Cambridge A Level Chemistry 9701 — 2007 Oct/Nov Paper 4 · Variant 1

This document consists of 15 printed pages and 1 blank page. SPA (KN) T34461/2 © UCLES 2007 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level and Advanced Level * 5 0 2 7 7 9 6 6 9 9 * CHEMISTRY 9701/04 Paper 4 Structured Questions October/November 2007 1 hour 45 minutes Candidates answer on the Question Paper. Additional Materials: Data Booklet 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. For Examiner’s Use 1 2 3 4 5 6 7 8 9 10 Total

2 9701/04/O/N/07 For Examiner’s Use © UCLES 2007 Section A Answer all questions in the spaces provided. 1 (a) Use the general formula of a carboxylic acid, RCO2H, to write equations to explain the following terms. (i) Ka … (ii) pKa … [2] (b) The pKa values of four carboxylic acids are listed in the table below. acid formula of acid pKa 1 CH3CH2CO2H 4.9 2 CH3CHCl CO2H 2.8 3 CH3CCl2CO2H 1.4 4 CH2Cl CH2CO2H 4.1 (i) Describe and explain the trend in acid strength shown by acids 1, 2 and 3. … … … … (ii) Suggest an explanation for the difference in the pKa values for acids 2 and 4. … … (iii) Calculate the pH of a 0.010 mol dm–3 solution of propanoic acid (acid 1). … … … [6]

3 9701/04/O/N/07 [Turn over For Examiner’s Use © UCLES 2007 (c) A good way of making synthetic amino acids uses chloro-acids as intermediates. Cl2 + trace of P I CH3CH2CO2H CH3CH(NH2)CO2H alanine NH3(excess) II CH3CHCl CO2H (i) Suggest the role that the trace of phosphorus plays in reaction I. … (ii) Write a fully balanced equation for reaction I. … (iii) State the type of mechanism of reaction II. … (iv) When 10.0 g of propanoic acid was used in this 2-stage synthesis, a yield of 9.5 g of alanine was obtained. Calculate the overall percentage yield. … [5] (d) In the solid state and in aqueous solutions, alanine exists as a zwitterion. Draw the structural formula of this zwitterion. [2] [Total: 15]

4 9701/04/O/N/07 For Examiner’s Use © UCLES 2007 2 (a) Describe and explain the trend in the solubilities of the sulphates of the Group II elements. … … … … …[4] (b) The salts formed by the Group II elements with other divalent anions show a similar trend in their solubilities, whereas most of their salts with monovalent anions are very soluble. Use this information to predict the identities of compounds A and B in the following description of some reactions of Group II compounds, and write balanced equations for the reactions. Magnesium hydroxide, Mg(OH)2, is almost insoluble in water. Stirring a mixture of magnesium hydroxide and aqueous ethanedioic acid, H2C2O4, produces a clear colourless solution containing A. When a solution of calcium nitrate, Ca(NO3)2, is added, a white precipitate of B is formed. identity of A … identity of B … equations … …[3] (c) The solubility product, Ksp, of magnesium hydroxide has a numerical value of 2.0 × 10–11. (i) Write an expression for the Ksp of magnesium hydroxide, stating its units. … (ii) Use the value of Ksp given to calculate the concentration of Mg(OH)2 in a saturated solution. … … (iii) Explain whether magnesium hydroxide would be more or less soluble in 0.1 mol dm–3 MgSO4(aq) than in water. … … [5] [Total: 12]

