Cambridge A Level Chemistry 9701 — 2018 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 2018  2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level This document consists of 18 printed pages and 2 blank pages. [Turn over IB18 06_9701_43/FP © UCLES 2018 *2624564396*

2 9701/43/M/J/18 © UCLES 2018 Answer all the questions in the spaces provided. 1 Sodium oxide, Na2O, is a white crystalline solid with a high melting point. (a) Write an equation for the reaction of sodium with oxygen, forming sodium oxide. Include state symbols. … [2] (b) Explain why sodium oxide has a high melting point. … … … … [2] (c) When sodium oxide reacts with water an alkaline solution is obtained. (i) Explain why the solution obtained is alkaline. You should use the Brønsted-Lowry theory of acids and bases in your answer. … … … … [2] (ii) Calculate the pH of the solution obtained when 3.10 g of sodium oxide are added to 400 cm3 of water. pH = … [3]

3 9701/43/M/J/18 © UCLES 2018 [Turn over (d) Use the data below, and other suitable data from the Data Booklet, to calculate the lattice energy of sodium oxide, Na2O(s). energy change value / kJ mol–1 standard enthalpy change of formation of sodium oxide, Na2O(s) – 416 standard enthalpy change of atomisation of sodium, Na(s) +109 electron affinity of O(g) –142 electron affinity of O–(g) +844 Na2O(s) = … kJ mol–1 [4] (e) State how Na2S(s) differs from Na2O(s). Indicate this by placing a tick () in the appropriate box in the table. Na2S(s) is more exothermic than Na2O(s) Na2S(s) is the same as Na2O(s) Na2S(s) is less exothermic than Na2O(s) Explain your answer. … … …  [2] [Total: 15]

4 9701/43/M/J/18 © UCLES 2018 2 Nitrogen monoxide, NO(g), reacts with hydrogen, H2(g), under certain conditions. 2NO(g) + 2H2(g) N2(g) + 2H2O(g) (a) Define the term rate of reaction. … … [1] (b) Identify a change in the reaction mixture that would enable the rate of this reaction to be studied. … [1] The rate equation for this reaction is given. rate = k [NO]2[H2] The result of an experiment in which NO reacted with H2 is shown in the table. initial [NO] / mol dm–3 initial [H2] / mol dm–3 initial rate of reaction / mol dm–3 s–1 2.50 × 10–3 2.50 × 10–3 1.27 × 10–3 (c) Use the data and the rate equation to calculate a value for the rate constant k. Give the units of k. k = … units = … [2] (d) A second experiment is performed at the same temperature. The initial concentration of H2(g) is 4.60 × 10–3 mol dm–3. The initial rate of the reaction is 2.31 × 10–3 mol dm–3 s–1. Calculate the initial concentration of NO(g). initial concentration of NO(g) = … mol dm–3 [1]

5 9701/43/M/J/18 © UCLES 2018 [Turn over (e) State the order of the reaction with respect to NO(g) and with respect to H2(g), and the overall order of the reaction. [NO] [H2] overall order [1] (f) The reaction is believed to proceed in three steps. 1 2NO N2O2 2 N2O2 + H2 N2O + H2O 3 N2O + H2 N2 + H2O (i) Deduce which of the three steps is the rate-determining step. … [1] (ii) Explain your answer to (i). … … … [1]

6 9701/43/M/J/18 © UCLES 2018 (g) A third experiment is performed under different conditions. A small amount of H2(g) of concentration 0.0200 mol dm–3 is mixed with a large excess of NO(g). The concentration of H2(g) is found to have a constant half-life of 2.00 seconds under the conditions used. (i) Define the term half-life. … … [1] (ii) Use the axes below to construct a graph of the variation in the concentration of H2(g) during the first 6 seconds under the conditions used. 0.02 0.01 0 0 1 2 3 time / s 4 5 6 [H2] / mol dm–3 [2] (h) NO(g) acts as a catalyst in the oxidation of atmospheric sulfur dioxide. (i) Give two equations to describe how NO(g) acts as a catalyst in this process. equation 1 … equation 2 … [1] (ii) Explain why NO(g) can be described as a catalyst in this reaction. … … [1] (iii) Describe, with the aid of an equation, an environmental consequence of the oxidation of atmospheric sulfur dioxide. … … [1] [Total: 14]

