Cambridge A Level Chemistry 9701 — 2013 May/June Paper 4 · Variant 3

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 soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fl uid. 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. Electronic calculators may be used. 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. CHEMISTRY 9701/43 Paper 4 Structured Questions May/June 2013 2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certifi cate of Education Advanced Level This document consists of 15 printed pages and 1 blank page. [Turn over IB13 06_9701_43/FP © UCLES 2013 *9673577343* For Examiner’s Use 1 2 3 4 5 6 7 8 Total

2 9701/43/M/J/13 © UCLES 2013 For Examiner’s Use Section A Answer all the questions in the spaces provided. 1 (a) What is meant by the term standard electrode potential, SEP? … … [2] (b) Draw a fully labelled diagram of the apparatus you could use to measure the SEP of the Fe3+ / Fe2+ electrode. [5] (c) The reaction between Fe3+ ions and I– ions is an equilibrium reaction. 2Fe3+(aq) + 2I–(aq) 2Fe2+(aq) + I2(aq) (i) Use the Data Booklet to calculate the for this reaction. … (ii) Hence state, with a reason, whether there will be more products or more reactants at equilibrium. … … (iii) Write the expression for Kc for this reaction, and state its units. Kc = units …

3 9701/43/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use An experiment was carried out using solutions of Fe3+(aq) and I–(aq) of equal concentrations. 100 cm3 of each solution were mixed together, and allowed to reach equilibrium. The concentrations at equilibrium of Fe3+(aq) and I2(aq) were as follows. [Fe3+(aq)] = 2.0 × 10–4 mol dm–3 [I2(aq)] = 1.0 × 10–2 mol dm–3 (iv) Use these data, together with the equation given in (c), to calculate the concentrations of Fe2+(aq) and I–(aq) at equilibrium. [Fe2+(aq)] = … mol dm–3 [I–(aq)] = … mol dm–3 (v) Calculate the Kc for this reaction. Kc = … [8] [Total: 15]

4 9701/43/M/J/13 © UCLES 2013 For Examiner’s Use 2 Ethyl ethanoate is hydrolysed slowly by water in the following acid-catalysed reaction. H+ CH3CO2CH2CH3 + H2O CH3CO2H + CH3CH2OH The concentration of ethyl ethanoate was determined at regular time intervals as the reaction progressed. Two separate experiments were carried out, with different HCl concentrations. The following graph shows the results of an experiment using [HCl ] = 0.1 mol dm–3. 0 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 0 20 40 60 80 100 120 time / min [CH3CO2CH2CH3] / mol dm–3 (a) When the experiment was carried out using [HCl ] = 0.2 mol dm–3, the following results were obtained. time / min [CH3CO2CH2CH3] / mol dm–3 0 0.200 10 0.160 25 0.115 50 0.067 75 0.038 100 0.022 125 0.013 (i) Plot these data on the axes above, and draw a line of best fi t.

5 9701/43/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (ii) Use one of the graphs to show that the reaction is fi rst order with respect to CH3CO2CH2CH3. Show all your working, and show clearly any construction lines you draw on the graphs. (iii) Use the graphs to calculate the order of reaction with respect to HCl. Show all your working, and show clearly any construction lines you draw on the graphs. (iv) Write the rate equation for this reaction, and calculate the value of the rate constant. rate = [7] (b) (i) Why is it not possible to determine the order of reaction with respect to water in this experiment? … … (ii) Although [CH3CO2CH2CH3] decreases during each experiment, [HCl ] remains the same as its initial value. Why is this? … … [2] [Total: 9]

