Cambridge A Level Chemistry 9701 — 2023 Oct/Nov Paper 4 · Variant 2

1 Propanone, CH3COCH3 , reacts with iodine, I2 , in the presence of an acid catalyst. CH3COCH3 + I2 CH3COCH2I + H+ + I– The rate equation for this reaction is shown. rate = k[CH3COCH3][H+] (a) Complete Table 1.1 to describe the order of the reaction. Table 1.1 order of the reaction with respect to [CH3COCH3 ] order of the reaction with respect to [I2] order of the reaction with respect to [H+] overall order of the reaction [2] (b) An experiment is performed using a large excess of CH3COCH3 and a large excess of H+(aq). The initial concentration of I2 is 1.00 × 10–5 mol dm–3. The initial rate of decrease in the I2 concentration is 2.27 × 10–7 mol dm–3 s–1. (i) Use the axes to draw a graph of [I2 ] against time for the first 10 seconds of the reaction. 1.00 × 10–5 0.90 × 10–5 0.80 × 10–5 0.70 × 10–5 0.60 × 10–5 [I2] / mol dm–3 0.50 × 10–5 0.40 × 10–5 0.30 × 10–5 0.20 × 10–5 0.10 × 10–5 0 0 1 2 3 4 5 6 7 8 9 10 time / s [1] (ii) State whether it is possible to calculate the numerical value of the rate constant, k, for this reaction from your graph. Explain your answer. ........................................................................................................................................... ..................................................................................................................................... [1] (c) The experiment is repeated at a different temperature. The initial concentrations of H+ ions, I2 and CH3COCH3 are all 0.200 mol dm–3. The value of k at this temperature is 2.31 × 10–5 mol–1 dm3 s–1. Calculate the initial rate of this reaction. rate = ..................................... mol dm–3 s–1 [1] (d) The experiment is repeated using an excess of H+(aq). The new rate equation is shown. rate = k1[CH3COCH3] (i) The value of k1 is 1.1 × 10–3 s–1. Calculate the value of the half-life, t12 . t1 = ...................................................... s [1] 2 (ii) Use your answer to (i) to draw a graph of [CH3COCH3] against time for this reaction. The initial value of [CH3COCH3 ] on your graph should be 0.200 mol dm–3. The final value of [CH3COCH3 ] on your graph should be 0.0250 mol dm–3. 0.200X 0.175 0.150 0.125 [CH3COCH3] 0.100 / mol dm–3 0.075 0.050 0.025 time / s [1] (e) A four-step mechanism is suggested for the overall reaction. CH3COCH3 + I2 CH3COCH2I + H+ + I– rate = k [CH3COCH3][H+ ] Part of this mechanism is shown. step 1: CH3COCH3 + H+ CH3C+(OH)CH3 step 2: CH3C+(OH)CH3 CH3C(OH)=CH2 + H+ step 3: step 4: CH3C+(OH)CH2I CH3COCH2I + H+ (i) Write an equation for step 3. ..................................................................................................................................... [1] (ii) Suggest the slowest step of the mechanism. Explain your answer. ........................................................................................................................................... ..................................................................................................................................... [1] (iii) Identify one conjugate acid-conjugate base pair in the mechanism. conjugate acid .......................................... conjugate base .......................................... [1] [Total: 10]

Mark scheme: Question Answer Marks 1(a) 1 2 0 [1] 1 2 [1] 1(b)(i) The line should start at 1  10–5 and decrease to 0.773  10–5 / 7.73  10–6 and be straight for 10 s [1] 1 1(b)(ii) its not possible AND the concentrations in the rate law are not known [1] 1 1(c) 9.24  10–7 [1] 1 1(d)(i) 630 [1] 1 1(d)(ii) Three halvings taking equal times, these times to agree with their answer to di.[1] 1 1(e)(i) CH3COH=CH2 + I2 → CH3C+(OH)CH2I + I– [1] 1 1(e)(ii) step 1 AND has (both) substances in rate law.[1] 1 1(e)(iii) CH3C+OHCH3 and CH3COCH3 OR 1 CH3C+(OH)CH3 and CH3C(OH)=CH2 OR CH3C+(OH)CH2I and CH3COCH2I [1]

