Cambridge A Level Chemistry 9701 — 2008 Oct/Nov Paper 4 · Variant 1

This document consists of 18 printed pages and 2 blank pages. SHW 00019 4/07 T63628/1 © UCLES 2008 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level and 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. * 8 2 6 8 3 7 4 4 3 6 * CHEMISTRY 9701/04 Paper 4 Structured Questions October/November 2008 1 hour 45 minutes Candidates answer on the Question Paper. Additional Materials: Data Booklet For Examiner’s Use 1 2 3 4 5 6 7 8 9 10 Total

9701/04/O/N/08 For Examiner’s Use © UCLES 2008 Section A Answer all questions in the spaces provided. 1 (a) Natural bromine consists of the two isotopes 79Br and 81Br in roughly equal proportions. The mass spectrum of bromine consists of 5 peaks. (i) Suggest the mass numbers for the 5 peaks and the identities of the species responsible for them. … … … (ii) Suggest the ratios of the relative abundances of • the three lines with the highest mass numbers, … • the two lines with the lowest mass numbers. … [4] Esters of 2,3-dibromopropan-1-ol with phosphoric acid are useful flame retardants used in plastics and fibres. 2,3-dibromopropan-1-ol can be made from propenal by the following two-stage process. CH2 CH2OH CH2 Br CHO C H II I propenal Br CH 2,3-dibromopropan-1-oI A 2

3 9701/04/O/N/08 [Turn over For Examiner’s Use © UCLES 2008 (b) (i) Draw the structure of the intermediate A in the box opposite. (ii) Suggest reagents and conditions for • reaction I, … • reaction II. … [3] (c) The mass spectrum of 2,3-dibromopropan-1-ol includes the following peaks. mass number relative abundance 31 100 106 44 108 45 185 0.3 187 0.6 189 0.3 (i) At what mass number would you expect the molecular ion to occur? … (ii) Identify the molecular formula (including isotopic composition where relevant) of these 6 peaks. mass number molecular formula 31 106 108 185 187 189 [5] [Total: 12]

4 9701/04/O/N/08 For Examiner’s Use © UCLES 2008 2 In the late 19th century the two pioneers of the study of reaction kinetics, Vernon Harcourt and William Esson, studied the rate of the reaction between hydrogen peroxide and iodide ions in acidic solution. H2O2 + 2I– + 2H+ 2H2O + I2 This reaction is considered to go by the following steps. step 1 H2O2 + I– IO– + H2O step 2 IO– + H+ HOI step 3 HOI + H+ + I– I2 + H2O The general form of the rate equation is as follows. rate = k[H2O2]a[I–]b[H+]c (a) Suggest how the appearance of the solution might change as the reaction takes place. … [1] (b) Suggest values for the orders a, b and c in the rate equation for each of the following cases. case numerical value a b c step 1 is the slowest overall step 2 is the slowest overall step 3 is the slowest overall [3] A study was carried out in which both [H2O2] and [H+] were kept constant at 0.05 mol dm–3, and [I–] was plotted against time. The following curve was obtained. 0.001 0.0009 0.0008 0.0007 0.0006 0.0005 0.0004 0.0003 0.0002 0.0001 0 0 30 60 90 120 150 180 210 240 270 300 time / s [I– ion] / mol dm–3

5 9701/04/O/N/08 [Turn over For Examiner’s Use © UCLES 2008 To gain full marks for the following answers you will need to draw relevant construction lines on the graph opposite to show your working. Draw them using a pencil and ruler. (c) Calculate the initial rate of this reaction and state its units. rate = … units … [2] (d) Use half-life data calculated from the graph to show that the reaction is first order with respect to [I–]. … [2] (e) Use the following data to deduce the orders with respect to [H2O2] and [H+], explaining your reasoning. [H2O2] / mol dm–3 [H+] / mol dm–3 relative rate 0.05 0.05 1.0 0.07 0.05 1.4 0.09 0.07 1.8 … … order with respect to [H2O2] = … order with respect to [H+] = … [2] (f) From your results, deduce which of the three steps is the slowest (rate determining) step. … [1] [Total: 11]

6 9701/04/O/N/08 For Examiner’s Use © UCLES 2008 3 (a) (i) Describe and explain the trend observed in the thermal stability of the carbonates of the Group II elements. … … … … … (ii) By quoting suitable data from the Data Booklet suggest how the thermal stabilities of • zinc carbonate and • lead carbonate might compare to that of calcium carbonate. … … … … … [6] (b) Malachite is an ore of copper. It contains the following percentages by mass. copper 57.7% oxygen 36.2% carbon 5.4% hydrogen 0.9% Malachite reacts with dilute H2SO4 producing a gas B that turns limewater milky and leaving a blue solution C. When heated in the absence of air, malachite produces gas B and steam, and leaves a black solid D. D reacts with dilute H2SO4 to produce the same blue solution C. Adding iron filings to C produces a pink solid E and a pale green solution F. (i) Calculate the empirical formula of malachite. …

