Cambridge A Level Chemistry 9701 — 2020 May/June Paper 4 · Variant 1

*4376886889* CHEMISTRY 9701/41 Paper 4 A Level Structured Questions May/June 2020  2 hours You must answer on the question paper. You will need: Data booklet INSTRUCTIONS ● ● Answer all questions. ● ● Use a black or dark blue pen. You may use an HB pencil for any diagrams or graphs. ● ● Write your name, centre number and candidate number in the boxes at the top of the page. ● ● Write your answer to each question in the space provided. ● ● Do not use an erasable pen or correction fluid. ● ● Do not write on any bar codes. ● ● You may use a calculator. ● ● You should show all your working, use appropriate units and use an appropriate number of significant figures. INFORMATION ● ● The total mark for this paper is 100. ● ● The number of marks for each question or part question is shown in brackets [ ]. IB20 06_9701_41/5RP © UCLES 2020 [Turn over This document has 20 pages. Blank pages are indicated. Cambridge International AS & A Level

2 9701/41/M/J/20 © UCLES 2020 Answer all the questions in the spaces provided. 1 (a) An aqueous solution of cobalt(II) contains the [Co(H2O)6]2+ complex ion. (i) Define the term complex ion. … … [1] (ii) Samples of [Co(H2O)6]2+ are reacted separately with aqueous sodium hydroxide and with an excess of aqueous ammonia. Give the following information about these reactions. ● ● the reaction of [Co(H2O)6]2+ with aqueous sodium hydroxide colour and state of the cobalt-containing species … ionic equation … type of reaction … ● ● the reaction of [Co(H2O)6]2+ with an excess of aqueous ammonia colour and state of the cobalt-containing species … ionic equation … type of reaction …  [6] (b) When concentrated hydrochloric acid is added to a solution containing [Co(H2O)6]2+, a blue solution of [CoCl 4]2– is formed and the following equilibrium is established. [Co(H2O)6]2+ + 4Cl – [CoCl 4]2– + 6H2O Use Le Chatelier’s principle to suggest the expected observations when silver nitrate solution is added dropwise to the blue solution of [CoCl 4]2–. Explain your answer. … … … [2]

3 9701/41/M/J/20 © UCLES 2020 [Turn over (c) The [Co(NH3)3Cl 3] complex shows stereoisomerism. Complete the three-dimensional diagrams to show the two isomers of [Co(NH3)3Cl 3]. Suggest the type of stereoisomerism. Co Co type of stereoisomerism …  [2] (d) Compound X, C6H18N4, is a tetradentate ligand. H2N NH2 N H X H N (i) Suggest why one molecule of X can form four dative bonds. … … … [1] (ii) C6H18N4 reacts with aqueous cobalt(II) ions, [Co(H2O)6]2+, in a 1 : 1 ratio to form a new complex ion. Construct an equation for this reaction. … [1]  [Total: 13]

4 9701/41/M/J/20 © UCLES 2020 2 (a) (i) Describe and explain the trend in the solubility of the Group 2 hydroxides down the group. … … … … … … [4] Group 2 hydroxides decompose on heating to give the corresponding metal oxide and water vapour. (ii) Suggest which of Mg(OH)2 and Sr(OH)2 will decompose at a lower temperature. Explain your answer. … … … … [2]  [Total: 6]

5 9701/41/M/J/20 © UCLES 2020 [Turn over 3 The overall reaction for photosynthesis is shown. 6CO2 + 6H2O → C6H12O6 + 6O2 Water is oxidised in this process according to the following half-equation. 2H2O → O2 + 4H+ + 4e– half-equation 1 (a) (i) Use these equations to deduce the half-equation for the reduction of carbon dioxide in this process.  [2] (ii) Draw a fully labelled diagram of the apparatus that should be used to measure the standard electrode potential, E o, of O2(g) in half-equation 1 under standard conditions. Include all necessary chemicals.  [4] (iii) For the cell drawn in (a)(ii), use the Data Booklet to calculate the and deduce which electrode is positive.  = … V  identity of the positive electrode = …  [1]  [Total: 7]

