HKDSE Chemistry Practice Paper: Organic Chemistry — Alkanes, Alkenes, Alcohols, Carboxylic Acids, Esters And Polymers (12 Questions, 100 Marks, 75 Minutes)

HKDSE Chemistry Practice Paper: Organic Chemistry — Alkanes, Alkenes, Alcohols, Carboxylic Acids, Esters and Polymers (12 questions, 100 marks, 75 minutes)

2026-11-04

HKDSE Chemistry: Organic Chemistry

Original HKDSE Chemistry paper on organic chemistry: alkanes, alkenes, alcohols, acids, esters, polymers, isomerism. 12 questions, 100 marks, 75 minutes, worked answers.

Download　Question paper PDF　|　Answer key PDF

Paper details　Level: S6 | Subject: Chemistry | Time: 75 minutes | Marks: 100 | Questions: 12

Section A (32 marks)

Hydrocarbons and functional groups: show the calculation and the formula.

#	Question	Answer and marking notes
1	An alkane contains 83.3% carbon by mass. (a) Using the general formula CnH2n+2, calculate its molecular formula. (b) Give the names of three structural isomers of C5H12. (c) Explain why C5H12 is a liquid while C20H42 is a solid at room temperature.	Answers: (a) Let the formula be CnH2n+2. Mass fraction of carbon = 12n/(14n+2) = 0.833, so 12n = 11.662n + 1.666 and n = 4.93, hence n = 5 and the formula is C5H12. (b) Pentane, 2-methylbutane and 2,2-dimethylpropane. (c) C20H42 has a long carbon chain and more electrons, so the van der Waals forces are stronger and it is a solid; C5H12 molecules are small with weak van der Waals forces, so it is a liquid. Marking points: equation 2, value of n 1, formula 1, three names 2, van der Waals explanation 2.
2	5.6 g of an alkene with molecular formula C4H8 reacts completely with bromine water. (a) Calculate the number of moles of the alkene. (b) Calculate the mass of bromine (Br2) required. (c) Explain why the bromine water is decolourised and state the type of reaction.	Answers: (a) Molar mass of C4H8 = 56 g/mol, so n = 5.6/56 = 0.10 mol. (b) One mole of alkene adds one mole of Br2, so 0.10 mol of Br2 is needed; mass = 0.10 x 160 = 16 g. (c) Bromine adds across the C=C bond to give colourless 1,2-dibromobutane, so the orange colour fades; the reaction is electrophilic addition. Marking points: molar mass 1, moles 2, 1:1 ratio 1, mass of Br2 2, explanation and reaction type 2.
3	23.0 g of ethanol is burned completely in excess oxygen. (a) Calculate the number of moles of ethanol. (b) Calculate the volume of oxygen required at room conditions (molar volume 24 dm3/mol). (c) Calculate the mass of carbon dioxide formed.	Answers: (a) Molar mass of ethanol = 46 g/mol, so n = 23.0/46 = 0.50 mol. (b) C2H5OH + 3O2 -> 2CO2 + 3H2O, so 1.5 mol of O2 is needed; volume = 1.5 x 24 = 36 dm3. (c) 1.0 mol of CO2 is formed, mass = 44 g. Marking points: moles 2, balanced equation 2, moles and volume of O2 2, mass of CO2 2.
4	0.50 mol of ethanoic acid is heated with excess ethanol and concentrated sulphuric acid. (a) Write the equation and name the organic product. (b) Calculate the theoretical yield of ethyl ethanoate. (c) Only 35.2 g is obtained; calculate the percentage yield. (d) State two functions of the concentrated sulphuric acid.	Answers: (a) CH3COOH + C2H5OH = CH3COOC2H5 + H2O with concentrated H2SO4 and heat; the product is ethyl ethanoate. (b) Theoretical yield = 0.50 x 88 = 44 g. (c) Percentage yield = 35.2/44 x 100% = 80%. (d) The acid catalyses the reaction and absorbs the water formed, shifting the equilibrium to the right. Marking points: equation and conditions 2, name 1, theoretical yield 2, percentage yield 2, two functions 2.

Section B (36 marks)

Isomerism, polymerisation, saponification and combustion analysis: set out the working and explain.

