Cambridge A Level Chemistry 9701 — 2011 May/June Paper 4 · Variant 1

This document consists of 18 printed pages and 2 blank pages. DC (CW/SW) 29521/4 © UCLES 2011 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen. You may use a pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. Section A Answer all questions. Section B Answer all questions. You may lose marks if you do not show your working or if you do not use appropriate units. A Data Booklet is provided. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. * 4 8 6 7 0 8 0 3 7 6 * CHEMISTRY 9701/41 Paper 4 Structured Questions May/June 2011 2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet For Examiner’s Use 1 2 3 4 5 6 7 8 9 Total

2 © UCLES 2011 9701/41/M/J/11 For Examiner’s Use Section A Answer all questions in the spaces provided. 1 Taken together, nitrogen and oxygen make up 99% of the air. Oxygen is by far the more reactive of the two gases, and most of the substances that react with air combine with the oxygen rather than with the nitrogen. (a) State one reason why the molecule of nitrogen, N2, is so unreactive. …[1] Despite the apparent lack of reactivity of N2, nitrogen atoms have been found to form bonds with almost all of the elements in the Periodic Table. Lithium metal reacts with nitrogen gas at room temperature to give lithium nitride, Li3N. Magnesium produces magnesium nitride, Mg3N2, as well as magnesium oxide, when heated in air. (b) Calculate the lattice energy of magnesium nitride using the following data, in addition to relevant data from the Data Booklet. enthalpy change value/kJ mol–1 atomisation of Mg(s) +148 total of electron affinities for the change N(g) N3–(g) +2148 enthalpy of formation of Mg3N2(s) –461 lattice energy = …kJ mol–1 [3]

3 © UCLES 2011 [Turn over 9701/41/M/J/11 For Examiner’s Use (c) Lithium reacts readily with nitrogen, and because of this Li3N has been considered as a possible intermediate in the ‘fixing’ of nitrogen to make ammonia-based fertilisers. + Li + H2O N2(g) Li3N NH3 + A (i) Construct an equation for the reaction between Li3N and H2O, and hence identify compound A. … (ii) Using your knowledge of the Haber process, consider one advantage and one disadvantage of using lithium as a means of fixing nitrogen, rather than the Haber process. advantage of the lithium method … disadvantage of the lithium method … [3] (d) Another possible advantage of Li3N is that it contains a large percentage by mass of nitrogen. Another fertiliser that contains a large percentage by mass of nitrogen is urea, NH2CONH2. (i) Calculate and compare the percentages by mass of nitrogen in Li3N and NH2CONH2. … … … (ii) What class of organic compound is urea? … (iii) Write an equation for the production of ammonia by the reaction between urea and water. … (iv) Urea can be applied directly to the soil either before or during the growing of crops. What would be a major disadvantage of using lithium nitride in this way? … … [5] [Total: 12]

4 © UCLES 2011 9701/41/M/J/11 For Examiner’s Use 2 (a) State briefly what is meant by the following terms. (i) reversible reaction … (ii) dynamic equilibrium … … [2] (b) Water ionises to a small extent as follows. H2O(l) H+(aq) + OH–(aq) ΔH = +58 kJ mol–1 (i) Write an expression for Kc for this reaction. … (ii) Write down the expression for Kw, the ionic product of water, and explain how this can be derived from your Kc expression in (i). … … (iii) State and explain how the value of Kw for hot water will differ from its value for cold water. … … [3] (c) Kw can be used to calculate the pH of solutions of strong and weak bases. (i) Use the value of Kw in the Data Booklet to calculate the pH of 0.050 mol dm–3 NaOH. pH = … Ammonia ionises slightly in water as follows. NH3(aq) + H2O(l) NH4 +(aq) + OH–(aq) The following expression applies to this equilibrium. [H2O] × Kc= [NH4 +][OH–] / [NH3] = 1.8 × 10–5 mol dm–3

