Cambridge A Level Chemistry 9701 — 2025 May/June Paper 4 · Variant 2

This document has 24 pages. Any blank pages are indicated. [Turn over Cambridge International AS & A Level CHEMISTRY 9701/42 Paper 4 A Level Structured Questions May/June 2025 2 hours You must answer on the question paper. No additional materials are needed. 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 and use appropriate units. INFORMATION ● The total mark for this paper is 100. ● The number of marks for each question or part question is shown in brackets [ ]. ● The Periodic Table is printed in the question paper. ● Important values, constants and standards are printed in the question paper. * 5 3 7 9 1 7 3 5 5 4 * DC (SL/SW) 342340/3 © UCLES 2025 ,










, * 0000800000001 * ¬OŠ> 4mHuOªEŠ`|6€W ¬=srUŸv¤_pwfI¡n¤¥‚ ¥e••u• 5eEEEE5e5uU

2 9701/42/M/J/25 © UCLES 2025 1 (a) (i) Calcium nitrate, Ca(NO3)2, decomposes on heating. Write an equation for the decomposition of calcium nitrate. … [1] (ii) Describe the trend in the decomposition temperature of the Group 2 nitrates. Explain your answer. … … … … … [3] (b) A sample of 0.333 g of strontium oxide, SrO, is completely dissolved in distilled water to form a solution of strontium hydroxide, Sr(OH)2. The resulting solution is added to a volumetric flask and made up to 250.0 cm3 with distilled water. Calculate the pH of this solution at 298 K. Give your answer to two decimal places. pH = … [4] [Total: 8] * 0000800000002 * ,










, ĬÍĊ¾Ġ´íÈõÏĪÅĊÝú¸þ× Ĭ½ññØīòĆØāāÉñě¾ÜĝĂ ĥõąĕµõÅÕĕąĕąÅõÅõÕÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

3 9701/42/M/J/25 © UCLES 2025 [Turn over 2 (a) When cobalt(II) sulfate, CoSO4, is dissolved in distilled water, solution A is formed. The reaction scheme in Fig. 2.1 shows some reactions of solution A. solution A solution C solid B NaOH(aq) excess NH3(aq) [CoCl4]2– Fig. 2.1 (i) Complete Table 2.1 to show the formula and colour of each of the cobalt-containing species present in A, B and C. Identify the type of reaction forming each of B and C. Table 2.1 formula of cobalt-containing species colour of cobalt-containing species type of reaction A B C [4] (ii) Suggest a suitable reagent for the formation of [CoCl4]2– from solution A. … [1] (b) The complex ion [CoCl4]2– has tetrahedral geometry. The 3d orbitals in an isolated Co2+ ion are degenerate. (i) Complete Fig. 2.2 to show the relative energies of the 3d orbitals in an isolated Co2+ ion and in Co2+ in a tetrahedral complex. energy isolated Co2+ ion Co2+ in a tetrahedral complex Fig. 2.2 [2] * 0000800000003 * ,










, ĬÏĊ¾Ġ´íÈõÏĪÅĊÝü¸þ× Ĭ½òòÐĝîöá÷ðĀåģĚÜčĂ ĥõõÕõĕåµąõÅąÅĕåµÅÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

4 9701/42/M/J/25 © UCLES 2025 (ii) Draw a three-dimensional diagram to show the structure of the complex ion [CoCl4]2–. [1] (c) H2O2 can act as an oxidising agent or a reducing agent when reacting with species that contain manganese. Table 2.2 shows electrode potentials, E o, for some electrode reactions. Table 2.2 electrode reaction E o / V MnO2 + 2H2O + 2e– Mn(OH)2 + 2OH– –0.04 MnO2 + 4H+ + 2e– Mn2+ + 2H2O +1.22 H2O2 + 2H+ + 2e– 2H2O +1.78 H2O2 + OH– + 2e– 3OH– +0.88 O2 + 2H+ + 2e– H2O2 +0.68 Use only the species listed in Table 2.2 to suggest: • one reaction in which H2O2 acts as an oxidising agent and • one reaction in which H2O2 acts as a reducing agent. Include the value of the standard cell potential, E o cell , and an overall equation for each reaction. H2O2 acting as an oxidising agent … … … … H2O2 acting as a reducing agent … … … … [4] * 0000800000004 * ,










, ĬÍĊ¾Ġ´íÈõÏĪÅĊßú¸Ā× Ĭ½òóÐħĀóÖù÷ćć¿¼ÌĥĂ ĥÅåÕµĕåĕĥÕµąąĕąµÕÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

5 9701/42/M/J/25 © UCLES 2025 [Turn over (d) Acidified manganate(VII) ions, MnO4 –, can be used to analyse the content of iron tablets by titration. Two identical iron tablets are crushed and dissolved in distilled water. The resulting solution is made up to 150.0 cm3 with distilled water. 25.0 cm3 of this solution requires 18.60 cm3 of 0.0500 mol dm–3 acidified MnO4 – to reach the end-point. All the Fe2+ ions are oxidised. The relevant half-equations are shown. MnO4 – + 8H+ + 5e– Mn2+ + 4H2O Fe2+ Fe3+ + e– (i) Describe the colour change observed at the end-point of this titration. from … to … [1] (ii) Calculate the mass, in mg, of iron in one tablet. Assume that all the iron in the tablets is Fe2+. Show your working. mass of iron in one tablet = … mg [4] [Total: 17] * 0000800000005 * ,