7 9701/04/O/N/08 [Turn over For Examiner’s Use © UCLES 2008 (ii) Suggest the formula of the ion responsible for the blue colour of solution C. … (iii) Identify the black solid D and calculate the mass of D that could be obtained by heating 10 g of malachite. … … … … (iv) Use data from the Data Booklet to identify the pink solid E and the solution F, and suggest an equation for the reaction producing them. … … … (v) What type of reaction is the reaction that produces E and F? … (vi) Describe and explain what you would see happen when dilute NH3(aq) is added slowly to the solution C until it is in an excess. … … … … … [13] [Total: 19]

8 9701/04/O/N/08 For Examiner’s Use © UCLES 2008 4 (a) The viscosity of engine oil can be improved by the addition of certain medium chain- length polymers. A portion of the chain of one such polymer is shown below. –CH2CH(CH2CH2CH3)CH2CH(CH2CH2CH3)CH2– On average, the molecules of the medium-chain polymer contain 40 carbon atoms. (i) Suggest the structure of the monomer. … (ii) How many monomer units are incorporated into the average molecule of the polymer? … [2] (b) Used car engine oil can be recycled for use as a fuel by the processes of distillation and cracking. (i) Assuming a typical molecule of engine oil has the formula C40H82, suggest an equation for a cracking reaction that could produce diesel fuel with the formula C16H34 and other hydrocarbons only. … (ii) What conditions are needed for this cracking reaction? … (iii) Considering only the bonds broken and the bonds formed during the reaction, use the Data Booklet to calculate the enthalpy change for the reaction you wrote in (b)(i). … … … (iv) Comment on how the conditions you described in (b)(ii) relate to the enthalpy change you calculated in (b)(iii). … … [4] [Total: 6]

9 9701/04/O/N/08 [Turn over For Examiner’s Use © UCLES 2008 5 (4-aminophenyl)ethanoic acid (4-APEA) and its derivatives are being investigated as possible drugs to treat chronic inflammation of the intestines. The synthesis of 4-APEA from methylbenzene is shown in the following scheme. CH3 I II CH2Cl G H3O+ HNO3 + H2SO4 NO2 NO2 V IV III H CH2CN NH2 4-APEA CH2CO2H (a) Draw the structures of the compounds G and H in the boxes above. [2] (b) Suggest reagents and conditions for the following steps. • step II … • step III … • step V … [3] [Total: 5]

10 9701/04/O/N/08 For Examiner’s Use © UCLES 2008 6 Suggest a test or simple reaction you could carry out on each of the following pairs of compounds to enable them to be distinguished. (a) CH3COCH2CH2CH3 CH3CH2COCH2CH3 J K (i) description of test or reaction … … (ii) observation with compound J … (iii) observation with compound K … [2] (b) NH2 NH2 L M (i) description of test or reaction … … (ii) observation with compound L … (iii) observation with compound M … [2]

11 9701/04/O/N/08 [Turn over For Examiner’s Use © UCLES 2008 (c) CH3CH2COCl CH3CH2CH2Cl N P (i) description of test or reaction … … (ii) observation with compound N … (iii) observation with compound P … [2] (d) CH3CH2CONH2 CH3CH2CH2NH2 Q R (i) description of test or reaction … … (ii) observation with compound Q … (iii) observation with compound R … [2] [Total: 8]

12 9701/04/O/N/08 For Examiner’s Use © UCLES 2008 7 (a) Explain briefly what is meant by the word protein. … … [1] (b) Describe how peptide bonds are formed between amino acids during the formation of a tripeptide. Include diagrams and displayed formulae in your answer. … [3] (c) Describe how proteins can be broken down into amino acids in the laboratory without the aid of enzymes. … [2] (d) When a small polypeptide S was broken down in this way, three different amino acids were produced according to the following reaction. CH2 S Mr = 165 2 NH2CHCO2H CH3 Mr = 89 2 NH2CHCO2H + Mr = 75 3 NH2CH2CO2H + (i) How many peptide bonds were broken during this reaction? … (ii) Calculate the Mr of the polypeptide S. Mr = … [3] [Total: 9]