7 9701/43/M/J/18 © UCLES 2018 [Turn over 3 (a) Complete the table, identifying the substance liberated at each electrode during electrolysis with inert electrodes. electrolyte substance liberated at the anode substance liberated at the cathode AgNO3(aq) concentrated NaCl (aq) CuSO4(aq) [3] (b) Molten calcium iodide, CaI2, is electrolysed in an inert atmosphere with inert electrodes. (i) Write ionic equations for the reactions occurring at the electrodes. ● … ● … [2] (ii) The electrolysis of molten CaI2 is a redox process. Identify the ion that is oxidised and the ion that is reduced, explaining your answer by reference to oxidation numbers. … … … [2] (iii) Describe two visual observations that would be made during this electrolysis. 1 … 2 … [1] (c) An oxide of iron dissolved in an inert solvent is electrolysed for 2.00 hours using a current of 0.800 A. The electrolysis products are iron and oxygen. The mass of iron produced is 1.11 g. Calculate the oxidation number of Fe in the oxide of iron. Show all your working.  oxidation number of Fe = … [3] [Total: 11]

8 9701/43/M/J/18 © UCLES 2018 4 (a) Describe what you would see when calcium and barium are heated separately with oxygen. calcium … barium … [2] (b) The decomposition temperatures of the Group 2 carbonates vary down the group. State and explain the variation in the decomposition temperatures. … … … … … … [3] (c) Magnesium carbonate was heated in an open test‑tube. It was difficult to see whether a thermal decomposition reaction took place. Explain why. … … … [2] [Total: 7]

9 9701/43/M/J/18 © UCLES 2018 [Turn over 5 Copper is a transition element with atomic number 29. (a) Complete the electronic configurations of a Cu atom and a Cu+ ion. Cu atom 1s22s22p6 … Cu+ ion 1s22s22p6 … [2] (b) Cu+ ions form a linear complex with Cl – ions, which are monodentate ligands. Draw the structure of this complex and include its overall charge. [2] (c) Cu2+ ions exist as [Cu(H2O)6]2+ complex ions in aqueous solution. Complete a three-dimensional diagram to show the shape of this complex. Name its shape. Label and state the value of one bond angle. Cu name of shape …  [2]

10 9701/43/M/J/18 © UCLES 2018 (d) When NH3(aq) is added to Cu2+(aq), dropwise at first and then in excess, two chemical reactions occur as shown. [Cu(H2O)6]2+ A dropwise NH3(aq) reaction 1 B excess NH3(aq) reaction 2 For each reaction, describe what you would see and write an equation. reaction 1 observation … … equation … reaction 2 observation … … equation … [4] (e) EDTA4– is a polydentate ligand. When a solution of EDTA4– is added to a solution containing [Cu(H2O)6]2+ a new complex is formed. The formula of this complex is [CuEDTA]2–. (i) Name the type of reaction occurring here. … [1] (ii) Write an expression for the stability constant, Kstab, of [CuEDTA]2– in this reaction. [1] (iii) The numerical value of the Kstab of [CuEDTA]2– is 6.3 × 1019 at 298 K. State what this tells us about the [CuEDTA]2– complex ion. … [1]

11 9701/43/M/J/18 © UCLES 2018 [Turn over (f) Ethanedioate ions, C2O4 2–, can act as a bidentate ligand. (i) Explain what is meant by the term bidentate ligand. … … … … [2] (ii) When ethanedioate ions are added to a solution of zirconium ions, Zr 4+, a complex ion containing four C2O4 2– ions and one Zr 4+ ion is formed. All four ethanedioate ions act as bidentate ligands in this complex. Give the formula of this complex ion and explain why this complex is not octahedral. … … … … [2] [Total: 17]

