi university, a technician finds four bottles whose labels have fallen off. Records show they contain hexane, hex-1-ene, ethanol and dilute ethanoic acid — but nobody knows which is which.
(a)[4]
Plan a sequence of simple chemical and physical tests that would identify all four liquids. For each test, state the result that identifies the liquid.
Model Answer -- 1(a)
Add aqueous bromine to a sample of each liquid: only hex-1-ene decolourises it from orange to colourless, because it is the only unsaturated compound [1]
Add sodium carbonate (or magnesium) to samples of the remaining three: only ethanoic acid gives effervescence, as carbon dioxide (or hydrogen) is released [1]
To separate the last two, add water: ethanol mixes completely with water, whereas hexane does not mix and floats as a separate layer [1]
Logical sequence that identifies all four liquids by elimination, with a correct result stated for each [1]
⚠ If you missed marks here: A plan must give the result as well as the reagent — "add bromine water" without saying which bottle decolourises it identifies nothing. Universal indicator alone cannot finish the job either: it separates the acid, but hexane, hexene and ethanol are all neutral, so a solubility or bromine test is still needed.
(b)[4]
A fifth bottle contains hydrocarbon X. X decolourises aqueous bromine, and its relative molecular mass is 56.
(i) Deduce the molecular formula of X, showing your working. (Ar: C = 12, H = 1) [2]
(ii) X could be one of several structural isomers. Name two straight-chain isomers of X and state how their structures differ. [2]
Model Answer -- 1(b)
(i) X decolourises bromine water, so it is an alkene with general formula CnH2n; each CH₂ unit contributes 14, so 14n = 56 and n = 4 [1]
(i) X is C₄H₈ [1]
(ii) But-1-ene and but-2-ene [1]
(ii) They differ in the position of the C=C double bond — between carbons 1 and 2 in but-1-ene, and between carbons 2 and 3 in but-2-ene [1]
⚠ If you missed marks here: The clue chain matters: bromine result → alkene → CnH2n → 14n = 56. Jumping straight to C₄H₈ without the alkene deduction drops the working mark. Note that the Mr of CnH2n is exactly 14n only because 12 + 2(1) = 14 per carbon.
(c)[4]
Liquid Y is produced slowly when a bottle of wine is left open to the air. Y turns blue litmus red, and a solution of Y gives effervescence with sodium carbonate.
(i) Identify Y and explain how it forms in the wine. [2]
(ii) Write a balanced symbol equation for the reaction of Y with sodium carbonate. [1]
(iii) A 0.1 mol/dm³ solution of Y has pH 3, while 0.1 mol/dm³ hydrochloric acid has pH 1. Explain this difference. [1]
Model Answer -- 1(c)
(i) Y is ethanoic acid, CH₃COOH [1]
(i) It forms when bacteria oxidise the ethanol in the wine using oxygen from the air [1]
(iii) Ethanoic acid is a weak acid, only partially dissociated, so its solution contains a lower concentration of H⁺ ions than the fully dissociated hydrochloric acid of the same concentration [1]
⚠ If you missed marks here: The wine clue points to bacterial oxidation — not fermentation, which makes ethanol rather than destroying it. In (iii) both solutions have the same concentration; only the degree of dissociation differs, and that word must appear in your answer.
Question 2 -- From Kerosene to Perfume: a Four-Step Synthesis
Total: 12 marks
A fragrance chemist in Mumbai explains that the fruity ester in a perfume can be traced all the way back to petroleum. The route has four steps:
Step 1: Dodecane from the kerosene fraction is cracked to give ethene and one other product.
(i) Write the balanced symbol equation for this cracking reaction, assuming one molecule of ethene forms per molecule of dodecane, and name the type of the second product. [2]
(ii) State the conditions needed. [1]
Model Answer -- 2(a)
(i) C₁₂H₂₆ → C₂H₄ + C₁₀H₂₂ — the atoms balance: 12 C and 26 H on each side [1]
(i) The second product, C₁₀H₂₂, fits CnH2n+2, so it is a shorter-chain alkane (decane) [1]
(ii) High temperature (about 600 °C) and a silica or alumina catalyst [1]
⚠ If you missed marks here: In any cracking equation the products must add up to the starting formula — write the carbon and hydrogen totals underneath to check. One product is always the alkene you want; the other is usually an alkane, and classifying it with the 2n+2 test is what the second mark rewards.
(b)[3]
Step 2: The ethene is converted to ethanol.
Write the balanced symbol equation for this step and give the full conditions of temperature, pressure and catalyst.