6 9701/43/M/J/13 © UCLES 2013 For Examiner’s Use 3 (a) (i) What is meant by the density of a substance? … (ii) Use data from the Data Booklet to explain why the density of iron is greater than that of calcium. … … … [3] (b) In general, reactions of the compounds of transition elements can be classifi ed under one or more of the following headings. acid-base ligand exchange precipitation redox Choose the most suitable heading to describe each of the following reactions, by placing a tick ( ) in the appropriate column in the table below. Only one tick should be placed against each reaction. reaction acid-base ligand exchange precipitation redox [Cu(H2O)6]2+ + 4NH3 → [Cu(NH3)4]2+ + 6H2O [Cu(H2O)6]2+ + 4HCl → [CuCl 4]2– + 4H+ + 6H2O 2FeCl 2 + Cl 2 → 2FeCl 3 [Fe(H2O)6]2+ + 2OH– → Fe(OH)2 + 6H2O 2Fe(OH)2 + ½O2 + H2O → 2Fe(OH)3 CrO3 + 2HCl → CrO2Cl 2 + H2O Cr(H2O)3(OH)3 + OH– → [Cr(H2O)2(OH)4]– + H2O [Cr(OH)4]– + 1½H2O2 + OH– → CrO4 2– + 4H2O [8] 

7 9701/43/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (c) Alloys of aluminium, titanium and vanadium are used in aerospace and marine equipment, and in medicine. When a powdered sample of one such alloy is heated with an excess of aqueous NaOH, only the aluminium reacts, according to the following equation. 2Al (s) + 2OH–(aq) + 6H2O(l) → 2[Al (OH)4]–(aq) + 3H2(g) Reacting 100 g of alloy in this way produced 8.0 dm3 of hydrogen, measured under room conditions. Calculate the percentage by mass of aluminium in the alloy. percentage = … % [3] [Total: 14]

8 9701/43/M/J/13 © UCLES 2013 For Examiner’s Use 4 Because of the lack of reactivity of the nitrogen molecule, extreme conditions need to be used to synthesise ammonia from nitrogen in the Haber process. (a) Suggest an explanation for the lack of reactivity of the nitrogen molecule, N2. … … [1] (b) Under conditions of high temperature, nitrogen and oxygen react together to give oxides of nitrogen. (i) Write an equation for a possible reaction between nitrogen and oxygen. … (ii) State two situations, one natural and one as a result of human activities, in which nitrogen and oxygen react together. … … (iii) What is the main environmental effect of the presence of nitrogen oxides in the atmosphere? … [4] (c) Describe and explain how the basicities of ethylamine and phenylamine compare to that of ammonia. … … … … … [4]

9 9701/43/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (d) Compound X is a useful intermediate in the synthesis of pharmaceuticals. X can be synthesised from chloromethylbenzene according to the following scheme. T W step 1 step 3 step 4 step 5 X CH2Cl step 2 CH2CN CH2CH2NH CH2 CH2CO2H O C (i) What type of reaction is each of the following? step 1 … step 2 … (ii) Suggest reagents and conditions for step 1, … step 2. … (iii) Draw the structures of the intermediates T and W in the boxes above. [6] [Total: 15]

10 9701/43/M/J/13 © UCLES 2013 For Examiner’s Use 5 (a) A series of experiments is carried out in which the reagent shown at the top of the column of the table is mixed, in turn, with each of the reagents at the side. Complete the following table by writing in each box the formula of any gas produced. Write x in the box if no gas is produced. The fi rst column has been completed as an illustration. H2O OH CO2H OH Na H2 KOH(aq) x Na2CO3(aq) x [5] (b) Compound C is responsible for the pleasant aroma of apples. It can be prepared from phenol by the following 3-step synthesis. OH OH C A CH3 H3C CH3 step 1 OH C B CH3 H3C CH3 step 2 O O C C CH3 CH3 H3C CH3 step 3 (i) The only by-product of step 1 is HCl. Suggest the reagent that was used to react with phenol to produce compound A. … (ii) What type of reaction is occurring in step 2? … (iii) What reagents and conditions are required for step 3? … (iv) State the reagent and conditions needed to convert C back to B, the reverse of step 3. … [5]

11 9701/43/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (c) (i) Either compound A or compound B, or both, react with the following reagents. For each reagent draw the structure of the organic product formed with A, and with B. If no reaction occurs, write ‘no reaction’ in the relevant box. reagent and conditions product with A product with B an excess of Br2(aq) heat with HBr pass vapour over heated Al 2O3 heat with acidifi ed K2Cr2O7 (ii) Choose one of the above reactions to enable you to distinguish between A and B. State below the observations you would make with each compound. reagent observation with A observation with B [7] [Total: 17]

