Cambridge A Level Chemistry 9701 — 2019 May/June 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 May/June 2019  2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet Cambridge Assessment International Education Cambridge International Advanced Subsidiary and Advanced Level This document consists of 23 printed pages and 1 blank page. [Turn over IB19 06_9701_42/3RP © UCLES 2019 *2283347273*

2 9701/42/M/J/19 © UCLES 2019 Answer all the questions in the spaces provided. 1 (a) (i) Complete the electronic configuration of the copper(II) ion. 1s22s22p6 … [1] (ii) State the colour of the solutions containing the following ions. ● [Cu(H2O)6]2+(aq) … ● [CuCl 4]2–(aq) …  [1] (iii) Octahedral complexes of Cu2+ with different ligands can have different colours. Explain why. … … … … … [2] (b) Copper(I) and silver(I) salts are colourless. Suggest why. … … … … [2]

3 9701/42/M/J/19 © UCLES 2019 [Turn over (c) Consider the following two equilibria and associated data values at 298 K. AgBr(s) Ag+(aq) + Br –(aq) equilibrium 1 Ksp = 5.0 × 10–13 mol2 dm–6 Ag+(aq) + 2NH3(aq) [Ag(NH3)2]+(aq) equilibrium 2 Kstab = 1.7 × 107 mol–2 dm6 The equilibrium constant for equilibrium 1 is the solubility product, Ksp, of AgBr(s). The equilibrium constant for equilibrium 2 is the stability constant, Kstab, for the formation of [Ag(NH3)2]+(aq). (i) Calculate the solubility of AgBr at 298 K in mol dm–3.  solubility of AgBr = … mol dm–3 [1] (ii) Use Le Chatelier’s principle as applied to equilibria 1 and 2 to suggest why AgBr(s) dissolves in concentrated NH3(aq). … … … … [2] (iii) Use equilibria 1 and 2 to construct an equation for the reaction of AgBr(s) with concentrated NH3(aq). This is equilibrium 3. … equilibrium 3 [1] (iv) Write an expression for the equilibrium constant of equilibrium 3, Keq3, in terms of Ksp for equilibrium 1 and Kstab for equilibrium 2. Keq3 =  [1]

4 9701/42/M/J/19 © UCLES 2019 (d) Define the term standard electrode potential, E o. … … … [1] (e) (i) Complete and label the diagram to show how the standard electrode potential, E o, of Ag+(aq) / Ag(s) could be measured under standard conditions.  [4] (ii) Use the Data Booklet to label the diagram in (e)(i) to show ● ● which is the positive electrode, ● ● the direction of electron flow in the external circuit when a current flows.  [1]  [Total: 17]

5 9701/42/M/J/19 © UCLES 2019 [Turn over 2 (a) Group 2 carbonates decompose when heated. Write an equation for the decomposition of the carbonate ion. … [1] (b) Describe and explain how the thermal stability of the Group 2 carbonates changes with increasing atomic number. … … … … … [3] (c) Lead(II) carbonate, PbCO3, and zinc carbonate, ZnCO3, decompose on heating in a similar way to calcium carbonate, CaCO3. State relevant data from the Data Booklet and use it to predict the order of thermal stability of these three carbonates. data … … (most stable) … > … > … (least stable)  [2] (d) Dolomite contains the double carbonate of calcium and magnesium, CaMg(CO3)2, and some impurities. When 0.642 g of a sample of dolomite reacts with an excess of hydrochloric acid, 125.0 cm3 of CO2 is formed under room conditions. Calculate the percentage of CaMg(CO3)2 in the sample of dolomite. Show all your working. Assume that none of the impurities react with HCl.  % of CaMg(CO3)2 in dolomite = … %  [3]  [Total: 9]

6 9701/42/M/J/19 © UCLES 2019 3 (a) Sketch the shape of a d orbital.  [1] (b) (i) Explain what is meant by the term transition element. … … [1] Transition elements can form complex ions which contain ligands. (ii) Name the type of bonding that occurs between a ligand and a transition element. … [1] (c) Give the formulae of two oxides of iron. State the oxidation number of iron in each compound. … … [1] (d) CO and CN– are monodentate ligands. Complete the table for the following two complexes. metal ion ligand co-ordination number formula of complex ion charge of complex ion Ni2+ CO 4 Fe3+ CN– 3-  [2]

