Cambridge A Level Chemistry 9701 — 2019 May/June Paper 4 · Variant 1

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 9701/41 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 20 printed pages. [Turn over IB19 06_9701_41/3RP © UCLES 2019 *2557393576*

2 9701/41/M/J/19 © UCLES 2019 Answer all the questions in the spaces provided. 1 (a) Aqueous solutions of copper(II) salts contain the blue-coloured [Cu(H2O)6]2+ complex ion. Separate portions of this blue solution react with aqueous sodium hydroxide and with concentrated hydrochloric acid. Give the following information for each of these reactions. ● ● reaction with aqueous sodium hydroxide ionic equation … type of reaction … colour and state of the copper-containing product … ● ● reaction with concentrated hydrochloric acid ionic equation … type of reaction … colour and state of the copper-containing product …  [6] (b) Chloride ions can be identified using aqueous silver nitrate, AgNO3(aq). Ag+(aq) + Cl –(aq) AgCl (s) 0.303 g of a chloride of sulfur is completely hydrolysed with water. All the chlorine atoms present in the chloride of sulfur are converted into chloride ions. The solution is diluted to 100.0 cm3. A 25.00 cm3 sample of this solution is titrated with 0.0500 mol dm–3 AgNO3(aq). The titration requires 22.40 cm3 of 0.0500 mol dm–3 AgNO3(aq). Calculate the empirical formula of the chloride of sulfur. Show all your working.  empirical formula of chloride of sulfur = … [3]  [Total: 9]

3 9701/41/M/J/19 © UCLES 2019 [Turn over 2 (a) Group 2 nitrates decompose when heated. Write an equation for the decomposition of strontium nitrate. … [1] (b) Describe and explain how the thermal stability of Group 2 nitrates changes with increasing atomic number. … … … … … [3] (c) The variation in the thermal stability of Group 2 amides is similar to that of Group 2 nitrates. (i) Suggest whether calcium amide, Ca(NH2)2, will decompose more or less readily than barium amide, Ba(NH2)2. Explain your answer. … … [1] (ii) Ba(NH2)2 decomposes when heated to form barium nitride, Ba3N2, and ammonia as the only products. Write an equation for this reaction. … [1] (d) Ba(NH2)2 contains the NH2 – ion. Predict the bond angle of NH2 –. Explain your answer using the qualitative model of electron‑pair repulsion. bond angle … explanation … … …  [3]  [Total: 9]

4 9701/41/M/J/19 © UCLES 2019 3 Chlorate(V) ions are powerful oxidising agents. (a) The reduction of chlorate(V) ions, Cl O3 –, with SO2 forms chlorine dioxide, Cl O2, and sulfate ions, SO4 2–, as the only products. Construct an equation for this reaction. … [1] (b) (i) Chlorine dioxide, Cl O2, disproportionates with hydroxide ions, OH–(aq), to form a mixture of Cl O2 – and Cl O3 – ions. 2Cl O2 + 2OH– Cl O2 – + Cl O3 – + H2O Explain, using this reaction as an example, what is meant by disproportionation. … … [1] (ii) Deduce the ionic half-equations for the reaction in (b)(i). … … [2] (c) A lithium-iodine electrochemical cell can be used to generate electricity for a heart pacemaker. The cell consists of a lithium electrode and an inert electrode immersed in body fluids. When current flows lithium is oxidised and iodine is reduced. (i) Use the Data Booklet to write half-equations for the reactions taking place at the two electrodes. Hence write the overall equation for when a current flows. ● … ● … overall equation …  [2] (ii) Use the Data Booklet to calculate the for this cell.  = … V [1]

5 9701/41/M/J/19 © UCLES 2019 [Turn over (iii) A current of 2.5 × 10–5 A is drawn from this cell. Calculate the time taken for 0.10 g of lithium electrode to be used up. Assume the current remains constant throughout this period.  time = … s [3]  [Total: 10]