5 © UCLES 2011 [Turn over 9701/41/M/J/11 For Examiner’s Use (ii) Calculate [OH–(aq)] in a 0.050 mol dm–3 solution of NH3. You may assume that only a small fraction of the NH3 ionises, so that [NH3] at equilibrium remains at 0.050 mol dm–3. [OH–(aq)] = … (iii) Use the value of Kw in the Data Booklet, and your answer in (ii), to calculate [H+(aq)] in 0.050 mol dm–3 NH3(aq). [H+(aq)] = … (iv) Calculate the pH of this solution. pH = … [6] [Total: 11]

6 © UCLES 2011 9701/41/M/J/11 For Examiner’s Use 3 (a) State and explain the variation in the oxidation numbers of the chlorides of the elements Na, Mg, Al and Si. … … … …[2] (b) Describe the reaction of phosphorus(V) chloride with water, and write an equation for the reaction. … …[2] (c) When microwave radiation is passed through phosphorus(III) chloride, PCl3, at low pressure, a new chloride of phosphorus, B, is formed. B contains 69.6% by mass of chlorine and 30.4% by mass of phosphorus, and its Mr is approximately 200. (i) Calculate the empirical and molecular formulae of B. … (ii) Assuming phosphorus and chlorine show their typical valencies, draw the displayed formula of B, showing all bonds and lone pairs. … (iii) Calculate the oxidation number of phosphorus in B. … (iv) One mole of B reacts with four moles of water. Suggest the structure of the phosphorus-containing product of this reaction. … [6] [Total: 10]

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8 © UCLES 2011 9701/41/M/J/11 For Examiner’s Use 4 The combustion of fuels in motor vehicles, trains, aeroplanes and power stations produces the pollutant gas NO2. (a) Write an equation to show how NO2 is formed in these situations. …[1] (b) (i) How is the NO2 removed from the exhaust gases of motor vehicles? … (ii) Write an equation for this process. … [2] (c) Suggest whether the production of the pollutant NO2 would be reduced if fossil fuels were replaced by hydrogen as a fuel for combustion. Explain your answer. … …[1] (d) In the atmosphere, NO2 acts as a catalyst for the oxidation of SO2 to SO3. NO2 SO2(g) + ½ O2(g) SO3(g) (i) What is the environmental significance of this reaction? … The oxidation takes place in two steps. The initial reaction is that between NO2 and SO2. reaction 1 NO2(g) + SO2(g) NO(g) + SO3(g) ΔH = –168 kJ mol–1 (ii) Write an equation to show how the NO2 is regenerated in the second step of the oxidation. … (iii) Write an expression for the equilibrium constant, Kp for reaction 1, stating its units. Kp = units … (iv) If equal amounts of NO2(g) and SO2(g) are allowed to react at room temperature, it is found that 99.8% of the gases have been converted into products at equilibrium. Calculate a value for Kp. Kp = …

9 © UCLES 2011 [Turn over 9701/41/M/J/11 For Examiner’s Use (v) The temperature of the atmosphere decreases with height. How will this affect the position of the equilibrium in reaction 1? Explain your answer. … … [7] [Total: 11]

10 © UCLES 2011 9701/41/M/J/11 For Examiner’s Use 5 (a) There are several ways of introducing chlorine atoms into organic molecules. State the reagents and conditions necessary to carry out the following transformations. transformation reagents + conditions C2H4 C2H5Cl C2H5OH C2H5Cl C2H6 C2H5Cl C2H4 C2H4Cl2 CH3CO2H CH3COCl CH3 CH3 Cl CH3 CH2Cl [6] (b) (i) When treated with concentrated HNO3 + H2SO4 at 55 °C, benzene produces nitrobenzene. Outline the mechanism of this reaction. You should include all charges, and use curly arrows to represent the movement of electron pairs.