12 9701/43/M/J/13 © UCLES 2013 For Examiner’s Use Section B Answer all the questions in the spaces provided. 6 There are two important polymerisations that occur within living organisms – protein synthesis and the formation of DNA. (a) Complete the table placing a tick ( ) in the correct column to indicate in which process each substance could be used. substance protein synthesis formation of DNA adenine alanine aspartate phosphate [3] (b) Proteins and DNA form different helical structures. Briefl y describe the bonding that maintains the shape of each of these helical structures. protein … … … DNA … … … [4] (c) Describe the differences in bonding in the primary and tertiary structures of proteins. Your answer should include reference both to the nature of the bonding and the types of amino acid causing it. … … … … … [3] [Total: 10] 

13 9701/43/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use 7 Modern methods of analysis have had far-reaching effects on a number of branches of science including medicine, forensic science, environmental monitoring and archaeology. (a) Outline, in simple terms, the technique of DNA fi ngerprinting. … … … … … … [4] (b) Complete the table by indicating whether the items can be used for DNA fi ngerprinting. Use a tick ( ) for items which can be used for DNA fi ngerprinting and a cross (x) for items which cannot. item for testing suitable for DNA fi ngerprinting human hair piece of a fl int tool piece of Iron Age pot piece of Roman leather [3] (c) Various forms of chromatography can be used to separate and analyse mixtures. HPLC (high performance liquid chromatography) can be used to separate each of the following mixtures. State another method of chromatography which would separate each mixture. insecticides in a sample of water … dyes present in a foodstuff … drug residue in an athlete’s urine … [3] [Total: 10] 

14 9701/43/M/J/13 © UCLES 2013 For Examiner’s Use 8 In recent years there has been a lot of interest in polymers in the form of gels that absorb aqueous materials. One of the largest uses of these polymers is in disposable nappies (diapers). The gel which is used in this case is a polymer of propenoic acid. O OH propenoic acid (a) (i) Draw a section of the polymer of propenoic acid showing two repeat units. (ii) By what type of chemical reaction is this polymer formed? … (iii) By what type of bonding is water held on the polymer? … [3] (b) For some disposable nappies (diapers), the monomer is a mixture of propenoic acid and sodium propenoate. The properties of the polymer are infl uenced by the proportion of sodium salt in the monomer mixture. (i) Suggest and explain how the difference in the structure of this polymer compared to one formed only from propenoic acid might affect the water absorbing properties of the polymer. … … … (ii) Suggest a property the polymer should have in order to be used in disposable products. … [3]

15 9701/43/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (c) A variation on the gel used for disposable nappies (diapers) containing more sodium propenoate has been used to treat soils contaminated by heavy metals such as lead (Pb2+) and cadmium (Cd2+). Suggest why the gel is effective. … … … [2] (d) Another variation on this type of polymer is used in hair gels. In these, the polymer chains are cross-linked by a compound known as pentaerythritol. OH OH HO HO pentaerythritol (i) By what type of chemical reaction are the cross-links in this polymer formed? … (ii) It is important that the gel should be easily washed out of hair. What is it about the structure of the polymer that allows this to happen? … [2] [Total: 10]

16 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. 9701/43/M/J/13 © UCLES 2013 BLANK PAGE