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2 Benzoic acid, C6H5COOH, is a weak acid. The Ka of benzoic acid is 6.31 × 10–5 mol dm–3 at 298 K. A 1.00 dm3 buffer solution is made at 298 K containing 1.00 g of C6H5COOH and a slightly greater mass of sodium benzoate, C6H5COO–Na+. This buffer solution has a pH of 4.15. (a) Define buffer solution. ................................................................................................................................................... ............................................................................................................................................. [1] (b) Write equations to show how this solution acts as a buffer solution when the named substances are added to it: (i) dilute aqueous sodium hydroxide ..................................................................................................................................... [1] (ii) dilute aqueous nitric acid. ..................................................................................................................................... [1] (c) Calculate the H+ concentration and the C6H5COOH concentration in the buffer solution described. Use the expression for the Ka of C6H5COOH to calculate the concentration of C6H5COO–Na+ in the buffer solution. Show your working and give each answer to a minimum of three significant figures. [H+] = .............................. mol dm–3 [C6H5COOH] = .............................. mol dm–3 [C6H5COO–Na+] = .............................. mol dm–3 [3] (d) A 10.0 cm3 sample of the buffer solution is mixed with 10.0 cm3 of 1.00 mol dm–3 KOH. Both solutions are at 298 K. A reaction is allowed to occur without stirring. Two observations are recorded: ● the temperature, after the reaction is complete, is fractionally above 298 K ● the pH, after the reaction, is greater than 13. Explain these two observations. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (e) Magnesium benzoate, Mg(C6H5COO)2, has a solubility in water of less than 1.00 g dm–3 at 298 K. × 10–7 at 298 K Ksp = [Mg2+][C6H5COO–]2 = 1.76 (i) Calculate the solubility of Mg(C6H5COO)2 in water at 298 K. Give your answer in g dm–3. Show your working. [Mr: Mg(C6H5COO)2 , 266.3] solubility = .............................. g dm–3 [2] (ii) An excess of Mg(C6H5COO)2 is added to a sample of 0.50 mol dm–3 MgSO4 at 298 K. State whether the equilibrium concentration of Mg(C6H5COO)2 is higher than, the same as, or lower than your answer to (i). Explain your answer. The concentration is ..................................................................... the concentration in (i). explanation ........................................................................................................................ ........................................................................................................................................... [1] [Total: 11]

Mark scheme: 2(a) resists pH change when small amount of acid or alkali is added [1] 1 2(b)(i) C6H5COOH + NaOH → C6H5COO–Na+ + H2O [1] 1 2(b)(ii) C6H5COO–Na+ + HNO3 → C6H5COOH + NaNO3 [1] 1 2(c) [H+] = 7.08  10–5 [1] 3 [C6H5COOH] = 8.20  10–3 [1] [C6H5COO–Na+] = 7.31  10–3 [1] 2(d) neutralisation is exothermic [1] 2 All of the C6H5COOH has reacted AND excess KOH 1] 2(e)(i) If [X] = [Mg2+] = [Mg(C6H5COO–)2] 2 4X3 = 1.76  10–7 so X = 3.53  10–3 [1] 0.940 g dm–3 [1] 2(e)(ii) lower than AND common ion effect [1] 1

