Cambridge A Level Chemistry 9701 — 2024 May/June Paper 2 · Variant 1

This document has 16 pages. Any blank pages are indicated. [Turn over Cambridge International AS & A Level CHEMISTRY 9701/21 Paper 2 AS Level Structured Questions May/June 2024 1 hour 15 minutes 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 60. ● 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. * 3 4 6 4 8 3 0 5 3 7 * DC (PQ/CT) 324892/3 © UCLES 2024

2 9701/21/M/J/24 © UCLES 2024 1 (a) The elements of Group 17 are called halogens. Complete Table 1.1. Table 1.1 halogen colour at 293 K chlorine bromine iodine [1] (b) State the trend in volatility of the halogens chlorine, bromine and iodine. Explain your answer. … … … … [3] (c) Iodine is made by reacting bromine with sodium iodide. (i) Construct an ionic equation for the reaction of bromine with sodium iodide. … [1] (ii) State the role of bromine in the reaction. Explain your answer. … … [1]

3 9701/21/M/J/24 © UCLES 2024 [Turn over (d) Concentrated sulfuric acid is added to separate samples containing equal amounts of NaCl , NaBr and NaI. All three samples initially react to produce the hydrogen halide. (i) Write an equation to describe the acid–base reaction that occurs when concentrated sulfuric acid reacts with NaBr. … [1] (ii) Deduce which sodium halide, NaCl , NaBr or NaI, produces the largest percentage yield of hydrogen halide when concentrated sulfuric acid is added. Explain your answer by considering the relative reactivity of the halide ions as reducing agents. identity of sodium halide … explanation … … … [3] [Total: 10]

4 9701/21/M/J/24 © UCLES 2024 2 (a) Sulfur chloride, SCl 2, is a liquid at room temperature. When SCl 2 is added to water, misty fumes are seen and a solution is made that turns universal indicator red. (i) Identify the type of reaction that occurs when SCl 2 is added to water. … [1] (ii) Name a chloride of a different Period 3 element that is also a liquid at room temperature and produces misty fumes when added to water. … [1] (b) A molecule of SCl 2 contains two S–Cl covalent bonds. (i) Complete the dot-and-cross diagram in Fig. 2.1 to show the arrangement of the outer electrons in a molecule of SCl 2. Use × to show electrons from the chlorine atoms. Use ● to show electrons from the sulfur atom. Cl Cl S Fig. 2.1 [2] (ii) Predict the shape of, and bond angle in, a molecule of SCl 2 by using VSEPR theory. shape … bond angle … [2] (c) Solid magnesium nitride, Mg3N2, is a crystalline solid. (i) Deduce the oxidation numbers of magnesium and nitrogen in magnesium nitride to complete Table 2.1. Table 2.1 oxidation number in Mg3N2 magnesium nitrogen [1]

5 9701/21/M/J/24 © UCLES 2024 [Turn over (ii) Magnesium nitride reacts with an excess of water to produce ammonia and magnesium hydroxide only. Construct an equation to describe this reaction. … [1] (iii) Explain why the solution produced in the reaction in (c)(ii) has a pH greater than 7. Refer to the products of the reaction in your answer. … … … [2] (d) Boron nitride is a white solid that melts above 2900 °C. Fig. 2.2 shows part of the lattice structure of a crystal of boron nitride. boron atom nitrogen atom van der Waals’ forces covalent bond Fig. 2.2 (i) Use Fig. 2.2 to deduce the empirical formula of boron nitride. … [1] (ii) Suggest the identity of another crystalline solid that has atoms arranged in layers similar to that of solid boron nitride. … [1] [Total: 12]

6 9701/21/M/J/24 © UCLES 2024 3 (a) Define Le Chatelier’s principle. … … … [2] (b) Reaction 1 describes the reversible reaction between yellow Fe3+(aq) and colourless SCN–(aq) to produce red FeSCN2+(aq). reaction 1 Fe3+(aq) + SCN–(aq) FeSCN2+(aq) ∆H = –x kJ mol–1 yellow colourless red A mixture of Fe3+(aq), SCN–(aq) and FeSCN2+(aq) is at equilibrium at 20 °C. The temperature of this mixture is then increased to 50 °C and allowed to reach equilibrium. Deduce the changes that occur, if any, in the equilibrium mixture at 50 °C compared to the equilibrium mixture at 20 °C. • change in appearance … • change in relative concentration of FeSCN2+(aq) … • change in value of the equilibrium constant, Kc … [3]