5 9701/04/O/N/07 [Turn over For Examiner’s Use © UCLES 2007 3 The following account describes the preparation of Péligot’s salt, named after the 19th century French chemist who first made it. Place 6.0 g of potassium dichromate(VI) in a 100 cm3 beaker and add 8.0 g of concentrated hydrochloric acid and 1.0 cm3 water. Warm the mixture gently; if carefully done the dichromate(VI) will dissolve without the evolution of chlorine. On cooling the beaker in an ice bath the solution will deposit long orange-red crystals of Péligotʼs salt. An analysis of Péligot’s salt showed that it contained the following percentages by mass: K, 22.4%; Cr, 29.8%; Cl, 20.3%; O, 27.5%. (a) Calculate the empirical formula of Péligot’s salt. [2] (b) Suggest a balanced equation for the formation of Péligot’s salt. …[1] (c) The instructions suggest that strong heating might cause chlorine to be evolved. (i) What type of reaction would produce chlorine in this system? … (ii) Use the Data Booklet to identify relevant half equations and E o values for the production of chlorine from the reaction between K2Cr2O7 and HCl. … … Use these equations to write the overall full ionic equation for this reaction. … (iii) The use of dilute HCl (aq) does not result in the production of chlorine. Suggest why this is so. … (iv) Use the Data Booklet to suggest a reason why it is not possible to prepare the bromine analogue of Péligot’s salt by using HBr(aq) instead of HCl (aq). … [6] [Total: 9]

6 9701/04/O/N/07 For Examiner’s Use © UCLES 2007 4 (a) By choosing the chlorides of two of the Group IV elements as examples, describe the trend in the reactions of these chlorides with water. Suggest an explanation for any differences, and write equations for any reactions that occur. … … … …[3] (b) The standard enthalpy changes of formation of lead(II) chloride and lead(IV) chloride are given in the following table. compound ∆H o f / kJ mol–1 PbCl2(s) –359 PbCl4(l) –329 Use these data, and also bond energy data from the Data Booklet, to calculate the enthalpy changes for the following two reactions. (i) CCl2(g) + Cl2(g) CCl4(g) ∆H o = … kJ mol–1 (ii) PbCl2(s) + Cl2(g) PbCl4(l) ∆H o = … kJ mol–1 (iii) Make use of your answers to parts (i) and (ii) to suggest how the relative stabilities of the two oxidation states vary down the Group. … … [3] [Total: 6]

7 9701/04/O/N/07 [Turn over For Examiner’s Use © UCLES 2007 5 Potassium manganate(VII) can be used to estimate the percentage of hydrogen peroxide in household bleach. The following unbalanced equation represents the reaction between them. … MnO– 4 + … H2O2 + … H+ … Mn2+ + … H2O + … O2 (a) Balance this equation by putting the appropriate numbers in the spaces above. [1] (b) Use data from the Data Booklet to calculate the E o cell for the reaction. …[1] (c) When 0.020 mol dm–3 KMnO4(aq) was added from a burette into an acidified 25.0 cm3 sample of H2O2, 15.0 cm3 of KMnO4 was required to reach the end-point. (i) Describe what you would see during this titration, and also at the end-point. … … (ii) Calculate the concentration of H2O2 in the sample. … … … [4] [Total: 6]

8 9701/04/O/N/07 For Examiner’s Use © UCLES 2007 6 The phenol 1-naphthol is a starting point for the manufacture of carbaryl, an insecticide and a plant growth inhibitor. OH 1-naphthol + NaOH O carbaryl C O + CH3NHCOCl CH3NH C (a) (i) Suggest a structure for the intermediate C and draw it in the box above. (ii) Name the functional groups in carbaryl. … … (iii) Suggest structures for the three products formed when carbaryl is hydrolysed. (iv) What reagents and conditions would you use for this hydrolysis? … [7] (b) Suggest reagents and conditions for converting 1-naphthol into each of the following compounds. (i) … OH Br Br (ii) … [2] OH NO2

9 9701/04/O/N/07 [Turn over For Examiner’s Use © UCLES 2007 (c) Compound D is an isomer of 4-nitro-1-naphthol. D is formed as a by-product during the reaction in b(ii). It can be converted into 2-amino-1-naphthol, E. OH I NH2 D E (i) Suggest the structural formula of the isomer D. (ii) Suggest reagents needed for reaction I. … (iii) Suggest the structural formula of the compound formed when compound E reacts with an excess of CH3COCl. [3] (d) When an alkaline solution of compound E is added to a solution containing Cu2+(aq) ions, a pale green-blue precipitate F forms. Analysis of F shows that its formula is Cu(C10H8NO)2(H2O)2. (i) Complete the following structural formula of F. O H2O Cu N H2 When an excess of concentrated NH3(aq) is added to F, the precipitate dissolves to form a deep blue solution. (ii) State the formula of the ion responsible for the deep blue colour. … (iii) What type of reaction is occurring here? … [3] [Total: 15]