12 9701/43/M/J/18 © UCLES 2018 6 (a) Benzene reacts with D in the presence of a suitable catalyst to give cumene and non-organic product E. This is an electrophilic substitution reaction. + D + E catalyst cumene (i) Name the reactant D and the non-organic product E. D … E … [2] (ii) Give the name of the type of aromatic electrophilic substitution reaction taking place. … [1] (b) Cumene undergoes substitution reactions with chlorine to give several different isomeric products with the formula C9H11Cl. The substitution can occur in the aromatic ring or in the side‑chain of cumene. (i) Describe the conditions that are used to ensure substitution takes place only in the aromatic ring. … [1] (ii) Draw the structures of the two major isomeric products of the reaction, formula C9H11Cl, when substitution takes place in the aromatic ring. [1]

13 9701/43/M/J/18 © UCLES 2018 [Turn over (iii) Describe the conditions that are used to ensure substitution takes place only in the side‑chain. … [1] (iv) Draw the structures of two isomeric products of the reaction, formula C9H11Cl, when substitution takes place in the side‑chain. [1] (c) Complete the following table to show the structures of the organic products formed when cumene reacts with each reagent. reagent structure of organic product hot KMnO4(aq) H2 + Ni, high pressure [2]

14 9701/43/M/J/18 © UCLES 2018 (d) Cumene can be nitrated using a mixture of concentrated nitric and sulfuric acids. The mechanism for this reaction is similar to the mechanism for the nitration of benzene. Complete the mechanism for this reaction. ● Include all relevant charges and curly arrows showing the movement of electron pairs. ● Draw the structure of the intermediate. ● You do not need to draw the products. cumene intermediate products [4] [Total: 13]

15 9701/43/M/J/18 © UCLES 2018 [Turn over 7 The three substances shown all have some acidic properties. O phenol propan-1-ol C H H H H O C H H C propanoic acid C H H H C C H H H O O H H H (a) Write an equation for the reaction between propan-1-ol and sodium metal. … [1] (b) (i) Give the order of the relative acidities of propanoic acid, propan‑1‑ol and phenol, stating the most acidic first. … [1] (ii) Explain your answer to (i). … … … … … … [2] (c) Methanoic acid, HCO2H, has a similar acid strength to propanoic acid. Describe a chemical test to distinguish between these two acids. Name the acid which gives a positive result in this test and describe the observations that would be made. … … … … [2]

16 9701/43/M/J/18 © UCLES 2018 (d) The ester phenyl propanoate, C2H5CO2C6H5, can be made from phenol and propanoic acid in a two-step synthesis. The first step produces an acyl chloride. For this two-step synthesis, ● draw the structure of the product of the first step, ● state the reagents and conditions needed for each step of the synthesis. … … … … [3] (e) An unknown compound, Z, is propan‑1‑ol, propanal or propanoic acid. The proton NMR spectrum of Z dissolved in CDCl 3 is shown. 12 10 8 6 4 2 0 δ / ppm –2 (i) From the proton NMR spectrum, identify Z. … [1] (ii) State one feature that would be seen, and why, in the proton NMR spectra of each of the two compounds that are not Z. … … … … [2] [Total: 12]

17 9701/43/M/J/18 © UCLES 2018 [Turn over 8 Abscisic acid, C15H20O4, is a plant hormone. abscisic acid, C15H20O4 OH O O OH (a) On the diagram of abscisic acid, use an asterisk (*) to label each chiral carbon atom. [1] (b) Abscisic acid is reacted with an excess of NaBH4. Give the molecular formula of the organic product formed. … [1] (c) If abscisic acid is treated with an excess of hot, concentrated, acidified KMnO4, three different carbon‑containing products are formed. (i) Draw the skeletal formula of the carbon-containing product with the largest molecular mass. [1] (ii) Identify the carbon‑containing product with the smallest molecular mass. Explain how this product arises. … … … [2] (iii) Identify the third carbon-containing product of this reaction by giving its displayed or structural formula.  [1]  [Total: 6]

