Cambridge A Level Chemistry 9701 — 2009 May/June Paper 4 · Variant 1

This document consists of 18 printed pages and 2 blank pages. SP (FF/DT) T69679/1 © 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. * 0 7 3 7 7 1 4 9 3 0 * CHEMISTRY 9701/04 Paper 4 Structured Questions May/June 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 9701/04/M/J/09 © UCLES 2009 For Examiner’s Use Section A Answer all questions in the spaces provided. 1 (a) Explain what is meant by the Bronsted-Lowry theory of acids and bases. … … … [2] (b) The Ka values for some organic acids are listed below. acid Ka /mol dm−3 CH3CO2H 1.7
10−5 Cl CH2CO2H 1.3
10−3 Cl 2CHCO2H 5.0
10−2 (i) Explain the trend in Ka values in terms of the structures of these acids. … … … … (ii) Calculate the pH of a 0.10 mol dm−3 solution of Cl CH2CO2H. pH = …

3 9701/04/M/J/09 © UCLES 2009 [Turn over For Examiner’s Use (iii) Use the following axes to sketch the titration curve you would obtain when 20 cm3 of 0.10 mol dm−3 NaOH is added gradually to 10 cm3 of 0.10 mol dm−3 Cl CH2CO2H. 0 0 5 10 15 20 7 pH volume of NaOH added / cm 3 14 [8] (c) (i) Write suitable equations to show how a mixture of ethanoic acid, CH3CO2H, and sodium ethanoate acts as a buffer solution to control the pH when either an acid or an alkali is added. … … (ii) Calculate the pH of a buffer solution containing 0.10 mol dm−3 ethanoic acid and 0.20 mol dm−3 sodium ethanoate. pH = … [4] [Total: 14]

4 9701/04/M/J/09 © UCLES 2009 For Examiner’s Use 2 (a) Describe the observations you would make when concentrated sulfuric acid is added to separate portions of NaCl (s) and NaBr(s). Write an equation for each reaction that occurs. NaCl (s): observation … … equation NaBr(s): observation … … equation [4] (b) By quoting relevant E o- data from the Data Booklet, explain how the observations you have described above relate to the relative oxidising power of the elements. … … … [2] (c) By referring to relevant E o- data choose a suitable reagent to convert Br2 into Br −. Write an equation and calculate the E o- for the reaction. … … … [3] [Total: 9]

5 9701/04/M/J/09 © UCLES 2009 [Turn over For Examiner’s Use 3 (a) Explain what is meant by the term transition element. … … [1] (b) Complete the electronic configuration of (i) the vanadium atom, 1s22s22p6 … (ii) the Cu2+ ion. 1s22s22p6 … [2] (c) List the four most likely oxidation states of vanadium. … [1] (d) Describe what you would see, and explain what happens, when dilute aqueous ammonia is added to a solution containing Cu2+ ions, until the ammonia is in an excess. … … … … … … … [5] (e) Copper powder dissolves in an acidified solution of sodium vanadate(V), NaVO3, to produce a blue solution containing VO2+ and Cu2+ ions. By using suitable half-equations from the Data Booklet, construct a balanced equation for this reaction. … [2] [Total: 11]

6 9701/04/M/J/09 © UCLES 2009 For Examiner’s Use 4 (a) The reaction between iodide ions and persulfate ions, S2O 8 2−, is slow. 2I− + S2O 8 2− I2 + 2SO 4 2− 1 The reaction can be speeded up by adding a small amount of Fe2+ or Fe3+ ions. The following two reactions then take place. 2I− + 2Fe3+ I2 + 2Fe2+ 2 2Fe2+ + S2O 8 2− 2Fe3+ + 2SO 4 2− 3 (i) What type of catalysis is occurring here? … (ii) The rates of reactions 2 and 3 are both faster than that of reaction 1. By considering the species involved in these reactions, suggest a reason for this. … … (iii) The following reaction pathway diagram shows the enthalpy profile of reaction 1. enthalpy progress of reaction Use the same axes to draw the enthalpy profiles of reaction 2 followed by reaction 3, starting reaction 2 at the same enthalpy level as reaction 1. [4]