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9701/04/O/N/08 For Examiner’s Use © UCLES 2008 Section B – Applications of Chemistry Answer all questions in the spaces provided. 8 (a) Enzymes play a vital role in all living organisms, helping chemical reactions to take place at body temperature. (i) The diagram below shows the reaction pathway of an enzyme-catalysed reaction without an enzyme present. On the diagram sketch the pathway if the enzyme was present. energy reaction pathway reactants products (ii) What type of molecule are most enzymes? … (iii) Why do many enzymes lose their catalytic effectiveness above 40 °C? … [3] (b) (i) Explain the difference between competitive and non-competitive inhibition of an enzyme. … … … 14

15 9701/04/O/N/08 [Turn over For Examiner’s Use © UCLES 2008 (ii) The graph below shows how the rate of an enzyme-catalysed reaction varies with substrate concentration in the absence of an inhibitor. For a given amount of enzyme, Vmax represents the rate when all of the active sites on the enzyme are being used. reaction rate substrate concentration Vmax Sketch on the diagram curves to show the effect on the rate of reaction of: I a competitive inhibitor; II a non-competitive inhibitor. Clearly label your curves. [4] (c) Heavy metal ions like Hg2+ can bind irreversibly to enzymes and this can result in poisoning. (i) Suggest to what atom or group Hg2+ ions bind. … (ii) Explain how this affects enzyme activity. … … … [3] [Total: 10]

16 9701/04/O/N/08 For Examiner’s Use © UCLES 2008 9 The technology of DNA fingerprinting has enormously advanced scientific identification techniques in medicine, crime detection and archaeology in recent years. (a) (i) In order to prepare a DNA sample for analysis, the DNA is treated with restriction enzymes. What do restriction enzymes do? … … (ii) What is the next stage in DNA analysis, after the treatment with restriction enzymes? … (iii) How are the DNA fragments made visible? … [3] (b) NMR and X-ray crystallography have made significant contributions to our knowledge of the structure of proteins and, in the pharmaceutical industry, how drugs react with target proteins. (i) Suggest an advantage of each technique in helping to determine protein structure. … … … (ii) MRI scanning is a medical technique based on NMR spectroscopy. It is particularly useful for looking for tumours in healthy tissue. Suggest how this technique can distinguish tumour tissue from healthy tissue. … … [3]

17 9701/04/O/N/08 [Turn over For Examiner’s Use © UCLES 2008 (c) A saturated molecule of formula CxHyNO was subjected to analysis by mass spectrometry and NMR spectroscopy. In the mass spectrum of the compound, the M peak was at m/e 73 and the ratio of the heights of the M:M+1 peak was 48 : 1.7. (i) Using the data from the mass spectrum, determine the values of x and y in the formula of the compound. (ii) Use the data from (i) together with the NMR spectrum below to deduce a structure for the compound, explaining how you arrive at your answer. 11 10 9 8 7 6 5 chemical shift,
/ ppm 4 3 2 1 0 [4] [Total: 10]

18 9701/04/O/N/08 For Examiner’s Use © UCLES 2008 10 (a) Silk from silkworms, used as a fabric shows a different secondary structure to that produced by spiders. silkworm silk spider dragline silk less ordered
sheets glycine rich strand highly ordered
sheets (i) What sort of bonding would you expect to occur between adjacent parts of the protein chains in each form of silk? silkworm … spider … (ii) Suggest two differences in properties that these forms of silk could have. Explain your answer. … … … … (iii) Spider dragline silk contains large amounts of the amino acid glycine. How does this affect the properties of the silk? … … [5]

19 9701/04/O/N/08 For Examiner’s Use © UCLES 2008 (b) Both forms of silk are condensation polymers. (i) Explain what is meant by a condensation polymer. … … (ii) Another type of polymer is called an addition polymer. Name an example of an addition polymer. … (iii) Suggest why condensation polymers such as proteins show a wider range of properties than addition polymers. … … … … … [5] [Total: 10]

