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Chemistry 9701/41 October/November 2024: question paper and mark scheme

Cambridge AS & A Level Chemistry (9701), Paper 4 A Level Structured Questions, October/November 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

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Paper
Paper 4 A Level Structured Questions
Total marks
100
Time
2 hours
Questions
8
Code on the paper
9701/41/O/N/24
9701_w24_ms_41.pdf

9701/41 mark scheme, question by question

Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.

Question 116 marks

Question

(a) Disodium phosphate, (Na+)2(HPO42–), reacts with an acid to form monosodium phosphate, Na+(H2PO4–). (i) Identify the ions that are a conjugate acid–base pair in this reaction, using the formulae of the species involved. conjugate acid conjugate base [1] (ii) Define buffer solution. …… [2] (iii) Write two equations to show how a mixture of (Na+)2(HPO42–) and Na+(H2PO4–) can act as a buffer solution. equation 1 …… equation 2 …… [2] (iv) Identify one inorganic ion that acts as a buffer in blood. …… [1] (b) Compound E is the hydroxide of a Group 2 element. Compound E is a strong alkali. 2.63 g of E is dissolved in water to make 250 cm3 of solution F. Solution F has a pH of 13.09 at 298 K. (i) Show that the concentration of hydroxide ions in solution F is 0.123 mol dm–3. [2] (ii) Explain why the concentration of compound E in solution F is 0.0615 mol dm–3. …… [1] (iii) Use the concentration given in (ii) to identify compound E. compound E …… [1] (c) Compound E is much more soluble than magnesium hydroxide. A saturated solution of magnesium hydroxide in water has a concentration of 1.40 × 10–4 mol dm–3 at 298 K. Calculate the solubility product, Ksp, of magnesium hydroxide. Include units. Ksp = …… units …… [3] (d) Explain why compound E is much more soluble than magnesium hydroxide. …… [3] [Total: 16]

Mark scheme

1(a)(i) conjugate acid conjugate base 1 H2PO4– HPO42– [1] BOTH 1(a)(ii) allows only small changes in pH / resists changes in pH [1] 2 when small amounts of acid or alkali/base / H+ or OH– are added [1] 1(a)(iii) Na+2HPO42– + H+ → Na+H2PO4– + Na+ 2 or HPO42– + H+ → H2PO4– [1] Na+H2PO4– + NaOH → Na+2HPO42– + H2O or H2PO4– + OH– → HPO42– + H2O [1] 1(a)(iv) HCO3– / hydrogen carbonate [1] 1 1(b)(i) either 10–13.09 or [H+]= 8.1 × 10–14 seen [1] 2 8.1x10–14 × [OH–] = 1 × 10–14 seen [1] 1× 10−14 OR [OH–] = seen 8.1× 10 −14 ALLOW alternative method pOH = 0.91 [1] [OH–] = 10–0.91 seen [1] 1(b)(ii) 2 moles of OH– in one mole of E / X(OH)2 1 OR there are two hydroxide ions in each formula unit of E / X(OH)2 [1] 1(b)(iii) 0.0615 × 0.25 = 0.0154 moles of E 1 RFM = 2.63 / 0.0154 = 170.8 / 171 OR RAM = 34 – 2.63 / 0.0154 = 137.1 /137 AND barium hydroxide / Ba(OH)2 [1] 1(c) Ksp = [Mg2+][OH–]2 OR Ksp = 1.4 × 10-4 × (2 × 1.4 × 10-4)2 [1] 3 Ksp = 1.10 × 10–11 [1] min 2sf correct answer [2] mol3 dm–9 [1] ecf from M1 1(d) M1: ΔHlatt and ΔHhyd less exothermic (down the group) 3 OR ΔHlatt/LE and ΔHhyd more exothermic/more negative for Mg [1] M2: ΔHlatt changes more than ΔHhyd (down the group) OR ΔHlatt changes more / faster for Mg [1] M3: ΔHsol becomes more exothermic (down the group) OR ΔHsol becomes less exothermic / less negative for Mg [1] mark independently