6 9701/41/M/J/20 © UCLES 2020 4 (a) The molecular formulae of three nitrogen‑containing compounds are given. S C6H5CONH2 T C6H5NH2 U C6H5CH2NH2 Describe and explain the relative basicities of S, T and U. … > … > … most basic least basic … … … … … …  [3] (b) Compound U can be prepared by two different methods as shown. NH2 NH2 Br O U reaction 1 reaction 2 (i) Suggest reagents and conditions for reaction 1 and for reaction 2. reaction 1 … reaction 2 …  [2] (ii) State the type of reaction in reaction 1 and name the mechanism in reaction 2. type of reaction in reaction 1 … mechanism of reaction 2 …  [2]  [Total: 7]

7 9701/41/M/J/20 © UCLES 2020 [Turn over 5 (a) Benzene reacts with bromine in the presence of an aluminium bromide catalyst, Al Br3, to form bromobenzene. This is a substitution reaction. No addition reaction takes place. (i) Explain why no addition reaction takes place. … … [1] Al Br3 reacts with bromine to generate an electrophile, Br+. (ii) Draw the mechanism of the reaction between benzene and Br+ ions. Include all relevant arrows and charges.  [3] (iii) Write an equation to show how the Al Br3 catalyst is reformed. … [1] (b) Suggest why bromination of phenol occurs more readily than bromination of benzene. … … … … … [2]

8 9701/41/M/J/20 © UCLES 2020 (c) (i) There are four different carbocations with the same formula, C4H9 +. One structure is given in the table. Suggest the structural formulae of the three other carbocations. structure 1 CH3CH2CH2CH2 + structure 2 structure 3 structure 4  [3] (ii) Benzene reacts with each of these carbocations in separate Friedel-Crafts alkylation reactions. In each reaction an organic compound with formula C10H14 is formed. The number of peaks observed in the carbon‑13 NMR spectrum of each compound is given. Suggest the structures for the three other compounds. number of peaks in carbon-13 NMR = 8 number of peaks in carbon-13 NMR = 6 number of peaks in carbon-13 NMR = 7 number of peaks in carbon-13 NMR = 8  [4]  [Total: 14]

9 9701/41/M/J/20 © UCLES 2020 [Turn over 6 (a) Compare and explain the relative acidities of 2‑chloropropanoic acid, 3‑chloropropanoic acid, and propanoic acid. Explain your answer. … > … > … most acidic least acidic explanation … … … … … …  [3] (b) (i) The numerical values of Ka for methanoic acid, HCO2H, and pyruvic acid, CH3COCO2H, are given. acid Ka HCO2H 1.78 × 10–4 CH3COCO2H 4.07 × 10–3 An equilibrium mixture containing the two acid-base pairs is formed. HCO2 – + CH3COCO2H HCO2H + CH3COCO2 – Use the Ka values to calculate the equilibrium constant, Keq, for this equilibrium.  Keq = … [1] (ii) Use your value of Keq to predict the position of this equilibrium. Indicate this by placing a tick () in the appropriate box in the table. Explain your answer. equilibrium lies to the left equilibrium lies in the middle equilibrium lies to the right … …  [1]

10 9701/41/M/J/20 © UCLES 2020 (iii) Ethanedioic acid, HO2CCO2H, has two dissociation constants, Ka1 and Ka2, whose pKa values are 1.23 and 4.19. Suggest equations to show the two dissociations that give rise to these pKa values. pKa1 1.23 … pKa2 4.19 …  [2] (iv) State the mathematical relationship between pKa and the acid dissociation constant Ka. … [1] (c) Three tests were carried out on separate samples of the organic acids shown in the table. The following results were obtained. = observed change = no observed reaction test reagent(s) and conditions HCO2H CH3COCO2H HO2CCO2H observed change 1 … … …    2 … … …    3 … … …    Complete the table with the reagent(s) and conditions and the observed change for each test. Assume these organic acids all have a similar acid strength. [5]

11 9701/41/M/J/20 © UCLES 2020 [Turn over (d) A sample of pyruvic acid, CH3COCO2H, is analysed by carbon‑13 NMR spectroscopy. Three peaks are observed. Complete the table by: ● ● circling the carbon atom responsible for the chemical shift ● ● stating the hybridisation of the circled carbon atom. chemical shift (δ) carbon atom responsible for chemical shift hybridisation of the circled carbon atom 27 O O C C O H H H C H 163 O O C C O H H H C H 192 O O C C O H H H C H  [2]