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the May/June 2013 series 9701 CHEMISTRY 9701/43 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 should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Page 2 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9701 43 © Cambridge International Examinations 2013 1 (a) The potential of an electrode compared to that of a standard hydrogen electrode (SHE) or the EMF of a cell composed of the test electrode and the SHE [1] all measurement concentrations of 1 mol dm–3 and 298 K / 1 atm pressure [1] [2] (b) H2 and good delivery system [1] Fe2+/Fe3+ solution labelled [1] platinum electrodes (both) [1] salt bridge and voltmeter [1] H+ or HCl or H2SO4 [1] (acid is not sufficient) [5] (c) (i) E ⦵ = 0.77 – 0.54 = 0.23 (V) [1] (ii) Since E ⦵ is positive/ E ⦵ >0 So more products / the equilibrium will be over to the right / forward reaction is favoured ecf from (c)(i) [1] (iii) Kc = [Fe2+]2[I2] / [Fe3+]2[I–]2 [1] units are mol–1 dm3 ecf on expression [1] (iv) ([Fe2+] must always be twice [I2], so) [Fe2+] = 0.02 (mol dm–3) [1] ([I–] must always be equal to [Fe3+], so) [I–] = 2 × 10–4 (mol dm–3) [1] (v) Kc = {(0.02)2 × 0.01} / {(2 x 10–4)2 × (2 × 10–4)2} correct expression [1] (allow ecf from incorrect expression in (c)(iii)) (allow ecf from (c)(iv)) = (4 × 10–6) / (1.6 × 10–1.5) = 2.5 × 109 (mol–1 dm3) [1] [8] [Total: 15]

Page 3 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9701 43 © Cambridge International Examinations 2013 2 (a) (i) plotting of points (–1 for any error – plotted to within ½ square) [1] a good best fit curve [1] (ii) construction lines for two half-lives and t½ ≈ 63 m or 32 m (±3 min) / t½ is constant or construction lines for two tangents and mention of two values / concentration doubled, rate doubled [1] (iii) either ratio of (initial) rates (slopes) or ratio of t½ = 2.0 [1] so reaction is first order w.r.t. [HCl] [1] (iv) rate = k[CH3CO2CH2CH3][HCl] conditional on (a)(iii) and ecf from (a)(iii) [1] (initial) rate = 0.2/95 or 0.2/47 ≈ 2.1 × 10–3 or 4.3 × 10–3 (mol dm–3 min–1) [1] k = 2.1 × 10–3 / (0.2 × 0.1) or 4.3 × 10–3 / (0.2 × 0.2) ≈ 0.11 (mol–1 dm3 min–1) [1] [8 max 7] (b) (i) because H2O is the solvent or its concentration cannot change [1] (ii) because HCl is a catalyst [1] [2] [Total: 9] 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0 20 40 60 80 100 120 [CH3CO2CH2CH3]/mol dm-3 time/min

Page 4 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9701 43 © Cambridge International Examinations 2013 3 (a) (i) density = mass per unit volume [1] (ii) mass per atom or Ar is larger (for Fe) Or Fe 55.8 and Ca 40.1 [1] Fe radii/volume of atom/ion is smaller or RFe = 0.116 nm whereas RCa = 0.197 nm [1] [3] (b) reaction acid- base ligand exchange precipitation redox [Cu(H2O)6]2+ + 4NH3 → [Cu(NH3)4]2+ + 6H2O ✓ [Cu(H2O)6]2+ + 4HCl→ [CuCl4]2– + 4H+ + 6H2O ✓ 2FeCl2 + Cl2 → 2FeCl3 ✓ [Fe(H2O)6]2+ + 2OH– → Fe(OH)2 + 6H2O ✓ ✓ 2Fe(OH)2 + ½O2 + H2O → 2Fe(OH)3 ✓ CrO3 + 2HCl → CrO2Cl2 + H2O ✓ ✓ Cr(H2O)3(OH)3 + OH– → [Cr(H2O)2(OH)4]– + H2O ✓ ✓ [Cr(OH)4]– + 1½H2O2 + OH– → CrO4 2– + 4H2O ✓ ✓ (Where more than one tick appears on a line in the table above – these are alternatives – but allow the mark if both are given). [8] (c) n(H2) = 8/24 = 0.33 mol [1] from equation, this is produced from 0.22 mol of Al ecf (× 2/3) [1] Ar(Al) = 27 thus mass of Al = 27 × 0.22 = 5.9 – 6 g hence 5.9–6.0% ecf (× 27) [1] [3] [Total: 14]