7 9701/42/M/J/19 © UCLES 2019 [Turn over (e) Transition element complexes can exhibit stereoisomerism. [Cu(H2O)4(NH3)2]2+ and Pt(NH3)2Cl 2 show the same type of isomerism. (i) Name this type of isomerism. … [1] (ii) Complete the three-dimensional diagrams of the two isomers for [Cu(H2O)4(NH3)2]2+ and Pt(NH3)2Cl 2. Cu Cu Pt Pt  [2]

8 9701/42/M/J/19 © UCLES 2019 (f) Copper can form complexes with the ligands ammonia and en, H2NCH2CH2NH2, as shown. [Cu(H2O)6]2+(aq) + en(aq) [Cu(H2O)4(en)]2+(aq) + 2H2O(l) Kstab = 3.98 × 1010 equilibrium 4 [Cu(H2O)6]2+(aq) + 2NH3(aq) [Cu(H2O)4(NH3)2]2+(aq) + 2H2O(l) Kstab = 5.01 × 107 equilibrium 5 (i) Write an expression for the stability constant, Kstab, for equilibrium 5. State its units. Kstab =  units = …  [2] (ii) The standard entropy change, ΔS o, for equilibrium 4 is +23 J K–1 mol–1 and for equilibrium 5 is –8.4 J K–1 mol–1. Suggest an explanation for this difference by reference to both equilibria. … … … [1] (iii) Of the three copper complexes in equilibria 4 and 5, state the formula of the copper complex that is the most stable and explain your choice. copper complex … explanation … …  [1]  [Total: 13]

9 9701/42/M/J/19 © UCLES 2019 [Turn over Question 4 starts on the next page.

10 9701/42/M/J/19 © UCLES 2019 4 The initial rate of reaction for propanone and iodine in acid solution is measured in a series of experiments at a constant temperature. H+ catalyst CH3COCH3 + I2 CH3COCH2I + HI The rate equation was determined experimentally to be as shown. rate = k [CH3COCH3][H+] (a) State the order of reaction with respect to ● CH3COCH3 … ● I2 … ● H+ … and state the overall order of this reaction. …  [2] (b) The rate of this reaction is 5.40 × 10–3 mol dm–3 s–1 when ● the concentration of CH3COCH3 is 1.50 × 10–2 mol dm–3 ● the concentration of I2 is 1.25 × 10–2 mol dm–3 ● the concentration of H+ is 7.75 × 10–1 mol dm–3. (i) Calculate the rate constant, k, for this reaction. State the units of k.  k = …  units = …  [2] (ii) Complete the table by placing one tick () in each row to describe the effect of decreasing the temperature on the rate constant and on the rate of reaction. decreases no change increases rate constant rate of reaction  [1]

11 9701/42/M/J/19 © UCLES 2019 [Turn over (c) From the results, a graph is produced which shows how the concentration of I2 changes during the reaction. 00 [I2] time Describe how this graph could be used to determine the initial rate of the reaction. … … … [2] (d) On the axes below, sketch a graph to show how the initial rate changes with different initial concentrations of CH3COCH3 in this reaction. 00 rate [CH3COCH3]  [1]

12 9701/42/M/J/19 © UCLES 2019 (e) The rate of a reaction between metal ions was studied. The following three-step mechanism has been suggested for this reaction. Step 1 is the rate-determining step. step 1 Ce4+ + Mn2+ Ce3+ + Mn3+ step 2 Ce4+ + Mn3+ Ce3+ + Mn4+ step 3 Mn4+ + Tl + Tl 3+ + Mn2+ (i) Explain the meaning of the term rate-determining step. … … [1] (ii) Use this mechanism to ● determine the overall equation for this reaction … ● suggest the role of Mn2+ ions in this mechanism. Explain your answer. … … …  [2]  [Total: 11]

13 9701/42/M/J/19 © UCLES 2019 [Turn over 5 (a) Complete the table by placing one tick () in each row to indicate the sign of each type of energy change under standard conditions. energy change always positive always negative either negative or positive lattice energy enthalpy change of neutralisation  [1] (b) Define, in words, the term enthalpy change of solution. … … [1] (c) The following enthalpy changes are given. enthalpy change value / kJ mol–1 standard enthalpy change of formation, , for K3PO4(s) –2035 standard enthalpy change, ∆H o, for P(s) + 2O2(g) + 3e– PO4 3–(aq) –1284 standard enthalpy change, ∆H o, for K(s) K+(aq) + e– –251 Determine the standard enthalpy change of solution of potassium phosphate, K3PO4(s). It may be helpful to draw a labelled energy cycle.  = … kJ mol–1 [3]