7 9701/04/M/J/09 © UCLES 2009 [Turn over For Examiner’s Use (b) The oxidation of SO2 to SO3 in the atmosphere is speeded up by the presence of nitrogen oxides. (i) Describe the environmental significance of this reaction. … (ii) Describe a major source of SO2 in the atmosphere. … (iii) By means of suitable equations, show how nitrogen oxides speed up this reaction. … … [4] [Total: 8]

8 9701/04/M/J/09 © UCLES 2009 For Examiner’s Use 5 (a) In the following boxes draw the structural formulae of three alcohols having straight (i.e. unbranched) chains, with the molecular formula C5H12O. A B C [2] Use the letters A, B or C as appropriate when answering the following questions. Each letter may be used once, more than once or not at all. (b) Which of the alcohols are chiral? …[1] (c) (i) Which of these alcohols react with alkaline aqueous iodine? … (ii) Describe the observation you would make during this reaction. … (iii) Draw the structural formulae of the products of this reaction. [4] (d) Draw the structural formula of the product obtained when each of the alcohols A, B and C is heated with an excess of acidified K2Cr2O7(aq). A B C [3]

9 9701/04/M/J/09 © UCLES 2009 [Turn over For Examiner’s Use (e) One of the many suggestions for converting biomass into liquid fuel for motor transport is the pyrolysis (i.e. heating in the absence of air) of cellulose waste, followed by the synthesis of alkanes. (i) In the first reaction, cellulose, (C6H10O5)n, is converted into a mixture of carbon monoxide and hydrogen. Some carbon is also produced. Complete and balance the equation for this reaction. (C6H10O5)n ————————— + ————————— + ————————— (ii) The second reaction involves the combination of CO and H2 to produce alkanes such as heptane. 7CO + 15H2 C7H16 + 7H2O heptane Using the value of 1080 kJ mol−1 as the value for the C
O bond energy in CO, and other relevant bond energies from the Data Booklet, calculate the ∆H for this reaction. ∆H = … kJ mol−1 [5] [Total: 15]

10 9701/04/M/J/09 © UCLES 2009 For Examiner’s Use 6 Phenol and chlorobenzene are less reactive towards certain reagents than similar non-aromatic compounds. Thus hexan-1-ol can be converted into hexylamine by the following two reactions, I II CH3(CH2)5OH CH3(CH2)5Cl CH3(CH2)5NH2 hexan-1-ol 1-chlorohexane hexylamine whereas neither of the following two reactions takes place. OH Cl NH2 (a) (i) Suggest reagents and conditions for reaction I, … , reaction II. … . (ii) What type of reaction is reaction II? … (iii) Suggest a reason why chlorobenzene is much less reactive than 1-chlorohexane. … … [4]

11 9701/04/M/J/09 © UCLES 2009 [Turn over For Examiner’s Use (b) Phenylamine can be made from benzene by the following two reactions. NH2 NO2 III IV (i) Suggest reagents and conditions for reaction III, … , reaction IV. … . (ii) State the type of reaction for reaction III, … , reaction IV. … . [5] (c) Suggest a reagent that could be used to distinguish phenylamine from hexylamine. reagent and conditions … observation with phenylamine … observation with hexylamine … [2]

12 9701/04/M/J/09 © UCLES 2009 For Examiner’s Use (d) Phenylamine is used to make azo dyes. In the following boxes draw the structural formula of the intermediate D and of the azo dye E. D E NH2 NaNO2 + HCl in NaOH(aq) T < 5°C OH CH3 CH3 [2] [Total: 13]

13 9701/04/M/J/09 © UCLES 2009 [Turn over For Examiner’s Use Section B Answer all questions in the spaces provided. 7 Metals play a vital part in biochemical systems. In this question you need to consider why some metals are essential to life, whilst others are toxic. (a) For each of the metals, state where it might be found in a living organism, and what its chemical role is. iron location in organism … role … … sodium location in organism … role … … zinc location in organism … role … … [6] (b) Heavy metals such as mercury are toxic, and it is important that these do not enter the food chain. (i) Give a possible source of mercury in the environment. … (ii) Describe and explain two reasons why mercury is toxic, using diagrams and/or equations to help your explanation. … … … [4] [Total : 10]

