Cambridge A Level Chemistry 9701 — 2009 Oct/Nov Paper 4 · Variant 1

This document consists of 19 printed pages and 1 blank page. DC (FF/DT) 12823/4 © UCLES 2009 [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. * 5 6 4 0 2 4 4 4 0 7 * CHEMISTRY 9701/41 Paper 4 Structured Questions October/November 2009 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 Total

2 9701/41/O/N/09 © UCLES 2009 For Examiner’s Use Section A Answer all questions in the spaces provided. 1 (a) The Group IV oxides CO2 and SiO2 differ widely in their physical properties. Describe these differences and explain them in terms of their structure and bonding. … … … … [3] (b) What are the properties of a ceramic material? Why is silicon(IV) oxide very suitable as a component of ceramics? … … … … [2] (c) Lead(II) oxide reacts with both acids and bases. (i) What is the name given to oxides that have this property? … (ii) Write a balanced equation for the reaction between PbO and NaOH. … [2]

3 9701/41/O/N/09 © UCLES 2009 [Turn over For Examiner’s Use (d) Tin forms an oxide, A, that contains the metal in both oxidation states II and IV. The formula of A can be found by the following method. • A sample of A was dissolved in H2SO4(aq), producing solution B, which was a mixture of tin(II) sulfate and tin(IV) sulfate. • A 25.0 cm3 sample of solution B was titrated with 0.0200 mol dm–3 KMnO4. 13.5 cm3 of KMnO4 was required to reach the end-point. • Another 25.0 cm3 sample of solution B was stirred with an excess of powdered zinc. This converted all the tin into tin(II). The excess of zinc powder was filtered off and the filtrate was titrated with 0.0200 mol dm-3 KMnO4, as before. This time 20.3 cm3 of KMnO4 was required to reach the end-point. The equation for the reaction occurring during the titration is as follows. 2MnO4 – + 16H+ + 5Sn2+ 2Mn2+ + 8H2O + 5Sn4+ (i) Write a balanced equation for the reaction between Zn and Sn4+. … (ii) Use the Data Booklet to calculate the E o- values for the reactions between • Zn and Sn4+, … • MnO4 – and Sn2+.. … (iii) Use the results of the two titrations to calculate • the number of moles of Sn2+ in the first titration sample, … … • the number of moles of Sn2+ in the second titration sample. … … (iv) Use the results of your calculation in (iii) to deduce the Sn2+/ Sn4+ ratio in the oxide A, and hence suggest the formula of A. … … … [8]

4 9701/41/O/N/09 © UCLES 2009 For Examiner’s Use (e) A major use of tin is to make ‘tin plate’, which is composed of thin sheets of mild steel electroplated with tin, for use in the manufacture of food and drinks cans. A tin coating of 1.0 3 10–5 m thickness is often used. (i) Calculate the volume of tin needed to coat a sheet of steel 1.0 m 3 1.0 m to this thickness, on one side only. … … (ii) Calculate the number of moles of tin that this volume represents. [The density of tin is 7.3 g cm–3.] … … … (iii) The solution used for electroplating contains Sn2+ ions. Calculate the quantity of electricity in coulombs needed to deposit the amount of tin you calculated in (ii). … … … [4] [Total: 19]

6 9701/41/O/N/09 © UCLES 2009 For Examiner’s Use 2 Calcium chloride, CaCl2, is an important industrial chemical used in refrigeration plants, for de-icing roads and for giving greater strength to concrete. (a) Show by means of an equation what is meant by the lattice energy of calcium chloride. … [1] (b) Suggest, with an explanation, how the lattice energies of the following salts might compare in magnitude with that of calcium chloride. (i) calcium fluoride, CaF2 … … (ii) calcium sulfide, CaS … … [3] (c) Use the following data, together with additional data from the Data Booklet, to calculate the lattice energy of CaCl2. standard enthalpy change of formation of CaCl2 –796 kJ mol–1 standard enthalpy change of atomisation of Ca(s) +178 kJ mol–1 electron affinity per mole of chlorine atoms –349 kJ mol–1 enthalpy Ca(s) + Cl 2(g) lattice energy = … kJ mol–1 [3]

7 9701/41/O/N/09 © UCLES 2009 [Turn over For Examiner’s Use (d) When a solution of CaCl2 is added to a solution of the dicarboxylic acid, malonic acid, the salt calcium malonate is precipitated as a white solid. The solid has the following composition by mass: Ca, 28.2 %; C, 25.2 %; H, 1.4 %; O, 45.2 %. (i) Calculate the empirical formula of calcium malonate from these data. (ii) Suggest the structural formula of malonic acid. [3] [Total: 10]