12 9701/41/M/J/20 © UCLES 2020 (e) An ester of pyruvic acid, F, is dissolved in CDCl 3 and analysed by proton NMR spectroscopy. C C H3C O F O CH2 O CH3 The proton NMR spectrum of F is shown. 5 4 3 2 1 0 δ / ppm Use the proton NMR spectrum of F to complete the table. chemical shift (δ) group responsible for the peak splitting pattern number of 1H atoms responsible for the peak 1.3 2.2 4.0  [3]

13 9701/41/M/J/20 © UCLES 2020 [Turn over (f) Deuterium oxide, D2O, where D is 2H, can be used as a solvent in proton NMR spectroscopy. The proton NMR spectrum of alanine in CDCl 3 has 4 peaks. The proton NMR spectrum of alanine in D2O has 2 peaks. H3C C NH2 H alanine COOH On the diagram of alanine, circle the protons that show peaks in both NMR spectra. Explain your answer. … … … [2] (g) The ionic product, Kw, for D2O has a value of 1.35 × 10–15 mol2 dm–6 at 298 K. (i) Write the expression for the Kw of D2O. Kw =  [1] (ii) Calculate the pH of pure, neutral D2O at 298 K. Assume [D+] is equivalent to [H+] for pH calculations.  pH = … [2]  [Total: 23]

14 9701/41/M/J/20 © UCLES 2020 7 (a) Silver carbonate, Ag2CO3, is sparingly soluble in water. The numerical value of the solubility product, Ksp, for silver carbonate is 6.3 × 10–12 at 25 °C. (i) Write an expression for the solubility product, Ksp, of Ag2CO3, and state its units. Ksp =  units = …  [2] (ii) Calculate the equilibrium concentration of Ag+ in a saturated solution of Ag2CO3 at 25 °C.  [Ag+] = … mol dm–3 [1] (iii) Solid Ag2CO3 is stirred at 25 °C with 0.050 mol dm–3 AgNO3 until no more Ag2CO3 dissolves. Calculate the concentration of carbonate ions, [CO3 2–], in this solution.  [CO3 2–] = … mol dm–3 [1] (iv) An electrochemical cell is set up to measure the electrode potential, E, for the Ag+ / Ag half‑cell using the saturated Ag2CO3(aq) with a standard hydrogen electrode. Use the Data Booklet, your answer to (a)(ii), and the Nernst equation to calculate the electrode potential, E, for this Ag+ / Ag half-cell.  E for Ag+ / Ag half-cell = … V [2]

15 9701/41/M/J/20 © UCLES 2020 [Turn over (b) Silver chloride, AgCl, is sparingly soluble in water. The equation for the enthalpy change of solution is shown. AgCl (s) → Ag+(aq) + Cl –(aq) = +65.5 kJ mol–1 Standard entropies are shown in the table. species AgCl (s) Ag+(aq) Cl –(aq) S o / J K–1 mol–1 +96.2 +72.7 +56.5 (i) Calculate the standard entropy change of solution, ∆S o.  ∆S o = … J K–1 mol–1 [1] (ii) Explain, with the aid of a calculation, why AgCl is insoluble in water at 25 °C. You should use data from this question and your answer to (b)(i). … … [3]  [Total: 10]

16 9701/41/M/J/20 © UCLES 2020 8 (a) Explain what is meant by the term buffer solution. … … … [2] (b) (i) Write an expression for the acid dissociation constant, Ka, for ammonium ions, NH4 +(aq). Ka =  [1] (ii) Write two equations to describe how a solution containing ammonium ions, NH4 +(aq), and ammonia, NH3(aq), can act as a buffer. … … [2] (iii) The numerical value of Ka for NH4 +(aq) is 5.6 × 10–10 at 298 K. A buffer solution was prepared by adding 0.80 dm3 of 0.25 mol dm–3 ammonia, an excess, to 0.20 dm3 of 0.20 mol dm–3 hydrochloric acid. Calculate the pH of the buffer solution formed at 298 K.  pH = … [3]  [Total: 8]