14 9701/04/M/J/09 © UCLES 2009 For Examiner’s Use 8 A large number of organic compounds are soluble in both water and non-aqueous solvents such as hexane. If such a compound is shaken with a mixture of water and the non-aqueous solvent, it will dissolve in both solvents depending on the solubility in each. (a) (i) State what is meant by the term partition coefficient. … … (ii) When 100 cm3 of an aqueous solution containing 0.50 g of an organic compound X was shaken with 20 cm3 of hexane, it was found that 0.40 g of X was extracted into the hexane. Calculate the partition coefficient of X between hexane and water. (iii) If two 10 cm3 portions of hexane were used instead of a single 20 cm3 portion, calculate the total amount of X extracted and compare this with the amount extracted using one 20 cm3 portion. [5]

15 9701/04/M/J/09 © UCLES 2009 [Turn over For Examiner’s Use (b) PCBs are highly toxic compounds released into the atmosphere when some plastics are burned at insufficiently high temperatures. In recent years PCB residues have been found in the breast milk of Inuit mothers in northern Canada. Foods, such as oily fish, seal and whale meat, which are high in fat, form an important part of the Inuit diet. (i) Suggest why berries and drinking water are not contaminated by PCBs in the same way that oily fish, seal and whale meat are. … … … (ii) Based on the information provided, what can you say about the partition coefficient between fat and water for PCB residues? … … … [3]

16 9701/04/M/J/09 © UCLES 2009 For Examiner’s Use (c) The diagram shows the result of two-way paper chromatography. X starting point solvent 1 solvent 2 (i) How many spots were there after the first solvent had been used? … (ii) Circle the spot that moved very little in solvent 2, but moved a greater distance in solvent 1. (iii) Draw a square around the spot that could be separated from the rest by using only solvent 1. [3] [Total: 11]

17 9701/04/M/J/09 © UCLES 2009 For Examiner’s Use 9 (a) Spider silk is a natural polymer which has an exceptional strength for its weight. Kevlar is a man-made polymer designed to have similar properties. It has a wide variety of uses from sporting equipment to bullet-proof vests. C O H N H N O C N H C O O C N H O C Kevlar (i) In Kevlar, the polymer strands line up to form strong sheets with bonds between the strands. On the diagram above, draw part of a second polymer chain showing how bonds could be formed between the chains. (ii) Suggest what type of bonds these are. … (iii) Draw two possible monomer molecules for making the polymer Kevlar. [5]

18 9701/04/M/J/09 © UCLES 2009 (b) The transport of oil by sea has resulted in a number of oil spills in recent years. As well as a waste of a valuable resource, these have caused major environmental problems. Traditional sorbent materials absorb water and sink. Researchers have developed new sorbent materials to help collect the spilled oil. The sorbent consists of a material called ‘hydrophobic aerogels’. This is a network of silicon(IV) oxide with some of the silicon atoms attached to fluorine-containing groups. —O—Si—CH2—CF3 The introduction of these fluorine-containing groups allows the oil to be absorbed but not the water. Tests show that these materials can absorb more than 200 times their mass of oil without sinking. (i) Suggest what the word hydrophobic means. … (ii) Suggest why the fluorine-containing groups allow oil to pass through but not water molecules. … … … … (iii) Suggest another important fluorine-containing polymer that repels water-containing materials. … [4] [Total: 9]

20 9701/04/M/J/09 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. BLANK PAGE

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2009 question paper for the guidance of teachers 9701 CHEMISTRY 9701/04 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. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the May/June 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 – May/June 2009 9701 04 © UCLES 2009 Section A 1 (a) acids are proton/H+ donors [1] bases are proton/H+ acceptors [1] [2] (b) (i) more Cl atoms produce a stronger acid or the larger the Ka the stronger the acid (NOT just “the more Cl atoms, the larger the Ka” – must refer to acid strength) [1] because the anion/RCO2 – is more stable or the O-H bond is weaker/polarised [1] due to the electronegativity/electron-withdrawing effect of Cl [1] (ii) [H+] = √(Ka.c) = 0.0114 (mol dm–3) [1] pH = 1.94 (allow 1.9) ecf from [H+] [1] (correct answer = [2]) (iii) start at pH = 1.94 (ecf from (ii) and goes up > 2 pH units before steep portion) [1] steep portion (over at least 3 pH units) at V = 10 cm3 [1] flattens off at pH 12–13 [1] [8] (c) (i) CH3CO2H + OH– → CH3CO2 – + H2O [1] CH3CO2 – + H+ → CH3CO2H [1] (ii) pKa = –log10(1.7 x 10–5) = 4.77 or [H+] = 8.5 x 10–6 (mol dm–3) [1] pH = pKa + log10(0.2/0.1) = 5.07 (allow 5.1) [1] (correct answer = [2]) [4] [Total: 14]