18 9701/43/M/J/18 © UCLES 2018 9 Noradrenaline is a hormone found in humans. HO NH2 OH HO noradrenaline (a) Give the molecular formula of noradrenaline. … [1] (b) State whether or not noradrenaline shows stereoisomerism. Explain your answer. … … [1] (c) HNO2(aq) is reacted at 5 °C with separate samples of noradrenaline and phenylamine. The reaction with phenylamine produces a stable diazonium ion. The reaction with noradrenaline produces an unstable diazonium ion. (i) Suggest why the diazonium ion produced with phenylamine is stable. … … [1] (ii) When one noradrenaline molecule reacts with one HNO2 molecule, the products are one water molecule, one molecule of an unreactive gas, and one molecule of an organic compound made up of carbon, hydrogen and oxygen only. Complete the chemical equation for this reaction. HO NH2 + HNO2 + … + H2O OH HO  [2]  [Total: 5]

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IGCSE™ is a registered trademark. This document consists of 13 printed pages. © UCLES 2018 [Turn over Cambridge Assessment International Education Cambridge International Advanced Subsidiary and Advanced Level CHEMISTRY 9701/43 Paper 4 A Level Structured Questions May/June 2018 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 International will not enter into discussions about these mark schemes. Cambridge International is publishing the mark schemes for the May/June 2018 series for most Cambridge IGCSE™, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

9701/43 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 2 of 13 Generic Marking Principles These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside the specific content of the mark scheme or generic level descriptors for a question. Each question paper and mark scheme will also comply with these marking principles. GENERIC MARKING PRINCIPLE 1: Marks must be awarded in line with: • the specific content of the mark scheme or the generic level descriptors for the question • the specific skills defined in the mark scheme or in the generic level descriptors for the question • the standard of response required by a candidate as exemplified by the standardisation scripts. GENERIC MARKING PRINCIPLE 2: Marks awarded are always whole marks (not half marks, or other fractions). GENERIC MARKING PRINCIPLE 3: Marks must be awarded positively: • marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond the scope of the syllabus and mark scheme, referring to your Team Leader as appropriate • marks are awarded when candidates clearly demonstrate what they know and can do • marks are not deducted for errors • marks are not deducted for omissions • answers should only be judged on the quality of spelling, punctuation and grammar when these features are specifically assessed by the question as indicated by the mark scheme. The meaning, however, should be unambiguous. GENERIC MARKING PRINCIPLE 4: Rules must be applied consistently e.g. in situations where candidates have not followed instructions or in the application of generic level descriptors.

9701/43 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 3 of 13 GENERIC MARKING PRINCIPLE 5: Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may be limited according to the quality of the candidate responses seen). GENERIC MARKING PRINCIPLE 6: Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or grade descriptors in mind.

9701/43 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 4 of 13 Question Answer Marks 1(a) 4Na(s) + O2(g) → 2Na2O(s) balanced with all formulae correct 1 state symbols 1 1(b) giant ionic 1 strong bond / attraction between AND positive and negative ions / anions and cations / Na+ and O2– / oppositely charged ions 1 1(c)(i) the reaction produces sodium hydroxide / hydroxide ions / OH– ions 1 the hydroxide ions can receive / accept H+ ions / protons 1 1(c)(ii) Calculation of Na2O moles 3.10 g / 62 OR 0.05 1 Calculation of [OH-] 0.05 × (2 / 0.400) = 0.25 mol dm–3 1 Calculation of pH –log 0.25 = 0.60 14 – 0.60 = 13.40 1 1(d) use of (2 × 109) or 218 and (2 × 494) or 988 1 use of (0.5 × 496) or 248 1 use of 416, 142, 844 1 evaluation of expression correctly ∆Hlat = –416 – (2 × 109) – (0.5 × 496) – (2 × 494) – (–142 + 844) = –2572 1 1(e) the lattice energy of Na2S is less exothermic 1 the sulfide ion is larger than the oxide ion / S2– larger than O2 / ionic radii quoted 0.184 nm and 0.140 nm AND less attraction (between the ions)/bonds are weaker 1