17 9701/41/M/J/20 © UCLES 2020 [Turn over 9 (a) Manganese(IV) oxide, MnO2, catalyses the decomposition of hydrogen peroxide, H2O2, as shown. MnO2 2H2O2(aq) 2H2O(l) + O2(g) The mechanism involves the formation of the intermediate species, Mn2+, in the first step which is subsequently used up in the second step. State and use relevant electrode potentials, E o, to construct two equations to show how MnO2 can catalyse this reaction. … … … … … equation 1 … equation 2 …  [3] (b) The equation for the decomposition of hydrogen peroxide without a catalyst is shown. 2H2O2(aq) → 2H2O(l) + O2(g) Under certain conditions this reaction is found to be first order with respect to hydrogen peroxide, with a rate constant, k, of 2.0 × 10–6 s–1 at 298 K. Calculate the initial rate of decomposition of a 0.75 mol dm–3 hydrogen peroxide solution at 298 K.  initial rate = … mol dm–3 s–1 [1]

18 9701/41/M/J/20 © UCLES 2020 (c) A four-step mechanism is suggested for the reaction between hydrogen peroxide and iodide ions in an acidic solution. step 1 H2O2 + I– → IO– + H2O step 2 H+ + IO– → HIO step 3 HIO + I– → I2 + OH– step 4 OH– + H+ → H2O Step 1 is the rate-determining step. (i) State what is meant by the term rate-determining step. … … [1] (ii) Use this mechanism to construct a balanced equation for this reaction. … [1] (iii) Deduce the order of reaction with respect to each of the following. H2O2 = … I– = … H+ = … [1]  [Total: 7]

19 9701/41/M/J/20 © UCLES 2020 [Turn over 10 (a) The electronic configuration of transition element Q is [Ar] 3d2 4s2. Predict the likely oxidation states of element Q in compounds. … [1] (b) Suggest why transition elements often show variable oxidation states in their compounds, but typical s‑block elements such as calcium do not. … … [1] (c) Many enzymes contain transition element complexes. Describe, with the aid of a suitably labelled diagram, how an enzyme catalyses the breakdown of a substrate molecule. … … … … [3]  [Total: 5]

20 9701/41/M/J/20 © UCLES 2020 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 15 printed pages. © UCLES 2020 [Turn over Cambridge International AS & A Level CHEMISTRY 9701/41 Paper 4 A Level Structured Questions May/June 2020 MARK SCHEME Maximum Mark: 100 Published Students did not sit exam papers in the June 2020 series due to the Covid-19 global pandemic. This mark scheme is published to support teachers and students and should be read together with the question paper. It shows the requirements of the exam. The answer column of the mark scheme shows the proposed basis on which Examiners would award marks for this exam. Where appropriate, this column also provides the most likely acceptable alternative responses expected from students. Examiners usually review the mark scheme after they have seen student responses and update the mark scheme if appropriate. In the June series, Examiners were unable to consider the acceptability of alternative responses, as there were no student responses to consider. Mark schemes should usually be read together with the Principal Examiner Report for Teachers. However, because students did not sit exam papers, there is no Principal Examiner Report for Teachers for the June 2020 series. Cambridge International will not enter into discussions about these mark schemes. Cambridge International is publishing the mark schemes for the June 2020 series for most Cambridge IGCSE™ and Cambridge International A & AS Level components, and some Cambridge O Level components.

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED MAY/JUNE 2020 © UCLES 2020 Page 2 of 15 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.

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED MAY/JUNE 2020 © UCLES 2020 Page 3 of 15 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.