Model Answer -- 2(b)
C₂H₄ + H₂O → C₂H₅OH — steam adds across the double bond in an addition reaction [1]
300 °C and 60 atm pressure [1]
Phosphoric acid catalyst [1]
⚠ If you missed marks here: All three conditions are needed for the two condition marks — 300 °C, 60 atm and phosphoric acid form a fixed set. If you wrote sulfuric acid, you have jumped ahead to Step 4: sulfuric acid catalyses esterification, phosphoric acid catalyses hydration.
(c)[3]
Step 3: The ethanol is oxidised to ethanoic acid in the laboratory.
Describe how this oxidation is carried out, the observation made, and name one other way ethanol can be oxidised.
Model Answer -- 2(c)
Warm (heat) the ethanol with acidified potassium manganate(VII) [1]
The purple colour of the manganate(VII) fades to colourless as ethanoic acid forms [1]
Ethanol is also oxidised by combustion (burning) or by bacterial oxidation with oxygen from the air [1]
⚠ If you missed marks here: The syllabus lists three oxidation routes for ethanol — combustion, bacterial oxidation and acidified potassium manganate(VII) — and this part asks for two of them. Combustion counts as oxidation; students often forget it because the product is CO₂ rather than ethanoic acid.
(d)[3]
Step 4: The ethanoic acid is warmed with propan-1-ol and a catalyst to make the perfume ester.
(i) Name the ester and the catalyst. [2]
(ii) Describe the structure of the ester linkage formed in this reaction. [1]
Model Answer -- 2(d)
(i) The ester is propyl ethanoate [1]
(i) The catalyst is concentrated sulfuric acid [1]
(ii) The ester linkage is –COO– : a carbon atom double-bonded to one oxygen and single-bonded to a second oxygen, which is bonded to the carbon of the alcohol part [1]
⚠ If you missed marks here: The ester name runs alcohol-first: propan-1-ol + ethanoic acid → propyl ethanoate. When describing the linkage, both oxygens must be placed — one in the C=O and one bridging to the alkyl group; "COO" written without explanation may not convince the examiner you know the structure.
Question 3 -- Petrol versus Ethanol: the Numbers
Total: 12 marks
Flex-fuel cars sold in many countries can burn either petrol or ethanol. An engineer models petrol as pure octane, C₈H₁₈, and compares the carbon dioxide released by each fuel.
(a)[4]
(i) Write a balanced symbol equation for the complete combustion of octane. [2]
(ii) Calculate the percentage by mass of carbon in octane. (Ar: C = 12, H = 1) [2]
Model Answer -- 3(a)
(i) 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O — correct formulae of all four species [1]
(ii) Mr of octane = (8 × 12) + (18 × 1) = 114, and the mass of carbon per mole is 96 g [1]
(ii) Percentage of carbon = 96 ÷ 114 × 100 = 84.2% (accept 84%) [1]
⚠ If you missed marks here: Octane's 2 : 25 balance defeats many candidates — double the fuel first so the odd number of oxygens disappears. In the percentage, both the 96 and the 114 must appear in your working; an unexplained 84.2% risks losing the method mark if a slip occurs elsewhere.
(b)[4]
A test engine burns 11.4 g of octane completely.
(i) Calculate the number of moles of octane burned. [1]
(ii) Calculate the mass of carbon dioxide produced. (Ar: O = 16) [2]
(iii) Hence calculate the mass of carbon dioxide released per gram of octane burned. [1]
Model Answer -- 3(b)
(i) Moles of octane = 11.4 ÷ 114 = 0.1 mol [1]
(ii) Each mole of octane gives 8 moles of CO₂ (16/2 from the equation), so moles of CO₂ = 8 × 0.1 = 0.8 mol [1]
(ii) Mass of CO₂ = 0.8 × 44 = 35.2 g [1]
(iii) Mass of CO₂ per gram of fuel = 35.2 ÷ 11.4 = 3.09 g (accept 3.1 g) [1]
⚠ If you missed marks here: The mole ratio is 8 CO₂ per octane — from the eight carbons, or from 16 ÷ 2 in the doubled equation; using 16 doubles your answer. Keep the final division to at least three significant figures so the comparison in part (c) stays fair.