10 9701/04/O/N/07 For Examiner’s Use © UCLES 2007 7 The nitration of benzene occurs in the following steps. + NO2 + H+ NO2 NO2 H + + (a) What reagents and conditions are needed for this reaction? …[2] (b) Write an equation showing how the electrophile NO+ 2 is formed from the reagents. …[1] (c) The nitration of methylbenzene produces mainly 2-nitromethylbenzene, whereas the nitration of benzoic acid produces mainly 3-nitrobenzoic acid. CH3 2-nitromethylbenzene NO2 CO2H 3-nitrobenzoic acid NO2 Use this information to suggest suitable intermediates G and H in the following two 2-stage syntheses of chlorobenzoic acids, and suggest suitable reagents for reactions I to IV. II I III G CO2H Cl CH3 IV H CO2H Cl reagents: reaction I … reaction II … reaction III … reaction IV … [4] [Total: 7]

11 9701/04/O/N/07 [Turn over For Examiner’s Use © UCLES 2007 Section B – Applications of Chemistry Answer all questions in the spaces provided. 8 (a) DNA carries the genetic code in living organisms and consists of a double helix. (i) Describe what is meant by a double helix. … … (ii) How are the strands of the double helix held together? … … [2] (b) In replicating the genetic code two RNA molecules, mRNA and tRNA, are used to perform functions called transcription and translation. Describe the role of the RNA molecules in these two functions. transcription … … … translation … … …[4] (c) When an egg is boiled, the protein changes from a viscous liquid to a solid. (i) Suggest what causes this change as the protein is heated. … … (ii) Why is there no change to the primary structure of the protein under these conditions? … … [2]

12 9701/04/O/N/07 For Examiner’s Use © UCLES 2007 (d) Describe in outline how energy is provided in animal cells. … … … … … …[3] [Total: 11] 9 (a) Explain with reference to energy states how 1H NMR can supply information about the structure of molecules. … … … …[3] (b) Nuclear magnetic resonance is used in magnetic resonance imaging scanners. These scanners are increasingly used in hospitals to detect tumours. Suggest why magnetic resonance techniques are better than X-rays. … … … …[2]

13 9701/04/O/N/07 [Turn over For Examiner’s Use © UCLES 2007 (c) The NMR spectrum shown below was obtained from a simple organic molecule, G, CxHyO2. When a sample of G was placed in a mass spectrometer, the ratio of the M : M+1 peaks for the molecule was 14.5 : 0.66. 11 absorbance 10 9 8 7 6 5 chemical shift,
4 3 3 3 2 2 1 0 (i) Calculate how many carbon atoms there are in the molecule. (ii) Use the NMR spectrum and the Data Booklet to work out the structure of G. [5] [Total: 10]

14 9701/04/O/N/07 For Examiner’s Use © UCLES 2007 10 Read the following article about the use of bacteria in mining, and then answer the questions that follow it. The discovery that bacteria could ʻmineʼ metals for us was made in Spain. The Rio Tinto mine, in the southwest corner of Spain, was originally mined for copper by the Romans some 2,000 years ago. In 1752, some mining engineers looked over the mine to see if it could possibly be re-opened. They noticed streams of a blue-green liquid running from spoil heaps of the processed rock that lay around the mine. When this blue-green liquid ran over iron, it coated the iron with a brown film. The brown film was metallic copper. There was still some copper left in the spoil heaps. At the time, everybody thought that the copper was being dissolved in the liquid through a simple chemical reaction. But in 1947, US scientists discovered that the copper was being ʻminedʼ by a bacterium called Thiobacillus ferrooxidans. The bacterium Thiobacillus ferrooxidans lives off the chemical energy trapped in metal sulphides. In the ore, the copper exists as copper sulphide. The bacteria gain energy by converting the copper sulphide to copper sulphate, which is then excreted. At the same time, they absorb the difference in energy in the chemical bonds. These bacteria can also obtain energy in similar reactions with ores of zinc, lead and uranium. (a) Use the Data Booklet to explain why the blue-green liquid coated the iron with copper. Write an equation for the reaction. … … [2] (b) Suggest two reasons why this method of extracting copper might be useful for ore containing only a small percentage of copper. (i) … … (ii) … … [2] (c) Suggest one disadvantage of using bacteria rather than traditional mining and smelting methods. … …[1]

