Cambridge A Level Chemistry 9701 — 2018 Feb/March Paper 4 · Variant 2

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 9701/42 Paper 4 A Level Structured Questions February/March 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 19 printed pages and 1 blank page. [Turn over IB18 03_9701_42/4RP © UCLES 2018 *8205941016*

2 9701/42/F/M/18 © UCLES 2018 Answer all the questions in the spaces provided. 1 (a) (i) State how the solubilities of the hydroxides of the Group 2 elements vary down the group. … [1] (ii) Explain the factors that are responsible for this variation. … … … … … [3] (b) The solubility of Sr(OH)2 is 3.37 × 10–2 mol dm–3 at 0 °C. (i) Write an expression for the solubility product of Sr(OH)2. Ksp =  [1] (ii) Calculate the value of Ksp at 0 °C. Include units in your answer.  Ksp = … units = …  [2]

3 9701/42/F/M/18 © UCLES 2018 [Turn over (c) Metal peroxides contain the –O–O– ion. The peroxides of the Group 2 elements, MO2, decompose on heating to produce a single gas and the solid oxide, MO, only. (i) Write an equation for the thermal decomposition of strontium peroxide, SrO2. … [1] (ii) Suggest how the temperature at which thermal decomposition of MO2 occurs varies down Group 2. Explain your answer. … … … … … [3] (d) (i) The ethanedioates of the Group 2 elements, MC2O4, decompose on heating to produce a mixture of two different gases and the solid oxide, MO, only. Complete the equation for the thermal decomposition of barium ethanedioate. BaC2O4 … + … + …  [1] (ii) Describe two observations you would make during the reaction when ethanedioic acid, H2C2O4, is warmed with acidified manganate(VII) ions. … … … … [2]  [Total: 14]

4 9701/42/F/M/18 © UCLES 2018 2 (a) Describe the trend in the reactivity of the halogens Cl 2, Br2 and I2 as oxidising agents. Explain this trend using values of E o (X2 / X–) from the Data Booklet. … … … … … [2] (b) (i) Write an equation for the reaction between chlorine and water. … [1] (ii) Use standard electrode potential, E o, data from the Data Booklet to calculate the for the following reaction. Cl 2 + 2OH– Cl – + Cl O– + H2O  = … V [2] (iii) The [OH–] was increased and the Ecell was measured. Indicate how the value of the Ecell measured would compare to the calculated in (ii) by placing one tick () in the table. Ecell becomes less positive than . Ecell stays the same as . Ecell becomes more positive than . Explain your answer. … … …  [2]

5 9701/42/F/M/18 © UCLES 2018 [Turn over (c) A half-equation involving bromate(V) ions, BrO3 –, and bromide ions is shown. BrO3 –(aq) + 3H2O(l) + 6e– Br –(aq) + 6OH–(aq) E o = +0.58 V (i) An alkaline solution of chlorate(I), Cl O–, can be used to oxidise bromide ions to bromate(V) ions. Use the Data Booklet and the half-equation shown to write an equation for this reaction. … [1] (ii) Calculate the for the reaction in (i).  = … V [1] (iii) When a concentrated solution of bromic(V) acid, HBrO3, is warmed, it decomposes to form bromine, oxygen and water only. Write an equation for this reaction. The use of oxidation numbers may be helpful. … [1]  [Total: 10]

6 9701/42/F/M/18 © UCLES 2018 3 (a) Hydrogen cyanide, HCN, is a weak acid in aqueous solution. HCN(aq) H+(aq) + CN–(aq) Ka = 6.2 × 10–10 mol dm–3 (i) Calculate the pH of 0.10 mol dm–3 HCN(aq).  pH = … [2] (ii) Draw a ‘dot-and-cross’ diagram to represent the bonding in the hydrogen cyanide molecule. Show the outer shell electrons only.  [1] (iii) State the hybridisation of the carbon and nitrogen atoms in hydrogen cyanide, and give the H–C–N bond angle. hybridisation of C … hybridisation of N … H–C–N bond angle …  [2] (iv) Suggest structures for the organic products A and B in the following reactions. Assume that HCN reacts in a similar way to RCN. HCN H2 + Ni heat with HCl (aq) A B  [2]