, ĬÏĊ¾Ġ´íÈõÏĪÅĊßü¸Ā× Ĭ½ñôØġĄăãÿĊÂÓ·ĠÌĕĂ ĥÅÕĕõõÅõõåĥąąõĥõÅÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

6 9701/42/M/J/25 © UCLES 2025 3 (a) Define entropy. … … [1] (b) Fig. 3.1 shows how the entropy, S, of a pure substance changes with temperature, T. temperature, T T1 T2 entropy, S Fig. 3.1 (i) Identify the process occurring at each of the temperatures T1 and T2. T1 … T2 … [1] (ii) Explain why the entropy change, ∆S, at T2 is bigger than the entropy change at T1. … … … [1] * 0000800000006 * ,










, ĬÑĊ¾Ġ´íÈõÏĪÅĊÞüµþ× Ĭ½òôÛīĬĉëČû»³¿ýĤĥĂ ĥĕõĕõÕÅÕµÅÅąÅõŵĕÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

7 9701/42/M/J/25 © UCLES 2025 [Turn over (c) The equation for the reduction of iron(III) oxide by carbon monoxide at 450 °C is shown. Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g) ∆G = –36.2 kJ mol–1 Table 3.1 shows the enthalpy of formation, ∆H f , and the entropy, S , for some substances. Table 3.1 Fe2O3(s) CO(g) Fe(s) CO2(g) ∆H f / kJ mol–1 –824.2 –110.5 0.0 –393.5 S / J K–1 mol–1 87.4 to be calculated 27.3 213.8 Use the data in Table 3.1 to calculate the entropy, S , of carbon monoxide at 450 °C. Show your working. S of CO(g) = … J K–1 mol–1 [3] (d) Iron(II) oxide can also be reduced to iron by carbon monoxide, as shown. FeO(s) + CO(g) Fe(s) + CO2(g) ∆H = –11.1 kJ mol–1 ∆S = –15.2 J K–1 mol–1 State the effect of increasing temperature on the feasibility of this reaction. Explain your answer. … … … … [2] [Total: 8] * 0000800000007 * ,










, ĬÓĊ¾Ġ´íÈõÏĪÅĊÞúµþ× Ĭ½ñóÓĝĨùÎîĆîħ·ÙĤĕĂ ĥĕąÕµµåµåµĕąÅĕåõąÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

8 9701/42/M/J/25 © UCLES 2025 4 (a) In aqueous solution, iodide ions react with acidified hydrogen peroxide, as shown in reaction 1. reaction 1 2I– + H2O2 + 2H+ I2 + 2H2O The rate equation for reaction 1 is shown. rate = k [I–][H2O2] (i) Explain what is meant by order of reaction. … … … [1] (ii) Complete Table 4.1. Table 4.1 the order of reaction with respect to [H+] the order of reaction with respect to [I–] the order of reaction with respect to [H2O2] overall order of the reaction [2] (iii) Sketch a line on Fig. 4.1 to show the relationship between [I–] and time. time 0 [I–] Fig. 4.1 [1] * 0000800000008 * ,










, ĬÑĊ¾Ġ´íÈõÏĪÅĊàüµĀ× Ĭ½ñòÓħĖðéôýõÅěûôĝĂ ĥåÕÕõµåĕÅĕĥąąĕąõĕÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

9 9701/42/M/J/25 © UCLES 2025 [Turn over (b) Nitrogen dioxide, NO2, reacts with ozone, O3, as shown in reaction 2. reaction 2 2NO2 + O3 O2 + N2O5 The rate equation for reaction 2 is shown. rate = k [NO2][O3] (i) Two experiments are carried out to measure the rate of reaction 2. In the first experiment, the initial rate is measured starting with known concentrations of NO2 and O3. In the second experiment, the concentrations of NO2 and O3 are both increased by a factor of four. Predict how the initial rate for reaction 2 would change. … [1] (ii) The reaction mechanism for reaction 2 has two steps. Define rate-determining step. … [1] (iii) Suggest equations for the two steps of the reaction mechanism for reaction 2. step 1 … step 2 … [2] * 0000800000009 * ,










, ĬÓĊ¾Ġ´íÈõÏĪÅĊàúµĀ× Ĭ½òñÛġĚĀÐĆô´đģßôčĂ ĥååĕµÕÅõÕĥµąąõĥµąÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