14 9701/42/M/J/19 © UCLES 2019 (d) Some lattice energy values are shown in the table. compound lattice energy value / kJ mol–1 CaBr2(s) –2176 KBr(s) –679 Suggest an explanation for why CaBr2 is more exothermic than KBr. … … … [2] (e) For a particular gas phase reaction the variation in standard Gibbs free energy change, ΔG o, with temperature is shown. Assume standard enthalpy change, ΔH o, and standard entropy change, ΔS o, remain constant with temperature. 0 350 ∆G o / kJ mol–1 T / K 550 (i) Write the equation that relates ΔG o to ΔH o and ΔS o. … [1] (ii) Use this equation to explain why ΔG o becomes less positive as temperature increases in this reaction. … … [1]  [Total: 9]

15 9701/42/M/J/19 © UCLES 2019 [Turn over Question 6 starts on the next page.

16 9701/42/M/J/19 © UCLES 2019 6 (a) By reference to the formation of σ and π bonds, describe and explain the shape of a benzene molecule, C6H6. … … … … … … … … [3] (b) 2,3-dimethylphenylamine can be prepared from 1,2-dimethylbenzene in two steps as shown. 1,2-dimethylbenzene CH3 CH3 M CH3 CH3 NO2 2,3-dimethylphenylamine CH3 CH3 NH2 HNO3 / H2SO4 step 1 step 2 Step 1 is catalysed by H2SO4. (i) Write an equation to show how H2SO4 generates the electrophile during step 1. … [1] (ii) Draw the mechanism of the reaction between this electrophile and 1,2-dimethylbenzene to form M. Include all relevant curly arrows and charges. intermediate CH3 CH3 M CH3 CH3 NO2  [3]

17 9701/42/M/J/19 © UCLES 2019 [Turn over (iii) Write an equation to show how the H2SO4 catalyst is reformed. … [1] (iv) For step 2, suggest the reagents and conditions and name the type of reaction. ● reagents and conditions … ● type of reaction …  [2] (c) The drug mefenamic acid can be made using 2,3-dimethylphenylamine in an excess of 2‑chlorobenzoic acid. 2,3-dimethylphenylamine CH3 CH3 2-chlorobenzoic acid mefenamic acid CO2H CO2H Cl + + HCl NH2 CH3 CH3 N H (i) Deduce the molecular formula of mefenamic acid. … [1] (ii) Name the functional groups, apart from the benzene ring, in mefenamic acid. … [1] (iii) Calculate the maximum mass of mefenamic acid that could be formed from 5.00 g of 2,3‑dimethylphenylamine in this reaction. Give your answer to three significant figures.  mass of mefenamic acid = … g [2]

18 9701/42/M/J/19 © UCLES 2019 (d) The position of substitution in the electrophilic substitution of arenes can be explained based on the stability of the intermediate cations formed in the first step. The example given involves the bromination of methylbenzene. methylbenzene CH3 CH3 + CH3 3-position 2-position 4-position Br H Br H CH3 + + Br H Use this information and your knowledge about the stability of cations to suggest why the CH3 group directs incoming electrophiles to the 2- and 4-positions in preference to the 3-position. … … … … [2]  [Total: 16]

19 9701/42/M/J/19 © UCLES 2019 [Turn over 7 (a) Amino acids can be separated by thin-layer chromatography. A mixture of amino acids is analysed using this technique. The chromatogram obtained is shown, drawn to scale. The table shows some Rf values for different amino acids in the solvent used. 5 4 3 2 1 0 cm baseline solvent front amino acid Rf value alanine 0.40 glutamic acid 0.29 leucine 0.71 valine 0.61 A B (i) Use the chromatogram and the Rf values to deduce the amino acid responsible for spot A and spot B. amino acid responsible for spot A … amino acid responsible for spot B …  [1] (ii) A second chromatogram of the same mixture is taken using a more polar solvent. Predict the effect on the Rf values of the amino acids. Explain your reasoning. … … … … [1] (b) Glycine, H2NCH2CO2H, is the simplest amino acid. (i) Complete the equations to show the acid-base properties of glycine. H2NCH2CO2H(aq) + HCl (aq) … H2NCH2CO2H(aq) + NaOH(aq) …  [2]