15 9701/04/O/N/07 For Examiner’s Use © UCLES 2007 (d) In conventional copper mining, the ore will typically contain 0.5 – 2.0% copper, which gives an idea of what a valuable resource copper is. (i) The ore from a particular mine contains 0.75% copper, and 150 000 tonnes of ore are mined each year. From this ore about 60% of the copper is extracted, and the remainder is left in the ‘spoil heaps’ of processed ore. What mass of copper is extracted each year? (ii) If the use of bacteria can recover a further 17% of copper from the spoil heaps, what is the extra mass of copper produced? [2] (e) Suggest why bacteria are unlikely to be used in the extraction of aluminium. … …[1] (f) Metals like copper and zinc from abandoned mines can contaminate ground-water. Suggest one way of removing these contaminants. … …[1] [Total: 9]

16 9701/04/O/N/07 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 2007 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 2007 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 2007 9701 04 © UCLES 2007 1 (a) (i) Ka = [H+][RCO2 –]/[RCO2H] [1] (ii) pKa = –log10Ka or –logKa or log [H+]2/[RCO2H] NOT ln; [1] [2] (b) (i) acid strength increases from no. 1 to no. 3 or down the table or as Cls increase [1] due to the electron-withdrawing effect/electronegativity of chlorine (atoms) [1] stabilising the anion or weakening the O-H bond NOT H+ more available [1] (ii) chlorine atom is further away (from O-H) in no. 4, so has less influence [1] (iii) either: pH = ½ (pKa – log10[acid]) or Ka = 10–pKa = 1.259 x 10–3 = ½ (4.9 + 2) [H+] = √(Ka. c) = 3.55 x 10–4 [1] = 3.4 (allow 3.5) pH = 3.4 ecf [1] ([1] for correct expression & values; [1] for correct working) [6] (c) (i) catalyst [1] (ii) CH3CH2CO2H + Cl2 → CH2CHClCO2H + HCl [1] (iii) nucleophilic substitution NOT addition/elimination [1] (iv) Mr (CH3CH2CO2H) = 74 Mr(CH2CH(NH2)CO2H) = 89 [1] ∴ 10.0 g should give 10 x 89/74 = 12.03 g ∴ percentage yield = 100 x 9.5/12.03 = 79% ecf [1] ([2] for correct answer) [5] (d) +NH3-CH(CH3)-CO2 – correct atoms [1] Allow charges on H of H3N, and –COO but not –C-O-O correct charges [1] [2] [Total: 15]

Page 3 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2007 9701 04 © UCLES 2007 2 (a) solubility decreases (down Group II) [1] lattice energy decreases [1] solvation/hydration energy (of cation) decreases [1] but more so than does lattice energy/is not able to overcome LE [1] ∆Hsoln becomes more endothermic/positive/less exothermic [1] [max 4] (b) identities of A and B 2 x [1] Mg(OH)2 + H2C2O4 → MgC2O4(aq) + 2H2O [1] (A) MgC2O4(aq) + Ca(NO3)2 → Mg(NO3)2 + CaC2O4(s) [1] (B) [max 3] (c) (i) (Ksp =) [Mg2+][OH–]2 [1] units are mol3dm–9 ecf from Ksp [1] (ii) (call [Mg(OH)2(aq)] = [Mg2+] = x) ∴ Ksp = 2 x 10–11 = 4x3 [1] ∴ x = 1.71 x 10–4 mol dm–3 ecf [1] (iii) less soluble because of the common ion effect or the equilibrium Mg(OH)2(s) ⇋ Mg2+(aq) + 2OH–(aq) is moved to the left [1] [5] [Total: 12]