Question 214 marks

Question

(a) Predict and explain the variation in enthalpy change of hydration for the ions F–, Cl –, Br – and I–. …… [2] (b) Fig. 2.1 shows an incomplete energy cycle involving calcium fluoride, CaF2. line B: Ca2+(g) + 2F–(g) process 1: enthalpy change of formation of Ca2+(g) plus process 2: enthalpy change of twice the enthalpy change of hydration of calcium ions plus formation of F–(g), twice the enthalpy change of ΔHf Ca2+(g) + 2ΔHf F–(g) hydration of fluoride ions, line A: Ca(s) + F2(g) ΔHhyd Ca2+(g) + 2ΔHhyd F–(g) process 3: enthalpy change of line D: formation of calcium fluoride, ΔHf CaF2(s) line C: CaF2(s) process 4: enthalpy change of solution of calcium fluoride, ΔHsol CaF2(s) Fig. 2.1 (i) Complete line D. Include state symbols. [1] (ii) The value of the enthalpy change for process 1 can be calculated using the values of five other enthalpy changes which are not referred to in Fig. 2.1. process 1: Ca(s) + F2(g) Ca2+(g) + 2F–(g) Identify these five other enthalpy changes, using either names or symbols. …… [2] (iii) Define lattice energy, ΔHlatt. …… [2] (iv) Complete the expression to give the mathematical relationship between ΔHlatt of calcium fluoride and the enthalpy changes for processes 1 and 3. ΔHlatt = [1] (c) Use data from Table 2.1 to calculate a value for the hydration energy, ΔHhyd, of fluoride ions, F–(g). Table 2.1 value / kJ mol–1 enthalpy change of solution of calcium fluoride, CaF2(s) +13 overall enthalpy change of process 1 in Fig. 2.1 +1395 enthalpy change of formation of calcium fluoride –1214 enthalpy change of hydration of Ca2+(g) –1650 ΔHhyd F–(g) = …… kJ mol–1 [2] (d) Define entropy. …… [1] (e) At 298 K, the Gibbs free energy change, ΔG, for the solution of compound T is +6.00 kJ mol–1. The enthalpy change of solution, ΔHsol, of compound T is +30.0 kJ mol–1 at 298 K. Calculate the value of the entropy change, ΔS, for the solution of compound T at 298 K. ΔS = …… J K–1 mol–1 [2] (f) Predict whether compound T becomes more or less soluble as the water is heated from 298 K to 360 K. Explain your answer. …… [1] [Total: 14]

Mark scheme

2(a) becomes less negative/less exothermic (down the group / from F to I) 2 AND due to increase in (ionic) radius / size [1] decreased attraction to water OR weaker ion-dipole force to water [1] 2(b)(i) CaF2(aq) OR Ca2+ (aq) + 2F– (aq) [1] 1 2(b)(ii) • atomisation energy of Ca / ΔHat(o)m)) 2 • atomisation energy of F(2) / ΔHat(o)m)) OR F-F bond energy / BE of F-F • first ionisation energy / IE1 of Ca / ΔHi1 • second ionisation energy / IE2 of Ca / ΔHi2 • (first) electron affinity /EA of F / ΔHea any three for [1] ALL five [2] 2(b)(iii) energy change / energy released 2 when one mole of an ionic compound is formed [1] from its gaseous ion(s) [1] 2(b)(iv) (ΔHlatt = ) (change) 3 – (change) 1 [1] 1 OR (ΔHlatt = ) ΔHf(CaF2(s)) – ΔHf(Ca2+(g)) – 2ΔHf(F–(g)) 2(c) expression involves four correct numbers (13, 1395, 1214, 1650) 2 AND 2 times [1] 1395 – 1650 + 2x = –1214+13 ΔHf(F–(g)) × = –473 [1] ecf correct answer [2] 2(d) number of possible arrangements of particles / energy in a system [1] 1 2(e) states or clearly uses ΔG = ΔH - TΔS [1] 2 OR 6000 = 30 000 – (298 × ΔS) ΔS = (+)80.5(4) [1] min 3sf ecf correct answer [2] 2(f) becomes more soluble 1 AND ΔS is positive / TΔS is positive / –TΔS is negative (as T inc) [1] (so ΔG becomes more negative)