#	Question	Answer and marking notes
5	A compound with molecular formula C4H9Br has several structural isomers. (a) Name all the structural isomers and state how many there are. (b) Identify the isomer with the lowest boiling point and explain your answer. (c) 1-bromobutane is heated with aqueous sodium hydroxide; state the type of reaction and the main product.	Answers: (a) Four isomers: 1-bromobutane, 2-bromobutane, 1-bromo-2-methylpropane and 2-bromo-2-methylpropane. (b) 2-bromo-2-methylpropane has the lowest boiling point: the molecule is nearly spherical with the most branching, so the contact surface area and van der Waals forces are smallest. (c) Nucleophilic substitution, giving butan-1-ol as the main product and sodium bromide. Marking points: number 1, four names 4, lowest boiling point 1, branching and van der Waals 2, reaction type and product 1.
6	A sample of poly(ethene) has a relative molecular mass of 56,000. (a) Give the name and formula of the monomer and the repeating unit. (b) Calculate the average degree of polymerisation. (c) Compare addition polymerisation with condensation polymerisation, giving one example and the by-product of each.	Answers: (a) The monomer is ethene, C2H4; the repeating unit is -CH2-CH2-. (b) Degree of polymerisation = 56,000/28 = 2,000. (c) Addition polymerisation uses monomers with C=C bonds and produces no by-product (e.g. poly(ethene)); condensation polymerisation joins two functional groups and releases a small molecule such as water or HCl (e.g. polyester, nylon). Marking points: monomer and repeating unit 3, degree of polymerisation 2, two polymerisations with examples 4.
7	1.00 kg of a triglyceride (relative molecular mass 890) is completely saponified with sodium hydroxide. (a) Calculate the number of moles of the triglyceride. (b) Calculate the mass of sodium hydroxide required. (c) Name the two types of product of saponification and give a use of each.	Answers: (a) n = 1,000/890 = 1.12 mol. (b) Three moles of NaOH are needed per mole of triglyceride: 1.12 x 3 = 3.37 mol, mass = 3.37 x 40 = 135 g (134.8 g). (c) Glycerol and sodium salts of fatty acids (soap); the soap is used as a cleaning agent and glycerol in cosmetics and medicines. Marking points: moles 2, 3:1 stoichiometry 2, moles of NaOH 2, mass 2, products and uses 1.
8	0.44 g of an organic compound containing only carbon, hydrogen and oxygen is burned completely, giving 0.88 g of carbon dioxide and 0.36 g of water. Its relative molecular mass is 88. (a) Calculate the mass of each element, the mole ratio and the empirical formula. (b) Determine the molecular formula. (c) Give two isomers with this molecular formula but different functional groups.	Answers: (a) Carbon: 0.88/44 = 0.020 mol, mass 0.24 g; hydrogen: 0.36/18 = 0.020 mol, mass 0.04 g; oxygen: 0.44 – 0.24 – 0.04 = 0.16 g. Mole ratio C:H:O = 0.020:0.040:0.010 = 2:4:1, so the empirical formula is C2H4O. (b) The empirical formula mass is 44 and 88/44 = 2, so the molecular formula is C4H8O2. (c) For example butanoic acid (a carboxylic acid) and ethyl ethanoate (an ester). Marking points: mass of C and H 2, mass of O 1, ratio and empirical formula 2, molecular formula 2, two isomers 2.

Section C (32 marks)

Reaction types, chemical tests and synthesis routes: write equations and observations.

#	Question	Answer and marking notes
9	Methane reacts with chlorine in ultraviolet light. (a) Name the reaction and state the three stages by which CH4 becomes CH3Cl. (b) Explain why the product is a mixture. (c) Explain why alkenes are more reactive than alkanes.	Answers: (a) Free radical substitution; the three stages are initiation (Cl2 is split by light into chlorine radicals), propagation (the radical reacts with methane to give CH3. and HCl, then CH3Cl) and termination (radicals combine). (b) Successive substitution gives CH3Cl, CH2Cl2, CHCl3 and CCl4, all of which remain in the mixture. (c) The C=C bond has loosely held pi electrons above and below the molecular plane, so it is attacked readily by electrophiles; an alkane has only sigma bonds that must first be broken, so it is less reactive. Marking points: name of reaction 1, three stages 3, mixture explanation 2, pi electrons and electrophilic attack 2.
10	Four unlabelled liquids are hexane, hex-1-ene, ethanol and ethanoic acid. Only bromine water, sodium hydrogencarbonate solution and sodium metal are available. (a) Design a test procedure and state the expected observation at each step. (b) State which test should be carried out first and explain why. (c) Write the equation for the reaction between ethanol and sodium.	Answers: (a) Add sodium hydrogencarbonate first: only ethanoic acid gives carbon dioxide bubbles. Then add bromine water: hex-1-ene decolourises the orange solution. Finally add sodium metal: ethanol gives flammable hydrogen gas and hexane does not react. (b) The sodium hydrogencarbonate test should be done first because it is specific to the carboxylic acid; if bromine water were added first the alkene would be used up and could not be identified later. (c) 2C2H5OH + 2Na -> 2C2H5ONa + H2. Marking points: three tests and observations 4, choice of first test 1, specificity explanation 2, equation 1.
11	100 g of glucose is fermented by yeast. (a) Calculate the number of moles of glucose. (b) Calculate the theoretical mass of ethanol formed. (c) Calculate the volume of carbon dioxide released at room conditions (molar volume 24 dm3/mol). (d) Explain why the actual yield is lower than the theoretical value.	Answers: (a) The molar mass of glucose is 180 g/mol, so n = 100/180 = 0.556 mol. (b) C6H12O6 -> 2C2H5OH + 2CO2, so 1.11 mol of ethanol is formed, mass = 51.1 g. (c) 1.11 mol of CO2 is released, volume = 26.7 dm3. (d) Some glucose is used for the growth and respiration of the yeast, the reaction is incomplete, and some ethanol evaporates or is oxidised to ethanoic acid. Marking points: moles 2, stoichiometry 2, mass of ethanol 1, volume of CO2 2, reason for lower yield 1.
12	Using ethene as the only organic starting material, design a three-step synthesis of ethyl ethanoate; each step has a yield of 80%. (a) State the reagents and conditions for the three steps. (b) Calculate the number of moles and the mass of ethyl ethanoate obtainable from 1.0 mol of ethene. (c) Suggest one way to raise the overall yield.	Answers: (a) (1) Ethene is hydrated with steam using concentrated sulphuric acid as catalyst and heat to give ethanol; (2) ethanol is oxidised with acidified potassium dichromate or acidified potassium manganate(VII) on heating to give ethanoic acid; (3) ethanoic acid is esterified with ethanol using concentrated sulphuric acid to give ethyl ethanoate. (b) Overall yield = 0.8 x 0.8 x 0.8 = 51.2%, so 0.512 mol of ethyl ethanoate is obtained, mass = 45.1 g. (c) Use excess ethanol and remove the water formed during the reaction so that the esterification equilibrium shifts to the right, or raise the temperature and use a more effective catalyst. Marking points: three steps with reagents and conditions 4, overall yield 2, mass 1, method to improve yield 1.

Find a tutor — free Tutor Registration →

A TutorZone practice paper; original questions and passages, not publisher material.