11 © UCLES 2011 [Turn over 9701/41/M/J/11 For Examiner’s Use In aromatic substitution of monosubstituted benzenes, the orientation of an incoming group depends on the nature of the group already attached to the ring. For example, using the same reagents and conditions as in (i), methylbenzene and benzoic acid produce the following nitro compounds. CH3 CO2H NO2 CO2H NO2 methylbenzene benzoic acid CH3 (ii) Using this information as an aid, suggest a structure for compound C in the following synthesis of 3-bromobenzoic acid. CH3 step 1 step 2 C Br CO2H (iii) Suggest reagents and conditions for steps 1 and 2. step 1 step 2 [6] [Total: 12]

12 © UCLES 2011 9701/41/M/J/11 For Examiner’s Use 6 (a) The reaction producing tri-iodomethane (iodoform) can be used as a test for the presence of certain groups within a molecule. (i) State the reagents and conditions used for this reaction. … (ii) Write the structural formula of one functional group that would give a positive result with this iodoform reaction. … (iii) What do you observe in a positive test? … … (iv) In the following table place a tick (✓) in the column against each compound that would give a positive result with this test, and a cross (✗) against each compound that would give a negative result. compound result CH3OH CH3CH2OH CH3CHO CH3CO2H CHO COCH3 [6]

13 © UCLES 2011 [Turn over 9701/41/M/J/11 For Examiner’s Use (b) The iodoform test can be used, along with other reactions, to work out the structures of unknown compounds. Use the information in the table below to deduce the structures of the compounds in the following scheme, and draw these structures in the boxes provided. hot concentrated acidified KMnO4 C7H12O C3H6O + C4H6O3 D E F Results of tests (✓ indicates a positive result; ✗ indicates a negative result) test results of tests with each compound D E F iodoform ✗ ✓ ✓ Fehling’s solution ✓ ✗ ✗ 2,4-dinitrophenyl- hydrazine reagent ✓ ✓ ✓ Na2CO3(aq) ✗ ✗ ✓ structures D(C7H12O) E(C3H6O) F(C4H6O3) [3] (c) Treatment of compound F with NaBH4 gives compound G, C4H8O3. Heating G with Al2O3 gives a mixture of three isomeric unsaturated carboxylic acids H, J and K, C4H6O2, two of which are stereoisomers of each other. Suggest structures for G, H, J, and K, and name the type of stereoisomerism shown. G Al2O3 H + J + K type of stereoisomerism …[5] [Total: 14]

14 © UCLES 2011 9701/41/M/J/11 For Examiner’s Use Section B Answer all questions in the spaces provided. 7 Enzymes are a special group of protein molecules present in large amounts in living organisms. Enzymes behave as catalysts but, unlike inorganic catalysts, they generally catalyse only one particular reaction. (a) Inorganic catalysts often work better on heating, but enzymes rarely work at temperatures much above 45°C. Explain why this is the case. … … …[2] (b) Using the shape below to represent an enzyme, sketch how an enzyme is specific to the breakdown of a particular substrate molecule enzyme + substrate enzyme-substrate complex enzyme + products [3]

15 © UCLES 2011 [Turn over 9701/41/M/J/11 For Examiner’s Use (c) Describe the effects of a competitive, and of a non-competitive inhibitor on the interaction between enzyme and substrate. … … … …[2] (d) (i) The diagram shown illustrates an enzyme-catalysed reaction. On the diagram sketch the graph that would be obtained if the same reaction was carried out in the presence of a non-competitive inhibitor. concentration of substrate / mol dm–3 initial reaction rate / mol dm–3 s–1 (ii) Explain why a non-competitive inhibitor has this effect on the reaction. … … [3] [Total: 10]

16 © UCLES 2011 9701/41/M/J/11 For Examiner’s Use 8 Chromatography is an important analytical technique in chemistry. There is a number of techniques under the general heading of chromatography. (a) Paper and gas chromatography rely on partition to separate the components in a mixture, whereas thin-layer chromatography uses adsorption. Explain what is meant by (i) partition and (ii) adsorption, in the context of chromatography. (i) partition … … … (ii) adsorption … … …[2] (b) In paper or thin-layer chromatography, better separation may be achieved by running the chromatogram in one solvent, then turning the paper at right angles and running it in a second solvent. The chromatogram below was produced in this way. X solvent 2 sample applied here solvent 1 (i) Ring the spot which was insoluble in solvent 1. (ii) Label as A and B the spots which were not resolved using solvent 1. [2]