7 9701/42/F/M/18 © UCLES 2018 [Turn over (b) Adding a measured quantity of KCN to a solution of NiCl 2 produces the complex [Ni(CN)2Cl 2]x. (i) Deduce the overall charge, x, on this complex.  x = … [1] The complex can exist as two separate isomers with the same geometry (shape) around the nickel ion. (ii) State the type of isomerism shown by these isomers. … [1] (iii) If bromide ions are present in the solution, the complex [Ni(CN)2Cl Br]x can form. Assuming that [Ni(CN)2Cl Br]x has the same geometry as [Ni(CN)2Cl 2]x, state the number of isomers of [Ni(CN)2Cl Br]x that could exist, and draw their structures in the box. ● number of isomers of [Ni(CN)2Cl Br]x … structures of the isomers of [Ni(CN)2Cl Br]x  [3] (c) An aqueous solution of KCN is gradually added to a solution of NiSO4 until the KCN is in excess. The following series of reactions takes place. NiSO4 more KCN KCN C green precipitate D yellow solution E red solution KCN in excess ● ● The oxidation state of nickel does not change during these reactions. ● ● None of C, D or E contains sulfur. ● ● C contains no potassium. ● ● The K : Ni ratio in D is 2 : 1. ● ● The K : Ni ratio in E is 3 : 1. Use the information to suggest the formulae of C, D and E. C … D … E …  [3]  [Total: 15]

8 9701/42/F/M/18 © UCLES 2018 4 (a) (i) State what is meant by the term partition coefficient. … … … [1] Ammonia is soluble in both water and organic solvents. An aqueous solution of ammonia is shaken with the immiscible organic solvent trichloromethane. The mixture is left to reach equilibrium. Samples are taken from each layer and titrated with dilute hydrochloric acid. ● ● A 25.0 cm3 sample from the trichloromethane layer requires 13.0 cm3 of 0.100 mol dm–3 HCl to reach the end-point. ● ● A 10.0 cm3 sample from the aqueous layer requires 12.5 cm3 of 0.100 mol dm–3 HCl to reach the end-point. (ii) Calculate the partition coefficient, Kpartition, of ammonia between trichloromethane and water.  Kpartition = … [2] (iii) Butylamine, C4H9NH2, is also soluble in both water and organic solvents. Suggest how the value of Kpartition of butylamine between trichloromethane and water would compare to the value of Kpartition calculated in (ii). Explain your answer. … … … [2]

9 9701/42/F/M/18 © UCLES 2018 [Turn over (b) (i) Explain why butylamine is basic. … … [1] (ii) Write an equation to show butylamine reacting as a base. … [1] (iii) State how the basicity of butanamide, C3H7CONH2, compares to that of butylamine. … [1] (iv) State a reagent for the conversion of butanamide into butylamine. … [1]  [Total: 9]

10 9701/42/F/M/18 © UCLES 2018 5 (a) (i) Complete the electronic configuration of a chromium atom. 1s22s22p63s2 … [1] (ii) State the two highest oxidation states of chromium commonly seen in its compounds. … [1] (b) Six different compounds or complexes, H, J, K, L, M and N, are formed when an excess of aqueous NH3, aqueous NaOH and concentrated aqueous HCl are separately added to separate solutions containing Cu2+(aq) or Co2+(aq). solution reagent an excess of NH3(aq) an excess of NaOH(aq) an excess of concentrated HCl (aq) Cu2+(aq) H J K Co2+(aq) L M N (i) State the colours of the following compounds or complexes. H … K … M …  [2] (ii) Write the formulae of the following compounds or complexes. L … N …  [2] (iii) State the appearance of compound J. … [1]  [Total: 7]

11 9701/42/F/M/18 © UCLES 2018 [Turn over 6 The apparatus shows a cell which can be used to determine a value of the Avogadro constant, L. X power supply silver electrodes electrolyte Y variable resistor (a) (i) Name component X. … [1] (ii) Suggest a suitable electrolyte Y. … [1] (b) In an experiment, a current of 0.200 A was passed through the cell for 40.0 minutes. The mass of the silver cathode increased by 0.500 g. The charge on the electron is –1.60 × 10–19 C. Calculate the: ● ● number of moles of silver deposited on the cathode ● ● number of coulombs of charge passed ● ● number of electrons passed ● ● number of electrons needed to deposit 1 mol of silver at the cathode.  [3]  [Total: 5]

12 9701/42/F/M/18 © UCLES 2018 7 (a) (i) Complete the equations to show the two types of polymerisation. Draw one repeat unit for each polymer. Include any other products. ● addition polymerisation n CH2=CHCH3(g) ● condensation polymerisation n HO2CCH2CO2H(s) + n HOCH2CH2OH(l)  [3] (ii) Suggest the sign of the entropy changes, ΔS o, for each of the two types of polymerisation. Explain your answers. ● ΔS o for addition polymerisation … … … ● ΔS o for condensation polymerisation … … …  [2]