Question 316 marks

Question

(a) A and B react together to give product AB. A+B AB When the concentrations of A and B are both 0.0100 mol dm–3, the rate of formation of AB is 7.62 × 10–4 mol dm–3 s–1. When the concentrations of A and B are both 0.0200 mol dm–3, the rate of formation of AB is 3.05 × 10–3 mol dm–3 s–1. (i) Complete the three possible rate equations that are consistent with these data. rate = …… rate = …… rate = …… [2] (ii) Choose one of the rate equations you have written in (i), and calculate the value of the rate constant, k. Include the units of k. k = …… units …… [2] (iii) Explain why it is not possible to calculate a value for the half-life, t 1 , of this reaction 2 using the value of the rate constant k calculated in (ii) and the equation k = 0.693 / t 1 . 2 …… [1] (b) Catalysts may be homogeneous or heterogeneous. (i) Identify two metals that act as heterogeneous catalysts in the removal of NO2 from the exhaust gases of car engines. …… and …… [1] (ii) Iron acts as a heterogeneous catalyst in the Haber process. Describe the mode of action of this iron catalyst. …… [2] (iii) Fe2+ ions act as a homogeneous catalyst in the reaction between I–(aq) and S2O82–(aq). Write equations for the two reactions that occur when Fe2+(aq) is added to a mixture of I–(aq) and S2O82–(aq). equation 1 S2O82– + …… equation 2 …… [2] (iv) Explain the difference between a homogeneous catalyst and a heterogeneous catalyst. …… [1] (c) Fe2+ ions can be oxidised to Fe3+ ions under alkaline conditions by suitable oxidising agents. (i) Iron is a transition element. Explain why iron forms stable compounds in both the +2 and the +3 oxidation states. …… [1] (ii) The half-equation for the reduction of Fe3+ under alkaline conditions, and its E o value, are shown. Fe(OH)3 + e– Fe(OH)2 + OH– E o = – 0.56 V Four more half-equations for reactions under alkaline conditions, and their E o values, are shown. Al (OH)4– + 3e– Al + 4OH– E o = –2.35 V Cl O– + H2O + 2e– Cl – + 2OH– E o = +0.89 V O2 + 2H2O + 4e– 4OH– E o = +0.40 V Zn(OH)42– + 2e– Zn + 4OH– E o = –1.22 V Select two oxidising agents that can oxidise Fe2+ ions to Fe3+ ions under alkaline conditions. o Write an equation, and give the E cell value, for each of the two reactions that occur. oxidising agent 1: …… equation: …… o E cell = …… V oxidising agent 2: …… equation: …… o E cell = …… V [4] [Total: 16]

Mark scheme

3(a)(i) rate = k[A][B] [1] 2 rate = k[A]2 / k[A]2[B]0 AND rate = k[B]2 / k[A]0[B]2 [1] BOTH 3(a)(ii) k = 7.62 / 7.625 / 7.63 [1] ecf from 3a(i) min 3sf 2 mol–1 dm3 s–1 [1] ecf from 3a(i) 3(a)(iii) reaction is 2nd order 1 OR reaction is not 1st order OR reaction does not have constant half-life [1] 3(b)(i) Pt, Pd, Rh any two [1] 1 3(b)(ii) • adsorption (of reactants) 2 / bond forms between the surface/catalyst and reactants • bond weakening (in reactants) • desorption (of reactants / products) / bond breaks between the surface/catalyst and products any two [1] all three [2] 3(b)(iii) S2O82– + 2Fe2+ → 2SO42– + 2Fe3+ [1] any multiple 2 2Fe3+ + 2I– → 2Fe2+ + I2 [1] any multiple 3(b)(iv) (heterogeneous) catalyst and reactants in different states / phases 1 AND (homogeneous) catalyst and reactants in same states / phases [1] BOTH 3(c)(i) similar energy of the (3)d and (4)s subshells / orbitals [1] 1 3(c)(ii) M1M2: 4 • Cl O– chosen • O2 chosen • E⦵cell values 1.45 (with Cl O–) • E⦵cell values 0.96 (with O2) any two [1] all four [2] M3: 2Fe(OH)2 + Cl O– + H2O → Cl– + 2Fe(OH)3 [1] M4: 4Fe(OH)2 + O2 + 2H2O → 4Fe(OH)3 [1]