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED MAY/JUNE 2020 © UCLES 2020 Page 4 of 15 Science-Specific Marking Principles 1 Examiners should consider the context and scientific use of any keywords when awarding marks. Although keywords may be present, marks should not be awarded if the keywords are used incorrectly. 2 The examiner should not choose between contradictory statements given in the same question part, and credit should not be awarded for any correct statement that is contradicted within the same question part. Wrong science that is irrelevant to the question should be ignored. 3 Although spellings do not have to be correct, spellings of syllabus terms must allow for clear and unambiguous separation from other syllabus terms with which they may be confused (e.g. ethane / ethene, glucagon / glycogen, refraction / reflection). 4 The error carried forward (ecf) principle should be applied, where appropriate. If an incorrect answer is subsequently used in a scientifically correct way, the candidate should be awarded these subsequent marking points. Further guidance will be included in the mark scheme where necessary and any exceptions to this general principle will be noted. 5 ‘List rule’ guidance (see examples below) For questions that require n responses (e.g. State two reasons …): • The response should be read as continuous prose, even when numbered answer spaces are provided • Any response marked ignore in the mark scheme should not count towards n • Incorrect responses should not be awarded credit but will still count towards n • Read the entire response to check for any responses that contradict those that would otherwise be credited. Credit should not be awarded for any responses that are contradicted within the rest of the response. Where two responses contradict one another, this should be treated as a single incorrect response • Non-contradictory responses after the first n responses may be ignored even if they include incorrect science.

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED MAY/JUNE 2020 © UCLES 2020 Page 5 of 15 6 Calculation specific guidance Correct answers to calculations should be given full credit even if there is no working or incorrect working, unless the question states ‘show your working’. For questions in which the number of significant figures required is not stated, credit should be awarded for correct answers when rounded by the examiner to the number of significant figures given in the mark scheme. This may not apply to measured values. For answers given in standard form, (e.g. a × 10n) in which the convention of restricting the value of the coefficient (a) to a value between 1 and 10 is not followed, credit may still be awarded if the answer can be converted to the answer given in the mark scheme. Unless a separate mark is given for a unit, a missing or incorrect unit will normally mean that the final calculation mark is not awarded. Exceptions to this general principle will be noted in the mark scheme. 7 Guidance for chemical equations Multiples / fractions of coefficients used in chemical equations are acceptable unless stated otherwise in the mark scheme. State symbols given in an equation should be ignored unless asked for in the question or stated otherwise in the mark scheme.

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED MAY/JUNE 2020 © UCLES 2020 Page 6 of 15 Question Answer Marks 1(a)(i) (a molecule or ion) formed by a (central) metal atom / ion surrounded by / bonded to (one or more) ligands 1 1(a)(ii) M1: blue ppt/solid M2: [Co(H2O)6]2+ + 2OH-  Co(OH)2 + 6H2O OR [Co(H2O)6]2+ + 2OH-  [Co(H2O)4(OH)2] + 2H2O M3: precipitation/ acid-base M4: blue solution M5: [Co(H2O)6]2+ + 6NH3  [Co(NH3)6]2+ + 6H2O M6: ligand exchange/displacement/substitution/replacement 6 1(b) • solution turns blue → pink • a white ppt. of AgCl forms • equilibrium shifts to the left / [Cl-] decreases Two correct responses = 1 mark Three correct responses = 2 marks 2 1(c) NH3 Co Cl Cl NH3 NH3 Cl NH3 Co Cl Cl NH3 Cl H3N geometric ALLOW cis-trans Two correct responses = 1 mark Three correct responses = 2 marks 2 1(d)(i) each nitrogen / the four nitrogen’s has a lone pair of electrons (to the metal ion) Two correct responses = 1 mark 1

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED MAY/JUNE 2020 © UCLES 2020 Page 7 of 15 Question Answer Marks 1(d)(ii) [Co(H2O)6]2+ + C6H18N4  [Co(C6H18N4)]2+ + 6H2O OR [Co(H2O)6]2+ + C6H18N4  [Co(C6H18N4)(H2O)2]2+ + 4H2O 1 Question Answer Marks 2(a)(i) M1 solubility increases down the group M2 ΔHlatt and ΔHhyd both become less exothermic / less negative M3 ΔHlatt changes more (than ΔHhyd as OH– being smaller than M2+) M4 ΔHsol becomes more exothermic / more negative 4 2(a)(ii) M1 Mg(OH)2 AND Mg2+ has a smaller ionic radii/ Mg2+ has a higher charge density M2 OH- ion is polarised/distorted more 2 Question Answer Marks 3(a)(i) 6CO2 + 24H+ + 24e-  C6H12O6 + 6H2O ALLOW 6CO2 + 12H+ + 12e-  C6H12O6 + 3O2 for both marks ALLOW one mark for an unbalanced equation showing the correct species of either equation 2