13 9701/42/F/M/18 © UCLES 2018 [Turn over (b) An amide bond forms when a carboxylic acid reacts with an amine. (i) Complete the equation by writing the products in the box. R–CO2H + H2N–R' +  [1] (ii) Use your answer to (i) to work out the bonds that are broken and the bonds that are formed during the reaction between a carboxylic acid and an amine. ● ● bonds that are broken … ● ● bonds that are formed …  [2] (iii) Use bond energy values from the Data Booklet to calculate the enthalpy change, ΔH o, when one mole of amide bonds is formed in the reaction in (i). ΔH o = … kJ [2] (c) Amide bonds can also be formed by reacting acyl chlorides with amines. The enthalpy change for this process, ΔH o, is – 6.00 kJ mol–1. Calculate the minimum entropy change, ΔS o, for this reaction to be spontaneous (feasible) at 298 K. ΔS o = … J K–1 mol–1 [2]

14 9701/42/F/M/18 © UCLES 2018 The repeat unit of the polyamide Kevlar is shown. O O N H N H (d) The monomers of Kevlar, benzene-1,4-dioic acid and 1,4-diaminobenzene, can be synthesised as follows. H3C CH3 HO2C CO2H benzene-1,4-dioic acid reaction 1 O2N NO2 H2N NH2 1,4-diaminobenzene reaction 2 State the reagents and conditions needed for: (i) reaction 1 … [1] (ii) reaction 2 … … [2] (e) Kevlar is both strong and rigid. Complete the table to identify two intermolecular forces and the groups involved which are responsible for these properties of Kevlar. intermolecular force group(s) involved  [2]  [Total: 17]

15 9701/42/F/M/18 © UCLES 2018 [Turn over Question 8 starts on the next page.

16 9701/42/F/M/18 © UCLES 2018 8 (a) Cumin is a spice used to flavour food. Two compounds responsible for its flavour are cuminaldehyde and cuminyl alcohol. O cuminaldehyde OH cuminyl alcohol (i) Deduce the number of peaks that would be present in the 13C NMR spectrum of cuminyl alcohol. number of peaks … [1] (ii) Identify two bonds that are responsible for the differences in the infra-red spectra of cuminaldehyde and cuminyl alcohol, and state their absorption ranges. absorption range in the infra-red spectrum / cm–1 bond responsible for the difference cuminaldehyde cuminyl alcohol  [2]

17 9701/42/F/M/18 © UCLES 2018 [Turn over (b) Cuminyl alcohol can be synthesised from benzene by the following route. OH cuminyl alcohol benzene step 4 CO2H step 3 step 2 step 1 O (i) Suggest reagents and conditions for steps 1– 4. step 1 … step 2 … step 3 … step 4 …  [4] (ii) Name the mechanism of step 2 and state the type of reaction in step 4. mechanism of step 2 … type of reaction in step 4 …  [2]  [Total: 9]

18 9701/42/F/M/18 © UCLES 2018 9 (a) Two molecules of compound F react together under suitable conditions to form compound G. Some information about compounds F and G is given. ● ● The mass spectrum of F has a peak due to the molecular ion at m / e = 106, and a peak at m / e = 107 with an abundance 8% of the 106 peak. ● ● The mass spectrum of G has a peak due to the molecular ion at m / e = 212, and major peaks at m / e = 91 and m / e = 105. ● ● Both F and G contain oxygen and are neutral compounds which are insoluble in water. ● ● The 1H NMR spectrum of F includes a singlet peak at δ = 10.0 due to one proton. ● ● The 1H NMR spectrum of G includes a singlet peak at δ = 5.1 due to two protons. ● ● When G is heated with dilute sulfuric acid, benzoic acid, C6H5CO2H, and phenylmethanol, C6H5CH2OH, are produced. Use this information to answer (i) – (vi). (i) Explain how the mass spectrum of F shows that a molecule of F contains seven carbon atoms. Show your working.  [1] (ii) Suggest the molecular formula of the fragment of G at m / e = 91. … [1] (iii) Suggest the molecular formulae of F and G. F … G …  [2] (iv) Suggest structures for compounds F and G. F G  [2] (v) On the structures you have drawn in (iv), circle the protons responsible for the 1H NMR peaks at δ = 10.0 in F and δ = 5.1 in G. [1]

19 9701/42/F/M/18 © UCLES 2018 [Turn over (vi) State the type of reaction that G undergoes when heated with dilute sulfuric acid. … [1] (b) Describe and explain the relative acidities of benzoic acid, phenylmethanol and 4-methylphenol. CO2H benzoic acid CH2OH H3C phenylmethanol OH 4-methylphenol … … … … … … [3] (c) The ester 4-methylphenyl benzoate is used in the manufacture of perfumes. O O 4-methylphenyl benzoate H3C Suggest a two-step route for the synthesis of 4-methylphenyl benzoate from 4-methylphenol and benzoic acid. Include reagents and conditions for each step, and the structure of the intermediate compound.  [3]  [Total: 14]

20 9701/42/F/M/18 © UCLES 2018 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. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge International Examinations Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cie.org.uk after the live examination series. 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.