Question 412 marks

Question

Transition metal atoms and transition metal ions form complexes by combining with ligands. (a) Explain why transition elements form complex ions. …… [1] (b) Co2+ ions form complex ion G. Each G ion contains two Co2+ ions, both of which are octahedrally coordinated. Each G ion contains one O2 molecule, which donates one pair of electrons to each Co2+ ion, and one NH2– ion, which donates one pair of electrons to each Co2+ ion. The remaining ligands are NH3 molecules. (i) Deduce the formula of complex ion G. Include its overall charge. formula of G …… [2] (ii) The d-orbitals of the Co2+ ions present in complex ion G are split. State the number of d-orbitals that are at a higher energy level and the number of d-orbitals that are at a lower energy level in each Co2+ ion. number of d-orbitals at a higher energy level number of d-orbitals at a lower energy level [1] (iii) Co2+ ions form a different complex ion, M. Each M ion contains two Co2+ ions, both of which are octahedrally coordinated, but the ligands are different from the ligands in G. Explain why G and M have different colours. …… [2] (c) Cadmium forms complex ion X, [Cd(NH3)4]2+. When a solution containing CN– ions is added to an aqueous solution of X, a ligand exchange reaction takes place, forming complex ion Y. Y contains no NH3 ligands and no H2O ligands. Y is in a much higher concentration in the mixture than X. The oxidation state and coordination number of cadmium do not change in this reaction. (i) Write an ionic equation for this reaction, using the formulae of the complex ions. …… [2] (ii) Cadmium forms complex ion Z in the same oxidation state and with the same coordination number as in X. All the ligands in Z are Cl – ions. When NaCl (aq) is added to a solution of X, very little Z forms. Write the three cadmium complexes, X, Y and Z, in order of increasing stability constant, Kstab. smallest value of Kstab largest value of Kstab [1] (d) Ethanedioate ions, C2O42–, form complexes with transition element ions. The concentration of C2O42– ions can be found by reaction with acidified Cr2O72– ions. C2O42– ions are protonated and form HOOCCOOH molecules which are oxidised by Cr2O72–. The half-equations are shown. Cr2O72– + 14H+ + 6e– 2Cr3+ + 7H2O 2CO2 + 2H+ + 2e– HOOCCOOH (i) Construct an equation for the reaction between acidified Cr2O72– and HOOCCOOH. …… [1] (ii) A 25.0 cm3 sample of a solution of Na2C2O4 reacts with exactly 16.20 cm3 of an acidified solution of 0.0500 mol dm–3 K2Cr2O7. Calculate the concentration of the solution of Na2C2O4. [Na2C2O4] = …… mol dm–3 [2] [Total: 12]

Mark scheme

4(a) empty (d) orbitals are energetically accessible 1 OR empty (d) orbitals can form dative bonds with ligands OR empty (d) orbitals can accept a lone pair from ligands [1] 4(b)(i) [Co2O2NH2(NH3)8]3+ scores [2] must have Co2 2 • Co2O2NH2 • 8 ammonia so both Co are octahedral (ecf bullet 1) • correct charge based on ligands present (ecf Co and NH2) any two [1] all three [2] 4(b)(ii) 2 1 3 BOTH [1] 4(b)(iii) ΔE different OR (d–d) energy gap different [1] 2 different frequency/wavelength/energy from visible light absorbed [1] 4(c)(i) [Cd(NH3)4]2+ + 4CN– → [Cd(CN)4]2– + 4NH3 2 X Y complex formed is [Cd(CN)4]2– / Na2[Cd(CN)4] [1] equation fully correct [1] ECF 4(c)(ii) [CdCl4]2– [Cd(NH3)4]2+ [Cd(CN)4]2– 1 OR Z X Y [1] 4(d)(i) Cr2O72– + 8H+ + 3HOOCCOOH → 2Cr3+ + 7H2O + 6CO2 [1] 1 4(d)(ii) • mol Cr2O72– = 0.0500 × 16.20 / 1000 = 8.10 × 10–4 2 • mol HOOCCOOH / C2O42– = 8.10 × 10–4 × 3 = 2.43 × 10–3 ecf • [C2O42–] = 2.43 × 10–3 × 40 = 0.0972 mol dm–3 min 2sf ecf ✓ ✓ [2]