Page 3 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – May/June 2009 9701 04 © UCLES 2009 2 (a) NaCl: steamy fumes [1] NaCl + H2SO4 → NaHSO4 + HCl (or ionic, i.e. without the Na+) or 2NaCl + H2SO4 → Na2SO4 + 2HCl [1] NaBr: orange/brown fumes [1] 2NaBr + 3H2SO4 → 2NaHSO4 + 2H2O + SO2 + Br2 or 2HBr + H2SO4 → 2H2O + SO2 + Br2 (ignore equations producing HBr) [1] [4] (b) relevant Eo quoted: Cl2/Cl–, 1.36; Br2/Br–, 1.07; (H2SO4/SO2, 0.17 – not required) [1] Br– is more easily oxidised because its Eo is more negative or Cl2 is more oxidising because its Eo is more positive [1] [2] (c) Allow almost any reducing agent from the Data Booklet (see below) with Eo less than 1.07 V. But do not allow reducing agents that require conditions that would react with Br2 in the absence of the reducing agent (e.g. NH3 or OH–), and also do not allow “reducing agents” that could produce, or act as, oxidising agents (e.g. MnO4 2– and H2O2) balanced equ. showing reduction of Br2 by the chosen reducing agent (either ionic or molecular) [1] Eo = 1.07 – (Eo of reductant) = x.xx (V) (see below) [1] [2] [Total: 8] List of acceptable reductants with resulting Eo cell values reductant Eo cell/V reductant Eo cell/V reductant Eo cell/V Ag 0.27 Fe⇒Fe2+ 1.51 Na 3.78 Al 2.73 Fe⇒Fe3+ 1.11 Ni 1.32 Ba 3.97 Fe2+ 0.30 Pb 1.20 Ca 3.94 H2 1.07 SO2 0.90 Co 1.35 I– 0.53 S2O3 2– 0.98 Cr ⇒ Cr2+ 1.98 K 3.99 Sn 1.21 Cr ⇒ Cr3+ 1.81 Li 4.11 Sn2+ 0.92 Cr2+ 1.48 Mg 3.45 V 2.27 Cu⇒Cu+ 0.55 Mn 2.25 V2+ 1.33 Cu⇒Cu2+ 0.73 NO2 0.26 V3+ 0.73 Cu+ 0.92 HNO2 0.13 VO2+ 0.07 NH4 + 0.20 Zn 1.83 e.g. for Sn2+: Sn2+ + Br2 → Sn4+ + 2Br– [1] Eo = 1.07 – 0.15 = 0.92 V [1] (or similarly for other suitable reagents)

Page 4 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – May/June 2009 9701 04 © UCLES 2009 3 (a) a (d-block) element forming stable ions/compounds/oxidation states with incomplete/ partially filled [NOT empty] d-orbitals [1] [1] (b) (i) (1s2 2s2 2p6) 3s2 3p6 3d3 4s2 [1] (ii) (1s2 2s2 2p6) 3s2 3p6 3d9 [1] [2] (c) (+)2, (+)3, (+)4, (+)5 or II, III, IV, V [1] [1] (d) (pale blue solution ⇒) blue/cyan solid/ppt.(or (s) in the formula) [1] (blue ppt. is) Cu(OH)2 or copper hydroxide [1] (then produces a) deep blue or purple solution [1] which contains [Cu(NH3)4]2+ or [Cu(NH3)4(H2O)2]2+ [1] formed by ligand replacement [1] [5] (e) 2VO3 – + 8H+ + Cu → 2VO2+ + 4H2O + Cu2+ or 2VO2 + + 4H+ + Cu → 2VO2+ + 2H2O + Cu2+ correct species [1] balancing [1] (award only [1] for just the two half-equations) [2] [Total: 11] 4 (a) (i) homogeneous [1] (ii) ions in 2 and 3 are oppositely charged ions (thus attract each other) or ions in 1 are similarly charged ions (thus repel each other) [1] (iii) two contiguous activation humps [1] both less than the original [1] starting and finishing at the same points as before [1] [5]