17 © UCLES 2011 [Turn over 9701/41/M/J/11 For Examiner’s Use (c) The mass spectrum shown was obtained from a compound of formula CpHqX, where X represents a halogen atom. 100 80 60 40 20 relative intensity 0 10 20 30 40 50 60 70 80 90 100 110 120 130 m / e R (i) Deduce the identity of X, giving a reason. X is … … (ii) If the relative heights of the M and M+1 peaks are 9 and 0.3 respectively, calculate the value of p. Use this value and the m / e value of the molecular ion to calculate the value of q, and hence the molecular formula of the compound. Show your working. (iii) Suggest a formula for the ion responsible for the peak labelled R. … [4] (d) In the fragmentation of alcohols which occurs in a mass spectrometer, small stable, neutral molecules are sometimes produced. Suggest the identity of two such molecules, each with an Mr less than 30. (i) … (ii) … [2] [Total: 10]

18 © UCLES 2011 9701/41/M/J/11 For Examiner’s Use 9 In today’s world, many traditional materials have been replaced by different sorts of polymers. This includes rigid polymers such as those used in car bodies to replace steel and flexible polymers like those used in textiles to replace cotton or wool. (a) (i) To form a polymer, what is the minimum number of functional groups that the monomer must possess? … (ii) Illustrate your answer to (i) with the structure of a possible monomer. [2] (b) State two differences between addition and condensation polymerisation. (i) … … (ii) … …[2] (c) The polymer formed from the co-polymerisation of the two monomers shown is known as Terylene. C benzene-1, 4-dicarboxylic acid ethane-1-2-diol CH2 H2C C O HO HO OH OH O (i) The two monomers react by condensation polymerisation. What other molecule is formed in this reaction? …

19 © UCLES 2011 9701/41/M/J/11 For Examiner’s Use (ii) Draw the structure of one repeat unit of Terylene. (iii) What is the name given to polymers containing the same functional group as Terylene? … [4] (d) The monomers ethene and but-1-ene can also co-polymerise to form a polyalkene, but this does not produce a regular alternating structure like Terylene. Explain why this is the case, drawing diagrams if you wish. … … … … [2] [Total: 10]

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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the May/June 2011 question paper for the guidance of teachers 9701 CHEMISTRY 9701/41 Paper 4 (A2 Structured Questions), maximum raw mark 100 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes must be read in conjunction with the question papers and the report on the examination. • Cambridge will not enter into discussions or correspondence in connection with these mark schemes. Cambridge is publishing the mark schemes for the May/June 2011 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 2 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9701 41 © University of Cambridge International Examinations 2011 1 (a) N≡N triple bond is (very) strong or the N2 molecule has no polarity [1] (b) 3Mg(s) → 3Mg2+(g) ∆H1 = 3 × 148 + 3 × 2186 = 7002 N2(g) → 2N3–(g) ∆H2 = 994 + 2 × 2148 = 5290 LE = –∆H1 – ∆H2 – 461 = –12,753 (kJ mol–1) (–[1] for each error) [3] (c) (i) Li3N + 3H2O → NH3 + 3LiOH (balanced equation) [1] (ii) advantage: no high pressure/temperature/catalyst needed/standard conditions used [1] disadvantage: Li is expensive or Li would need to be recycled/removed or LiOH by-product is corrosive/strongly basic or this would be a batch, rather than continuous process [1] (d) (i) Li3N: 100 × 14/35 = 40% N [1] urea: 100 × 28/60 = 47% N [1] (ii) amide [1] (iii) NH2CONH2 + H2O → 2NH3 + CO2 or → NH2CO2H + NH3 or NH2CONH2 + 2H2O → 2NH3 + H2CO3 [1] (iv) The LiOH would be strongly alkaline or would increase the pH of the soil or would ‘burn’ the crops/reduce plant growth/stunt plants or would contaminate the environment [1] [Total: 12]