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED MAY/JUNE 2020 © UCLES 2020 Page 8 of 15 Question Answer Marks 3(a)(ii) salt bridge (indicated) voltmeter / V labelled O2 good delivery system H2 good delivery system Pt electrode H+ / HCl / H2SO4 solution labelled (at least once) 1 atm 1 mol dm–3 quoted Every two correct responses = 1 mark 4 3(a)(iii) Eocell = (+) 1.23 V AND positive electrode = O2 half-cell identified 1 Question Answer Marks 4(a) M1 phenylmethanamine / U > phenylamine / T > benzamide / S [1] any two from: • alkyl group is electron donating so lone pair more able to accept a proton • lone pair on N overlaps with delocalised system so less able to accept a proton • presence of electron-withdrawing oxygen / carbonyl group means lone pair is not available to accept a proton OR amides are neutral 3 4(b)(i) reaction 1 LiAlH4 reaction 2 heat NH3 under pressure/ heat NH3 in a sealed tube 2 4(b)(ii) reaction 1 reduction reaction 2 nucleophilic substitution 2

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED MAY/JUNE 2020 © UCLES 2020 Page 9 of 15 Question Answer Marks 5(a)(i) The substitution product is stabilised by delocalisation of (6)π-electrons OR The addition product is not stabilised by delocalisation of (6)π-electrons [1] 1 5(a)(ii) Br Br+ + Br H + H+ [1] intermediate [1] curly arrow + H+ lost [1] • first curly arrow • intermediate • 2nd curly arrow, product and H+ formed / lost 3 5(a)(iii) AlBr4- + H+  AlBr3 + HBr 1 5(b) lone pair of oxygen is delocalised into the ring any one from: • phenol has a higher electron density in the ring • phenol can polarise/induce a dipole in Br2 2 5(c)(i) CH3CH2CH+CH3 (CH3)2CHCH2+ (CH3)3C+ Each correct structure = 1 mark 3

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED MAY/JUNE 2020 © UCLES 2020 Page 10 of 15 Question Answer Marks 5(c)(ii) number of peaks in carbon-13 NMR = 8 number of peaks in carbon-13 NMR = 6 number of peaks in carbon-13 NMR = 7 number of peaks in carbon-13 NMR = 8 Two correct organic products = 1 mark three correct organic products = 2 marks all products linked correctly to NMR = 2 marks 4 Question Answer Marks 6(a) M1 2-chloropropanoic acid > 3-chloropropanoic acid > propanoic acid [1] M2 CH3CHClCO2H / ClCH2CH2CO2H (are more acidic) as they contain an electronegative Cl atom so weaken O-H bond / stabilise carboxylate anion [1] M3 CH3CHClCO2H (is more acidic than ClCH2CH2CO2H) as the Cl atom is closer to CO2H so weaken O-H bond more / stabilise carboxylate anion more [1] 3

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED MAY/JUNE 2020 © UCLES 2020 Page 11 of 15 Question Answer Marks 6(b)(i) Keq = 4.07x10-3/1.78x10-4 = 22.9 1 6(b)(ii) 3rd box ticked [to the right] AND as the Keq is greater than one ecf on Keq 1 6(b)(iii) pKa 1.23 HO2CCO2H + H2O ⇌ HO2CCO2- + H3O+ OR HO2CCO2H ⇌ HO2CCO2- + H+ pKa 4.19 HO2CCO2- + H2O ⇌ -O2CCO2- + H3O+ OR HO2CCO2- ⇌ -O2CCO2- + H+ 2 6(b)(iv) pKa = – logKa 1 6(c) reagents and conditions observed change test 1 M1 Tollen’s reagent, warm OR Fehling’s solution, warm silver mirror (brick)-red ppt. test 2 M2 aqueous alkaline iodine OR 2,4-DNPH yellow ppt. orange ppt. test 3 M3 acidified MnO4- , warm decolourises (and bubbles) Two correct observations = 1 mark Three correct observations = 2 marks 5