9701/43 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 5 of 13 Question Answer Marks 2(a) change in amount / mass / concentration of reactant / product per time 1 2(b) decrease in volume or pressure 1 2(c) 8.13 × 104 / 81280 / 81300 1 mol–2 dm6 s–1 1 2(d) √(0.00231 / (0.0046 × 81280)) = 2.49 × 10–3 1 2(e) 2, 1, 3 1 2(f)(i) 2 1 2(f)(ii) the total of steps 1 and 2 / the components of 2 are two NO and one H2 1 2(g)(i) time for amount or mass or concentration to halve 1 2(g)(ii) 0.02 at start and 0.01 after 2 seconds 1 0.005 after 4 seconds and 0.0025 after 6 seconds 1 2(h)(i) NO + ½ O2 → NO2 or NO + O2 → NO2 + ½ O2 AND NO2 + SO2 → NO + SO3 1 2(h)(ii) (NO is) regenerated / reformed 1 2(h)(iii) SO3 + H2O → H2SO4 AND acid rain or consequence of this described 1

9701/43 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 6 of 13 Question Answer Marks 3(a) anode cathode AgNO3 (aq) oxygen / O2 silver / Ag saturated NaCl (aq) chlorine / Cl2 hydrogen / H2 CuSO4 (aq) oxygen / O2 copper / Cu 3 3(b)(i) 2I– → I2 + 2e– 1 Ca2+ + 2e– → Ca 1 3(b)(ii) • Ca / Calcium reduced and I / iodine oxidised • Oxidation number of calcium decreases from 2 to 0 • Oxidation number of iodine increases from –1 to 0 2 points = 1 mark 3 points = 2 marks 2 3(b)(iii) • metal / grey / silvery • purple AND vapour / gas / fumes • amount of melt decreases any 2 points for 1 mark 1 3(c) 2 × 60 × 60 × 0.8 = 5760 C AND 5760 / 96500 = 0.060 (0.0597) F 1 1.11 / 55.8 = 0.020 (0.0199) mol of Fe 1 0.06 / 0.02 = 3 ∴ Fe3+ or +3 or 3 1

9701/43 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 7 of 13 Question Answer Marks 4(a) calcium – red flame 1 barium – green flame 1 4(b) • the temperature increases down the group • ionic radius increases / charge density decreases down the group • decreasing distortion / polarisation or decreasing weakening of bonds • of the anion / the CO3 2– ion 2 points = 1 mark 3 points = 2 marks 4 points = 3 marks 3 4(c) the gas is colourless / looks like air / cannot be seen 1 appearance of solid doesn’t change / white solid becomes white solid 1 Question Answer Marks 5(a) 3s23p63d104s1 1 3s23p63d10 1 5(b) Cl-Cu-Cl 1 one minus charge 1 5(c) 1 octahedral and 90o or 180o labelled correctly on diagram as appropriate 1

9701/43 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 8 of 13 Question Answer Marks 5(d) reaction 1: blue ppt / blue solid 1 [Cu(H2O)6]2+ + 2OH– → Cu(OH)2 + 6H2O or [Cu(H2O)6]2+ + 2OH– → Cu(OH)2(H2O)4 + 2H2O or [Cu(H2O)6]2+ + 2NH3 → Cu(OH)2(H2O)4 + 2NH4 + 1 reaction 2: deep / dark / royal blue solution 1 Cu(OH)2 + 4NH3 + 2H2O → [Cu(NH3)4(H2O)2]2+ + 2OH– or Cu(OH)2 + 4NH3 → [Cu(NH3)4]2+ + 2OH– or Cu(OH)2(H2O)4 + 4NH3 → [Cu(NH3)4(H2O)2]2+ + 2OH– + 2H2O 1 5(e)(i) ligand exchange / displacement / replacement / substitution 1 5(e)(ii) Kstab =[ [CuEDTA]2– ] / [[Cu(H2O)6]2+ ][[EDTA]4–] 1 5(e)(iii) stable / more stable than [Cu(H2O)6]2+ 1 5(f)(i) donates lone pairs / forms dative / co-ordinate bonds to (central) metal atom / metal ion 1 donates two lone pairs / forms two (dative or coordinate) bonds 1 5(f)(ii) [Zr(C2O4)4]4– 1 not octahedral because 8 dative bonds to Zr or not octahedral because not 6 dative bonds to Zr or not octahedral because co-ordination number is 8 / is not 6 1