20 9701/42/M/J/19 © UCLES 2019 (ii) In aqueous solution, amino acids exist as zwitterions. Draw the zwitterionic structure of glycine. Explain how the zwitterion for glycine is formed. … … …  [2] (c) Apart from glycine, all naturally occurring amino acids have a chiral centre and exhibit stereoisomerism. Draw the two stereoisomers of alanine, CH3CH(NH2)CO2H. C C  [1] (d) The amino acid alanine can be synthesised from 2‑chloropropanoic acid, CH3CHCl CO2H. (i) State the reagents and conditions and name the mechanism for this reaction. reagents and conditions … name of mechanism …  [2] (ii) State and explain the relative acidities of trichloroethanoic acid, chloroethanoic acid and ethanoic acid. … … … … … [3]

21 9701/42/M/J/19 © UCLES 2019 [Turn over (e) Serine, HOCH2CH(NH2)CO2H, can react with alanine, CH3CH(NH2)CO2H, to form three different structural isomers, each with the molecular formula C6H12N2O4. Draw the structures of these three structural isomers. isomer 1 (C6H12N2O4) isomer 2 (C6H12N2O4) isomer 3 (C6H12N2O4)  [3]  [Total: 15]

22 9701/42/M/J/19 © UCLES 2019 8 Compound R is shown. CH3 Cl HO R CH3 (a) State the systematic name of compound R. … [1] (b) (i) R is dissolved in CDCl 3 and analysed using carbon‑13 and proton NMR spectroscopy. ● Predict the number of peaks that are seen in the carbon-13 NMR spectrum of R. … ● Predict the number of peaks that are seen in the proton NMR spectrum of R. …  [2] (ii) A separate sample of R is dissolved in D2O. The proton NMR spectrum of this solution shows one less peak than is obtained in CDCl 3. Explain why. … … [1]

23 9701/42/M/J/19 © UCLES 2019 [Turn over (c) Compound R reacts separately with the four reagents shown in the table. Complete the table by ● drawing the structures of the organic products formed, ● stating the type of reaction. reagent organic product structure type of reaction Na CH3COCl Br2(aq) + N N Cl –  [6]  [Total: 10]

24 9701/42/M/J/19 © UCLES 2019 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 Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cambridgeinternational.org after the live examination series. Cambridge Assessment International Education is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which itself is a department of the University of Cambridge.

This document consists of 13 printed pages. © UCLES 2019 [Turn over Cambridge Assessment International Education Cambridge International Advanced Subsidiary and Advanced Level CHEMISTRY 9701/42 Paper 4 A Level Structured Questions May/June 2019 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 2019 series for most Cambridge IGCSE™, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

9701/42 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 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.

9701/42 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 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.

9701/42 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 4 of 13 Question Answer Marks 1(a)(i) 3s23p63d9 [1] 1 1(a)(ii) [Cu(H2O)6]2+ (pale) blue CuCl42- yellow both [1] 1 1(a)(iii) M1 energy gap / ∆E is different (for the ligands) [1] M2 different frequency / wavelength of light absorbed / transmitted / reflected [1] 2 1(b) M1 (Cu+/Ag+) d-shell is full / complete OR d-orbitals are full [1] M2 no electrons can be promoted [1] 2 1(c)(i) solubility = √5.0 × 10–13 = 7.1 × 10–7 (mol dm–3) [1] min 2sf 1 1(c)(ii) M1 (in conc. NH3) [NH3] increases and equilibrium 2 shifts to the right [1] M2 [Ag+] decreases and equilibrium 1 shifts to the right [1] 2 1(c)(iii) AgBr + 2NH3 ⇌ [Ag(NH3)2]+ + Br– [1] 1 1(c)(iv) Keq3 = Ksp × Kstab [1] ALLOW Keq3= [Ag(NH3)2+][Br–]/[NH3]2 1 1(d) The potential difference when a half-cell is connected to a (standard) hydrogen electrode under standard conditions [1] OR the potential difference / voltage / EMF between a hydrogen electrode and another half-cell under standard conditions [1] 1 1(e)(i) salt bridge y voltmeter / V y Ag y Ag+ (or soluble silver salt) y Pt y H2 (and delivery correct) + H+ (or named strong acid) y 1 atm. (pressure) y 1 mol dm–3 (and 298 K) y mark as y 9 y 9 y 9 y 9 [4] 4 1(e)(ii) Ag electrode labelled and arrow (in the external circuit moving towards this electrode) [1] 1

