Cambridge A Level Chemistry 9701 — 2016 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 an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. Answer all questions. Electronic calculators may be used. 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. CHEMISTRY 9701/43 Paper 4 A Level Structured Questions May/June 2016 2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet This document consists of 17 printed pages and 3 blank pages. [Turn over IB16 06_9701_43/FP © UCLES 2016 *1516755053* Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level

2 9701/43/M/J/16 © UCLES 2016 Answer all the questions in the spaces provided. 1 (a) Limewater is a saturated solution of Ca(OH)2 in water. It is used to test for the presence of CO2 in a gaseous mixture. (i) Write an equation for the reaction between limewater and CO2. … [1] A saturated solution of Ba(OH)2 can be used instead of Ca(OH)2 to test for CO2. A saturated solution of Mg(OH)2 cannot be used for this test. (ii) Explain why a saturated solution of Ba(OH)2 can be used to test for CO2. … … [1] (iii) Explain why a saturated solution of Mg(OH)2 cannot be used to test for CO2. … … … [2] (b) Describe and explain the trend in the thermal stabilities of the carbonates down Group 2. … … … … … [3] (c) Nickel carbonate, NiCO3, decomposes on heating. Use the Data Booklet to explain whether NiCO3 will decompose more or less readily than CaCO3. … … … … [2] [Total: 9]

3 9701/43/M/J/16 © UCLES 2016 [Turn over 2 (a) (i) Complete the following electronic configurations. • the cobalt atom, Co 1s22s22p6 … • the cobalt(II) ion, Co2+ 1s22s22p6 … [1] (ii) State the colours you would observe when concentrated HCl (aq) is added to an aqueous solution of cobalt(II) nitrate, Co(NO3)2. Give the formulae and geometry of the complexes formed. … … … … … [5] (b) There are two isomers with the formula Ni(R3P)2I2, but only one structure with the formula Co(R3P)2I2. (R = alkyl, R3P is a monodentate ligand) Draw diagrams showing the structure of Co(R3P)2I2 and the two isomers of Ni(R3P)2I2. Ni(R3P)2I2 Ni(R3P)2I2 Co(R3P)2I2 [3] [Total: 9]

4 9701/43/M/J/16 © UCLES 2016 3 One method of producing hydrogen from natural gas is the reaction between hydrogen sulfide and methane. 2H2S(g) + CH4(g) CS2(g) + 4H2(g) (a) Write the expression for Kp for this reaction, and state its units. Kp = units … [2] (b) The initial partial pressures of the two gases in a mixture at 1000 K are recorded. H2S(g) 200 atm CH4(g) 100 atm The mixture is left to reach equilibrium. It is found that the equilibrium partial pressure of CS2(g) is 2 atm and that of the remaining CH4(g) is 98 atm. (i) Calculate the equilibrium partial pressures of H2S(g) and H2(g). p(H2S) = … atm p(H2) = … atm [2] (ii) Calculate the value of Kp at this temperature. Kp = … [1]

5 9701/43/M/J/16 © UCLES 2016 [Turn over (c) (i) Predict the sign of ∆S o for this reaction. Explain your answer. 2H2S(g) + CH4(g) CS2(g) + 4H2(g) ∆H o = +241 kJ mol–1 … … [1] The free energy change, ∆G o, for this reaction at 1000 K is +51 kJ mol–1. (ii) Calculate the value of ∆S o for this reaction, stating its units. ∆S o = … units … [2] (d) How would the value of ∆G o, and hence the spontaneity (feasibility) of this reaction change as the temperature increases? Explain your answer. … … … [2] [Total: 10]

6 9701/43/M/J/16 © UCLES 2016 4 (a) (i) Define the term standard cell potential, . … … … [1] The following incomplete diagram shows the apparatus that can be used to measure the for a cell composed of the Fe3+ / Fe2+ and Ag+ / Ag half-cells. A B C D (ii) Complete the diagram, labelling the components you add. [1] (iii) Identify the components A-D. A … B … C … D … [3] (b) (i) Use E o values to write an equation for the cell reaction that takes place if the two electrodes in (a) are connected by a wire and the circuit is completed. … … [1] (ii) Another electrochemical cell was set up using 0.31 mol dm–3 Ag+(aq) instead of the standard Ag solution. Use the Nernst equation, E = E o + 0.059 log [Ag+(aq)], and the relevant E o values to calculate the new E cell in this experiment. E cell = … V [2] [Total: 8]