Page 3 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9701 41 © University of Cambridge International Examinations 2011 2 (a) (i) One that can go in either direction. [1] (ii) both forward & reverse reactions are going on at the same time, but the concentrations of all species do not change (owtte) or rate of forward = rate of backward reaction [1] (b) (i) Kc = [H+][OH–]/[H2O] [1] (ii) Kw = [H+][OH–] [1] rearrangement of equation in (i) gives Kc[H2O] = [H+][OH–] & Kw = Kc[H2O] (owtte) or the [H2O] is contained within Kw [1] (iii) Kw will be higher in hot water because reaction is endothermic [1] (c) (i) [OH–] = 5 × 10–2; [H+] = (1 × 10–14)/ 5 × 10–2 = 2 × 10–13 [1] pH = –log10[H+] = 12.7 (correct ans = [2]) ecf [1] (ii) [NH4 +] = [OH–] (= x) [1] x2 = 1.8 × 10–5 × 0.05 ⇒ x (= [OH–]) = 9.49 × 10–4 (mol dm–3) (correct ans = [2]) [1] (iii) [H+] = Kw/[OH–] = (1 × 10–14)/ 9.49 × 10–4 = 1.05 × 10–11(mol dm–3) ecf [1] (iv) pH = 11.0 ecf [1] [Total: 12 max 11] 3 (a) (+)1; (+)2; (+)3; (+)4 [1] O.N. corresponds to the no. of electrons in outer/valence shell/lost [1] (b) PCl5 fizzes or white/misty fumes or heat evolved [1] PCl5 + 4H2O → H3PO4 + 5HCl or PCl5 + 3H2O → HPO3 + 5HCl (allow partial hydrolysis: PCl5 + H2O → POCl3 + 2HCl) [1] (c) (i) P = 30.4/31 = 0.98 Cl = 69.6/35.5 = 1.96 [1] Thus E.F = PCl2 [1] Mr(PCl2) = 102, so 2 × PCl2 = 204 ≈ 200, so M.F. = P2Cl4 [1] (ii) [1] (iii) O.N. = (+)2 [1] (iv) (HO)2P-P(OH)2 or H(HO)P(=O)-P(=O)(OH)H ecf from structure in (ii) [1] Allow HO-P-OH or HO-P=O │ H [Total: 10] Cl P P Cl Cl Cl (ignore lone pairs on Cl)

Page 4 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9701 41 © University of Cambridge International Examinations 2011 4 (a) N2 + 2O2 → 2NO2 (or via NO) or 2NO + O2 → 2NO2 [1] (b) (i) catalytic converter and passing the exhaust gases over a catalyst/Pt/Rh [1] (ii) NO2 + 2CO → ½ N2 + 2CO2 or similar [1] Allow 2NO2 + CH4 → CO2 + N2 + 2H2O (c) No, it wouldn’t be reduced. Because the reaction in (a) does not presuppose a particular fuel (owtte) [1] Allow formed from N2 and O2 in air during combustion (d) (i) SO3 produces acid rain [1] (ii) NO + ½ O2 → NO2 [1] (iii) Kp = ( 3 SO NO.p p )/( 2 SO 2 NO .p p ) [1] units: dimensionless/none (don’t accept just a blank!) [1] (iv) Kp = 99.82/0.22 = 2.5 × 105 [1] (v) It will shift to the right (owtte) [1] because the reaction is exothermic. NOT just Le Chatelier argument [1] [Total: 11] 5 (a) transformation reagent + conditions C2H4 → C2H5Cl HCl, no light or catalyst C2H5OH → C2H5Cl conc HCl + ZnCl2 or SOCl2 or PCl5 or PCl3 and heat C2H6 → C2H5Cl Cl2 + light C2H4 → C2H4Cl2 Cl2, no light or catalyst CH3CO2H → CH3COCl SOCl2 or PCl5 or PCl3 and heat H3C H3C Cl Cl2 + AlCl3 CH3 CH2Cl Cl2 + light or heat [6]