9701/42 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 5 of 13 Question Answer Marks 2(a) CO32– → O2– + CO2 [1] 1 2(b) Increases (with increasing atomic number / implied) [1] cationic radius / ion size increases (down the group) [1] less polarisation / distortion of anion/CO32– [1] 3 2(c) (Pb2+) 0.120 nm; (Ca2+) 0.099 nm; (Zn2+) 0.075 nm [1] (most stable) PbCO3 > CaCO3 > ZnCO3 (least stable) [1] ECF from atomic radii 2 2(d) amount of CO2 = 125 / 24000 = 5.21 × 10–3 mol [1] CaMg(CO3)2 :CO2 1:2 amount of carbonate = 2.60(4) × 10–3 mol [1] ECF mass of carbonate = 184(.4) × 2.60(4) × 10–3 = 0.480 g % of CaMg(CO3)2 = 100 × 0.480 / 0.642 = 74.8 % [1] ECF 3 Question Answer Marks 3(a) any diagram [1] 1 3(b)(i) (elements) forming one or more (stable) ions with incomplete / partially filled d orbital(s) / sub-shell [1] 1 3(b)(ii) dative covalent / coordinate [1] 1

9701/42 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 6 of 13 Question Answer Marks 3(c) FeO and +2 Fe2O3 and +3 all [1] ALLOW Fe3O4 and +3 and +2 1 3(d) metal ion ligand co-ordination number formula of complex ion charge of complex ion Ni2+ CO 4 Ni(CO)4 2+ Fe3+ CN- 6 Fe(CN)6 3- mark as y 9 y 9 [2] 2 3(e)(i) cis-trans isomerism [1] ALLOW geometric(al) 1 3(e)(ii) one correct pair [1] two correct pairs [2] 2 3(f)(i) Kstab = 2 2 4 3 2 2 2 2 6 3 [Cu(H O) (NH ) ] [Cu(H O) ][NH ] + + [1] units = dm6 mol–2 [1] ecf from M1 2 3(f)(ii) equilibrium 4 has a (net) increase in moles of product / 2 moles goes to 3 moles whereas equilibrium 5 has same number of moles of reactants and products / 3 moles vs 3 moles [1] 1

9701/42 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 7 of 13 Question Answer Marks 3(f)(iii) [Cu(H2O)4(en)]2+ and (equilibrium) constant / Kstab is the largest / highest [1] ALLOW [Cu(H2O)4(en)]2+ and constant / Kstab of eqm 4 is greater / higher 1 Question Answer Marks 4(a) CH3COCH3 = 1 I2 = 0 H+ = 1 overall order = 2 M1 3 orders [1] M2 overall order based on their M1 [1] 2 4(b)(i) k = 5.40 × 10–3 / (1.50 × 10–2 × 7.75 × 10–1) k = 0.46(452) [1] dm3 mol–1 s–1 [1] 2sf min 2 4(b)(ii) decreases no change increases rate constant 9 rate of reaction 9 both [1] 1 4(c) draw a tangent at time, t=0 [1] measure the gradient / slope of the tangent [1] 2 4(d) straight line graph starting at 0,0 and showing rate α [CH3COCH3] [1] 1 4(e)(i) slowest step / reaction (in the mechanism) [1] 1 4(e)(ii) 2Ce4+ + Tl+ → Tl3+ + 2Ce3+ [1] catalyst and (used in step 1 and) regenerated / reformed in step 3 / end of the reaction [1] 2

9701/42 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 8 of 13 Question Answer Marks 5(a) energy change always positive always negative either negative or positive lattice energy 9 enthalpy of neutralisation 9 both [1] 1 5(b) (energy change) when 1 mole of solute is dissolved in an infinite amount of water to form a dilute solution 1 5(c) calculation of ∆Hosol with –251, –1284 and –2035 only and two correct signs [1] calculation of ∆Hosol with –251, –1284 and –2035 only and correct signs OR calculation of ∆Hosol with (–251 × 3), –1284 and –2035 only and two correct signs [2] ∆Hosol = (3 × –251) + (–1284) – (–2035) = –2 (kJ mol–1) [3] 3 5(d) Ca2+ have a higher charge / greater charge density [1] ora stronger electrostatic forces between Br– and Ca2+ [1] 2 5(e)(i) ∆Go = ∆Ho – T∆So [1] 1 5(e)(ii) T∆S is more positive OR –T∆S becomes more negative [1] 1