10 9701/42/M/J/25 © UCLES 2025 (c) Dinitrogen pentoxide, N2O5, can decompose to NO2 and O2. The rate equation for this decomposition is shown. rate = k [N2O5] Fig. 4.2 shows the graph of rate against [N2O5] for this decomposition. 0.0 0.0000 0.0100 0.0200 0.0300 [N2O5] / mol dm–3 rate / 10–4 mol dm–3 s–1 0.0400 0.0500 2.0 4.0 6.0 12.0 14.0 16.0 8.0 10.0 Fig. 4.2 (i) Use Fig. 4.2 to calculate a value for the rate constant, k, for this decomposition. k = … [1] (ii) Use your answer to (c)(i) to calculate the half-life, t 1 2, in seconds, for this decomposition. t 1 2 = … s [1] [Total: 10] * 0000800000010 * ,










, ĬÑĊ¾Ġ´íÈõÏĪÅĊÝü·þ× Ĭ½ôòÚĥĎĪçûĆÏéęďÌĕĂ ĥõµĕõµąĕĥåĥÅąµąõåÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

11 9701/42/M/J/25 © UCLES 2025 [Turn over 5 (a) Chloric(I) acid, HCl O, is a weak Brønsted–Lowry acid. When chloric(I) acid is added to aqueous sodium hydroxide, an acid–base reaction takes place, as shown. HCl O + NaOH NaCl O + H2O (i) Identify the two conjugate acid–base pairs in this reaction. acid I HCl O conjugate base of acid I … acid II … conjugate base of acid II … [1] (ii) The value for the acid dissociation constant, Ka, of HCl O(aq) is 3.70 × 10–8. Calculate the concentration, in mol dm–3, of HCl O(aq) at a pH of 4.51. concentration of HCl O = … mol dm–3 [2] (iii) When a solution of HCl O(aq) is heated, chloric(V) acid and a strong acid not containing oxygen are formed as the only products. Write an equation for this reaction. … [1] (b) (i) Define a buffer solution. … … [1] (ii) Suggest a substance that could be added to aqueous ethanoic acid to form a buffer solution. Explain your answer. … … … [1] * 0000800000011 * ,









, ĬÓĊ¾Ġ´íÈõÏĪÅĊÝú·þ× Ĭ½óñÒģĒĚÒýûĚíġËÌĥĂ ĥõÅÕµÕĥõõÕµÅąÕĥµµÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

12 9701/42/M/J/25 © UCLES 2025 (c) Some fertilisers contain calcium dihydrogenphosphate, Ca(H2PO4)2. An aqueous solution containing dihydrogenphosphate ions, H2PO4 –, can act as a buffer solution. Write two equations to show how H2PO4 – ions can act as a buffer. equation 1 … equation 2 … [2] (d) The solubility of calcium phosphate, Ca3(PO4)2, is 1.14 × 10–7 mol dm–3 at 25 °C. (i) The expression for the solubility product, Ksp, of Ca3(PO4)2 is shown. Ksp = [Ca2+]3[PO4 3–]2 Calculate the value of Ksp for Ca3(PO4)2. Include units. Ksp = … units … [3] (ii) Some solid sodium phosphate is added to a saturated solution of Ca3(PO4)2. Predict the effect, if any, on the solubility of Ca3(PO4)2. Explain your answer. … … … [1] [Total: 12] * 0000800000012 * ,









, ĬÑĊ¾Ġ´íÈõÏĪÅĊßü·Ā× Ĭ½óôÒĩĤďåăôđϽĩÜčĂ ĥÅĕÕõÕĥÕĕõÅÅÅÕŵåÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

13 9701/42/M/J/25 © UCLES 2025 [Turn over 6 (a) Methylbenzene reacts readily with nitronium ions, NO2 +. NO2 + ions are generated by the reaction between concentrated nitric acid and concentrated sulfuric acid. (i) Write an equation for the formation of the NO2 + ion. … [1] (ii) Complete the mechanism in Fig. 6.1 for the nitration of methylbenzene to form 1-methyl-2-nitrobenzene. Include all relevant curly arrows and charges. NO2 + organic intermediate NO2 CH3 1-methyl-2-nitrobenzene + … CH3 Fig. 6.1 [3] (b) Phenol can be nitrated with dilute nitric acid. Explain why the nitration of phenol occurs under milder conditions than the nitration of benzene. … … … … … [2] (c) A sample of 2-nitrophenol is reacted with sodium. Complete the equation in Fig. 6.2 for the reaction of 2-nitrophenol with sodium. NO2 OH + Na Fig. 6.2 [1] * 0000800000013 * ,









, ĬÓĊ¾Ġ´íÈõÏĪÅĊßú·Ā× Ĭ½ôóÚğĠğÔõýØċµ­ÜĝĂ ĥÅĥĕµµąµąąĕÅŵåõµÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

14 9701/42/M/J/25 © UCLES 2025 (d) 2-nitrophenol can undergo different reactions as shown in Fig. 6.3. NH2 OH OH NO2 CH3COCl reaction 1 2-nitrophenol reaction 2 Fig. 6.3 (i) Suggest reagents and conditions for reaction 1. … [1] (ii) Name the type of reaction for reaction 1. … [1] (iii) Reaction 2 is carried out at room temperature. Draw the structure of the organic product from reaction 2 in Fig. 6.3. [1] (iv) Name the mechanism for reaction 2. … [1] (e) The NO2 group in 2-nitrophenol is electron withdrawing. Suggest the relative acidities of ethanol, 2-nitrophenol, phenol and water. Explain your answer. … … … … most acidic least acidic … … … … … … [4] * 0000800000014 * ,