6 9701/41/M/J/19 © UCLES 2019 4 (a) Sketches of the shapes of some atomic orbitals are shown. Identify the type of orbital, s, p, or d. x y z x y z x y z shape of orbital type of orbital  [1] (b) Cadmium forms the two ions, Cd2 2+ and Cd2+. The electronic configuration of cadmium in these ions is shown. ● ● [Kr] 4d105s1 ● ● [Kr] 4d10 Use this information to explain why cadmium is not a transition element. … … [1] (c) Methylamine, CH3NH2, is a monodentate ligand. (i) State what is meant by the term monodentate in this context. … … [1]

7 9701/41/M/J/19 © UCLES 2019 [Turn over In the presence of aqueous methylamine, [Cd(H2O)6]2+ reacts to form a mixture of two isomeric octahedral complexes. equilibrium 1 [Cd(H2O)6]2+ + 4CH3NH2 [Cd(CH3NH2)4(H2O)2]2+ + 4H2O = –57 kJ mol–1 (ii) Complete the three-dimensional diagrams to show the isomers of [Cd(CH3NH2)4(H2O)2]2+. Use L to represent CH3NH2 in your diagrams. Cd Cd  [2] (d) (i) State what is meant by the term stability constant. … … [1] (ii) Complete the table by placing one tick () in each row to suggest how increasing temperature will affect Kstab and the equilibrium concentration of the cadmium complex, [[Cd(CH3NH2)4(H2O)2]2+], for equilibrium 1. Explain your answer. decreases no change increases Kstab [[Cd(CH3NH2)4(H2O)2]2+] explanation … …  [2]

8 9701/41/M/J/19 © UCLES 2019 EDTA4– is a polydentate ligand. When a solution of EDTA4– is added to [Cd(H2O)6]2+ a new complex [CdEDTA]2– is formed. The values for the stability constants for two Cd2+ complexes are shown. Kstab [Cd(CH3NH2)4(H2O)2]2+ 4.0 × 106 [CdEDTA]2– 4.0 × 1016 (iii) A solution containing equal numbers of moles of CH3NH2 and EDTA is added to [Cd(H2O)6]2+. Predict which complex is formed in the larger amount. Explain your answer. … … [1] (e) Methylamine is a Brønsted-Lowry base. Write an equation showing how methylamine dissolves in water to give an alkaline solution. … [1] (f) Methylamine is a useful reagent in organic chemistry. (i) Write an equation for the reaction of ethanoyl chloride with methylamine. … [2] (ii) Methylamine also reacts with propanone to form compound P as shown. H3C H3C C O H2N CH3 CH3 + H2O + H3C H3C C N P Deduce the type of reaction shown here. … [1]  [Total: 13]

9 9701/41/M/J/19 © UCLES 2019 [Turn over 5 (a) Chlorate(I) ions undergo the following reaction under aqueous conditions. 2NH3 + Cl O– N2H4 + Cl – + H2O A series of experiments was carried out at different concentrations of Cl O– and NH3. The table shows the results obtained. experiment [Cl O–] / mol dm–3 [NH3] / mol dm–3 initial rate / mol dm–3 s–1 1 0.200 0.100 0.256 2 0.400 0.200 2.05 3 0.400 0.400 8.20 (i) Use the data in the table to determine the order with respect to each reactant, Cl O– and NH3. Show your reasoning. … … … … … … [2] (ii) Write the rate equation for this reaction. rate = … [1] (iii) Use the results of experiment 1 to calculate the rate constant, k, for this reaction. Include the units of k.  k = …  units = …  [2]

10 9701/41/M/J/19 © UCLES 2019 (iv) On the axes sketch a graph to show how the value of k changes as temperature is increased. rate constant, k temperature  [1] (b) In another experiment, the reaction between chlorate(I) ions and iodide ions in aqueous alkali was investigated. A solution of iodide ions in aqueous alkali was added to a large excess of chlorate(I) ions and [I–] was measured at regular intervals. (i) Describe how the results of this experiment can be used to confirm that the reaction is first-order with respect to [I–]. … … … [2] A three-step mechanism for this reaction is shown. step 1 Cl O– + H2O HCl O + OH– step 2 I– + HCl O HIO + Cl – step 3 HIO + OH– H2O + IO– (ii) Use this mechanism to deduce the overall equation for this reaction. … [1] (iii) Identify a step that involves a redox reaction. Explain your answer. … … [1]  [Total: 10]