8 9701/41/O/N/09 © UCLES 2009 For Examiner’s Use 3 One major difference between the properties of compounds of the transition elements and those of other compounds is that the compounds of the transition elements are often coloured. (a) Explain in detail why many transition element compounds are coloured. … … … … … [3] (b) The following graph shows the absorption spectrum of two complexes containing copper. absorbance [Cu(NH3)4(H2O)2]2+ [Cu(H2O)6]2+ wavelength / nm 400 500 600 700 800 900 1000 blue green yellow red infra-red (i) State the colours of the following complex ions. [Cu(H2O)6]2+ … [Cu(NH3)4(H2O)2]2+ … (ii) Using the spectra above give two reasons why the colour of the [Cu(NH3)4(H2O)2]2+ ion is deeper (more intense) than that of the [Cu(H2O)6]2+ ion. … … … (iii) Predict the absorption spectrum of the complex [Cu(NH3)2(H2O)4]2+, and sketch this spectrum on the above graph. [6]

9 9701/41/O/N/09 © UCLES 2009 [Turn over For Examiner’s Use (c) Copper forms a complex with chlorine according to the following equilibrium. Cu2+(aq) + 4Cl –(aq) [CuCl4]2–(aq) (i) Write an expression for the equilibrium constant, Kc , for this reaction, stating its units. Kc = units … … (ii) The numerical value of Kc is 4.2 3 105. Calculate the [[CuCl4] 2–] / [Cu2+ ] ratio when [Cl – ] = 0.20 mol dm–3. … … [3] [Total: 12]

10 9701/41/O/N/09 © UCLES 2009 For Examiner’s Use 4 Cyclohexanol and phenol are both solids with low melting points that are fairly soluble in water. OH cyclohexanol phenol OH (a) Explain why these compounds are more soluble in water than their parent hydrocarbons cyclohexane and benzene. … … … [2] (b) Explain why phenol is more acidic than cyclohexanol. … … … [2]

11 9701/41/O/N/09 © UCLES 2009 [Turn over For Examiner’s Use (c) For each of the following reagents, draw the structural formula of the product obtained for each of the two compounds. If no reaction occurs write no reaction in the box. reagent product with cyclohexanol product with phenol Na(s) NaOH(aq) Br2(aq) I2(aq) + OH–(aq) an excess of acidified Cr2O2 7 –(aq) [7] (d) Choose one of the above five reagents that could be used to distinguish between cyclohexanol and phenol. Describe the observations you would make with each compound. reagent … observation with cyclohexanol … observation with phenol … [2] [Total: 13]

12 9701/41/O/N/09 © UCLES 2009 For Examiner’s Use 5 Kevlar is a tough polyamide used in bullet-proof vests and high-specification bicycle tyres. It can be manufactured by the following process. I II H3C H2N NH2 CH3 HO2C C D Kevlar CO2H ClOC COCl (a) (i) Suggest reagents and conditions for reaction I, … reaction II. … (ii) Draw the structural formula of one repeat unit of Kevlar in the box above. [4] (b) The di-acid chloride C reacts with a variety of reagents. Suggest the structural formulae of the products of the reaction of C with (i) CH3NH2, (ii) HOCH2CH2OH. [3]

13 9701/41/O/N/09 © UCLES 2009 [Turn over For Examiner’s Use (c) The diamine D also reacts with a variety of reagents. Suggest the structural formulae of the products of the reaction of D with (i) HCl (aq), (ii) Br2(aq). [3] (d) 4-aminobenzoic acid, E, is a useful intermediate for making dyes. III IV HO2C OH a dye NH2 HO2C N N Cl - HO2C N=N + E Suggest reagents and conditions for reaction III, … reaction IV. … [4] (e) 4-aminobenzoic acid, E, forms a zwitterion. (i) What is meant by the term zwitterion? … … (ii) Draw the structural formula of the zwitterion formed from 4-aminobenzoic acid. [2] [Total: 16]

14 9701/41/O/N/09 © UCLES 2009 For Examiner’s Use Section B Answer all questions in the spaces provided. 6 (a) The diagram shows part of one strand of DNA. Draw the complementary strand, labelling the bonds formed to the original strand, and labelling the components of the strand you draw. phosphate sugar T phosphate sugar A phosphate sugar C [3] (b) Briefly describe the roles of each of the following in protein synthesis. (i) tRNA … … … (ii) the ribosome … … … [4]