® IGCSE is a registered trademark. This document consists of 10 printed pages. © UCLES 2018 [Turn over Cambridge Assessment International Education Cambridge International Advanced Subsidiary and Advanced Level CHEMISTRY 9701/42 Paper 4 A Level Structured Qeustions March 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 March 2018 series for most Cambridge IGCSE®, Cambridge International A and AS Level components and some Cambridge O Level components.

9701/42 Cambridge International AS/A Level – Mark Scheme PUBLISHED March 2018 © UCLES 2018 Page 2 of 10 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.

9701/42 Cambridge International AS/A Level – Mark Scheme PUBLISHED March 2018 © UCLES 2018 Page 3 of 10 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.

9701/42 Cambridge International AS/A Level – Mark Scheme PUBLISHED March 2018 © UCLES 2018 Page 4 of 10 Question Answer Marks 1(a)(i) (solubility) increases (down the group) 1 1(a)(ii) down the group: lattice energy or hydration energy decrease lattice energy decreases more than hydration energy enthalpy change of solution becomes more negative / exothermic 3 1(b)(i) Ksp = [Sr2+][OH–]2 1 1(b)(ii) Ksp = (3.37 × 10–2) × (6.74 × 10–2)2 = 1.5 × 10–4 units: mol3 dm–9 2 1(c)(i) 2 SrO2 → 2SrO + O2 1 1(c)(ii) temperature will increase (down the group) charge density of cation decreases (down the group) this means less polarisation of the O2 2– ion or weakens the O-O bond less 3 1(d)(i) BaC2O4 →) BaO + CO + CO2 1 1(d)(ii) the KMnO4 would decolourise bubbles / gas evolution would be seen 2 Question Answer Marks 2(a) the Eo for X2 / X– becomes less positive / decrease down the group so the halogens are less reactive (as oxidants) down the group 2 2(b)(i) Cl2 + H2O → HCl + HClO 1 2(b)(ii) Cl2/Cl– = +1.36 V and ClO– / (Cl– + OH–) = +0.89 V so Eo cell = 1.36 – 0.89 = (+) 0.47 V 2

9701/42 Cambridge International AS/A Level – Mark Scheme PUBLISHED March 2018 © UCLES 2018 Page 5 of 10 Question Answer Marks 2(b)(iii) box three ticked Le Chatelier argument, more OH– / increase reactant concentration so equilibrium shifts right or an argument based on the half cell with OH– 2 2(c)(i) Br – + 3ClO – → BrO3 – + 3Cl – 1 2(c)(ii) Eo cell = 0.89 – 0.58 = + 0.31 V 1 2(c)(iii) 4HBrO3 → 2Br2 + 5O2 + 2H2O 1 Question Answer Marks 3(a)(i) [H+] = √(Ka. c) = √(6.2 × 10–10 × 0.1) [H+] = 7.9 × 10–6 pH = –log[H+] = 5.1(0) 2 3(a)(ii) H C N 1 3(a)(iii) C: sp and N: sp angle 180° 2 3(a)(iv) A is CH3NH2 B is HCO2H 2 3(b)(i) 2- 1 3(b)(ii) geometrical / cis-trans 1

9701/42 Cambridge International AS/A Level – Mark Scheme PUBLISHED March 2018 © UCLES 2018 Page 6 of 10 Question Answer Marks 3(b)(iii) 2 isomers Br Ni CN CN Cl Br Ni CN Cl NC (must be clearly square planar) 3 3(c) C is Ni(CN)2 D is K2Ni(CN)4 E is K3Ni(CN)5 3 Question Answer Marks 4(a)(i) (ratio of the) concentrations of a solute in two solvents / liquids at equilibrium 1 4(a)(ii) [NH3]aq = 0.1 × 12.5 / 10 = 0.125 mol dm–3 [NH3]CHCl3 = 0.1 × 13 / 25 = 0.052 mol dm–3 ratio = Kpartition = 0.052 / 0.125 = 0.416 2 4(a)(iii) Kpartition will be larger for butylamine than for ammonia butylamine contains a hydrophobic / non-polar (C4) group 2 4(b)(i) nitrogen has a lone pair which can accept a proton or can be donated to a proton 1 4(b)(ii) e.g. C4H9NH2 + HCl ⇌ C4H9NH3 + + Cl – etc. 1 4(b)(iii) butanamide is non-basic / neutral or (much) less basic than butylamine 1 4(b)(iv) LiAlH4 (in dry ether) 1