11 9701/41/M/J/19 © UCLES 2019 [Turn over 6 (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 bond energy enthalpy change of formation  [1] (b) Explain what is meant by the term enthalpy change of atomisation. … … [1] (c) The overall reaction for the atomisation of liquid bromine molecules, Br2(l), is shown. Br2(l) 2Br(g) This happens via a two-step process. ● ● Construct a labelled energy cycle to represent this atomisation process, including state symbols. ● ● Use your cycle and relevant data from the Data Booklet to calculate the enthalpy change of vaporisation of Br2(l), . The enthalpy change of atomisation of bromine, ∆Hat, = +112 kJ mol–1.  = … kJ mol–1 [3] (d) Suggest how the of iodine, I2(l), would compare to that of bromine, Br2(l). Explain your answer. … … [1]

12 9701/41/M/J/19 © UCLES 2019 (e) (i) Explain what is meant by the term enthalpy change of hydration. … … [1] (ii) Suggest why the enthalpy change of hydration of Br –(g) is more exothermic than that of I–(g). … … … [2]  [Total: 9]

13 9701/41/M/J/19 © UCLES 2019 [Turn over 7 (a) Benzene can be converted into cyclohexane. (i) For this reaction name the type of reaction and identify the reagent and conditions needed. type of reaction … reagent and conditions …  [2] (ii) State the bond angles in benzene and cyclohexane. bond angle in benzene … bond angle in cyclohexane … Explain your answers. … … …  [2] (b) When benzene reacts with SO3, benzenesulfonic acid is produced. SO3H benzenesulfonic acid + SO3 The mechanism of this reaction is similar to that of the nitration of benzene. Concentrated H2SO4 is used in an initial step to generate the SO3H+ electrophile as shown. SO3 + H2SO4 SO3H+ + HSO4 – (i) Draw a mechanism for the reaction of benzene with SO3H+ ions. Include all necessary curly arrows and charges. SO3H benzenesulfonic acid SO3H+  [3] (ii) Write an equation to show how the H2SO4 catalyst is reformed. … [1]

14 9701/41/M/J/19 © UCLES 2019 (c) 3-dodecylbenzenesulfonic acid can be prepared from benzenesulfonic acid. 3-dodecylbenzenesulfonic acid SO3H SO3H C12H25 benzenesulfonic acid Suggest the reagents and conditions and name the mechanism for this reaction. reagents and conditions … mechanism …  [2] (d) When concentrated sulfuric acid is added to water, dissociation takes place in two stages. stage 1 H2SO4 H+ + HSO4 – stage 2 HSO4 – H+ + SO4 2– Ka2 = 1.0 × 10–2 mol dm–3 Ka2 is the acid dissociation constant for stage 2. (i) Write the expression for the acid dissociation constant Ka2. Ka2 =  [1] (ii) H2SO4 is considered a strong acid whereas HSO4 – is considered a weak acid. Suggest how the magnitude of the acid dissociation constant for stage 1 compares to Ka2. … [1] (e) Benzoic acid, C6H5CO2H, is a weak acid. A solution of 0.0250 mol dm–3 benzoic acid has a pH of 2.90. Calculate the Ka of benzoic acid.  Ka = … mol dm–3  [2]  [Total: 14]

15 9701/41/M/J/19 © UCLES 2019 [Turn over 8 (a) The mass spectrum of compound X, C5H10O2, is recorded. The peak heights of the M and M+1 peaks are 22.65 and 1.25 respectively. (i) Use these data to show that there are five carbon atoms present in one molecule of X. Show your working.  [1] (ii) The mass spectrum has a peak at m/e = 57. Complete the equation to show the fragmentation of X to produce this peak. [C5H10O2]+ … + …  [2] (b) State the use of TMS and CDCl 3 in NMR spectroscopy. ● TMS … ● CDCl 3 …  [1]