9701/42 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 9 of 13 Question Answer Marks 6(a) any three points from: 3 × [1] • bond angle = 120° and shape is (hexagonal ring) planar / (trigonal) planar • carbons are sp2 hybridised • contains delocalised electrons in the π bonds / system • sp2 orbitals between C-H / C-C overlap to form σ bonds • a p orbital from each carbon atom overlap sideways with each other above and below the ring forming π bonds ALLOW labelled diagrams for bullets 1–5 3 6(b)(i) HNO3 + H2SO4 → HSO4- + H2O + NO2+ or HNO3 + 2H2SO4 → 2HSO4- + H3O+ + NO2+ [1] 1 6(b)(ii) first curly arrow to N of NO2+ [1] correct intermediate [1] 2nd curly arrow and H+ formed / lost [1] 3 6(b)(iii) HSO4- + H+ → H2SO4 [1] 1 6(b)(iv) Sn + conc. HCl (+ heat) [1] reduction [1] IGNORE redox 2 6(c)(i) C15H15NO2 [1] 1 6(c)(ii) amine and carboxylic acid both [1] 1

9701/42 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 10 of 13 Question Answer Marks 6(c)(iii) amount of 2,3-dimethylphenylamine = 5.00 / 121 = 0.0413 mol [1] amount of mefenamic acid = 0.0413 mol mass of mefenamic acid = 0.0413 × 241 = 9.96 / 9.95 g 3sf required [1] ECF 2 6(d) 3o carbocations are more stable than 2o carbocations [1] due to the methyl group acting as an electron donating group (leading to an increase in electron density on the cation stabilising it) [1] 2 Question Answer Marks 7(a)(i) A= leucine B= glutamic acid both [1] 1 7(a)(ii) greater and more soluble in the solvent / mobile phase OR greater and form more H-bonds with the solvent [1] 1 7(b)(i) H2NCH2CO2H + HCl → Cl–H3N+CH2CO2H [1] H2NCH2CO2H + NaOH → H2NCH2CO2–Na+ + H2O [1] 2 7(b)(ii) H3N+CH2CO2– [1] Proton is transferred from the CO2H group to the NH2 group [1] 2 7(c) CO2H C H2N CH3 H CO2H C NH2 H3C H two non-superimposable mirror images for alanine drawn [1] 1

9701/42 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 11 of 13 Question Answer Marks 7(d)(i) NH3 (in ethanol) heat in a sealed tube [1] nucleophilic substitution [1] 2 7(d)(ii) acidity of Cl3CCO2H > ClCH2CO2H > CH3CO2H [1] any two of: Cl is electronegative / electron withdrawing group AND Cl3CCO2H has more / 3 Cl groups [1] weakens O-H bond so more likely to ionise / dissociate OR negative charge on anion is more stabilised OR charge / electron density on COO– decreases so anion is (more) stabilised [1] CH3 is electron donating so O-H bond is stronger so less likely to ionise in CH3CO2H OR CH3CO2H has no -I group so O-H bond is stronger and less likely to ionise [1] 3 7(e) H2N HO O H N CO2H H2N O H N CO2H OH O O H2N CO2H NH2 One mark for each structure. [1] [1] [1] 3

9701/42 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 12 of 13 Question Answer Marks 8(a) 4-chloro-3,5-dimethylphenol OR 3,5-dimethyl-4-chlorophenol [1] ALLOW 2,6-dimethyl-4-hydroxychlorobenzene and 2-chloro-5-hydroxy-1,3-dimethylbenzene 1 8(b)(i) carbon-13 NMR = 5 peaks [1] proton NMR = 3 peaks [1] 2 8(b)(ii) OH proton had disappeared due to proton exchange with D / D2O [1] ALLOW OH + D2O → OD + HOD 1

9701/42 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 13 of 13 Question Answer Marks 8(c) reagent organic product structure type of reaction Na redox CH3COCl Condensation Br2(aq) (electrophilic) substitution (electrophilic) substitution each structure [1] × 4 type of reaction y 9 y 9 [2] 6