, ĬÍĊ¾Ġ´íÈõÏĪÅĊàù¶Ă× Ĭ½óñÛġēėâóĊûçĠ¹ĬĥĂ ĥĥĥÕõÕÅĕĥĕĕąąõąµĥÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

15 9701/42/M/J/25 © UCLES 2025 [Turn over (f) Salbutamol is a pharmaceutical drug that contains a phenol functional group. OH salbutamol H N HO HO Fig. 6.4 (i) Name and classify the three other functional groups in salbutamol in Table 6.1. Table 6.1 name of functional group classification of functional group [2] (ii) Salbutamol reacts with Br2(aq) to form organic product X. Draw the structure of X. [1] (iii) Salbutamol reacts with an excess of SOCl 2 to form organic product Y. The molecular formula of Y is C13H19Cl 2NO. Draw the structure of Y. [1] [Total: 19] * 0000800000015 * ,









, ĬÏĊ¾Ġ´íÈõÏĪÅĊàû¶Ă× Ĭ½ôòÓħďħ׹÷®óĘĝĬĕĂ ĥĥĕĕµµåõõĥÅąąĕĥõõÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

16 9701/42/M/J/25 © UCLES 2025 7 (a) 2-aminobutane, CH3CH(NH2)CH2CH3, exists as a mixture of two enantiomers. Define enantiomers. … … [1] (b) (i) Explain why an aqueous solution of 2-aminobutane has a pH greater than 7. Include an equation in your answer. … … … [2] (ii) A 0.10 mol dm–3 solution of diethylamine, (CH3CH2)2NH, has a higher pH than a 0.10 mol dm–3 solution of 2-aminobutane, CH3CH(NH2)CH2CH3. Suggest why. … … … … [2] (iii) (CH3CH2)2NH reacts with ethanoic acid, CH3COOH. Complete the equation for this reaction. (CH3CH2)2NH + CH3COOH … [1] (iv) (CH3CH2)2NH reacts with ethanoyl chloride, CH3COCl. Complete the equation for this reaction. (CH3CH2)2NH + CH3COCl … [1] * 0000800000016 * ,









, ĬÍĊ¾Ġ´íÈõÏĪÅĊÞù¶Ą× Ĭ½ôóÓĝĝĢäċðµÑ¼¿üĝĂ ĥÕÅĕõµåÕĕŵąÅĕÅõĥÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

17 9701/42/M/J/25 © UCLES 2025 [Turn over (c) Table 7.1 shows monomers that can undergo polymerisation. (i) Complete Table 7.1. Table 7.1 monomer type of polymerisation CH2C(CH3)COOH OH Cl O CH3CH(NH2)COOH [1] (ii) Ethanedioic acid, HOOCCOOH, can react with propane-1,3-diamine, H2NCH2CH2CH2NH2, to form polymer W. Draw a section of polymer W showing only one repeat unit. The new functional group formed should be displayed. [2] (iii) Poly(alkenes) biodegrade very slowly. Explain why. … … [1] [Total: 11] * 0000800000017 * ,









, ĬÏĊ¾Ġ´íÈõÏĪÅĊÞû¶Ą× Ĭ½óôÛīġĒÕíāôąÄěüčĂ ĥÕµÕµÕŵąµĥąÅõåµõÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

18 9701/42/M/J/25 © UCLES 2025 8 (a) Compound A is analysed by carbon-13 NMR and proton (1H) NMR spectroscopy. O A O O Fig. 8.1 State the reference substance and a solvent that can be used in NMR spectroscopy. reference … solvent … [1] (b) Predict the number of peaks in the carbon-13 NMR spectrum of A. … [1] (c) The proton (1H) NMR spectrum of A shows peaks in four different chemical environments. Complete Table 8.1 for the proton (1H) NMR spectrum of A. Table 8.1 chemical shift δ / ppm splitting pattern number of protons on adjacent carbon atoms number of 1H atoms responsible for the peak 1.10 1.50 6 2.85 3.75 [4] [Total: 6] * 0000800000018 * ,









 , ĬÍĊ¾Ġ´íÈõÏĪÅĊßù¸Ă× Ĭ½ñóÚğĥøÞĄ÷ď­ºëÄĕĂ ĥąåÕõµąÕµõµÅŵÅõÕÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

19 9701/42/M/J/25 © UCLES 2025 [Turn over Table 8.2 environment of proton example chemical shift range δ / ppm alkane –CH3, –CH2–, >CH– 0.9–1.7 alkyl next to C=O CH3–C=O, –CH2–C=O, >CH–C=O 2.2–3.0 alkyl next to aromatic ring CH3–Ar, –CH2–Ar, >CH–Ar 2.3–3.0 alkyl next to electronegative atom CH3–O, –CH2–O, –CH2–Cl 3.2–4.0 attached to alkene =CHR 4.5–6.0 attached to aromatic ring H–Ar 6.0–9.0 aldehyde HCOR 9.3–10.5 alcohol ROH 0.5–6.0 phenol Ar–OH 4.5–7.0 carboxylic acid RCOOH 9.0–13.0 alkyl amine R–NH– 1.0–5.0 aryl amine Ar–NH2 3.0–6.0 amide RCONHR 5.0–12.0 * 0000800000019 * ,