7 9701/21/M/J/24 © UCLES 2024 [Turn over (c) In another experiment, equimolar amounts of Fe3+(aq) and SCN–(aq) are mixed together and allowed to reach equilibrium. The total volume of the mixture is 25.0 cm3. reaction 1 Fe3+(aq) + SCN–(aq) FeSCN2+(aq) At equilibrium the mixture contains: • [SCN–] = 1.30 × 10–3 mol dm–3 • [FeSCN2+] = 0.300 × 10–3 mol dm–3. (i) Calculate the initial amount, in mol, of Fe3+(aq) added to SCN–(aq) to produce this mixture. initial amount of Fe3+(aq) = … mol [2] (ii) Calculate Kc for reaction 1 and state its units. Show your working. Kc = … units … [2] [Total: 9]

8 9701/21/M/J/24 © UCLES 2024 4 (a) Define enthalpy change of formation. … … … [2] (b) Iron is made when iron(III) oxide is heated with carbon monoxide, as shown by reaction 2. reaction 2 Fe2O3 + 3CO 2Fe + 3CO2 Table 4.1 shows enthalpy change of formation data measured at 298 K and 101 kPa. Table 4.1 substance equation value for ∆H f o / kJ mol–1 Fe2O3 –824.2 CO –110.5 CO2 C(s) + O2(g) CO2(g) –393.5 (i) Complete Table 4.1 by adding equations with relevant state symbols to represent: • standard enthalpy change of formation for Fe2O3 • standard enthalpy change of formation for CO. [2] (ii) Use the data in Table 4.1 to calculate the enthalpy change of reaction, ∆Hr , in kJ mol–1, for reaction 2. Show your working. ∆Hr = … kJ mol–1 [2] [Total: 6]

9 9701/21/M/J/24 © UCLES 2024 [Turn over 5 Hydrocarbon molecules contain covalent bonds. (a) Define covalent bond. … … [1] (b) A C=C bond in an alkene is made from a σ bond and a π bond. (i) Identify the hybridisation of the carbon atoms in a C=C bond in an alkene. … [1] (ii) Draw labelled diagrams to show, in terms of orbital overlap, how the σ and π bonds are made in a C=C bond. σ bond π bond [2] (c) In electrophilic reactions involving alkenes the π bond of C=C is broken. (i) Suggest one difference between σ and π bonds that explains why the π bond is broken in electrophilic addition reactions involving alkenes. … … [1] (ii) Complete Fig. 5.1 to show the mechanism for the electrophilic addition of hydrogen bromide to 2-methylpropene to produce the major organic product. Include charges, dipoles, lone pairs of electrons and curly arrows, as appropriate. CH3 H CH3 H C C H Br Fig. 5.1 [4] [Total: 9]

10 9701/21/M/J/24 © UCLES 2024 6 (a) V shows stereoisomerism. V O O O Fig. 6.1 (i) Explain what is meant by stereoisomerism. … … … [1] (ii) Deduce the number of stereoisomers of V. Explain your reasoning. … … … [2] (iii) Deduce the molecular formula of V. … [1] (iv) Name all the functional groups present in V. … … [1]

11 9701/21/M/J/24 © UCLES 2024 [Turn over (b) Fig. 6.2 shows two reactions involving V. W X V + reagent T reaction 1 + reagent U reaction 2 O O O OH O OH O O O Fig. 6.2 (i) Identify the role of reagent T for each functional group that reacts in reaction 1. … … [1] (ii) Suggest the identity of reagent U in reaction 2. … [1]

12 9701/21/M/J/24 © UCLES 2024 (c) Both functional groups in one molecule of Y react with an inorganic reagent to form one molecule of Q and one molecule of methanol, CH3OH, as shown in Fig. 6.3. inorganic reagent Y Q + CH3OH O O O Fig. 6.3 (i) Part of the mass spectrum for Q is shown in Fig. 6.4. Only peaks with m / e greater than 198 are shown. 198 199 200 100 x 201 m / e relative abundance Fig. 6.4 Calculate the relative abundance, x, of the peak at m / e = 201. Show your working. x = … [2] (ii) Q contains only hydroxyl functional groups. Complete Table 6.1 to show the observations that occur when 2,4-dinitrophenylhydrazine (2,4-DNPH reagent) is added to separate samples of Y and Q. Table 6.1 observation on addition of 2,4-DNPH reagent Y Q [1]