Question 56 marks

Question

The shapes of four different complexes, P, Q, R and S, are shown in Table 5.1. The symbol J represents an atom or ion of a transition element. The symbol L is used to represent a monodentate ligand. Table 5.1 P Q L L L J L J L L L L R S L L L J L L J L L L (a) Label one bond angle on each of complexes P, Q, R and S, and identify the size of the angle in degrees. [2] (b) Identify the shapes of complexes P, Q, R and S. P …… Q …… R …… S …… [2] (c) Two L ligands are exchanged with two different monodentate ligands X and Y in each of complexes P, Q, R and S. Identify all the complexes which form new complexes that show geometrical isomerism. …… [1] (d) Three L ligands are exchanged with three different monodentate ligands X, Y and Z in each of complexes P, Q and R. Identify all the complexes which form new complexes that show optical isomerism. …… [1] [Total: 6]

Mark scheme

5(a) bond angle must go from bond to bond 2 P 109 OR 109.5° Q 90° (or 180° if different angle labelled) R 90° (or 180° if different angle labelled) S 180° any two [1] all four [2] 5(b) P tetrahedral 2 Q square planar R octahedral S linear any two [1] all four [2] 5(c) Q AND R [1] 1 5(d) P AND R [1] 1

Question 614 marks

Question

Benzene, C6H6, reacts with chloroethane, C2H5Cl, in the presence of a suitable catalyst to form ethylbenzene, C6H5C2H5. In the presence of the catalyst, the ion C2H5+ is formed. This ion reacts with benzene. (a) Complete the equation for the reaction of C2H5Cl with this catalyst to form C2H5+ as one product. C2H5Cl + …… C2H5+ + …… [1] (b) Ethylbenzene reacts with more C2H5Cl, forming a mixture containing 1,2-diethylbenzene and 1,4-diethylbenzene. (i) Draw the structures of 1,2-diethylbenzene and 1,4-diethylbenzene. 1,2-diethylbenzene 1,4-diethylbenzene [1] (ii) Explain why there is very little 1,3-diethylbenzene in the product mixture. …… [1] (c) 1,2-diethylbenzene can be oxidised to benzene-1,2-dioic acid, C6H4(COOH)2. benzene-1,2-dioic acid COOH COOH (i) State the reagent and conditions used for this reaction. …… [1] (ii) Complete the overall equation for this reaction. An atom of oxygen from the oxidising agent is represented as [O]. All of the atoms in the two ethyl groups are fully oxidised in this reaction. …… + …… [O] C6H4(COOH)2 + …… + …… (1,2-diethylbenzene) [2] (iii) Predict the number of peaks in the carbon-13 NMR spectrum of benzene-1,2-dioic acid. …… [1] (d) The proton (1H) NMR spectra of ethylbenzene, C6H5C2H5, in CDCl 3 and of benzene-1,2-dioic acid, C6H4(COOH)2, in CDCl 3 are shown. They have not been identified. 11 10 9 8 7 6 5 4 3 2 1 0 δ / ppm Fig. 6.1 16 14 12 10 8 6 4 2 0 δ / ppm Fig. 6.2 (i) Explain the use of CDCl 3, instead of CHCl 3, as the solvent when obtaining these spectra. …… [1] (ii) Identify the substance shown by the spectrum in Fig. 6.1, and complete Table 6.1. substance …… Table 6.1 peak at δ = 1.2 peak at δ = 2.6 name of splitting pattern group responsible for peak explanation of splitting pattern [3] (iii) Identify the substance shown by the spectrum in Fig. 6.2, and complete Table 6.2. substance …… Table 6.2 peak at δ = 7.8 peak at δ = 13.1 group responsible for peak [1] (iv) When D2O is used as a solvent, the spectrum obtained is different from the spectrum in Fig. 6.2. Describe this difference and explain your answer. …… [1] (e) Benzene-1,2-dioic acid can be used to produce K. heat K benzene-1,2-dioic acid O COOH C O COOH C O Suggest the name of this type of reaction. …… [1] [Total: 14]