15 9701/41/O/N/09 © UCLES 2009 [Turn over For Examiner’s Use (c) Some diseases, such as sickle cell anaemia, are caused by a single mutation in the DNA for a particular gene. This causes the haemoglobin produced to change the shape of red blood cells, reducing their efficiency in carrying oxygen. (i) What is meant by a mutation? … (ii) Explain why such a mutation could alter the bonding in haemoglobin. … … … … [4] [Total: 11]

16 9701/41/O/N/09 © UCLES 2009 For Examiner’s Use 7 This question is about the modern techniques of analysis which may be used to determine molecular structures. (a) In X-ray crystallography X-rays are diffracted by the electron clouds surrounding individual atoms in the structure. (i) What useful information is provided by X-ray crystallography? … … (ii) Why cannot hydrogen atoms in a structure be detected by this technique? … [2] (b) Suggest how structures of complex molecules such as enzymes, derived from X-ray crystallography, can help explain their biochemical behaviour. … … … [2] (c) NMR spectroscopy, in contrast to X-ray crystallography, is frequently used to examine protons in organic molecules. (i) What feature of protons enables their detection by NMR spectroscopy? …

17 9701/41/O/N/09 © UCLES 2009 [Turn over For Examiner’s Use (ii) The NMR spectrum below was obtained from a compound X, CxHyOz. In the mass spectrum of the compound, the M : M+1 ratio was found to be 25:2. Determine the values of x, y and z in the formula of X and deduce a possible structure for the compound, explaining how you arrive at your conclusion. 11 10 9 8 7 6 5 4 4 3 1 δ/ppm 3 2 1 0 … … … … … … … … Possible structure of X [6] [Total:10]

18 9701/41/O/N/09 © UCLES 2009 For Examiner’s Use 8 A new method of making very light, flexible batteries using nanotechnology was announced in August 2007. Read the passage and answer the questions related to it. Researchers have developed a new energy-storage device that could easily be mistaken for a simple sheet of black paper. The nano-engineered battery is lightweight, ultra-thin and completely flexible. It is geared towards meeting the difficult design and energy requirements of tomorrow’s gadgets, such as implantable medical devices and even vehicles. Researchers soaked ‘paper’ in an ionic liquid electrolyte which carries the charge. They then treated it with aligned carbon nanotubes, which give the device its black colour. The nanotubes act as electrodes and allow the storage devices to conduct electricity. The device, engineered to function as both a battery and a supercapacitor, can provide the long, steady power output comparable to a conventional battery, as well as a supercapacitor’s quick burst of high energy. The device can be rolled, twisted, folded, or cut into shapes with no loss of strength or efficiency. The ‘paper’ batteries can also be stacked, like a pile of printer paper, to boost the total power output. 1. Conventional batteries produce electrons through a chemical reaction between electrolyte and metal. 2. Chemical reaction in the ‘paper’ battery is between electrolyte and carbon nanotubes. 3. Electrons collect on the negative terminal of a battery. 4. Electrons must flow from the negative terminal, through the external circuit to the positive terminal for the chemical reaction to continue. nanotube

19 9701/41/O/N/09 © UCLES 2009 [Turn over For Examiner’s Use For Examiner’s Use © UCLES 2009 (a) From your knowledge of the different structures of carbon, suggest which of these is used to make nanotubes. … [1] (b) Suggest a property of this structure that makes it suitable for making nanotubes. … … [1] (c) Carbon in its bulk form is brittle like most non-metallic solids. Suggest why the energy storage device described can be rolled into a cylinder. … … [1] (d) Name an example of an ‘ionic liquid electrolyte’ (not a solution). … [1] [Total: 4]

