HKDSE Chemistry Practice Paper: Organic Chemistry, Functional Groups And Polymers (12 Questions, 100 Marks, 75 Minutes, With Answers)

HKDSE Chemistry Practice Paper: Organic Chemistry, Functional Groups and Polymers (12 Questions, 100 Marks, 75 Minutes, with Answers)

2026-11-06

HKDSE Chemistry: Organic Chemistry and Polymers

S5 chemistry practice paper, 12 questions, 100 marks, 75 minutes: alkanes and combustion, cis-trans isomerism, electrophilic addition, structural isomers, addition polymerisation, oxidation

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Section A: Alkanes, Alkenes and Isomerism (24 marks)

Each question carries 6 marks; show your working or reasoning.

#	Question	Answer and marking notes
1	An alkane has the molecular formula C6H14. (a) State the general formula of the alkane homologous series and deduce the molecular formula of the alkane with eight carbon atoms. (b) Write the balanced equation for the complete combustion of C6H14 and calculate the number of moles of oxygen needed to burn 1 mol of C6H14 completely. (c) Explain why alkanes mainly undergo substitution rather than addition.	(a) CnH2n+2; the alkane with eight carbon atoms is C8H18. (2 marks) (b) C6H14 + 9.5O2 -> 6CO2 + 7H2O; 1 mol of C6H14 needs 9.5 mol of oxygen and gives 6 mol of carbon dioxide and 7 mol of water. (2 marks) (c) Every bond in an alkane is a single bond and each carbon already uses four bonds, so the molecule is saturated and has no double bond to add across; under light or heat a hydrogen atom can be replaced by a halogen atom, which is substitution. (2 marks)
2	2-butene and 1-butene both have the molecular formula C4H8. (a) Write the structural formulae of 1-butene and cis-2-butene. (b) Explain the structural difference between cis-2-butene and trans-2-butene and state whether they are isomers. (c) Explain why 1-butene shows no cis-trans isomerism.	(a) 1-butene: CH2=CH-CH2-CH3; cis-2-butene: CH3-CH=CH-CH3 with both methyl groups on the same side of the double bond. (2 marks) (b) The C=C bond cannot rotate freely, so the two methyl groups are locked on the same side (cis) or on opposite sides (trans); the two have the same molecular formula but different spatial arrangements, so they are stereoisomers. (2 marks) (c) One carbon of the double bond in 1-butene carries two identical hydrogen atoms, so interchanging the groups gives no different arrangement. (2 marks)
3	Ethene reacts with hydrogen bromide by addition to give bromoethane (C2H5Br). (a) Write the equation and name the reaction type. (b) 0.25 mol of ethene reacts with excess hydrogen bromide; calculate the theoretical yield in grams. (c) If only 21.8 g of bromoethane is obtained, calculate the percentage yield.	(a) C2H4 + HBr -> C2H5Br; electrophilic addition. (2 marks) (b) Mr of bromoethane = 2 x 12 + 5 x 1 + 80 = 109; mass = 0.25 mol x 109 g/mol = 27.25 g. (2 marks) (c) Percentage yield = 21.8 / 27.25 x 100% = 80.0%. (2 marks)
4	Pentane (C5H12) has three structural isomers. (a) Write the structural formula and IUPAC name of each. (b) Arrange them in order of decreasing boiling point and explain. (c) How many structural alkene isomers have the molecular formula C4H8? Name them.	(a) The three isomers are pentane, 2-methylbutane and 2,2-dimethylpropane. (2 marks) (b) Boiling point: pentane > 2-methylbutane > 2,2-dimethylpropane. More branching makes the molecule more compact, so the contact surface area between molecules decreases and the van der Waals forces become weaker. (2 marks) (c) Three: but-1-ene, but-2-ene (which has cis and trans forms) and 2-methylpropene. (2 marks)

Section B: Functional Groups, Alcohols, Acids and Esters (36 marks)

Each question carries 9 marks; note the marks for each part.