 , ĬÏĊ¾Ġ´íÈõÏĪÅĊßû¸Ă× Ĭ½òôÒĩĩĈÛöĊÚĩÂïÄĥĂ ĥąÕĕµÕĥµåąĥÅÅÕåµÅÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

20 9701/42/M/J/25 © UCLES 2025 9 (a) The structures of the amino acids serine and lysine are shown in Fig. 9.1. CH2OH serine lysine H2N COOH C H CH2CH2CH2CH2NH2 H2N COOH C H Fig. 9.1 Draw the structure for the dipeptide, ser–lys, with molecular formula C9H19N3O4. The peptide functional group formed should be displayed. [2] (b) The isoelectric point of serine is 5.7 and of lysine is 9.7. (i) State what is meant by isoelectric point. … … [1] (ii) A mixture of serine, lysine and ser–lys is analysed by electrophoresis using a buffer at pH 5.7. – + mixture applied here Fig. 9.2 Draw and label three spots on Fig. 9.2 to indicate the predicted position of each of these three species, serine, lysine and ser–lys, after electrophoresis. Explain your answer. … … … … [3] * 0000800000020 * ,










, ĬÍĊ¾Ġ´íÈõÏĪÅĊÝù¸Ą× Ĭ½òñÒģěāàüāÑËĞÍÔčĂ ĥµąĕõÕĥĕÅåĕÅąÕąµÕÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

21 9701/42/M/J/25 © UCLES 2025 (c) Fig. 9.3 shows the synthesis of compound R from compound P. OH O N H step 1 step 2 R P Q, C6H13NO3 O O NH2 Fig. 9.3 (i) Draw the structure of Q in Fig. 9.3. [1] (ii) State the reagents and conditions for steps 1 and 2 in Fig. 9.3. step 1 … step 2 … [2] [Total: 9] * 0000800000021 * ,









, ĬÏĊ¾Ġ´íÈõÏĪÅĊÝû¸Ą× Ĭ½ñòÚĥėñÙþðĘďĖĉÔĝĂ ĥµõÕµµąõÕÕÅÅąµĥõÅÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

22 9701/42/M/J/25 © UCLES 2025 BLANK PAGE * 0000800000022 * ,









, ĬÑĊ¾Ġ´íÈõÏĪÅĊàûµĂ× Ĭ½òòÕğïûÑĉýĝïĞĬüčĂ ĥĥÕÕµĕąÕĕµĥÅŵŵĕÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

23 9701/42/M/J/25 © UCLES 2025 Important values, constants and standards molar gas constant R = 8.31 J K–1 mol–1 Faraday constant F = 9.65 × 104 C mol–1 Avogadro constant L = 6.02 × 1023 mol–1 electronic charge e = –1.60 × 10–19 C molar volume of gas Vm = 22.4 dm3 mol–1 at s.t.p. (101 kPa and 273 K) Vm = 24.0 dm3 mol–1 at room conditions ionic product of water Kw = 1.00 × 10–14 mol2 dm–6 (at 298 K (25 °C)) specific heat capacity of water c = 4.18 kJ kg–1 K–1 (4.18 J g–1 K–1) * 0000800000023 * ,









, ĬÓĊ¾Ġ´íÈõÏĪÅĊàùµĂ× Ĭ½ññÍĩóċèïôìëĖ°üĝĂ ĥĥåĕõõĥµąÅµÅÅÕåõąÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