Page 4 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2007 9701 04 © UCLES 2007 3 (a) K = 22.4/39.1 = 0.573 thus ratio is: 1 Cr = 29.8/52.0 = 0.573 1 Cl = 20.3/35.5 = 0.572 1 O = 27.5/16.0 = 1.719 3 or KCrClO3 (scores 2) [1] [1] [2] (b) K2Cr2O7 + 2HCl → 2KCrClO3 + H2O [1] [1] (c) (i) redox or oxidation [1] (ii) Eo data and half equations: Cr2O7 2– + 14H+ + 6e– → 2Cr3+ + 7H2O Eo = 1.33 V [1] Cl2 + 2e– → 2 Cl– Eo = 1.36 V [1] overall ionic equation: Cr2O7 2– + 6Cl– + 14H+ → 2Cr3+ + 3Cl2 + 7H2O [1] (iii) (dilution will) lower Eo for Cr2O7 2–/Cr3+ or raise Eo for Cl2/Cl–   [1] or lower [Cl–] or [H+] will shift equilibrium in eqn to the left hand side  (iv) Br2/Br– = +1.07 V, so Cr(VI) would oxidise Br– (easily) [1] [6] [Total: 9]

Page 5 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2007 9701 04 © UCLES 2007 4 (a) CCl4 is unreactive. (The rest react (with increasing vigour)) [1] no d-orbitals or available/low-lying empty orbitals in carbon or unable to expand octet [1] e.g. SiCl4 + 2H2O → SiO2 + 4HCl (or GeCl4 etc) or Si(OH)2Cl2 or Si(OH)4 (allow balanced equations for partial hydrolysis) [1] [3] (b) (i) E(Cl-Cl) = 244 kJ mol–1; 2 E(C-Cl) = 2 x 340 = 680 kJ mol–1 ∴ ∆H = –436 (kJ mol–1) [1] (ii) ∆H = 359 – 329 = +30 (kJ mol–1) [1] (iii) since reaction (ii) is endothermic, the +4 oxidation state is less stable or the +2 oxidation state is more stable (down the group) [1] [3] [Total: 6] 5 (a) 2 MnO4 – + 5 H2O2 + 6 H+ → 2 Mn2+ + 8 H2O + 5 O2 [1] [1] (b) Ecell = 1.52 – 0.68 = +0.84 (V) [1] [1] (c) (i) (as KMnO4 is added), colour changed (from purple) to colourless – NOT pink or effervescence/bubbles (of O2) are produced [1] at end-point, change is to (first) pink [1] (ii) n(MnO4 –) = 0.02 x 15/1000 = 3 x 10–4 [1] since H2O2 : MnO4 – = 5:2, ⇒ n(H2O2) = (5/2) x 3 x 10–4 = 7.5 x 10–4 in 25 cm3 ∴ [H2O2] = 7.5 x 10–4 x 1000/25 = 3.0 x 10–2 mol dm–3 [1] [4] [Total: 6] o

Page 6 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2007 9701 04 © UCLES 2007 6 (a) (i) C is O Na allow ONa but no covalent O-Na bond [1] (ii) amide, ester 2 x [1] (iii) CO2 or H2CO3 or Na2CO3 [1] CH3NH2 or CH3NH3 +Cl– [1] OH [1] (iv) H3O+ and heat >80° or OH–(aq) and heat >80° [1] [7] (b) (i) Br2(aq) (or other suitable solvent) [1] (ii) dilute/aqueous HNO3 [1] [2] (c) (i) OH NO2 D is [1] (ii) tin/Fe + HCl NOT LiAlH4 [1] (iii) OCOCH3 NHCOCH3 mark each side chain separately 2 x [1] [4] (d) (i) (allow any orientation of groups) N H2 Cu O H2 N H2O OH2 O penalise missing H on NH2 [1] (ii) [Cu(NH3)4]2+ or [Cu(NH3)4(H2O)2]2+ NOT [Cu(NH3)6]2+ [1] (iii) ligand substitution/exchange [1] [3] [Total: max 15]

Page 7 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2007 9701 04 © UCLES 2007 7 (a) HNO3 + H2SO4 [1] at 50 – 60°C (or ≤ 60°C) not dilute or (aq) [1] [2] (b) 2H2SO4 + HNO3 → 2HSO4 – + H3O+ + NO2 + (allow equ. with only one H2SO4, giving H2O) [1] [1] (c) G is H is CH3 Cl CO2H [1] + [1] reaction I: Cl2 + AlCl3/accept other halogen carriers NOT aq, nor u.v. reaction II: KMnO4 + H+ NOT HCl nor HNO3 reaction III: KMnO4 + H+ NOT HCl nor HNO3 reaction IV: Cl2 + AlCl3/accept other halogen carriers NOT aq, nor u.v. both I + IV [1] both II + III [1] [4] [Total: 7]