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – May/June 2009 9701 04 © UCLES 2009 (b) (i) SO3 produces acid rain or SO3 + H2O → H2SO4 or a consequence of acid rain, e.g. lower pH of lakes; leaches aluminium from soils; kills fish/plants/rainforests; dissolves/corrodes/damages buildings (NOT global warming) [1] (NOT asthma etc – since this is not environmental) (ii) the burning of fossil fuels/coal/oil/petrol/gas/diesel/fuel or car exhausts or roasting of sulphide ores or cement manufacture or volcanoes [1] (iii) SO2 + NO2 → SO3 + NO [1] NO + ½O2 → NO2 [1] [4] [Total: 9] 5 (a) CH3CH2CH2CH2CH2OH CH3CH2CH2CH(OH)CH3 CH3CH2CH(OH)CH2CH3 A B C all three (any order) [2] (2 only = [1]) [2] (b) B above (may be different letter) ([0] if more than one compound stated) [1] [1] (c) (i) B above (may be different letter) ([0] if more than one compound stated) [1] (ii) (pale) yellow ppt. [1] (iii) CHI3 + CH3CH2CH2CO2Na or anion (no credit for the acid, RCO2H) [1] + [1] [4] (d) A → CH3CH2CH2CH2CO2H [1] B → CH3CH2CH2COCH3 [1] C → CH3CH2COCH2CH3 (letters may differ) [1] [3]

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – May/June 2009 9701 04 © UCLES 2009 (e) (i) (C6H10O5)n → 5n H2 + 5n CO + n C correct species and the 5:5:1 ratio [1] (allow n5 instead of 5n) balancing, i.e. multiplying by n [1] (ii) ∆H = 7(1080) + 15(436) – 6(350) – 16(410) – 14(460) = –1000 kJ mol–1 4 correct values from DB (in bold italics above) [1] correct multipliers [1] correct signs and arithmetic [1] (correct answer = [3]) Some ecf values for [2] marks (i.e. 1 error): for [1] mark (i.e. 2 errors): +1000 (signs reversed) –1350 (7 x (C-C) instead of 6) +1350 +2220 (7 x O-H instead of 14) –2220 –1410 (17 C-H instead of 16) +1410 The omission of a type of bond (C-C is the most common one that is omitted) forfeits 2 marks, in addition to any other errors there may be. [5] [Total: 15] 6 (a) (i) I: SOCl2 or PCl5 or HCl + ZnCl2 or PCl3 + heat or Cl2 + P + heat [NOT NaCl + H2SO4] [1] (mention of aq negates mark) II: NH3 (ignore any conditions stated) [1] (ii) nucleophilic substitution or SN or SN1 or SN2 [1] (iii) delocalisation of lone pair on Cl over benzene ring produces a stronger C-Cl bond [1] [4] (b) (i) III: HNO3 + H2SO4 [1] both conc., and at T < 60oC [1] IV: Sn + conc HCl [NOT LiAlH4 or H2 + Ni] [1] (ii) III: electrophilic substitution [1] IV: reduction or redox [1] [5] (c) e.g. add bromine water or Br2(aq) (a solvent is needed for the mark) [1] or add UI solution phenylamine decolorises the bromine or gives a white ppt., hexylamine does not [1] or hexylamine turns UI blue, with phenylamine it stays green [2]

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – May/June 2009 9701 04 © UCLES 2009 (d) N N N N Cl CH3 CH3 OH (phenylazo group must be at 4-position to -OH) (N=N must be double bond, not triple) (allow + charge on either N) (allow double or triple bond) [1] [1] [2] [Total: 13] Section B 7 (a) For each element, award [1] mark for each column in one particular line in the table below. The [2] marks awardable for each element are not conditional on each other, but don’t take the location from one line and the role from another. element location role red blood cells/haemoglobin to bind to/carry/transfer oxygen (to cells) or CO2 (away from cells) muscle (cells)/myoglobin to bind to/carry/transfer oxygen (to muscles) or CO2 (away from muscles) in mitochondria/cytochromes to aid redox reactions or to help oxidise NADH etc in iron-sulphide proteins to aid redox reactions iron in ferrodoxin to aid redox reactions in nerve cells/nerves/nervous system/neurones or in cell membranes/phospholipid bilayers Na+/K+ pump or ion pump or active transport or transmission/regulation of nerve impulses sodium in kidneys to help re-absorb glucose in blood (“cells” not needed, but “plasma” negates) or carbonic anhydrase as an enzyme co-factor/prosthetic group or to help the hydration/removal of CO2 or production of H2CO3/HCO3 – in the gut/carboxypeptidase as an enzyme co-factor/prosthetic group or to help hydrolyse polypeptides zinc in the liver/alcohol dehydrogenase as an enzyme co-factor/prosthetic group or to help oxidise/break down alcohol [1] + [1] for each element [6]