(c)[4]
The same engine burns 9.2 g of ethanol, C₂H₅OH, completely. (Mr of ethanol = 46)
(i) Calculate the mass of carbon dioxide produced and the mass of carbon dioxide per gram of ethanol. [2]
(ii) The engineer concludes: "Per gram of fuel, ethanol releases less carbon dioxide, and the ethanol is carbon-neutral anyway." Comment on both parts of this conclusion. [2]
Model Answer -- 3(c)
(i) Moles of ethanol = 9.2 ÷ 46 = 0.2 mol, and each mole gives 2 moles of CO₂, so 0.4 × 44 = 17.6 g of CO₂ [1]
(i) Mass of CO₂ per gram of ethanol = 17.6 ÷ 9.2 = 1.91 g, which is indeed less than the 3.09 g for octane [1]
(ii) The first claim is supported by the calculation (ethanol contains oxygen and less carbon per gram, so it releases less CO₂ per gram) [1]
(ii) The carbon-neutral claim is only partly true: the CO₂ released was absorbed by the sugar-cane crop during photosynthesis, but fossil-fuel energy is used in farming and distillation, so overall emissions are not zero [1]
⚠ If you missed marks here: "Comment on" means judge each claim against evidence — agree with the arithmetic for the first, and challenge the absolute word "neutral" in the second. A one-sided answer that simply accepts or rejects both claims together cannot reach full marks.
Question 4 -- Supply, Demand and the Cracker
Total: 12 marks
A refinery in Rotterdam analyses what its crude oil contains (supply) against what its customers want to buy (demand).
fraction
% in crude oil (supply)
% of customer demand
refinery gas + gasoline
18
34
naphtha + kerosene
21
24
diesel oil
19
23
fuel oil + residues
42
19
(a)[4]
(i) Use the table to identify the fraction group in greatest surplus and the group in greatest shortage. [2]
(ii) Explain how cracking allows the refinery to solve both problems at once. [2]
Model Answer -- 4(a)
(i) Fuel oil and residues are in greatest surplus — 42% supplied but only 19% demanded [1]
(i) Refinery gas and gasoline are in greatest shortage — 18% supplied against 34% demanded [1]
(ii) Cracking breaks the surplus long-chain fuel-oil molecules into short-chain molecules in the gasoline range, using up the surplus and filling the shortage simultaneously [1]
(ii) It also produces alkenes, which the refinery can sell as feedstock for polymers and other chemicals [1]
⚠ If you missed marks here: Quote the numbers — "42% supply against 19% demand" turns an assertion into evidence and secures the mark. In (ii) the clever phrase is "solves both problems": the same reaction removes the surplus and supplies the shortage, and your answer must say so explicitly.
(b)[4]
The refinery lists four unlabelled fractions W, X, Y and Z with these properties:
W: boils above 350 °C, almost solid at room temperature, dark and sticky
X: boils at 40–100 °C, runny, evaporates quickly
Y: boils at 250–350 °C, thick oily liquid
Z: gas at room temperature
(i) Match W, X, Y and Z to: refinery gas, gasoline, fuel oil, bitumen. [2]
(ii) Explain, in terms of molecular size and the forces between molecules, why W and X behave so differently. [2]
Model Answer -- 4(b)
(i) W = bitumen and X = gasoline [1]
(i) Y = fuel oil and Z = refinery gas [1]
(ii) W contains very long-chain molecules with strong attractive forces between them, so it barely flows and needs a very high temperature to boil [1]
(ii) X contains small molecules with weak intermolecular attractions, so it flows easily, evaporates readily and boils at a low temperature [1]
⚠ If you missed marks here: Data-matching questions are decided by boiling point: the higher the range, the longer the chains and the lower the fraction sits in the column. The explanation must connect size → strength of intermolecular attraction → property; skipping the middle link costs the mark.
(c)[4]
In the cracker, tetradecane from fraction Y reacts as follows:
C₁₄H₃₀ → C₈H₁₈ + 2X
(i) Deduce the molecular formula of X, showing your working, and name X. [3]
(ii) State one industrial use for X. [1]
Model Answer -- 4(c)
(i) Carbons: 14 − 8 = 6 shared between two molecules, so 3 each; hydrogens: 30 − 18 = 12, so 6 each [1]
(i) X is C₃H₆ [1]
(i) X is propene, an alkene [1]
(ii) Propene is polymerised to make poly(propene) (accept: converted to other chemicals / plastics manufacture) [1]
⚠ If you missed marks here: Divide by the coefficient before writing the formula — C₆H₁₂ is the total of both molecules, not one of them. C₃H₆ fits CnH2n, confirming an alkene, and its everyday destiny is poly(propene) — an answer of "fuel" gets the use mark only if qualified sensibly.
Question 5 -- Deductions from Relative Molecular Mass
Total: 10 marks
A chemistry olympiad in Singapore sets a round of molecular puzzles. Each compound must be identified from its relative molecular mass and a few clues.