Page 8 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2007 9701 04 © UCLES 2007 8 (a) (i) Two interlinked spirals or chains or strands woven round each other [1] (ii) By hydrogen bonds between bases [1] [2] (b) Transcription – (1)DNA/RNA/nucleic acid unravels [1] – (2)strand is used as a template [1] – (3)mRNA reads the sequence on this strand/ produces complementary strand [1] Translation – (4)mRNA binds to the ribosome [1] – (5)tRNA translates the codon from mRNA [1] – (6)tRNA carries amino acids to ribosome/adds a.a. to chain [1] [max 4] (c) (i) Disruption of the secondary/tertiary/quaternary/3D structure of the protein (could be answered in terms of bonds e.g. hydrogen bonds break) [1] (ii) The covalent/peptide bonds in the (protein) chain are too strong [1] [2] (d) Energy is provided by the breakdown/hydrolysis of adenosine triphosphate (ATP) [1] ATP (+ H2O) → ADP + Pi (+ energy) or in words [1] ATP is produced during respiration/Krebs cycle/oxidation of glucose, fats or proteins/ in mitochondria/ADP is recycled [1] [3] [Total: 11]

Page 9 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2007 9701 04 © UCLES 2007 9 (a) Suitable diagram showing origin of two energy states/or description [1] Needs to mention applied magnetic field/electron transfer negates Indication that energy difference is in the radio frequency range [1] Indication that frequency of absorption or gap between the 2 energy states depends on the nature of nearby atoms or the chemical environment of the 1H [1] [3] (b) They do not damage tissues/X-rays harmful/NMR of lower energy [1] They are not obscured by bones/skeleton [1] They can be tuned to examine particular tissues/tumours/organs/protons [1] [max 2] (c) (i) M : M+1 = 100/(1.1n) n = = = × × = 14 . 4 95 . 15 66 1 . 1 5 . 14 200 66 . 0 4 carbon atoms [1] Check for 1.1 in divisor, if missing, penalise (ii) Singlet at δ 2 suggests methyl adjacent to C=O [1] Quartet at δ 4 suggests a –CH2- group (adjacent to a –methyl group) [1] (allow –OCH2- ) Triplet at δ 1.2 suggests a methyl group (adjacent to a –CH2–) [1] G is ethyl ethanoate (or structure)/if methyl propanoate given here cannot score first marking point [1] [5] [Total: 10]

Page 10 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2007 9701 04 © UCLES 2007 10 (a) Iron is higher in the reactivity series than copper (owtte)/allow use of Eo [1] Cu2+(aq) + Fe(s) → Cu(s) + Fe2+(aq) [1] If conversion to Fe3+ given, Ecell is –0.38 [2] (b) It does not require investment in machinery/labour [1] It requires little energy [1] accept it produces little/no pollution/noise [1] Do not accept comparison with electrolytic method [max 2] (c) The process takes a long time/requires smaller workforce [1] [1] (d) (i) 0.75% is 7.5 kg in every tonne of ore Hence 150,000 tonnes of ore yield 1000 150000 5 . 7 × tonnes or 1,125 tonnes Cu 1125 x 0.6 = 675 tonnes (accept 680) [1] (ii) 450 x 0.17 = 76.5 tonnes (accept 77) [1] or 1125 x 0.17 = 191.25 tonnes (accept 191) – this is an ecf if 675 not in (i) [2] (e) Aluminium is too high in the reactivity series/very reactive/aluminium forms bonds with oxygen which are too strong/aluminium ore doesn’t exist as sulphide /Fe unable to displace Al [1] [1] (f) Control the pH (greater than pH 6.0) [1] Bioremediation/growth of special plants (to remove heavy metals) Other reasonable suggestions such as displacement by a more reactive metal/ precipitation/ion exchange [1] [2] [Total: 9]