Page 8 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – May/June 2009 9701 04 © UCLES 2009 (b) (i) manufacture of NaOH or manufacture of batteries or manufacture of felt or gold extraction or (mercury) fungicides or (mercury) compounds used in timber preservation [1] (ii) In each case below, a balanced equation is worth [2] marks breaks disulphide bonds/linkages or Hg bonds to S-H groups (or in an unbalanced equation) [1] –CH2-S-S-CH2- + 4Hg+ → 2 –CH2-S–Hg + 2Hg2+ or R-S-S-R + 4Hg+ → 2 R-S-Hg + 2Hg2+ or R-S-S-R + Hg+ → 2 R-S-Hg+ or R-SH + Hg+ → R-SHg + H+ or R-SH + Hg2+ → R-S-Hg+ + H+ or 2 R-SH + Hg2+ → (R-S)2Hg + 2 H+ etc [1] bonds to carboxyl side chains (in amino acids) (or in an unbalanced equation) [1] –CO2H + Hg+ → –CO2Hg + H+ or 2 RCO2H + Hg2+ → (RCO2)2Hg + 2H+ [1] [5] [11 max 10] 8 (a) (i) Partition coefficient (PC) is an equilibrium constant representing the distribution of a solute between two solvents. or PC = ratio of the concentrations of the solute in the two solvents or PC = [X]a/[X]b [1] (ii) If 0.4 g has been extracted, 0.1 g remain in the aqueous layer. the concentration in the hexane layer = 20 0.4 = 0.02 g cm–3 the concentration in the aqueous layer = 100 0.1 = 0.001 g cm–3 Kpc = 0.02/0.001 = 20 [1] (iii) 1st extraction: hexane x/10 g cm–3 water (0.50-x)/100 g cm–3 Kpc = 20 x)/100 - (0.5 x/10 = hence x/10 = (10 – 20x)/100 100x = 10(10 – 20x) or 100x = 100 – 200x x = 0.33 g [1] 2nd extraction: hexane y/10 g cm–3 water (0.17 – y)/100 g cm–3 Kpc = 20 y)/100 - (0.17 y/10 = hence y/10 = (3.4 – 20y)/100 100y = 10(3.4 – 20y) or 100y = 34 – 200y y = 0.11 g [1] total extracted = 0.44 g, or difference = 0.04 g or 10% more (is extracted) [1] (correct answer = [3]) [5]

Page 9 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – May/June 2009 9701 04 © UCLES 2009 (b) (i) berries are aqueous media [1] PCBs are insoluble/sparingly soluble in water or more fat-soluble [1] (ii) partition coefficient or [fat]/[water] is greater than 1 [1] [3] (c) (i) 4 (four) [1] (ii) correct spot circled [1] correct spot squared [1] [in each case, more than one spot circled or squared negates the mark] [3] [Total: 11] 9 (a) (i) correct diagram showing at least one monomer unit, and at least one N-H and C=O. i.e. –NH-C6H2-NH-CO– or –CO-C6H4-CO-NH– (no mark for this, but apply a penalty of –[1] if candidate’s diagram does NOT show these points correctly) one H-bond between N-H of original chain and C=O group of new chain [1] one H-bond between C=O of original chain and N-H group of new chain [1] (ii) hydrogen bonds or H-bonds (in words; can be written on diagram) (ignore ref to v d W) [1] (iii) HO2C CO2H ClOC COCl NH2 H2N or allow HO2C- HOOC- HOCO- allow ClCO- allow NH2- [5] [1] [1]

Page 10 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – May/June 2009 9701 04 © UCLES 2009 (b) (i) Water-hating/fearing/repelling/resistant or can’t form bonds with water (molecules) [1] [NOT insoluble or does not dissolve in water, also NOT “non-polar”] (ii) Fluorine-containing groups form van der Waals bonds (with the oil molecules)... [1] …but cannot form hydrogen bonds (with the water molecules) [1] (iii) Teflon/PTFE [1] [4] [Total: 9]