20 9701/04/O/N/08 BLANK PAGE Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. University of Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2008 question paper 9701 CHEMISTRY 9701/04 Paper 4 (Theory 2), 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. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2008 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 2 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2008 9701 04 © UCLES 2008 1 (a) (i) 162 (81Br – 81Br+) for molecular species [1] 160 (81Br – 79Br+) for atomic species [1] 158 (79Br – 79Br+) ignore missing charges for 5 masses [1] 81 (81Br+) 79 (79Br+) (ii) 158:160:162 =1:2:1 [1] 79:81 =1:1 [1] (b) (i) either BrCH2CHBr-CHO or CH2=CH-CH2OH (double bond needed) [1] (ii) reaction I: Br2(aq or in CCl4 etc.), light negates – solvent not needed [1] reaction II: NaBH4 or H2/Ni etc. (but not if A is CH2=CH-CH2OH) allow LiAlH4 or Na/ethanol [1] (reactions can be reversed) (c) (i) C3H6OBr2 = 216, 218 and 220 (any one) [1] (ii) 31 is CH2OH+/CH3O+ 106 is C2H3 79Br+ 108 is C2H3 81Br+ 185 is C2H3 79Br2 + ignore missing charges 187 is C2H3 79Br81Br+ 6 correct [4] 189 is C2H3 81Br2 + 5 correct [3] etc if no mass numbers given – [1] only [4] [Total: 13 max 12] 2 (a) solution will turn brown/purple [1] (b) table: case a b c 1 1 1 0 2 1 1 1 3 1 2 2 each horizontal row scores [1] if no marks scored, a correct vertical row can score [1] [3 max] (c) rate = 6.5–7.5 × 10–6 [1] units are mol dm–3 s–1 [1] (d) half-life measured and quoted as ≅ 90–94 s [1] evidence of two half-lives measured [1]

Page 3 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2008 9701 04 © UCLES 2008 (e) lines 1 and 2: as [H2O2] increases by 0.07/0.05 = 1.4, so does rate so order w.r.t. [H2O2] = 1 [1] lines 1 and 3: increase in rate (1.8) is also the increase in [H2O2], so rate is independent of [H+] (or zero order) [1] a description can be accepted here if both orders are correct but no working/explanation given score [1] (f) the first step/or the relevant equation [1] [Total: 11] 3 (a) (i) carbonates become more stable down the Group/higher decomposition temperature [1] cation/M2+ radius/size increases down the group/M2+ charge density decreases [1] anion/carbonate ion/CO3 2– suffers less polarisation/distortion [1] (ii) ionic radii quoted: Ca2+: 0.099 nm Zn2+: 0.074 nm Pb2+: 0.120 nm [1] thus we expect ZnCO3 to be less stable, but PbCO3 to be more stable [1] if candidate states PbCO3 is more stable than ZnCO3 (or converse) with no reference to CaCO3 give [1] as salvage. (b) (i) Cu = 57.7/63.5 = 0.91 ratios correct scores [1] O = 36.2/16 = 2.26 C = 5.4/12 = 0.45 H = 0.9/1 = 0.90 hence Cu2O5CH2 [1] (ii) Cu2+(aq) or [Cu(H2O)6]2+ NOT [Cu(H2O)4]2+ [1] (iii) D is CuO / copper(II) oxide [1] Cu2O5CH2 → 2CuO + CO2 + H2O [1] 221 → 159 (Mrs) [1] ∴ 10 → 10 × 159/221 = 7.2 g (7.19) if candidate thinks only CO2 is lost, answer will be 8.0 g [1] (iv) E is copper; F is Fe2+ / Fe SO4 [1] Fe + Cu2+ → Fe2+ + Cu (or molecular) [1] (v) redox/displacement [1] (vi) blue ppt./solid formed [1] (dissolves to give) dark blue/purple colour [1] blue ppt. is Cu(OH)2(s) [1] deep blue is [Cu(NH3)4]2+ (allow [Cu(NH3)4(H2O)2]2+ NOT [Cu(NH3)6]2+ [1] [Total: 19]

Page 4 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2008 9701 04 © UCLES 2008 4 (a) (i) CH2=CH–CH2CH2CH3 accept C3H7 on RHS [1] (ii) 8 [1] (b) (i) e.g. C40H82 → C16H34 + 2 C12H24 OR C24H48 [1] (ii) heat + catalysts/SiO2/Al2O3/Pt/ceramic/pumice/zeolite etc [1] if temp given >500 °C (iii) bonds broken: 4(C–C) = 4 × 350 = 1400 kJ mol–1 bond formed: 2 (C=C) = 2 × 610 = 1220 kJ mol–1 ∴∆H = +180 kJ mol–1 [1] from eqn in (i) : +90 kJ mol –1 for each C=C formed (could be multiples of 90) (iv) endothermic reactions ∆H > 0 [1] [Total: 6] 5 (a) G is 4-nitromethylbenzene [1] H is 4-nitrophenylethanoic acid [1] (b) step II: Cl2 + light or heat (T~100 °C) (AlCl3 or aq. negates) [1] step III: KCN (in ethanol) + heat (T~75 °C) (HCN negates) [1] step V: Sn or Fe + HCl (+ heat) [1] [Total: 5] 6 (a) alkaline aqueous iodine (NaOH/I2) (allow NaOI) [1] J gives yellow ppt; K gives no reaction [1] (b) aqueous bromine / Cu2+ aq / diazotisation with phenol [1] L gives no change; M decolourises/gives white ppt. with Cu2+ L goes blue, M goes green with diazotisation L gives no reaction, M a coloured compound [1] (c) drop of water [1] N fizzes/gives off steamy fumes; P has no reaction [1] or add AgNO3(aq) [1] N gives rapid ppt.; P gives ppt. very slowly [1] or add NH3/RNH2 [1] N gives off fumes; P has no reaction [1] or add alcohol/phenol [1] N produces sweet-smelling liquid, P gives no reaction [1] (d) Universal Indicator solution/litmus [1] Q shows no change; R will turn solution blue (alkaline) [1] [Total: 8]