9701/42 Cambridge International AS/A Level – Mark Scheme PUBLISHED March 2018 © UCLES 2018 Page 7 of 10 Question Answer Marks 5(a)(i) 1s22s22p63s2 3p63d54s1 1 5(a)(ii) (+)3 and (+)6 1 5(b)(i) H is deep / dark / royal and blue (solution) K is yellow / yellow-green M is blue (precipitate) (allow pink) 2 5(b)(ii) L is [Co(NH3)6]2+ N is [CoCl4]2– 2 5(b)(iii) (pale) blue precipitate 1 Question Answer Marks 6(a)(i) X is an ammeter 1 6(a)(ii) Y is AgNO3 or AgF or AgClO4 1 6(b) n(Ag) = 0.500 / 107.9 = 4.6(34) × 10–3 n(C) = 0.200 × 40 × 60 = 480 C n(e–) = 480/1.60 × 10–19 = 3(.00) × 1021 n(e–)/n(Ag) = 3.00 × 1021 / 4.634 × 10–3 = 6.474 × 1023 (6.5 × 1023) 3

9701/42 Cambridge International AS/A Level – Mark Scheme PUBLISHED March 2018 © UCLES 2018 Page 8 of 10 Question Answer Marks 7(a)(i) CH3 O O O + H2O [1] [1] [1] O 3 7(a)(ii) • for addition polymerisation: ∆S will be negative, as many gas molecules are combining to form one (large) molecule • for condensation polymerisation: ∆S likely to be positive, (as each pair of monomer molecules join to chain, two molecules of) water forms 2 7(b)(i) (RCO2H + H2NR' →) RCONHR' + H2O 1 7(b)(ii) broken: C-O, N-H formed: C-N, O-H 2 7(b)(iii) bonds formed: 305 + 460 or 765 bonds broken: 360 + 390 or 750 (both) ∆H = 750 – 765 = –15 (kJ) 2 7(c) (If ∆G = 0, then) ∆H = T∆S ∆S = ∆H / T = –6000 / 298 = –20.1 (J mol–1 K–1) 2 7(d)(i) heat with (conc.) KMnO4 1 7(d)(ii) Sn and HCl heat + conc. (then add NaOH) 2

9701/42 Cambridge International AS/A Level – Mark Scheme PUBLISHED March 2018 © UCLES 2018 Page 9 of 10 Question Answer Marks 7(e) intermolecular force group(s) involved hydrogen bonding N-H and C=O (of amide) induced dipole / van der Waals benzene rings 2 Question Answer Marks 8(a)(i) 7 peaks 1 8(a)(ii) C=O 1670–1740 OH 3200–3600 Or C-O 1000–1260 2 8(b)(i) step 1 heat with Al Cl3 + (CH3)2CHCl or CH3CH=CH2 step 2 heat with Al Cl3 + CH3COCl step 3 NaOH + I2 (or Cl2) (then H+) step 4 LiAl H4 (in dry ether) 4 8(b)(ii) step 2 electrophilic (aromatic) substitution step 4 reduction 2 Question Answer Marks 9(a)(i) n = (100 / 1.1)(8 / 100) = 7.3 (⇒ 7 C atoms) 1 9(a)(ii) C7H7 + 1 9(a)(iii) F is C7H6O G is C14H12O2 2

9701/42 Cambridge International AS/A Level – Mark Scheme PUBLISHED March 2018 © UCLES 2018 Page 10 of 10 Question Answer Marks 9(a)(iv) O O O F G H H H 2 9(a)(v) O O O F G H H H δ 10.0 δ 5.1 1 9(a)(vi) hydrolysis 1 9(b) benzoic acid > methylphenol > phenylmethanol methylphenoxide anion has delocalisation of the lone pair on oxygen over the ring benzoic acid has an (extra) electronegative oxygen or electron withdrawing C=O 3 9(c) step 1 treat benzoic acid with SOCl2 or PCl5 to make the acyl chloride formula is C6H5COCl step 2 dissolve the methylphenol in NaOH(aq) (and shake with the benzoyl chloride) 3