#	Question	Answer and marking notes
5	Poly(ethene) is made from ethene by addition polymerisation. (a) Write the equation for the polymerisation, stating the reaction type and the conditions required. (b) A sample of poly(ethene) has an average relative molecular mass of 42,000; calculate its average degree of polymerisation. (c) Compare the monomers of poly(ethene) and PVC, and explain why PVC is harder than poly(ethene) at room temperature.	(a) n CH2=CH2 -> -(CH2-CH2)n-; addition polymerisation, carried out at high pressure with a catalyst such as a Ziegler-Natta catalyst. (3 marks) (b) The repeating unit is C2H4 with Mr = 28; average degree of polymerisation = 42,000 / 28 = 1,500. (3 marks) (c) The monomer of poly(ethene) is ethene and that of PVC is chloroethene (vinyl chloride, CH2=CHCl). The chlorine atoms are larger and polar, so dipole-dipole attractions act between chains in addition to van der Waals forces; the chains pack more closely and slide less easily, so PVC is harder. (3 marks)
6	Ethanol can be oxidised to ethanal and then to ethanoic acid. (a) Write the equation for each oxidation step and state the reagent and apparatus used. (b) 23 g of ethanol is fully oxidised; calculate the theoretical mass of ethanoic acid formed. (c) Only 24 g of ethanoic acid is obtained; calculate the percentage yield and suggest two ways to improve it.	(a) Step one: CH3CH2OH + [O] -> CH3CHO + H2O, using acidified potassium dichromate(VI) with distillation to collect the ethanal; step two: CH3CHO + [O] -> CH3COOH, using acidified potassium dichromate(VI) under reflux. (3 marks) (b) Mr of ethanol = 46, so n = 23 / 46 = 0.5 mol; Mr of ethanoic acid = 60; theoretical mass = 0.5 x 60 = 30 g. (3 marks) (c) Percentage yield = 24 / 30 x 100% = 80%. Improvements: use excess oxidising agent, reflux for longer, or remove the acid as it forms to reduce the reverse reaction (catalytic oxidation with careful temperature control is an alternative). (3 marks)
7	Ethanol and ethanoic acid react with concentrated sulphuric acid as catalyst to give ethyl ethanoate. (a) Write the balanced equation and state the conditions. (b) 0.5 mol of ethanol reacts with 0.5 mol of ethanoic acid; calculate the theoretical mass of ethyl ethanoate. (c) Only 0.35 mol of ethyl ethanoate is obtained; calculate the percentage yield, explain why the yield is below 100% and state how industry improves it.	(a) CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O, with concentrated sulphuric acid as catalyst under heating and reflux. (3 marks) (b) Mr of ethyl ethanoate = 4 x 12 + 8 x 1 + 2 x 16 = 88; theoretical mass = 0.5 x 88 = 44 g. (3 marks) (c) Percentage yield = 0.35 / 0.5 x 100% = 70%. Esterification is reversible, so the equilibrium mixture still contains reactants and conversion is incomplete. Industry removes the water formed in a separator, uses an excess of the cheaper reactant and distils the ester out as it forms. (3 marks)
8	An organic compound contains only carbon, hydrogen and oxygen, with percentage composition by mass C 62.1%, H 10.3% and O 27.6%. Its relative molecular mass is 116. (a) Calculate the empirical formula, showing your working. (b) Deduce the molecular formula. (c) The compound releases carbon dioxide from sodium carbonate solution; name one structural isomer that fits and state its functional group.	(a) Taking 100 g: C 62.1 / 12 = 5.18 mol, H 10.3 / 1 = 10.3 mol, O 27.6 / 16 = 1.73 mol. Dividing each by 1.73 gives C 3, H 6 and O 1, so the empirical formula is C3H6O with formula mass 58. (4 marks) (b) 116 / 58 = 2, so the molecular formula is C6H12O2. (2 marks) (c) Hexanoic acid (or 2-methylpentanoic acid) fits; its functional group is the carboxyl group, -COOH. (3 marks)

Section C: Polymers, Industrial Use and Environmental Evaluation (40 marks)

Each question carries 10 marks; give full explanations or evaluations.