20 9701/41/O/N/09 © UCLES 2009 9 In recent years a great deal of research has been carried out into finding different anti-cancer drugs. Tumours, which are often symptoms of cancer, are produced when cells replicate uncontrollably. This in turn is brought about by the replication of DNA in these cells. Two anti-cancer agents are mechlorethamine and cis-platin. They work by binding to the DNA and preventing replication. CH3 NH3 NH3 Pt N Cl Cl Cl Cl CH3 N G DNA DNA mechlorethamine crosslinked DNA cis-platin G NH3 NH3 Pt G G C C (a) (i) What type of bonding attaches both anti-cancer agents to the DNA? … (ii) Suggest how each of the anti-cancer agents prevents replication of the DNA. … … … … … [5] [Total: 5] © UCLES 2009 For Examiner’s Use 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 2009 question paper for the guidance of teachers 9701 CHEMISTRY 9701/41 Paper 41 (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. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2009 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/AS LEVEL – October/November 2009 9701 41 © UCLES 2009 1 (a) CO2 is a gas (at room temperature); SiO2 is a high melting solid [1] CO2: simple / discrete molecular / covalent [1] SiO2: giant covalent or macromolecular / giant molecular [1] [3] (b) (a substance that is..) hard, high melting, electrical insulator any two [1] SiO2 has strong covalent bonds (can be in (a)) [1] [2] (c) (i) amphoteric [1] (ii) 2NaOH + PbO → Na2PbO2 + H2O [1] (or NaOH + PbO + H2O → NaPb(OH)3 etc.) [2] (d) (i) Zn + Sn4+ → Zn2+ + Sn2+ [1] (ii) Eθ = 0.15 – (–0.76) = 0.91 V [1] Eθ = 1.52 – 0.15 = 1.37 V [1] (iii) n(Sn2+) = 0.02 × 13.5/1000 × 5/2 = 6.75 × 10–4 mol use of the 5/2 ratio [1] correct rest of working [1] n(Sn2+) = 0.02 × 20.3/1000 × 5/2 = 1.02 × 10–3 mol [1] (iv) n(Sn4+) = 1.02 × 10–3 – 6.75 × 10–4 = 3.45 × 10–4 mol [1] ∴ ratio = 6.75/3.45 = 1.96:1 ≈ 2:1 ∴ formula is 2SnO + SnO2 ⇒ Sn3O4 (condl on calculation, but allow ecf) [1] [8] (e) (i) volume = 1 × 1 × 1 × 10–5 = 1 × 10–5 m3 or 10 cm3 [1] (ii) mass = vol × density = 10 × 7.3 = 73 g ecf [1] moles = mass/Ar = 73/119 = 0.61 mol ecf [1] (iii) Q = nFz = 0.61 × 9.65 × 104 × 2 = 1.18 (1.2) × 105 coulombs ecf [1] [4] [Total: 19]

Page 3 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – October/November 2009 9701 41 © UCLES 2009 2 (a) Ca2+(g) + 2Cl–(g) → CaCl2(s) [1] [1] (b) CaF2 and CaS both have larger lattice energies (than CaCl2) [1] (i) F– is smaller than Cl– [1] (ii) S2– is more highly charged than Cl– [1] [3] (c) LE = –[178 + 590 + 1150] – [244 – 2 × 349] – 796

signs
= –2260 (kJ mol–1) [3] [3] (d) (i) Ca = 28.2/40.1 = 0.703 ⇒ 1 C = 25.2/12 = 2.10 ⇒ 3 H = 1.4/1 = 1.4 ⇒ 2 (1 mark for initial step of calc’n) O = 45.1/16 = 2.82 ⇒ 4 formula is CaC3H2O4 (1) [2] (ii) malonic acid must be C2H4O4, i.e. CH3(CO2H)2 (must be structural) [1] [3] [Total: 10] 3 (a) d-orbitals split into two / different levels light is absorbed electron is promoted from a lower to a higher level colour observed is the complement of the colour absorbed E = hf any 3 points [3] [3] (b) (i) [Cu(H2O)6]2+ is pale blue [1] [Cu(NH3)4(H2O)2 ]2+ is deep / dark blue or purple [1] (ii) because it has a larger absorbance peak or a larger εo value [1] because λmax is in the visible region (hence more visible light is absorbed) [1] (iii) curve will have λmax between >600 nm and 800 nm [1] with maximum εo in between the other two [1] [6] (c) (i) Kc = [CuCl4 2–]/([Cu2+][Cl–]4) units are mol–4 dm12 [1] + [1] (ii) [CuCl4 2–]/[Cu2+] = Kc[Cl–]4 = 672 (no units) [1] [3] [Total: 12]