Mark scheme

6(a) C2H5Cl + AlCl3 → C2H5+ + AlCl4– [1] 1 OR C2H5Cl + FeCl3 → C2H5+ + FeCl4– 6(b)(i) 1 AND BOTH [1] 6(b)(ii) alkyl / ethyl group is 2,4 directing 1 OR ethyl group is electron donating / positive inductive effect [1] 6(c)(i) hot alkaline KMnO4 / MnO4– (followed by acid) [1] 1 6(c)(ii) C10H14 / C6H4(C2H5)2 + 12 [O] → C6H4(COOH)2 + 2CO2 + 4H2O 2 all species correct formulae C10H14 + CO2 + H2O [1] correct balancing [1] 6(c)(iii) 4 [1] 1 6(d)(i) CDCl3 does not cause a peak 1 OR does not interfere with spectrum / peaks [1] 6(d)(ii) • ethylbenzene / C6H5C2H5 3 • triplet • CH3 • 2H on neighbouring C / next to CH2 – quartet / quadruplet – CH2 – 3H on neighbouring C / next to CH3 / coupling by CH3 any three [1] any five [2] all seven [3] 6(d)(iii) • benzene-1,2-dioic acid / C6H4(COOH)2 1 • H on the benzene / C6H4 – COOH / carboxylic acid ALL three [1] 6(d)(iv) • COOH peak disappears OR remove peak at 13.1 (ecf (d)(iii) if δ value used) 1 AND • as proton / hydrogen exchanges with deuterium / D BOTH [1] 6(e) dehydration / elimination / (auto)condensation [1] 1

Question 715 marks

Question

A reaction scheme is shown in Fig. 7.1. The reagents needed for reaction 2 and reaction 3 are stated. Reaction 5 takes place when C2H5NH2 is mixed with compound V. No special conditions are required. reaction 1 compound U CH3COOH reaction 2 reaction 3 reagent = LiAl H4 reagent = SOCl2 compound W C2H5NH2 compound V reaction 4 reaction 5 reaction 6 C2H5NHC2H5 C2H5NHCOCH3 Fig. 7.1 (a) Identify compound U which contains only three elements. …… [1] (b) Describe the reagents and conditions for reaction 1. …… [1] (c) Identify compound V. …… [1] (d) Complete the equation for reaction 3. CH3COOH + SOCl2 …… [1] (e) Identify compound W. …… [1] (f) Describe the conditions for reaction 4. …… [1] (g) Suggest the reagent needed for reaction 6. …… [1] (h) Complete Table 7.1 by adding the reaction numbers, 1, 2, 3, 4, 5 and 6, to the right-hand column. Use the reaction numbers given in Fig. 7.1. Each of the numbers 1, 2, 3, 4, 5 and 6 should be used once only. Table 7.1 type of reaction reaction number(s) hydrolysis addition reduction substitution [4] (i) Compare the basicities of C2H5NHCOCH3, C2H5NHC2H5 and NH3. Explain your answer. most basic least basic [4] [Total: 15]

Mark scheme

7(a) CH3CN / ethanenitrile [1] 1 7(b) dilute acid / HCl (aq) 1 AND hot / heat [1] three components required (acid / aq / heat) 7(c) CH3COCl / ethanoyl chloride [1] 1 7(d) CH3COOH + SOCl2 → CH3COCl + HCl + SO2 [1] 1 7(e) C2H5Br / C2H5Cl / bromoethane / chloroethane [1] 1 7(f) heat in ethanol AND 1 under pressure / in sealed tube [1] 7(g) LiAlH4 / lithium aluminium hydride [1] 1 7(h) Examiner checks first that each of 1–6 is only used once and 4 IGNOREs any repeated unless a CON hydrolysis 1 addition 2 reduction 2, 6 substitution 3, 4, 5 two correct [1] four correct [2] five correct [3] six correct [4] 7(i) M1 C2H5NHC2H5 ammonia C2H5NHCOCH3 [1] 4 M2 (basicity linked to) lone pair/ p orbital on N AND being able accept / donate to a proton/H+ [1] M3 (amine stronger because of) +ve inductive effect / electron donating alkyl / R / ethyl group [1] M4 (amide weaker because) lone pair / p orbital on N is delocalised into C=O group [1]