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED MAY/JUNE 2020 © UCLES 2020 Page 12 of 15 Question Answer Marks 6(d) chemical shift (δ) environment of the carbon atom hybridisation of the carbon atom 27 CH3 circled sp3 163 COOH circled sp2 192 C=O(COOH) circled sp2 Award one mark for each correct column 2 6(e) chemical shift (δ) group responsible for the peak splitting pattern number of 1H atoms responsible for the peak 1.3 alkane / CH / CH3 triplet 3 2.2 CH3CO or alkyl / CH next to C=O singlet 3 4.0 CH2O or alkyl / CH next to electronegative atom / C=O quartet / quadruplet 2 Award one mark for every three correct responses. 3 6(f) CH AND CH3 circled these protons do not exchange with D2O OR OH and NH protons exchange with D2O 2 6(g)(i) Kw = [D+][DO-] 1 6(g)(ii) M1 [D+] = √1.35 x 10-15 = 3.67 x10-8 M2 pH = -log [D+] = 7.4(3) min 2sf 2

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED MAY/JUNE 2020 © UCLES 2020 Page 13 of 15 Question Answer Marks 7(a)(i) M1 Ksp = [Ag+]2[CO32-] M2 units = mol3 dm-9 2 7(a)(ii) x = 3√6.3 x 10-12/4 = 1.16 x 10-4 (mol dm-3) [Ag+] = 1.16 x 10-4 x 2 = 2.33 x 10-4 (mol dm-3) min 2sf 1 7(a)(iii) 6.3 x 10-12 = [0.05]2[CO32-] [CO32-] = 2.52 x 10-9 (mol dm-3) min 2sf 1 7(a)(iv) M1 E=Eo + 0.059log[Ag+] M2 E=0.80 + 0.059log(1.2 x 10-4) = 0.57 V ecf from (a)(ii) min 2sf 2 7(b)(i) ΔSo = 72.7 + 56.5 -96.2 = +33.0 J K-1 mol-1 1 7(b)(ii) M1 ΔG = ΔHo - TΔSo M2 ΔG = (65.5)- (298x0.033) = +55.7 kJ mol-1 min 3sf M3 ΔG = positive so not feasible/spontaneous 3 Question Answer Marks 8(a) M1 a solution that resists changes in pH M2 when small amounts of acid and alkali are added to it 2 8(b)(i) Ka = [NH3][H+] [NH4+] 1 8(b)(ii) M1 NH4+ + OH-  NH3 + H2O M2 NH3 + H3O+  NH4+ + H2O 2

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED MAY/JUNE 2020 © UCLES 2020 Page 14 of 15 Question Answer Marks 8(b)(iii) M1 moles NH3(initial) = 0.25 x 0.80 = 0.200 AND moles HCl = 0.20 x 0.20 = 0.040 (= moles NH4+eqm) M2 moles NH3(eqm) = 0.20 – 0.04 = 0.160 [H+] = (5.6 x10-10 x 0.04)/(0.16) = 1.4 x 10-10 (mol dm-3) ecf on M1 M3 pH = -log(1.4 x 10-10) = 9.85 ecf on M2 min 2sf 3 Question Answer Marks 9(a) M1 data seen H2O2/H2O +1.77V and MnO2/Mn2+ +1.23 V and O2/H2O2 +0.68 V OR Ecell= 0.55 V (first step) and 0.54 V (second step) M2 MnO2 + H2O2 + 2H+  Mn2+ + O2 + 2H2O M3 Mn2+ + H2O2  MnO2 + 2H+ 3 9(b) rate = 2.0 x 10-6 x 0.75 = 1.5 x 10-6 1 9(c)(i) slowest step in overall reaction 1 9(c)(ii) H2O2 + 2H+ + 2I-  I2 + 2H2O OR H2O2 + 2HI  I2 + 2H2O 1 9(c)(iii) H2O2 = 1 AND I- = 1 AND H+ = 0 1 Question Answer Marks 10(a) +4 and any of +1, +2, +3 1 10(b) close similarity of energy of the 4s and 3d sub-shells 1

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED MAY/JUNE 2020 © UCLES 2020 Page 15 of 15 Question Answer Marks 10(c) diagrams M1 (can be in words or diagram) substrate shape is complementary to active site M2 (can be in words or diagram) the substrate bind / bonds / fits into the active site M3 (can be in words or diagram) products are released M4 (words) lower EA / bonds weakened (in substrate) Any three points 3