7 9701/43/M/J/16 © UCLES 2016 [Turn over 5 Dicarboxylic acids dissociate in stages. stage 1 stage 2 HO2C(CH2)nCO2H HO2C(CH2)nCO2 – + H+ –O2C(CH2)nCO2 – + 2H+ (a) The pKa values for stage 1 and stage 2 for some dicarboxylic acids are listed below. n in HO2C(CH2)nCO2H pKa(1) for stage 1 pKa(2) for stage 2 1 2.83 5.69 2 4.16 5.61 3 4.31 5.41 For comparison, the pKa of ethanoic acid, CH3CO2H, is 4.76. (i) State the mathematical relationship between pKa and the acid dissociation constant Ka. … [1] (ii) With reference to the table above, suggest why the pKa(1) values • are all smaller than the pKa of ethanoic acid, … … … • become larger as n increases. … [3] (iii) Suggest why all the pKa(2) values in the table above are larger than the pKa of ethanoic acid. … … [1] (b) The monosodium salts of edible dicarboxylic acids are added to some foodstuffs as buffers. (i) Explain what is meant by the term buffer solution. … … [2] (ii) Write two equations to show how monosodium butanedioate, HO2CCH2CH2CO2Na, acts as a buffer. … … [2] [Total: 9]

8 9701/43/M/J/16 © UCLES 2016 6 Nitrobenzene, C6H5NO2, can be reduced to phenylamine, C6H5NH2, in acid solution in a two step process. (a) (i) Balance the half-equation for this reaction to work out how many moles of electrons are needed to reduce one mole of nitrobenzene. C6H5NO2 + …e– + …H+ → C6H5NH2 + …H2O [1] (ii) The reducing agent normally used is granulated tin and concentrated hydrochloric acid. In the first step, the reduction of nitrobenzene to phenylammonium chloride can be represented by the equation shown. Use oxidation numbers or electrons transferred to balance this equation. You might find your answer to (i) useful. …C6H5NO2 + …HCl + …Sn → …C6H5NH3Cl + …SnCl 4 + …H2O [2] (b) When 5.0 g of nitrobenzene was reduced in this reaction, 4.2 g of phenylammonium chloride, C6H5NH3Cl, was produced. Calculate the percentage yield. percentage yield of phenylammonium chloride = … % [2] (c) Following the reaction in (b), an excess of NaOH(aq) was added to liberate phenylamine from phenylammonium chloride. (i) Calculate the mass of phenylamine, C6H5NH2, produced when 4.20 g of phenylammonium chloride reacts with an excess of NaOH(aq). mass of phenylamine = … g [1] The final volume of the alkaline solution of phenylamine in (i) was 25.0 cm3. The phenylamine was extracted by addition of 50 cm3 of dichloromethane. After the extraction, the dichloromethane layer contained 2.68 g of phenylamine. (ii) Use the data to calculate the partition coefficient, K partition, of phenylamine between dichloromethane and water. K partition = … [2]

9 9701/43/M/J/16 © UCLES 2016 [Turn over (d) How does the basicity of phenylamine compare to that of ethylamine? Explain your answer. … … … [2] (e) Phenol can be synthesised from phenylamine in two steps. NH2 OH + N2 E step 1 step 2 (i) State the reagents and conditions for steps 1 and 2. step 1 … step 2 … [2] (ii) Draw the structure of the intermediate compound E in the box above. [1] [Total: 13]

10 9701/43/M/J/16 © UCLES 2016 7 (a) (i) Use the Data Booklet to draw the structure of the tripeptide Ala-Ser-Gly showing its peptide bonds in full. Ala-Ser-Gly [2] (ii) Calculate the relative molecular mass, Mr, of Ala-Ser-Gly. Mr = … [1] (b) Electrophoresis can be used to separate mixtures of amino acids and peptides. A mixture of the tripeptide Ala-Ser-Gly and its three constituent amino acids was subjected to electrophoresis in a buffer at pH 11. (i) Draw the structure of serine at pH 11. [1]

11 9701/43/M/J/16 © UCLES 2016 [Turn over At the end of the experiment the following results were seen. mixture applied here P G H Q R S I F (ii) Identify the components labelled F-I in the above diagram. F … G … H … I … [4] (iii) Suggest the identities of the species responsible for spot P, … spot S. … Explain your answers. … … [3] (c) (i) State the reagents and conditions needed for converting the tripeptide into its three constituent amino acids. … [1] (ii) Name the type of reaction in (i). … [1] [Total: 13]