16 9701/41/M/J/19 © UCLES 2019 (c) The proton NMR spectrum of compound X, C5H10O2, is shown. 5 4 3 δ / ppm 2 1 0 2 2 3 3 (i) By considering both the relative peak areas and their δ values, use the Data Booklet to ● deduce the part of the molecule that produces the peak at δ 2.2, … ● deduce the part of the molecule that produces the peaks at δ 1.2 and δ 3.5, … ● deduce the part of the molecule that produces the peak at δ 4.0. …  [3] (ii) When reacted with aqueous alkaline iodine, X produces a yellow precipitate. Use this information and your answers to (c)(i) to suggest a structure for X.  [1]

17 9701/41/M/J/19 © UCLES 2019 [Turn over (d) Compound W is an ester with the molecular formula C5H10O2. The proton NMR spectrum of W contains only two peaks. The relative areas of these two peaks are in the ratio 9 : 1. Suggest a structure for this ester, W.  [1] (e) Compound V is a carboxylic acid which contains a chiral centre. It also has the molecular formula C5H10O2. (i) Explain what is meant by the term chiral centre. … … [1] (ii) Suggest a structure for V.  [1]  [Total: 11]

18 9701/41/M/J/19 © UCLES 2019 9 (a) Organochlorine compounds can undergo hydrolysis. R–Cl + H2O R–OH + HCl State and explain the relative rates of hydrolysis of the following compounds. CH3CH2Cl CH3COCl C6H5Cl … … … … … [3] (b) Epibatidine is a naturally occurring organochlorine compound. Cl N epibatidine N H (i) Epibatidine is a weak base. State what is meant by the term weak base. … … [1] A molecule of epibatidine contains two nitrogen atoms, both of which can act as a base. (ii) Epibatidine reacts with HCl (aq). Complete the structure to suggest the product formed in this reaction. Cl N N H  [1]

19 9701/41/M/J/19 © UCLES 2019 [Turn over (c) Polyamides, such as nylon-6, can be prepared from a monomer that contains both an amine and an acyl chloride functional group. nylon-6 monomer H2N (CH2)5 C O Cl (i) When the nylon-6 monomer is hydrolysed, bonds are broken and formed. By considering the two steps in the mechanism of the reaction, complete the table by placing one tick () in each row to indicate the types of bonds broken and formed during the mechanism. σ bonds only π bonds only both σ and π bonds bonds broken bonds formed  [1] (ii) Draw two repeat units of nylon-6. The amide bond should be shown fully displayed.  [2] (d) An addition polymer made from two different alkene monomers is called a co-polymer. A section of a polyalkene co-polymer is shown. C CH3 Cl C H H H C C2H5 C H C CH3 CH3 Cl C H H Draw the structure of the two alkene monomers which produce this co-polymer.  [2]

20 9701/41/M/J/19 © UCLES 2019 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. (e) Explain why polyamides normally biodegrade more readily than polyalkenes. … … [1] (f) The alkene phenylethene can be prepared from benzene in three steps. step 1 step 3 step 2 O H phenylethene (i) Deduce the identity of compound H and draw its structure in the box. [1] (ii) Suggest reagents and conditions for each of the steps 1–3. step 1 … step 2 … step 3 …  [3]  [Total: 15]

This document consists of 14 printed pages. © UCLES 2019 [Turn over Cambridge Assessment International Education Cambridge International Advanced Subsidiary and Advanced Level CHEMISTRY 9701/41 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.

9701/41 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 2 of 14 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 2019 © UCLES 2019 Page 3 of 14 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 2019 © UCLES 2019 Page 4 of 14 Question Answer Marks 1(a) M1 [Cu(H2O)6]2+ + 2OH– → Cu(OH)2 + 6H2O M2 precipitation M3 blue precipitate M4 [Cu(H2O)6]2+ + 4Cl – → CuCl42– + 6H2O M5 ligand exchange / displacement / substitution / replacement M6 yellow solution 6 1(b) M1 amount of Ag+ = 0.050 × 0.0224 = 1.12 × 10–3 mol (in 25 cm3) amount of Ag+ = 1.12 × 10–3 × 4 = 4.48 × 10–3 mol (in 100 cm3) M2 amount of Cl – = 4.48 × 10–3 mol (in 100 cm3) mass of Cl – = 4.48 × 10–3 × 35.5 = 0.159 g (in 100 cm3) mass of S = 0.303 – 0.159 = 0.144 g (in 100 cm3) ecf M3 moles of S = 0.144 / 32.1 = 4.49 × 10–3 molar ratio S : Cl 1:1 → SCl ecf 3 Question Answer Marks 2(a) Sr(NO3)2 → SrO + 2NO2 + ½O2 1 2(b) M1 increases M2 cationic radius / ion size increases (down the group) M3 less polarisation/distortion of anion / nitrate ion / NO3– / nitrate group 3 2(c)(i) more readily and Ca2+ has a smaller ionic radius or more readily and Ca2+ has a greater charge density 1