24 9701/42/M/J/25 © UCLES 2025 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. Group The Periodic Table of Elements 1 H hydrogen 1.0 2 He helium 4.0 1 2 13 14 15 16 17 18 3 4 5 6 7 8 9 10 11 12 3 Li lithium 6.9 4 Be beryllium 9.0 atomic number atomic symbol Key name relative atomic mass 11 Na sodium 23.0 12 Mg magnesium 24.3 19 K potassium 39.1 20 Ca calcium 40.1 37 Rb rubidium 85.5 38 Sr strontium 87.6 55 Cs caesium 132.9 56 Ba barium 137.3 87 Fr francium – 88 Ra radium – 5 B boron 10.8 13 Al aluminium 27.0 31 Ga gallium 69.7 49 In indium 114.8 81 Tl thallium 204.4 6 C carbon 12.0 14 Si silicon 28.1 32 Ge germanium 72.6 50 Sn tin 118.7 82 Pb lead 207.2 22 Ti titanium 47.9 40 Zr zirconium 91.2 72 Hf hafnium 178.5 104 Rf rutherfordium – 23 V vanadium 50.9 41 Nb niobium 92.9 73 Ta tantalum 180.9 105 Db dubnium – 24 Cr chromium 52.0 42 Mo molybdenum 95.9 74 W tungsten 183.8 106 Sg seaborgium – 25 Mn manganese 54.9 43 Tc technetium – 75 Re rhenium 186.2 107 Bh bohrium – 26 Fe iron 55.8 44 Ru ruthenium 101.1 76 Os osmium 190.2 108 Hs hassium – 27 Co cobalt 58.9 45 Rh rhodium 102.9 77 Ir iridium 192.2 109 Mt meitnerium – 28 Ni nickel 58.7 46 Pd palladium 106.4 78 Pt platinum 195.1 110 Ds darmstadtium – 29 Cu copper 63.5 47 Ag silver 107.9 79 Au gold 197.0 111 Rg roentgenium – 30 Zn zinc 65.4 48 Cd cadmium 112.4 80 Hg mercury 200.6 112 Cn copernicium – 114 Fl flerovium – 116 Lv livermorium – 7 N nitrogen 14.0 15 P phosphorus 31.0 33 As arsenic 74.9 51 Sb antimony 121.8 83 Bi bismuth 209.0 8 O oxygen 16.0 16 S sulfur 32.1 34 Se selenium 79.0 52 Te tellurium 127.6 84 Po polonium – 9 F fluorine 19.0 17 Cl chlorine 35.5 35 Br bromine 79.9 53 I iodine 126.9 85 At astatine – 10 Ne neon 20.2 18 Ar argon 39.9 36 Kr krypton 83.8 54 Xe xenon 131.3 86 Rn radon – 113 Nh nihonium – 115 Mc moscovium – 117 Ts tennessine – 118 Og oganesson – 21 Sc scandium 45.0 39 Y yttrium 88.9 57–71 lanthanoids 89–103 actinoids 57 La lanthanum 138.9 89 Ac lanthanoids actinoids actinium – 58 Ce cerium 140.1 90 Th thorium 232.0 59 Pr praseodymium 140.9 91 Pa protactinium 231.0 60 Nd neodymium 144.2 92 U uranium 238.0 61 Pm promethium – 93 Np neptunium – 62 Sm samarium 150.4 94 Pu plutonium – 63 Eu europium 152.0 95 Am americium – 64 Gd gadolinium 157.3 96 Cm curium – 65 Tb terbium 158.9 97 Bk berkelium – 66 Dy dysprosium 162.5 98 Cf californium – 67 Ho holmium 164.9 99 Es einsteinium – 68 Er erbium 167.3 100 Fm fermium – 69 Tm thulium 168.9 101 Md mendelevium – 70 Yb ytterbium 173.1 102 No nobelium – 71 Lu lutetium 175.0 103 Lr lawrencium – * 0000800000024 * ,









, ĬÑĊ¾Ġ´íÈõÏĪÅĊÞûµĄ× Ĭ½ñôÍģāþÓñûãĉºĎĬĕĂ ĥÕõĕµõĥĕĥĥÅÅąÕąõĕÕ DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN

This document consists of 17 printed pages. © Cambridge University Press & Assessment 2025 [Turn over Cambridge International AS & A Level CHEMISTRY 9701/42 Paper 4 A Level Structured Questions May/June 2025 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 2025 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 May/June 2025 © Cambridge University Press & Assessment 2025 Page 2 of 17 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 descriptions 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 2025 © Cambridge University Press & Assessment 2025 Page 3 of 17 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. 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 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.

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 4 of 17 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.

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 5 of 17 Annotations guidance for centres Examiners use a system of annotations as a shorthand for communicating their marking decisions to one another. Examiners are trained during the standardisation process on how and when to use annotations. The purpose of annotations is to inform the standardisation and monitoring processes and guide the supervising examiners when they are checking the work of examiners within their team. The meaning of annotations and how they are used is specific to each component and is understood by all examiners who mark the component. We publish annotations in our mark schemes to help centres understand the annotations they may see on copies of scripts. Note that there may not be a direct correlation between the number of annotations on a script and the mark awarded. Similarly, the use of an annotation may not be an indication of the quality of the response. The annotations listed below were available to examiners marking this component in this series. Annotations Annotation Meaning Correct point or mark awarded Incorrect point or mark not awarded Unclear Information missing or insufficient for credit Benefit of the doubt given Contradiction in response otherwise markworthy, mark not given Part of the correct answer has been seen. Full credit has not been awarded. Error carried forward applied Incorrect or insufficient point ignored while marking the rest of the response Benefit of the doubt not applied in this instance

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 6 of 17 Annotation Meaning Rounding error Repetition Blank page or part of script seen Error in number of significant figures Transcription error