9701/43 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 9 of 13 Question Answer Marks 6(a)(i) D 2-chloropropane 1 E hydrogen chloride 1 6(a)(ii) (Friedel-Crafts) alkylation 1 6(b)(i) AlCl3 or FeCl3 1 6(b)(ii) 1 6(b)(iii) sunlight or UV OR T>100 °C 1 6(b)(iv) 1

970 © U Que 6 6 01/43 CLES 2018 estion 6(c) reactio reactio 6(d) attacki curly a correc 2nd cu on with hot KMnO on with H2 + Ni, h ing species is N arrow starting wi t intermediate urly arrow from C NO2 + Ca O4(aq) high pressure O2 + thin hexagon an C-H bond into rin H ambridge Interna nd going to NO2 + ng + NO2 -H+ ational AS/A Lev PUBLISHED Page 10 of 13 Answer + Products vel – Mark Scheme May/June 2 Ma 018 arks 1 1 1 1 1 1

9701/43 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 11 of 13 Question Answer Marks 7(a) 2C3H7OH + 2Na → 2C3H7ONa + H2 1 7(b)(i) propanoic acid, phenol, propan-1-ol 1 7(b)(ii) • propan-1-ol: O-H bond strengthened by positive inductive effect of alkyl group OR propoxide ion is destabilised by positive inductive effect of alkyl group • phenol: O-H bond weakened by negative inductive effect of ring OR phenoxide ion is stabilised by delocalisation of oxygen lone pair into ring • propanoic acid: O-H bond weakened by negative inductive effect of C=O OR propanoate ion is stabilised by delocalisation of minus charge by C=O 1 mark for a correct explanation, max 2 marks 2 7(c) Tollens’ reagent or Fehling’s reagent 1 methanoic acid gives a silver mirror/solid with Tollen’s reagent OR red / orange ppt / solid with Fehlings’ reagent 1 7(d) PCl5 or PCl3 (+heat) or SOCl2 (added to propanoic acid) 1 product of first step: 1 add product of first step to phenol in NaOH 1 7(e)(i) propanoic acid 1 7(e)(ii) propan-1-ol would have peak at 0.5–6.0 because of OH group 1 propanal would have peak at 9.3–10.5 because of CHO / aldehyde 1

970 © U Que 8 8 8 8( 8( 01/43 CLES 2018 estion 8(a) 8(b) C15H22 (c)(i) (c)(ii) CO2 oxidati (c)(iii) CH3CO O 2O4 ion / oxidative cle OCO2H correct ch O * Ca eavage hiral centre labe OH O ambridge Interna elled only OH C15H22O4 ational AS/A Lev PUBLISHED Page 12 of 13 Answer vel – Mark Scheme May/June 2 Ma 018 arks 1 1 1 1 1 1

970 © U Que 9 9 9 9( 01/43 CLES 2018 estion 9(a) C8H11O 9(b) yes, as (c)(i) (pheny (c)(ii) N2 O3N s it has a chiral C yldiazonium ion OH Ca C atom is stabilised bec ambridge Interna cause) positive c ational AS/A Lev PUBLISHED Page 13 of 13 Answer charge is deloca vel – Mark Sche alised by ring / po me ositive charge is spread over rin May/June 2 Ma ng 018 arks 1 1 1 1 1