(a)[3]
Puzzle 1: compound A is an alkane with a relative molecular mass of 72.
(i) State why the relative molecular masses of consecutive alkanes differ by 14. [1]
(ii) Deduce the molecular formula of A, showing your working. (Ar: C = 12, H = 1) [2]
Model Answer -- 5(a)
(i) Consecutive members of a homologous series differ by one CH₂ unit, and CH₂ has a mass of 12 + 2 = 14 [1]
(ii) For an alkane CnH2n+2, Mr = 12n + 2n + 2 = 14n + 2; setting 14n + 2 = 72 gives n = 5 [1]
(ii) A is C₅H₁₂ (pentane, the five-carbon alkane) [1]
⚠ If you missed marks here: The alkane formula gives Mr = 14n + 2, not 14n — forgetting the +2 turns the answer into n = 5.14 and no whole number. Whenever a deduction gives a non-integer, go back and check the general formula before rounding.
(b)[4]
Puzzle 2: compound E has a relative molecular mass of 88 and a sweet, fruity smell. E is made by warming ethanol with ethanoic acid and an acid catalyst. (Ar: C = 12, H = 1, O = 16)
(i) Name E and write its structural formula. [2]
(ii) Show that the relative molecular mass of E is 88. [1]
(iii) Describe the displayed formula of E around the ester linkage. [1]
Model Answer -- 5(b)
(i) E is ethyl ethanoate — the ester of ethanol and ethanoic acid [1]
(i) Structural formula CH₃COOC₂H₅ (accept CH₃COOCH₂CH₃) [1]
(iii) The central carbon is double-bonded to one oxygen and single-bonded to a second oxygen; that oxygen is bonded to the CH₂ of the ethyl group, while the CH₃ of the ethanoate part sits on the other side of the central carbon [1]
⚠ If you missed marks here: Count the atoms of CH₃COOC₂H₅ carefully — four carbons, eight hydrogens, two oxygens — before summing masses. In (iii) the two oxygens play different roles (one C=O, one bridging C–O–C); describing them as equivalent is the standard error.
(c)[3]
Puzzle 3: compounds A (an alkane), B (an alcohol) and C (a carboxylic acid) all contain three carbon atoms.
Explain why A, B and C have very different chemical reactions even though their chain lengths are the same, and why B reacts in the same way as ethanol despite their different chain lengths.
Model Answer -- 5(c)
A, B and C belong to different homologous series because each contains a different functional group (none, –OH, –COOH) [1]
The functional group determines the chemical reactions, so the three compounds react very differently — for example only C is acidic [1]
B and ethanol both contain the –OH group, so they undergo the same reactions (combustion, oxidation to a carboxylic acid, esterification); chain length changes only physical properties [1]
⚠ If you missed marks here: The whole question hangs on one principle: the functional group controls chemistry; the chain length controls physical properties. State it in both directions — different groups mean different reactions, same group means same reactions — to collect all three marks.
Question 6 -- Choosing an Ethanol Route: Two Countries
Total: 12 marks
Two companies each plan an ethanol plant. Company P is in a region of India with vast sugar-cane plantations; company Q is in a Gulf state with cheap ethene from its petrochemical industry. Their consultants prepare this comparison:
fermentation
ethene + steam
rate of production
slow (days)
fast (seconds)
process type
batch
continuous
product purity
dilute aqueous solution
essentially pure ethanol
raw material
sugars from crops
ethene from petroleum
reaction conditions
25–35 °C, 1 atm
300 °C, 60 atm
(a)[4]
Before choosing, each company reviews the chemistry of both routes.
(i) For fermentation, name the catalyst source and explain why the temperature must stay within 25–35 °C. [2]
(ii) For the ethene route, write the symbol equation and name the catalyst. [2]
Model Answer -- 6(a)
(i) The enzymes that catalyse fermentation come from yeast [1]
(i) Below about 25 °C the reaction becomes too slow; above about 35 °C the enzymes are denatured and fermentation stops [1]
(ii) C₂H₄ + H₂O → C₂H₅OH [1]
(ii) The catalyst is phosphoric acid [1]
⚠ If you missed marks here: The temperature window needs both halves: too cold = too slow, too hot = enzymes denatured. One-sided answers earn half. Note that yeast itself is not the catalyst — the enzymes in the yeast are, and precise wording protects the mark.
(b)[4]
Using the table and your own knowledge, recommend a route for each company. Justify both recommendations with evidence.