12 9701/43/M/J/16 © UCLES 2016 8 (a) Chromyl chloride, CrO2Cl 2, can be prepared by heating a mixture of potassium dichromate(VI) and potassium chloride with concentrated sulfuric acid. K2Cr2O7 + 4KCl + 3H2SO4 → 2CrO2Cl 2 + 3K2SO4 + 3H2O Use the following data to complete the Hess’ Law cycle and calculate the enthalpy change of the reaction, ∆Hr. compound enthalpy change of formation, / kJ mol–1 K2Cr2O7 –2061 KCl –437 H2SO4 –814 CrO2Cl 2 –580 K2SO4 –1438 H2O –286 K2Cr2O7 + 4KCl + 3H2SO4 ∆Hr 2CrO2Cl 2 + 3K2SO4 + 3H2O elements ∆Hr = … kJ mol–1 [2]

13 9701/43/M/J/16 © UCLES 2016 [Turn over (b) There are two isomeric complex ions with the formula [Cr(NH3)4Cl 2]+. One is green and the other is violet. (i) Suggest the type of isomerism shown by these complex ions. … [1] (ii) Explain why these two complex ions • are coloured, … … … … • have different colours. … … … … [4] [Total: 7]

14 9701/43/M/J/16 © UCLES 2016 9 The anti-inflammatory drug ketoprofen can be synthesised from benzene via the following five steps. O O Br step 1 step 5 step 2 step 3 T, (C15H14O) U step 4 O CO2H ketoprofen (a) Suggest the structures of compounds T and U and draw them in the boxes above. [2] (b) Suggest reagents and conditions for steps 1-5. step 1 … step 2 … step 3 … step 4 … step 5 … [5] (c) What types of reaction are steps 1 and 5? step 1 … step 5 … [2] [Total: 9]

15 9701/43/M/J/16 © UCLES 2016 [Turn over 10 (a) Ethanedioic acid, C2O4H2, occurs in many vegetables. The amount that occurs in spinach can be estimated as follows. • 40.0 g of spinach leaves are crushed and mixed with distilled water, using a mortar and pestle. • The mixture is filtered, and the leaves are washed with a little more water. • The combined filtrate and washings are made up to 100.0 cm3 with water. • A 25.0 cm3 portion of the resulting solution is added to a conical flask, along with an excess of dilute sulfuric acid. • The acidified solution is warmed, and then titrated with 0.0200 mol dm–3 KMnO4. The equation for the reaction between ethanedioic acid and acidified manganate(VII) ions is shown. 2MnO4 – + 6H+ + 5C2O4H2 → 2Mn2+ + 10CO2 + 8H2O In the titration, 15.20 cm3 of KMnO4 was required to reach the end-point. Calculate the percentage by mass of ethanedioic acid in the spinach leaves. percentage of ethanedioic acid = … % [3]

16 9701/43/M/J/16 © UCLES 2016 (b) Ethanedioic acid can be converted into ethanedioyl chloride: HO2CCO2H → Cl OCCOCl (i) State a suitable reagent for this reaction. … [1] (ii) For the reactions of ethanedioyl chloride below, suggest the structures of compounds J and K and draw them in the boxes. CH3OH H2NCH2CH2NH2 K (C4H6N2O2) J (C4H6O4) COCl COCl [2] (c) When ethanedioyl chloride is reacted with silver ethanedioate, AgO2CCO2Ag, in ethoxyethane at –30 °C, an oxide of carbon, L, is formed. The molecule of L has no overall dipole and has molecular formula C4O6. The carbon-13 NMR spectrum of a solution of L in ethoxyethane, CH3CH2OCH2CH3, is shown below. 150 100 50 0 δ (ppm) (i) Use the Data Booklet to state in the boxes below the δ values for the peaks in the spectrum which are due to the carbon atoms in ethoxyethane. CH3 CH3 CH2 CH2 O δ values [2] (ii) Explain what the rest of the carbon-13 NMR spectrum indicates about the structure of L. … [1]