13 9701/21/M/J/24 © UCLES 2024 (iii) Under certain conditions, 0.0020 mol of Q reacts with an excess of sodium to produce a total of 44.8 cm3 of gas at s.t.p. Calculate the number of hydroxyl groups present in a molecule of Q. Show your working. number of hydroxyl groups = … [2] (iv) Use Table 6.2 to describe and explain two differences between the infrared spectrum of Y and Q in the region above 1500 cm–1. … … … [2] Table 6.2 bond functional groups containing the bond characteristic infrared absorption range (in wavenumbers) / cm–1 C–O hydroxy, ester 1040–1300 C=C aromatic compound, alkene 1500–1680 C=O amide carbonyl, carboxyl ester 1640–1690 1670–1740 1710–1750 C≡N nitrile 2200–2250 C–H alkane 2850–2950 N–H amine, amide 3300–3500 O–H carboxyl hydroxy 2500–3000 3200–3650 [Total: 14]

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15 9701/21/M/J/24 © UCLES 2024 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 = 6.022 × 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)

16 9701/21/M/J/24 © UCLES 2024 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 –

This document consists of 12 printed pages. © Cambridge University Press & Assessment 2024 [Turn over Cambridge International AS & A Level CHEMISTRY 9701/21 Paper 2 AS Level Structured Questions May/June 2024 MARK SCHEME Maximum Mark: 60 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 2024 series for most Cambridge IGCSE, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

9701/21 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 2 of 12 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.

9701/21 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 3 of 12 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.

9701/21 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 4 of 12 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.

9701/21 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 5 of 12 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.

9701/21 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 6 of 12 Question Answer Marks 1(a) halogen colour at 293 K chlorine green bromine brown iodine (dark) grey All three correct for one mark 1 1(b) M1 (it / volatility) decreases M2 more electrons in molecules M3 (so) increased strength of instantaneous dipole – induced dipole forces (of attraction) 3 1(c)(i) Br2 + 2I– → 2Br – + I2 1 1(c)(ii) (Br2 / bromine is an) oxidising agent AND removes electron(s) from iodide (ions) I– OR increases the oxidation number of I– / iodide (ion) 1 1(d)(i) NaBr + H2SO4 → HBr + NaHSO4 OR 2NaBr + H2SO4 → 2HBr + Na2SO4 1 1(d)(ii) M1 sodium chloride M2 Cl – is not a strong enough reducing agent to reduce H2SO4 M3 Br– & I– are strong enough reducing agents to react (further) with (the S in) H2SO4 Alternative M2 and M3 M2 ability to behave as reducing agents increases down the group M3 EITHER Cl – is not strong enough to reduce H2SO4 OR Br – and I– react (further) with (the S in) H2SO4 3

9701/21 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 7 of 12 Question Answer Marks 2(a)(i) hydrolysis 1 2(a)(ii) silicon(IV) chloride / silicon tetrachloride ALLOW phosphorus(III) chloride / phosphorus trichloride 1 2(b)(i) M1 1 shared pair of electrons between each S-Cl shown as  M2 6 non-bonding electrons around each Cl shown as  AND 4 non-bonding electrons around S shown as  2 2(b)(ii) M1 shape non-linear M2 bond angle 103–105(°) 2 2(c)(i) oxidation number in Mg3N2 magnesium (+)II / 2(+) nitrogen –III/ 3 – Both correct for one mark 1 2(c)(ii) Mg3N2 + 6H2O  3Mg(OH)2 + 2NH3 1 2(c)(iii) M1 due to OH–(aq) M2 EITHER acid-base reaction of water by ammonia OR dissociation of (some) Mg(OH)2 2 2(d)(i) BN 1 2(d)(ii) graphite 1