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 7 of 17 Question Answer Marks 1(a)(i) Ca(NO3)2 → CaO + 2NO2 + ½O2[1] 1 1(a)(ii) M1: temperature increases down the group M2: ionic radius OR size of cation / M(2)+ AND increases M3: less polarisation / less distortion AND of anion / nitrate ion OR less weakening of NO bond / harder to break NO bond 3 1(b) M1: moles of SrO = 0.333  103.6 = 3.214  10-3 AND moles of Sr(OH)2 = 3.214  10-3 OR 9 / 2800 M2: moles of OH– in 250 cm3 = 3.214  10-3  2 = 6.429  10-3 AND moles of OH– in 1 dm3 = 6.429  10-3  4 = 0.0257 M3: [H+] = 1.00  10–14  0.0257 = 3.89  10–13 OR pOH = –log[OH–] = 1.59 M4: pH = –log(3.89  10–13) = 12.41 must be to 2dp OR pH = 14 –1.59 = 12.41 must be 2 dp 4 2(a)(i) A = [Co(H2O)6]2+ pink B = [Co(H2O)4(OH)2] / Co(OH)2 blue C = [Co(NH3)6]2+ straw / yellow-brown B = precipitation / deprotonation / acid-base C = ligand exchange Any two [1] any four [2] any six [3] all eight [4] 4 2(a)(ii) concentrated HCl 1 2(b)(i) • five (3d) orbitals (lines, boxes) in the isolated Co2+ ion of same energy • splitting three higher and two lower d orbitals • energy of all d orbitals in the complex clearly higher than all d orbitals in isolated ion any two [1] all three [2] 2

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 8 of 17 Question Answer Marks 2(b)(ii) 1 2(c) M1: H2O2 acting as oxidising agent with Mn(OH)2 / Mn2+ M2: Mn2+ + H2O2 → MnO2 + 2H+ AND E ocell = +0.56 OR Mn(OH)2 + H2O2 → MnO2 + 2H2O AND E ocell = +0.92 M3: H2O2 acting as reducing agent with MnO2 M4: MnO2 + 2H+ + H2O2 → Mn2+ + 2H2O + O2 AND Eocell = +0.54 4 2(d)(i) colourless to pale pink 1 2(d)(ii) • moles MnO4– = 0.0500  18.60 / 1000 = 9.30  10-4 in 25.0 cm3 • moles Fe2+ = 5  9.30  10-4 = 4.65  10-3 in 25.0 cm3 • moles Fe2+ = 6 x 4.65  10-3 = 2.79  10-2in 150.0 cm3 M1 / M2: any two of the above bullets [1] all three [2] OR • moles MnO4– = 0.0500  18.60 / 1000 = 9.30  10-4 in 25.0 cm3 • moles MnO4– = 6  9.30  10-4 = 5.58  10-3 in 150.0 cm3 • moles Fe2+ = 5  5.58  10-3 = 2.79  10-2in 150.0 cm3 M1 / M2: any two of the above bullets [1] all three [2] M3: mass Fe = 55.8  2.79  10-2 = 1.55682 (g) in two tablets M4: mass Fe = 0.5  1.55682 = 0.77841 (g) in one tablet AND mass Fe = 7.7841  10–2  1000 = 778.4 4

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 9 of 17 Question Answer Marks 3(a)(i) number of possible arrangements of particles/molecules AND energy in a system 1 3(b)(i) T1 melting OR solid to liquid T2 boiling OR liquid to gas 1 3(b)(ii) change in disorder from a liquid to a gas is much bigger than solid to liquid OR change in intermolecular distance between liquid & gas is much bigger than solid to liquid 1 3(c) M1: ΔHor = (3  –393.5) – ((–824.2) + (3  –110.5)) = –24.8 (kJ mol–1) M2: ΔGo = ΔHo – TΔSo AND T= 723 used M3: (24800 – 36200) ÷ 723 = – [3  213.8 + 2  27.3) – (87.4) – (3x)] –15.8 = –608.6 + 3x, so 3x = 592.8, so x = 197.6 / 198 (J K–1 mol–1) min 3sf 3 3(d) • (as temperature increases) the reaction is less feasible • as ΔGo becomes less negative / (more) positive • due to TS becoming more negative / –TS (becoming more) positive M1 / M2: any two of the above bullets [1] all three [2] 2 Question Answer Marks 4(a)(i) the power/exponent to which a concentration AND of a reactant is raised in the rate equation 1 4(a)(ii) Any two [1] all three [2] 2