17 9701/43/M/J/16 © UCLES 2016 [Turn over (d) When pure L is reacted with an excess of CH3OH, a mixture of three compounds is formed. L and CH3OH → M and N and O (C4O6) (C2H2O4) (C3H4O4) (C4H6O4) M is formed as one of the products when either N or O is heated with aqueous acid. The table gives information of the peaks recorded in the carbon-13 NMR spectra of M, N and O. compound peaks recorded in carbon-13 NMR spectrum M δ 162 N δ 53 δ 160 δ 162 O δ 53 δ 160 (i) Suggest the structures of M, N and O. M, (C2H2O4) N, (C3H4O4) O, (C4H6O4) [3] (ii) Suggest a structure for L that fits all the data given in (c) and (d). L, (C4O6) [1] [Total: 13]

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® IGCSE is the registered trademark of Cambridge International Examinations. This document consists of 8 printed pages. © UCLES 2016 [Turn over Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level CHEMISTRY 9701/43 Paper 4 A Level Structured Questions May/June 2016 MARK SCHEME Maximum Mark: 100 Published 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 2016 series for most Cambridge IGCSE®, Cambridge International A and AS Level components and some Cambridge O Level components.

Page 2 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2016 9701 43 © Cambridge International Examinations 2016 Question Answer Marks 1 (a) (i) Ca(OH)2 + CO2 → CaCO3 + H2O [1] (ii) Ba(OH)2 is soluble, OR BaCO3 is insoluble [1] (iii) Mg(OH)2 is insoluble / not very soluble will not form ppt. of MgCO3 [1] [1] (b) carbonates are more stable down the group due to increase in cationic size / radius (causing) less polarisation of CO3 2– ion [1] [1] [1] (c) radius of Ni2+ = 0.070 nm; radius of Ca2+ = 0.099 nm so NiCO3 decomposes more readily than CaCO3 [1] [1] [Total: 9] 2 (a) (i) Co: ...3s23p63d74s2 Co2+: ...3s23p63d7 [1] (ii) solution starts pink turns blue pink is [Co(H2O)6]2+ blue is [CoCl4]2– this complex is tetrahedral [1] [1] [1] [1] [1]

Page 3 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2016 9701 43 © Cambridge International Examinations 2016 Question Answer Marks (b) R3P Ni R3P I I I Ni R3P I PR3 R3P I Co PR3 I [1] [1] [1] [Total: 9] 3 (a) Kp = {p(CS2) × (p(H2))4} / {(p(H2S))2 × p(CH4)} units: atm2 OR Pa2 [1] [1] (b) (i) p(H2S) = 196 atm p(H2) = 8 atm [1] [1] (ii) Kp = (2 × 84) / (1962 × 98) = 2.176 × 10–3 [1] (c) (i) ∆So will be positive, because more gas moles on the RHS / products [1] (ii) ∆So = (∆Ho – ∆Go) / T = (241 – 51)/1000 = 0.19 OR 190 kJ mol–1 K–1 OR J mol–1 K–1 [1] [1] (d) ∆Go will become less positive/more negative as T increases, …because ∆So is positive (or –T ∆So is more negative) …therefore the reaction becomes more feasible / spontaneous as T increases [2] [Total: 10]

Page 4 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2016 9701 43 © Cambridge International Examinations 2016 Question Answer Marks 4 (a) (i) SCP is the EMF / potential of a cell composed of two electrodes (OR half cells) under standard conditions (OR at 289 K OR 1 mol dm–3) [1] (ii) voltmeter and salt bridge [1] (iii) A is Ag B is Ag+(aq) or AgNO3(aq) C is Pt D is Fe2+(aq) and Fe3+(aq) (combination of A and B can be reversed with combination of C and D) [3] (b) (i) Ag+ + Fe2+ → Ag + Fe3+ [1] (ii) E = Eo + 0.059log [Ag+] = 0.80 – 0.03 = 0.77 V so Ecell = 0.77 – 0.77 = 0.0 V [1] [1] [Total: 8] 5 (a) (i) pKa = –log Ka [1] (ii) diacids are more acidic than CH3CO2H HO2C– group is electron-withdrawing, stabilising the monoanion OR HO2C– group is electron-withdrawing, weakening the O–H bond OR monoanion is stabilised by H–bonding as n increases, the electron–withdrawing group is further away from the ionising CO2H group OR the (intervening) alkyl groups destabilise the anion [1] [1] [1] (iii) removing H+ from an anion is not electrostatically favourable [1] (b) (i) a solution which resists changes in pH when small amounts of H+ or OH– are added [1] [1]