Page 5 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9701 43 © Cambridge International Examinations 2013 4 (a) (due to the) strong N≡N bond [1] [1] (b) (i) Any balanced equation forming a stable nitrogen oxide e.g. N2 + O2 → 2NO or N2 + 2O2 → 2NO2 [1] (ii) in lightning [1] in an engine/combustion of fuels (or a specific example) [1] (iii) (NOx produces) acid rain or forms (photochemical) smog [1] [4] (c) (base is a) proton acceptor [1] basicities: ethylamine > NH3 > phenylamine [1] ethylamine (more basic) due to electron donating ethyl group [1] phenylamine (less basic) due to lone pair being delocalised into the ring [1] [4] (d) (i) step 1: nucleophilic substitution [1] step 2: hydrolysis [1] (ii) step 1: KCN (in ethanol) and reflux [1] step 2: H3O+ / aqueous acid and reflux [1] (iii) T is NH2 [1] W is Cl O [1] [6] [Total: 15]

Page 6 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9701 43 © Cambridge International Examinations 2013 5 (a) H2O OH CO2H OH Na H2 H2 H2 H2 KOH(aq) X X X X Na2CO3(aq) X X CO2 X [5] (b) (i) (CH3)3 C–Cl (any unambiguous structure or name) [1] (ii) reduction or hydrogenation [1] (iii) either CH3CO2H and heat with (conc) H2SO4 or CH3COCl [1] (iv) reflux [1] dilute HCl [1] [5]

Page 7 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9701 43 © Cambridge International Examinations 2013 (c) (i) reagent and conditions product with A product with B Br2(aq) OH Br Br C(CH3)3 no reaction heat with HBr no reaction Br C(CH3)3 pass vapour over heated Al2O3 no reaction C(CH3)3 heat with acidified K2Cr2O7 no reaction O C(CH3)3 [6] (ii) either: Cr2O7 2–/H+: no observation with A and goes from orange to green with B. or: Br2(aq): white ppt. with A and no observation/ppt with B [1] [7] [Total: 17]

Page 8 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9701 43 © Cambridge International Examinations 2013 6 (a) substance protein synthesis formation of DNA adenine ✓ alanine ✓ aspartate ✓ phosphate ✓ [3] [3] (b) protein : hydrogen bonds [1] between –NH and C=O groups on different (peptide) groups [1] DNA : hydrogen bonds [1] between bases / A & T / C & G on different chains [1] [4] (c) primary: covalent bonds between (successive) amino acids [1] tertiary : hydrogen bonds between –COOH / –OH and –NH2 (in side chains) ionic bonds between –NH3 + and –CO2 – (in side chains) disulfide bonds between cysteine molecules / residues / –SH groups (in side chains) van der Waals/VDW forces between alkyl groups / non-polar residues (in side chains) any two rows [2] [3] [Total: 10]

Page 9 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9701 43 © Cambridge International Examinations 2013 7 (a) Any four from: • extract DNA • use restriction enzymes (to break DNA into fragments) • use polymerase chain reaction (to increase concentration of fragments) • place samples on (agarose) gel • carry out electrophoresis • label fragments (transferred to a membrane) with radioactive isotope [4 × 1] [4] (b) item for testing suitable for DNA fingerprinting human hair ✓ piece of a flint tool ✗ piece of Iron Age pot ✗ piece of Roman leather ✓ [3] [3] (c) insecticides: gas-liquid or thin-layer chromatography [1] dyes : paper or thin-layer chromatography [1] drugs: gas-liquid or thin-layer chromatography [1] [3] [Total: 10]

Page 10 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9701 43 © Cambridge International Examinations 2013 8 (a) (i) CH CH2 CH CH2 CO2H CO2H [1] (ii) Addition [1] (iii) Hydrogen bonding [1] [3] (b) (i) more / increase water absorbing properties (allow attracts water more) [1] more polar(ity)/more hydrophilic / has ionic side-chains (as well as hydrophilic ones) [1] (ii) It should be biodegradable/decompose [1] [3] (c) idea of ion exchange / replacement of Na+ for Cd2+/Pb2+ [1] (the metal ions) will be attracted to the carboxylate ions [1] [2] (d) (i) condensation [1] (ii) OH/alcohol groups so highly soluble / able to form hydrogen bonds [1] [2] [Total: 10]