9701/41 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 5 of 14 Question Answer Marks 2(c)(ii) 3Ba(NH2)2 → Ba3N2 + 4NH3 1 2(d) M1 bond angle 104–105° M2 explanation two lone pairs and two bonding pairs M3 lone pairs repel more 3 Question Answer Marks 3(a) 2ClO3– + SO2 → 2ClO2 + SO42– 1 3(b)(i) Cl in ClO2 gets both oxidised and reduced or Cl goes from +4 → +5 and +4 → +3 1 3(b)(ii) M1 ClO2 + 2OH– → ClO3– + H2O + e– M2 ClO2 + e– → ClO2– 2 3(c)(i) M1 Li → Li+ + e– and I2 + 2e– → 2I– M2 2Li + I2 → 2Li+ + 2I– 2 3(c)(ii) Eocell = 0.54 –(–3.04) = +3.58 V [1] 1 3(c)(iii) M1 amount of Li = 0.10 / 6.9 = 1.45 × 10–2 mol [1] M2 Q needed = 96500 × 1.45 × 10–2 = 1399 (1398.55) C [1] ecf M3 t = 1399 / (2.5 × 10–5) = 5.6 × 107 s [1] ecf 2sf min 3

9701/41 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 6 of 14 Question Answer Marks 4(a) All shapes required for mark p s d 1 4(b) both cadmium ions have full d subshells 1 4(c)(i) donates one lone pair to the central metal ion 1 4(c)(ii) M1 one 3D diagram of [Cd(CH3NH2)4(H2O)2]2+ M2 cis and trans structures 2 4(d)(i) equilibrium constant for the formation of a complex ion in solution / solvent [1] 1

9701/41 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 7 of 14 Question Answer Marks 4(d)(ii) decreases no change increases Kstab 9 [[Cd(CH3NH2)4(H2O)2]2+] 9 M1 both ticks correct [1] M2 equilibrium moves to the left as the (forward) reaction is exothermic [1] 2 4(d)(iii) [CdEDTA]2– and larger Kstab value 1 4(e) CH3NH2 + H2O ⇌ CH3NH3+ + OH– 1 4(f)(i) CH3COCl + CH3NH2 → CH3CONHCH3 + HCl M1 Correct formulae of CH3COCl or CH3CONHCH3 M2 rest of the equation 2 4(f)(ii) condensation or addition-elimination 1 Question Answer Marks 5(a)(i) M1: using expt 2 and 3, [NH3] × 2, rate × 4 so order with respect to [NH3] = 2 M2: using expt 1 and 2, [ClO–] × 2 and [NH3] × 2, as rate × 8 (=22 * x) so order with respect to [ClO–] = 1 2 5(a)(ii) rate = k[NH3]2[ClO–] 1 5(a)(iii) M1: k = 0.256 / (0.200 × 0.1002) k = 128 M2: Units dm6 mol–2 s–1 2

9701/41 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 8 of 14 Question Answer Marks 5(a)(iv) curve / line showing k increasing as temperature increases 1 5(b)(i) M1: plot a graph of [I–] against time M2: constant half-lives 2 5(b)(ii) ClO– + I– → IO– + Cl – 1 5(b)(iii) step 2 and Cl is reduced / oxid no. decreases / oxid no. +1 → –1 or step 2 and I is oxidised / oxid no. increases / oxid no. –1 → +1 1