Page 5 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2016 9701 43 © Cambridge International Examinations 2016 Question Answer Marks (ii) HO2CCH2CH2CO2Na + H+ → HO2CCH2CH2CO2H + Na+ HO2CCH2CH2CO2Na + NaOH → NaO2CCH2CH2CO2Na + H2O [1] [1] [Total: 9] 6 (a) (i) C6H5NO2 + 6e– + 6H+ → C6H5NH2 + 2H2O [1] (ii) 2C6H5NO2 + 14HCl + 3Sn → 2C6H5NH3Cl + 3SnCl4 + 4H2O [2] (b) (Mr values: C6H5NO2 = 123 C6H5NH3Cl = 129.5) theoretical yield = 5.0 × 129.5/123 = 5.26 g percentage yield = 100 × 4.2/5.26 = 79.8% (80%) [1] [1] (c) (i) C6H5NH2 = 93 yield of phenylamine = 4.2 × 93/129.5 = 3.016 g [1] (ii) mass left in water = 3.016 – 2.68 = 0.336 g Kpart = (2.68/50) / (0.336/25) = 3.99 [1] [1] (d) phenylamine is less basic that ethylamine the lone pair on N is delocalised over the ring… …making it less available for reaction with a proton / δ+ H [2] (e) (i) step 1: HNO2 OR (NaNO2 + HCl ) at T ⩽ 10 °C step 2: boil / heat in water [1] [1] (ii) N N (Cl-) E is [1] [Total: 13]

Page 6 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2016 9701 43 © Cambridge International Examinations 2016 Question Answer Marks 7 (a) (i) NH2 CH C H N CH C N H CH2 CO2H CH3 O CH2OH O [2] (ii) Mr = 233 [1] (b) (i) NH2CH(CH2OH)CO2 – [1] (ii) F is a DC power supply G is the anode OR positive electrode I is the cathode OR negative electrode H is filter paper (OR gel) soaked in buffer solution [4] (iii) P is NH2CH2CO2 – or NH2CH2CO2 H or glycine S is [ala–ser–gly](–) glycine is the smallest, so travels fastest; tripeptide is the largest, so travels slowest [1] [1] [1] (c) (i) heat with H3O+ OR heat with OH–(aq) [1] (ii) hydrolysis [1] [Total: 13] 8 (a) ∆H = [2(–580) + 3(–286) + 3(–1438)] – [–2061 + 4(–437) + 3(–814)] = –81 kJ mol–1 [2] (b) (i) cis–trans OR geometrical [1]

Page 7 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2016 9701 43 © Cambridge International Examinations 2016 Question Answer Marks (ii) in a complex the d–orbitals are split into 2 energy levels colour is due to absorption of light (in visible region) electron promotion to higher orbital absorbs a photon the d–d energy gap is different for the two complexes, hence different colours [1] [1] [1] [1] [Total: 7] 9 (a) O O CN T is U is [1] [1] (b) step 1: C6H5COCl + AlCl3 (+ heat) step 2: CH3CH2Cl + AlCl3 (+ heat) step 3: Br2 + light (or heat) step 4: KCN + heat (in ethanol) step 5: H3O+ OR H+ in H2O OR HCl (aq) etc AND heat / boil / reflux [1] [1] [1] [1] [1] (c) step 1: electrophilic substitution OR nucleophilic substitution step 5: hydrolysis OR nucleophilic substitution [1] [1] [Total: 9] 10 (a) n(MnO4 –) = 0.02 × 15.2 / 1000 = 3.04 × 10–4 mol n(C2O4H2) = 3.04 × 10–4 × 5 / 2 = 7.6 × 10–4 (in 25 cm3) = 3.04 × 10–3 mol in 100 cm3 Mr = 24 + 64 + 2 = 90 mass of C2O4H2 = 3.04 × 10–3 × 90 = 0.2736 g (0.274) percentage = 0.2736 × 100/40 = 0.68% [1] [1] [1] (b) (i) SOCl2 or PCl5 or PCl3 [1]

Page 8 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2016 9701 43 © Cambridge International Examinations 2016 Question Answer Marks (ii) J is CH3OCO–COOCH3 K is HN NH O O [1] [1] (c) (i) CH3 at δ 15 CH2O at δ 65 [1] [1] (ii) Only one peak, so only one type / environment of C atom [1] (d) (i) M is HO2C–CO2H N is CH3OCO–CO2H O is CH3OCO–COOCH3 [3] (ii) C C O C C O O O O O L is [1] [Total: 13]