Page 4 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – October/November 2009 9701 41 © UCLES 2009 4 (a) (cyclohexanol & phenol) hydrogen bonding to (solvent) water molecules [1] due to OH group [1] [2] (b) phenoxide anion is more stable (than cyclohexoxide) / OH bond is weaker [1] due to delocalisation of charge / lone pair over the ring [1] [2] (c) reagent product with cyclohexanol product with phenol Na(s) RONa or RO–Na+ ArONa or ArO–Na+ NaOH(aq) no reaction ArONa or ArO–Na+ Br2(aq) no reaction tribromophenol I2(aq) + OH–(aq) no reaction no reaction an excess of acidified Cr2O7 2–(aq) cyclohexanone no reaction five correct products 5 × [1] five correct “no reaction”s [2] (4 correct = [1]; 3 correct = [0]) [7] (d) either Br2(aq): no reaction with cyclohexanol; decolourises or white ppt with phenol or Cr2O7 2– + H+: turns from orange to green with cyclohexanol; no reaction with phenol correct reagent chosen and the correct “no reaction” specified [1] correct positive observation [1] [2] [Total: 13]

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – October/November 2009 9701 41 © UCLES 2009 5 (a) (i) I: KMnO4 [1] heat with H+ or OH– [1] II: SOCl2 or PCl5 or PCl3 (NOT aq) [1] (ii) -[-CO-C6H4-CO-NH-C6H4-NH-]- (Peptide bond must be displayed for minm) [1] [4] (b) (i) CH3NHCO-C6H4-CONHCH3 (1 mark for each end) [1] + [1] (ii) HOCH2CH2O-CO-C6H4-CO-OCH2CH2OH for [1] or the polymer -[- OCH2CH2O-CO-C6H4-CO-]- for [2] [4 max 3] (c) (i) Cl– +NH3-C6H4-NH3 + Cl- (1 mark for each end) [1] + [1] (ii) H2N-C6H2Br2-NH2 or H2N-C6H2Br3-NH2 or H2N-C6Br4-NH2 [1] [3] (d) I: HNO2 (or NaNO2 + HCl/H2SO4) [1] at T < 10oC [1] II: m-prop-2-yl phenol, (CH3)2CH-C6H4OH [1] + NaOH(aq) [1] [4] (e) (i) A species having positive and negative ionic centres / charges, with no overall charge [1] (ii) -O2C-C6H4-NH3 + [1] [2] [Total: 16]

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – October/November 2009 9701 41 © UCLES 2009 6 (a) All three amino acids correctly paired (2) Two amino acids correctly paired (1) One labelled H-bond between strands (1) [3] (b) (i) tRNA – each amino acid has its own specific / appropriate tRNA (1) – carry amino acids to ribosomes / mRNA (1) – contains a triplet code / anticodon (1) (ii) ribosome – attaches / moves along / binds to mRNA (1) – assemble amino acids in correct sequence for / synthesises protein (1) [5] (c) (i) Base miscopied / deleted (1) (ii) Sequence of bases is changed (1) This may result in different amino acid sequence – different protein (1) Can affect shape / tertiary structure of protein (1) [Max 3] [Total: 12 max 11]

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – October/November 2009 9701 41 © UCLES 2009 7 (a) (i) Positions of atomic nuclei / atoms (1) (ii) Insufficient electrons / electron density / electron cloud (around H atom) (1) [2] (b) X-ray crystallography can show the geometry of the arrangement of atoms / bonding between atoms / shape of atoms (1) This can help explain how e.g. enzymes work (any reasonable example) (1) [2] (c) (i) Nuclear spin (1) (ii) (If M : M+1 gives a ratio 15 : 2) Then x = 25 1.1 2 100 × × = 7 (1) Single peak at 3.7 δ due to –O-CH3 (1) Single peak at 5.6 δ due to phenol / OH (1) 1,2,1 peak at 6.8 δ due to hydrogens on benzene ring (1) Pattern suggests 1,4 subsitution (1) (x = 7,) y = 8, z = 2 (1) Compound is 4-methoxylphenol (1) Max 5 [6] [Total: 10]

Page 8 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – October/November 2009 9701 41 © UCLES 2009 8 (a) Graphite / graphene (1) (b) They do not exist as sheets / layers of carbon atoms (1) (c) The lengths of nanotubes are much shorter than the curvature of the paper / they are so small that they are not effected by rolling (1) (d) Any molten ionic salt (or plausible organic ionic compounds) (1) [Total: 4] 9 (a) (i) Covalent / co-ordinate (1) (ii) Mechlorethamine – binds the two chains together (1) – prevents unravelling (1) Cis-platin – binds to two Gs / bases in one chain (1) – so they are not available for base pairing (1) [Total: 5]