9701/41 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 9 of 14 Question Answer Marks 6(a) energy change always positive always negative either negative or positive bond energy 9 enthalpy of formation 9 both ticks correct 1 6(b) (energy change) when 1 mole of gaseous atoms are formed (from an element in its standard state) 1 6(c) M1: correct cycle: formulae and state symbols M2: use of 1 × 193 and 2 × (112) M3: for the correct sum and answer ecf from M2 ∆Hovap (= (2 × 112) – (193) ) = +31 kJ mol–1 [scores M2 and M3] 3 6(d) more endothermic and greater Van der Waals / London / induced dipole-dipole forces both 1 6(e)(i) (energy change) when 1 mole of gaseous ions is dissolved in (an excess of) water 1 6(e)(ii) M1: Br- has a smaller ionic radii M2: stronger (ion-dipole) attractions with water molecules 2

9701/41 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 10 of 14 Question Answer Marks 7(a)(i) M1: reduction / hydrogenation M2: H2 + Ni / Pt catalyst 2 7(a)(ii) M1: benzene (120°) and cyclohexane (109.5°) M2: as π-bonds are transformed into σ-bonds 2 7(b)(i) M1: first curly arrow to the sulfur atom M2: intermediate shown M3: 2nd curly arrow and H+ formed / lost 3 7(b)(ii) HSO4– + H+ → H2SO4 1 7(c) M1: C12H25Br and halogen carrier e.g. Al Br3 (+ heat) M2: electrophilic substitution 2 7(d)(i) + − − = 2 4 a2 4 [H ][SO ] K [HSO ] 1 7(d)(ii) Ka of H2SO4 is larger than Ka2 1

9701/41 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 11 of 14 Question Answer Marks 7(e) M1: [H+] = 10–2.90 = 1.26 × 10–3 M2: Ka = [1.26 × 10–3]2 / 0.025 = 6.3 × 10–5 (mol dm–3) 2

9701/41 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 12 of 14 Question Answer Marks 8(a)(i) no. of carbons = 100 × 1.25 / (22.65 × 1.1) (= 5.02) 1 8(a)(ii) M1: C2H5O M2: C3H5O+ (positive sign required for m / e = 57 fragment) 2 8(b) TMS: Reference CDCl3: Solvent 1 8(c)(i) M1: CH3CO M2: CH3CH2O M3: (CO)CH2O 3 8(c)(ii) CH3COCH2OCH2CH3 1 8(d) HCO2C(CH3)3 1 8(e)(i) this is a (carbon) atom which has four different atoms or groups attached to it 1 8(e)(ii) CH3CH2CH(CH3)COOH 1 Question Answer Marks 9(a) M1: CH3COCl > CH3CH2Cl > C6H5Cl M2 & M3 any two from: • in C6H5Cl (no hydrolysis) C-Cl bond is part of delocalised system OR p-orbital on Cl overlaps with π system OR electrons from Cl overlap with π system • CH3COCl carbon in C-Cl bond is more electron deficient since it is also attached to an oxygen atom (ora) or C-Cl bond strength is weakest in CH3COCl (ora) • CH3CH2Cl carbon in C-Cl bond strengthened by positive inductive effect of alkyl group 3

9701/41 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 13 of 14 Question Answer Marks 9(b)(i) partially ionised and proton acceptor 1 9(b)(ii) or 1 9(c)(i) σ-bonds only π-bonds only both σ- and π-bonds bonds broken 9 bonds formed 9 Both ticks correct 1 9(c)(ii) M1: amide link M2: rest of the structure 2 9(d) or CH3CCl=CH2 and C2H5CH=CHCH3 each correct structure scores one mark 2

9701/41 Cambridge International AS/A Level – Mark Scheme PUBLISHED May/June 2019 © UCLES 2019 Page 14 of 14 Question Answer Marks 9(e) C-C bonds are non-polar / polyalkenes cannot be hydrolysed and polyamides can be broken down by hydrolysis 1 9(f)(i) 1 9(f)(ii) M1: step 1: CH3COCl + AlCl3 [1] M2: step 2: NaBH4 / LiAlH4 [1] M3: step 3: conc. H2SO4, heat [1] 3