Page 5 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2008 9701 04 © UCLES 2008 7 (a) protein: polymer of amino acids / amino acids are monomers. [1] (b) diagram of at least two amino acids joining by the loss of water [1] at least one peptide bond drawn out in full [1] correct formula of the tripeptide [1] (c) acid/H+/HCl etc. or alkali/OH–/NaOH NOT conc H2SO4 or any HNO3 [1] heat/boil/reflux if temp given >90 °C [1] (d) (i) six [1] (ii) Mr = 3 × 75 + 2 × 89 + 2 × 165 – 6 × 18 [1] = 625 [1] (allow [1] for Mr = 733) (also ecf from (i)) [Total: 9] 8 (a) (i) dotted line must start and end at same points [1] (ii) protein/polypeptide NOT polymer/polyamide [1] (iii) they are denatured/lose their 2°/3° structure/or H-bonds/vdW [1] (b) (i) competitive inhibitor resembles the substrate OR competes for the active site of the enzyme [1] non-competitive inhibitor can bind to a different site on the enzyme OR forms a covalent bond/bonds permanently with the enzyme [1]

Page 6 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2008 9701 04 © UCLES 2008 (ii) mark for each line NB lines must cross to score mark for II [2 × 1] (c) (i) –S–H groups (allow sulphide/S/cysteine residue) [1] (ii) this inhibits/reduces/decreases the enzyme activity/stops normal function [1] the bonding disrupts the 3-dimensional structure of the enzyme [1] [Total: 10] 9 (a) (i) cut DNA into sections / fragments / minisatellites [1] (ii) these undergo electrophoresis OR are placed on agarose gel [1] (iii) radioactive phosphorus / 32P OR darkens photographic film [1] (b) (i) NMR can be done in solution / in vivo / shows labile protons / shows positions of protons and/or carbon atoms [1] X-ray crystallography shows the positions of most atoms in structure / allows measurement of bond length [1] (ii) different types of tissue have protons in different chemical environments / tumour and healthy tissue absorb differently / allow at different frequencies [1] (c) (i) M : M+1 = 48 : 1.7 x = 100 × 1.7 = 3.2 hence there are 3 carbon atoms in the compound [1] 1.1 × 48 NB if calculation shown 1.1 divisor MUST be present since the compound has an m/e of 73 and contains 3 carbon atoms, 1 nitrogen atom and 1 oxygen atom, y = 73–(36 +14+16) = 7 [1] (ii) the NMR spectrum shows a quartet, triplet pattern characteristic of an ethyl group [1] the other broad peak must be due to N–H protons [1] thus the structure of the compound is likely to be CH3CH2CONH2 [1] [Total: 11 max 10]

Page 7 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2008 9701 04 © UCLES 2008 10 (a) (i) silkworm – hydrogen bonds [1] spider – van der Waals’ OR hydrogen bonds [1] (ii) spider silk is more elastic/flexible/less rigid than silkworm silk/has a lower density [1] silkworm silk absorbs water more easily [1] (iii) this increases the elasticity/hydrophobic nature of the silk [1] (b) (i) a polymer formed with the elimination/formation of a small molecule (or example) [1] (ii) any addition polymer e.g. poly(ethene), PVC, etc. [1] (iii) 3 from: addition polymers have a limited range of bonds/monomers [1] addition polymers are non-polar/have fewer/no H-bonds [1] condensation polymers/proteins have a range of combinations of amino acids which give a wide range of properties [1] condensation polymers/proteins have more functional groups/sidechains [1] different sequences of amino acids result in different 2°/3° structure [1] [Total: 12 max 10]