9701/21 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 8 of 12 Question Answer Marks 3(a) M1 (if a change in conditions occurs) the equilibrium shifts M2 to minimise the change in conditions 2 3(b) M1 change in appearance paler red / more yellow / less orange M2 change in relative concentration of FeSCN2+(aq) lower M3 change in value of the equilibrium constant, Kc lower 3 3(c)(i) M1 deduce the concentration of SCN– present in initial mixture [SCN– eq’m] = [Fe3+ eq’m] = [original SCN-] – [FeSCN2+ produced] 1.30  10-3 + 0.300  10-3 = 1.60  10-3 mol dm-3 M2 find initial amount of Fe3+ initially M1  25 / 1000 = amount of Fe3+ added initially = 4.00  10-5 (mol) 2 3(c)(ii) M1 value for Kc = 178 M2 units = mol-1 dm3 2

9701/21 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 9 of 12 Question Answer Marks 4(a) M1 (enthalpy change / energy change) when one mole of a compound / substance is formed M2 from its elements in their standard states 2 4(b)(i) equation for reaction ∆Hformation Fe2O3 2Fe (s) + ଷ ଶO2 (g) → Fe2O3 (s) –824.2 CO C (s) + ଵ ଶO2 (g) → CO (g) –110.5 CO2 C (s) + O2 (g)  CO2 (g) –393.5 2 4(b)(ii) ∆Hreaction = –∆HFe2O3 – (3  –∆HCO) + (3  –∆HCO2) = –(–824.2) – (3  –110.5) + (3  –393.5) = –24.8 kJ mol-1  evidence of use of correct expression and / or any rearrangement of only these bond energy values  use correct stoichiometry for these bond energy values only  calculate their expression correctly provided the expression uses only addition / subtraction and all 3 values 2

9701/21 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 10 of 12 Question Answer Marks 5(a) (electrostatic) attraction between nuclei of two atoms and shared pair of electrons 1 5(b)(i) sp2 1 5(b)(ii) M1 𝜎 bond – diagram showing how 𝜎 bond is made from direct overlap of orbitals i.e. OR M2 𝜋 bond - diagram showing how 𝜋 bond is made from sideways overlap of 2 p– orbitals i.e. OR OR 2 5(c)(i) EITHER (pair of) electrons in 𝜋 bond are further away from the nuclei so weaker attraction OR (pair of) electrons in 𝜎 bond are closer to the two nuclei so stronger attraction 1

9701/21 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 11 of 12 Question Answer Marks 5(c)(ii) C H3 C C CH3 H H Br H C H3 C + C CH3 H H H Br – C H3 C C CH3 H H Br H M1 arrow from = of C=C to H of H-Br M2 correct dipole on H-Br AND arrow from H-Br bond to Br M3 correct intermediate AND 2-bromopropane M4 arrow from lone pair on Br — to C+ of their intermediate (with +ve or + charge) 4 Question Answer Marks 6(a)(i) (molecules with the) same structural formula (and same molecular formula) with different arrangement of atoms / groups in space 1 6(a)(ii) M1 number of stereoisomers = 8 M2 reasoning: 2 chiral centres AND 1 C=C (producing a cis-and trans or geometrical pair) 2 6(a)(iii) C13H20O3 1 6(a)(iv) ester AND carbonyl / ketone AND C=C (bond) / alkene 1 6(b)(i) reducing agent for C=O AND C=C 1 6(b)(ii) sodium borohydride / NaBH4 1 6(c)(i) M1 deduce n = 12 (from Y – 1C (in methanol)) M2 (100  x) ÷ (1.1  100) = 12 (so) x = 13.2 2

9701/21 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 12 of 12 Question Answer Marks 6(c)(ii) observation on addition of 2,4–DNPH Y orange precipitate Q no precipitate Both correct for one mark 1 6(c)(iii) ROH + Na → RONa + 1 / 2H2 (0.002 mol Q produced 0.002 mol H2 gas so) 2 OH groups M1 answer indicates that OH group(s) in Q react with Na to produce the H2 in the ratio 1 mol OH : ½ mol H2 M2 uses data to show 2OH groups 2 6(c)(iv) M1 Y will have absorption / peak / trough between 1670–1740 due to C=O (Q will not) M2 Q will have absorption / peak / trough between 3200–3600 due to O-H (Y will not) 2