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9701 41 © University of Cambridge International Examinations 2011 (b) (i) production of NO2 +: 2H2SO4 + HNO3 → 2HSO4 – + H3O+ + NO2 + [1] (accept H2SO4 + HNO3 → HSO4 – + H2O + NO2 +) NO2 H NO2 curly arrow from ring to NO2 + and from C-H bond to ring [1] correct intermediate, including charge in the right place Note charge area must be more than half ring [1] (ii) C is C6H5CO2H [1] (iii) step 1: reagent is hot acidified or alkaline KMnO4 [1] step 2: reagent is Br2 + FeBr3/AlCl3 etc. (H2O or light negates) [1] (If C is given as 3-bromotoluene, then allow the last [2] marks if steps 1 and 2 are reversed.) [Total: 12] 6 (a) (i) aqueous alkaline iodine or I2 + OH–(aq) allow NaClO + KI [1] (ii) CH3CO- or CH3CH(OH)- [1] (iii) Pale yellow ppt. or antiseptic smell [1] (iv) compound result CH3OH
CH3CH2OH ✓ CH3CHO ✓ CH3CO2H
CHO
COCH3 ✓ •✓•✓•✓ [3]

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9701 41 © University of Cambridge International Examinations 2011 (b) (allow displayed, skeletal or structural formulae) CHO O + CO2H O D E F for D allow CHO (no mark for CO2H O ) (D + E + F): 3 × [1] (c) (allow displayed, skeletal and structural formulae) Must be consistent with F CO2H CO2H CO2H CO2H OH G H J K allow CO2H OH (only allow mark for H if G is the 3-OH acid) for G (N.B. letters H, J, K can be swapped around) (G + H + J + K): 4 × [1] geometrical or cis-trans isomerism [1] [Total: 14]

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9701 41 © University of Cambridge International Examinations 2011 7 (a) The tertiary/3-dimensional structure/shape is held together by hydrogen/ionic/van der Waals bonds [1] These break (relatively) easily/are weak/break at/above 45 °C [1] (b) (or similar diagrams) Enzyme + substrate Enzyme-substrate complex Enzyme + products 3 × [1] (c) a competitive inhibitor combines with the enzyme’s active site (so preventing the substrate from binding) [1] non-competitive inhibitor bonds with the enzyme away from the active site/at an allosteric site [1] this changes the shape of the active site [1] Also allow competitive inhibition can be overcome by increasing [substrate] or non-competitive inhibition cannot be removed by increasing [substrate] for the 3rd mark (d) (i) Line must be of similar shape to original but level out below original line [1] (ii) Inhibitor reduces the number of enzymes with ‘working’ active sites (owtte) [1] [Total: 10]

Page 8 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9701 41 © University of Cambridge International Examinations 2011 8 (a) partition – separation due to the different solubilities of compounds in two solvents/phases [1] adsorption – separation due to the different attractions between the compounds and the stationary phase, relative to their solubility in the solvent [1] Note, if candidates do not refer to different solubilities and different attractions max 1 (b) Ring: [1] A + B: [1] (c) (i) X is bromine – M and (M+2) peaks almost same height [1] (ii) 0.3 100 n 9 1.1 100 1 M M = × = + 1.1 × n Hence 3.03 9 1.1 0.3 100 n = × × = p = 3 (answer + working) [1] (If the mass peak is at 122 and the compound contains Br and 3 C atoms then Q = (122 – 79 – 36)) thus Q = 7 ecf from (ii) [1] (The compound is C3H7Br) (iii) (R is at m/e 43), hence C3H7 + [1] (d) Any two from H2, H2O, CO, C2H4, C2H2, CH4 2 × [1] [Total: 10] X - sample A B

Page 9 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9701 41 © University of Cambridge International Examinations 2011 9 (a) (i) One [1] (ii) Any alkene (or allow a cyclic amide, as in caprolactam) [1] (b) Any TWO from: addition needs unsaturated/double bonds/alkene condensation eliminates a small molecule condensation needs a molecule other than a hydrocarbon empirical formula of addition polymer is the same as that of its monomer condensation needs two different functional groups (NOT – “condensation needs two different monomers”) 2 × [1] (c) (i) Water [1] (ii) O CH2 CH2 O O O Correct ‘ester’ bond [1] ‘sticks’ to rest of molecule [1] Note : candidates need only show ‘brackets’ if more than one repeat unit shown (iii) Polyesters [1] (d) Monomers in Terylene have to alternate in order to condense out water (owtte) [1] Alkenes can link in any order (and still form a polyalkene) (or diagram showing this) [1] [Total: 10]