Question 87 marks

Question

(a) An aqueous solution of phenol, C6H5OH, is acidic at 298 K. Explain why phenol is more acidic than water. …… [2] (b) (i) Name the two products formed when phenol reacts with an excess of Br2(aq). …… and …… [1] (ii) Draw the structures of the two isomeric organic products, with Mr = 139, that are formed when phenol reacts with HNO3(aq) at room temperature. [1] (iii) Write the equation for the reaction between phenol, C6H5OH, and sodium metal. …… [1] (c) Phenol can be produced from phenylamine in a two-step synthesis. step one step two phenylamine intermediate compound phenol Describe the reagents and conditions needed in each step. step one: reagents …… conditions …… step two: reagents …… conditions …… [2] [Total: 7] 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) The Periodic Table of Elements Group 1 2 13 14 15 16 17 18 1 2 H He hydrogen helium Key 1.0 4.0 3 4 atomic number 5 6 7 8 9 10 Li Be atomic symbol B C N O F Ne lithium beryllium name boron carbon nitrogen oxygen fluorine neon 6.9 9.0 relative atomic mass 10.8 12.0 14.0 16.0 19.0 20.2 11 12 13 14 15 16 17 18 Na Mg Al Si P S Cl Ar sodium magnesium aluminium silicon phosphorus sulfur chlorine argon 3 4 5 6 7 8 9 10 11 12 23.0 24.3 27.0 28.1 31.0 32.1 35.5 39.9 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 at www.cambridgeinternational.org after the live examination series. K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr potassium calcium scandium titanium vanadium chromium manganese iron cobalt nickel copper zinc gallium germanium arsenic selenium bromine krypton 39.1 40.1 45.0 47.9 50.9 52.0 54.9 55.8 58.9 58.7 63.5 65.4 69.7 72.6 74.9 79.0 79.9 83.8 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 24 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe rubidium strontium yttrium zirconium niobium molybdenum technetium ruthenium rhodium palladium silver cadmium indium tin antimony tellurium iodine xenon 85.5 87.6 88.9 91.2 92.9 95.9 – 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3 55 56 57–71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 lanthanoids Cs Ba Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn caesium barium hafnium tantalum tungsten rhenium osmium iridium platinum gold mercury thallium lead bismuth polonium astatine radon 132.9 137.3 178.5 180.9 183.8 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 – – – 87 88 89–103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 actinoids Fr Ra Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv Ts Og francium radium rutherfordium dubnium seaborgium bohrium hassium meitnerium darmstadtium roentgenium copernicium nihonium flerovium moscovium livermorium tennessine oganesson – – – – – – – – – – – – – – – – – 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 lanthanoids La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu lanthanum cerium praseodymium neodymium promethium samarium europium gadolinium terbium dysprosium holmium erbium thulium ytterbium lutetium 138.9 140.1 140.9 144.2 – 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.1 175.0 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 actinoids Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr actinium thorium protactinium uranium neptunium plutonium americium curium berkelium californium einsteinium fermium mendelevium nobelium lawrencium – 232.0 231.0 238.0 – – – – – – – – – – – Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge

Mark scheme

8(a) M1: p orbital / lone pair on the oxygen / O 2 AND is delocalised into the ring / overlaps with delocalised ring [1] M2: O-H weakened (in phenol) OR anion / phenoxide ion / conjugate base is stabilised/more stable [1] 8(b)(i) 2,4,6-tribromophenol (name) AND hydrogen bromide / HBr [1] 1 8(b)(ii) 1 AND 8(b)(iii) 2C6H5OH + 2Na → 2C6H5ONa + H2 [1] 1 8(c) step one: HNO2 (+ HCl) OR NaNO2 + HCl 2 AND T⩽10 °C step two: H2O AND warm / T> 10 °C any two [1] all four [2]

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Chemistry 9701/41 for October/November 2024 is out of 100 marks, with 2 hours to answer it, across 8 questions.

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