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 10 of 17 Question Answer Marks 4(a)(iii) negative gradient curve 1 4(b)(i) rate = 4  4 = 16 times 1 4(b)(ii) slowest step / slow step in the mechanism / multi-step reaction 1 4(b)(iii) M1: step 1 O3 + NO2 → NO3 + O2 M2: step 2 NO3 + NO2 → N2O5 AND sum of step 1 and 2 is consistent with equation for reaction 2 2 4(c)(i) Use of graph k = 12  10–4 / 0.04 = 0.03 1 4(c)(ii) k = 0.693 / t1/2 t1/2 = 0.693 / 0.030 = 23.1 (s) min 2sf 1 Question Answer Marks 5(a)(i) acid I HClO conjugate base of acid I ClO– acid II H2O conjugate base of acid II OH– 1 (a)(ii) M1: [H+] = 10–4.51 = 3.09  10–5 M2: [HClO] = (3.09  10–5)2  (3.70  10–8) = 0.0258 min 2sf 2 (a)(iii) 3 HClO → HClO3 + 2 HCl 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 11 of 17 Question Answer Marks (b)(i) (a solution that) resists / minimises changes pH AND when small amounts of acid / H+ and base / OH– are added to it 1 5(b)(ii) hydroxide ions / NaOH / OH– AND it reacts with the acid to form the conjugate base / ethanoate ions OR ethanoate ions / sodium ethanoate AND as it is the conjugate base (of ethanoic acid) 1 5(d)(i) M1: Use of [Ca2+] = [PO43–] = 1.14  10–7 in the expression OR Ksp = (3.42  10–7)3  (2.28  10–7)2 M2: Ksp = (3  1.14  10–7)3  (2  1.14  10–7)2 Ksp = 2.079  10–33 OR 2.08  10–33 OR 2.1  10–33 min 2sf M3: units = mol5 dm–15 3 5(d)(ii) solubility decreases AND due to the common ion effect 1 Question Answer Marks 6(a)(i) HNO3 + H2SO4 → NO2+ + HSO4– + H2O OR HNO3 + 2H2SO4 → NO2+ + 2HSO4– + H3O+ 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 12 of 17 Question Answer Marks 6(a)(ii) M1: curly arrow 1 from inside the hexagon to nitrogen of NO2+ M2: intermediate, usual rules for horseshoe and + charge M3: curly arrow 2 from C-H bond into hexagon AND H+ 3 6(b) M1: p-orbital/lone pair from O / oxygen AND is delocalised / overlaps AND into the ring /  system M2: so electron density of the ring is increased OR polarises electrophiles / NO2+ better 2 6(c) 1 6(d)(i) (hot) tin / Sn AND concentrated AND HCl 1 6(d)(ii) reduction 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 13 of 17 Question Answer Marks 6(d)(iii) 1 6(d)(iv) (nucleophilic) addition – elimination 1 6(e) M1: 2-nitrophenol  phenol  water  ethanol M2: weakens O—H bond / anion stabilised AND proton/H+ (more easily) lost / proton / H+ (more easily) donated M3: (for 2-nitrophenol / phenol), a p-orbital / lone pair from O / oxygen AND is delocalised / overlaps AND into the ring /  system (and increases electron density in the ring) M4: (for ethanol), positive inductive effect / electron donating of alkyl group 4 6(f)(i) name of functional group classification of functional group alcohol / hydroxyl primary alcohol / hydroxyl secondary amine secondary Any three correct [1] all six correct [2] 2

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 14 of 17 Question Answer Marks 6(f)(ii) 1 6(f)(iii) 1 Question Answer Marks 7(a) rotate the plane of polarised light equally in the opposite direction 1 7(b)(i) M1: nitrogen / N can accept / bond to a proton / H+ ion OR donation of lone pair/electrons on the nitrogen / N to a proton / H+ ion M2: CH3CH(NH2)CH2CH3 + H2O → CH3CH(NH3+)CH2CH3 + OH– 2 7(b)(ii) M1: more electron donating alkyl / ethyl group on N / NH (in diethylamine) OR two electron donating alkyl / R / alkane / ethyl group on N / NH (in diethylamine) M2: increase electron density on N OR make lone pair on N more available (to accept H+) 2 7(b)(iii) (CH3CH2)2NH + CH3COOH → (CH3CH2)2NH2+ CH3COO– 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 15 of 17 Question Answer Marks 7(b)(iii) (CH3CH2)2NH + CH3COOH → (CH3CH2)2NH2+ CH3COO– 1 7(b)(iv) (CH3CH2)2NH + CH3COCl → CH3CON(CH2CH3)2 + HCl 1 7(c)(i) monomer type of polymerisation addition condensation CH3CH(NH2)COOH condensation 1 7(c)(ii) M1: correct displayed amide linkage with an adjacent C=O (to CO) and C(H2) to NH M2: rest of the structure correct (only one repeat unit) with continuation bonds 2 7(c)(iii) they are chemically inert / difficult to hydrolyse / C-C bonds are non-polar 1 Question Answer Marks 8(a) reference: TMS / tetramethylsilane / (CH3)4Si AND solvent: D2O / CDCl3 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 16 of 17 Question Answer Marks 8(b) six / 6 peaks 1 8(c) chemical shift  / ppm splitting pattern number of protons on adjacent carbon atoms number of 1H atoms responsible for the peak 1.10 triplet 2 3 1.50 doublet 1 6 2.45 quartet / quad ruplet 3 2 3.75 multiplet 6 1 Any three [1] any six [2] any nine [3] all eleven [4] 4 Question Answer Marks 9(a) M1: correct displayed peptide bond with an adjacent C(H) (to CO) and C(H) to NH M2: rest of the dipeptide correct 2 9(b)(i) pH at which a molecule has no overall charge / no net charge 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 17 of 17 Question Answer Marks 9(b)(ii) M1: correct relative positions of the spots drawn M2: ser not moved AND as it is a zwitterion / neutral OR ser-lys / lys AND move towards negative (pole) as they are positively charged M3: lys moves the furthest / fastest AND as it has the lower Mr / smaller size (than ser-lys) 3 9(c)(i) 1 9(c)(ii) M1: step 1 aqueous AND HCl / H2SO4 AND heat / reflux[1] M2: step 2 conc. H2SO4 catalyst AND high temp / heat 2