Model Answer -- 6(b)
Company P should choose fermentation, because its raw material — sugar from cane — is abundant, cheap and renewable locally [1]
P's mild conditions (25–35 °C, 1 atm) need little energy and simpler, cheaper equipment [1]
Company Q should choose the ethene route, because it has cheap ethene on hand and the continuous, fast process gives a much higher output [1]
Q's product is essentially pure ethanol, avoiding the cost of fractional distillation that fermentation would require [1]
⚠ If you missed marks here: Each recommendation needs two distinct pieces of evidence, and they must fit the country — recommending fermentation to Q because "it is renewable" ignores the scenario. Answers that copy the table without linking a row to a company score the choice but not the justification.
(c)[4]
Company P's fermented mixture contains about 12% ethanol in water.
(i) Name the method used to concentrate the ethanol and explain why it works for this mixture. [2]
(ii) A director claims the ethene route will still be running "in a hundred years". Comment on this claim. [2]
Model Answer -- 6(c)
(i) Fractional distillation is used to concentrate the ethanol [1]
(i) It works because ethanol boils at 78 °C, lower than water at 100 °C, so ethanol vapour rises and is collected first [1]
(ii) The claim is doubtful: ethene is made by cracking petroleum fractions, and petroleum is a finite resource that is being used up [1]
(ii) In the long term the renewable fermentation route (or ethene from renewable sources) is more sustainable, so P's route is more secure over a century [1]
⚠ If you missed marks here: Simple distillation is not enough credit at Extended level — fractional distillation with the two boiling points quoted is the full answer. For (ii), tie the doubt to the finite nature of petroleum; "technology might change" is speculation, not chemistry.
Question 7 -- Polymer Forensics
Total: 10 marks
A materials laboratory in Chennai receives two unidentified polymer samples and must deduce the monomers from which each was made.
(a)[4]
Sample 1 has a carbon backbone with this repeating pattern:
–CH₂–CH(CH₃)–CH₂–CH(CH₃)–CH₂–CH(CH₃)–
(i) Identify the repeat unit and deduce the monomer, giving its name and structural formula. [3]
(ii) Name the type of polymerisation and justify your choice. [1]
Model Answer -- 7(a)
(i) The repeat unit is –CH₂–CH(CH₃)– : two backbone carbons, one carrying a methyl side group [1]
(i) Restoring the C=C double bond between the two backbone carbons gives the monomer CH₂=CHCH₃ [1]
(i) The monomer is propene [1]
(ii) Addition polymerisation — the backbone is unbroken carbon and no small molecule is lost, so the monomers simply added together as their double bonds opened [1]
⚠ If you missed marks here: To go from chain to monomer, cut out one two-carbon repeat and restore the double bond — the CH₃ side group stays attached, so the monomer is propene, not butene. An all-carbon backbone is the fingerprint of addition polymerisation; linkages containing oxygen or nitrogen would signal condensation.
(b)[3]
Sample 2 is a fibre whose chain contains regularly repeating –CO–NH– groups between hydrocarbon blocks.
(i) Identify the type of polymer and name a synthetic example. [1]
(ii) Name the two types of monomer used to make it. [1]
(iii) Explain why this polymerisation is described as condensation. [1]
Model Answer -- 7(b)
(i) The –CO–NH– group is an amide linkage, so sample 2 is a polyamide; the synthetic example is nylon [1]
(ii) The monomers are a dicarboxylic acid and a diamine [1]
(iii) Each time a linkage forms, a small molecule — water — is eliminated, which is the defining feature of condensation polymerisation [1]
⚠ If you missed marks here: Read the linkage like a barcode: –CO–NH– means amide and therefore polyamide/nylon; –COO– would mean ester and polyester/PET. The condensation mark always needs the loss of water stated explicitly.
(c)[3]
The laboratory also analyses a protein fibre from silk and finds –CO–NH– linkages, yet the two fibres behave differently.
(i) State what this shows about proteins. [1]
(ii) The city wants to ban one of: PET recycling, plastic incineration without filters, or sanitary landfill. Recommend which to ban and give two reasons. [2]
Model Answer -- 7(c)
(i) Proteins are natural polyamides — the same amide linkage joins their amino acid monomers [1]
(ii) Ban incineration without filters, because burning plastics releases toxic gases such as carbon monoxide (and acidic gases) directly into the air people breathe [1]
(ii) The other two options are defensible: recycling PET conserves crude oil and reduces waste, and even landfill contains the waste rather than dispersing toxins into the atmosphere [1]
⚠ If you missed marks here: A recommendation question is marked on the quality of the reasons, not the choice alone. Naming a specific toxic product of combustion, and explaining why the alternatives are less harmful, is what separates a two-mark answer from a bare opinion.
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