#	Question	Answer and marking notes
9	A city uses 24,000 tonnes of poly(ethene) each year, of which 12% is recycled. (a) Calculate the mass of poly(ethene) not recycled each year, in tonnes. (b) The calorific value of poly(ethene) is 43 MJ/kg; calculate the energy released if the amount in (a) were completely incinerated, in MJ to three significant figures. (c) Give one environmental and one economic reason for recycling rather than incinerating, and state one practical difficulty in recycling poly(ethene).	(a) 24,000 x (1 – 0.12) = 21,120 tonnes. (3 marks) (b) 21,120 tonnes = 2.112 x 10^7 kg; energy = 2.112 x 10^7 x 43 = 9.08 x 10^8 MJ. (4 marks) (c) Environmentally, recycling saves the petroleum feedstock and avoids the carbon dioxide and possible toxic gases produced by burning; economically, the recyclate can be made into new products and sold, whereas incineration recovers energy only once. Difficulty: mixed plastics are hard to sort, washing is costly and the reprocessed material is of lower quality. (3 marks)
10	Oleic acid (C18H34O2), which contains one C=C bond, is hydrogenated to make margarine. (a) Write the equation, stating the reaction type and the catalyst. (b) 5.64 kg of oleic acid is fully hydrogenated; calculate the mass of hydrogen needed. (c) Explain why hydrogenation raises the melting point and state the health concern about partially hydrogenated vegetable oils.	(a) C18H34O2 + H2 -> C18H36O2; addition (hydrogenation) with a nickel catalyst under heat and pressure. (3 marks) (b) Mr of oleic acid = 18 x 12 + 34 x 1 = 250 + 32 = 282; 5.64 kg = 5,640 g; n = 5,640 / 282 = 20 mol, so 20 mol of hydrogen is needed, i.e. 20 x 2 = 40 g. (4 marks) (c) Hydrogenation turns the bent cis unsaturated chains into straight saturated chains, which pack more closely, increasing the contact area and the van der Waals forces, so the melting point rises. Partial hydrogenation also forms trans fats, which raise the risk of cardiovascular disease. (3 marks)
11	A saturated monocarboxylic acid, CnH2n+1COOH, has a relative molecular mass of 74. (a) Calculate n and give the name and molecular formula of the acid. (b) The acid reacts with ethanol with concentrated sulphuric acid as catalyst; name the product and give its molecular formula. (c) Compare the boiling points of the acid and the alcohol with the same number of carbon atoms, and explain the difference. (d) Explain why the acid is stronger than the alcohol.	(a) 14n + 46 = 74, so n = 2; the acid is propanoic acid, C2H5COOH (C3H6O2). (3 marks) (b) The product is ethyl propanoate, C5H10O2. (2 marks) (c) The acid has the higher boiling point because its molecules form dimers through two hydrogen bonds, giving much stronger intermolecular forces than the single hydrogen bonds between alcohol molecules. (3 marks) (d) The carboxylate ion left after losing a proton is stabilised by resonance, with the negative charge shared over two oxygen atoms, so the proton is released more readily; the alkoxide ion from an alcohol has no such delocalisation. (2 marks)
12	A triglyceride has a relative molecular mass of 890, and 1 mol of it requires 3 mol of sodium hydroxide for complete saponification. (a) Write a word equation for the saponification. (b) 1,000 g of the triglyceride is completely saponified; calculate the mass of sodium hydroxide needed. (c) Explain why soap cleans less well in hard water and how synthetic detergents avoid this problem.	(a) triglyceride + 3 sodium hydroxide -> glycerol + 3 sodium salts of fatty acids (soap). (3 marks) (b) n(triglyceride) = 1,000 / 890 = 1.124 mol; sodium hydroxide = 3 x 1.124 = 3.371 mol; mass = 3.371 x 40 = 134.8 g, about 135 g. (4 marks) (c) Hard water contains calcium and magnesium ions, which form insoluble scums with fatty acid ions, wasting the soap and giving no lather. Synthetic detergents are sulphonates whose calcium and magnesium salts are soluble, so they still clean in hard water. (3 marks)

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