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3 Some electrode potentials are shown in Table 3.1. Table 3.1 electrode reaction E o– / V V2+ + 2e– V –1.20 V3+ + e– V2+ –0.26 VO2+ + 2H+ + e– V3+ + H2O +0.34 VO2+ + 2H+ + e– VO2+ + H2O +1.00 Fe2+ + 2e– Fe –0.44 Fe3+ + 3e– Fe –0.04 Fe3+ + e– Fe2+ +0.77 2H+ + 2e– H2 0.00 Cl O– + H2O + 2e– Cl – + 2OH– +0.89 (a) (i) Complete the diagram to show a standard hydrogen electrode. Label your diagram. Identify all substances. You do not need to state standard conditions. [1] (ii) An electrochemical cell is set up using an Fe3+ / Fe2+ electrode and a standard hydrogen electrode. Identify the positive electrode in the electrochemical cell and the direction of electron flow in the external circuit. positive electrode .............................................................................................................. Electrons flow from the .............................. electrode to the .............................. electrode. [1] (b) The vanadium-containing species in the electrode reactions given in Table 3.1 are V, V2+, V3+, VO2+ and VO2+. (i) Identify one vanadium-containing species that does not react with Fe2+ ions under standard conditions. Use data from Table 3.1 to explain your answer. ........................................................................................................................................... ..................................................................................................................................... [1] (ii) Identify all the vanadium-containing species that will react with Fe2+ ions under standard conditions. ..................................................................................................................................... [1] (iii) Write an equation for one of the possible reactions identified in (ii). ..................................................................................................................................... [1] (c) Another electrochemical cell is set up using an Fe3+ / Fe2+ electrode and an alkaline Cl O– / Cl – electrode. The concentration of Fe3+ is 1000 times greater than the concentration of Fe2+ in the Fe3+ / Fe2+ electrode. All other conditions are standard. (i) Use the Nernst equation to calculate the E value of the Fe3+ / Fe2+ electrode. Show your working. E = .............................. V [2] (ii) Write an equation for the reaction that occurs in the cell, under these conditions. ..................................................................................................................................... [1] (d) Another electrochemical cell is set up using an Fe2+ / Fe electrode and an alkaline Cl O– / Cl – electrode under standard conditions. Calculate the value of ΔG o– for the cell. ΔGo – = .............................. kJ mol–1 [3] (e) A solution of iron(II) sulfate, FeSO4(aq) is electrolysed with iron electrodes. Under the conditions used, no gas is evolved at the cathode. A current of 0.640 A is passed for 17.0 minutes. The mass of the cathode increases by 0.185 g. Use these results to calculate an experimental value for the Avogadro constant, L. Show your working. L = .............................. mol–1 [3] (f) Iron(II) chloride, FeCl2, is oxidised by chlorine to form iron(III) chloride, FeCl 3, under standard conditions. 2FeCl 2(s) + Cl 2(g) 2FeCl 3(s) ΔH o– = –128 kJ mol–1 Table 3.2 species S o– / J K–1 mol–1 Cl 2(g) 223 FeCl 2(s) 120 FeCl 3(s) 142 (i) Use Table 3.2 and other data to calculate the Gibbs free energy change, ΔG o–, for this reaction. Show your working. ΔG o– = .............................. kJ mol–1 [3] (ii) Predict whether this reaction becomes more or less feasible at a higher temperature. Explain your answer. The reaction becomes .............................. feasible. explanation ........................................................................................................................ ........................................................................................................................................... [1] [Total: 18]

Mark scheme: 3(a)(i) hydrogen, delivery system, H+, platinum, [1] 1 3(a)(ii) iron . . . . . hydrogen . . . . . iron [1] 1 3(b)(i) (for specified V2+, V3+ or VO2+) Eo is more positive than / above –0.44 AND more negative than / below 0.77 V [1] 1 3(b)(ii) V and VO2+ [1] 1 3(b)(iii) V + Fe2+ → V2+ + Fe OR 1 VO2+ + 2H+ + Fe2+ → VO2+ + H2O + Fe3+ [1] 3(c)(i) Nernst: E = 0.77 + (0.059 / z)log[ox] / [red] [1] 2 0.947 [1] 3(c)(ii) 2Fe3+ + Cl– + 2OH– → 2Fe2+ + ClO– + H2O [1] 1 3(d) Eocell = 1.33 V [1] 3 Go = –nEocellF [1] –257 [1] 3(e) 0.64  17  60 = 653 / 652.8 Coulombs [1] 3 652.8 ÷ 1.6x10–19 = 4.08  1021 (number of electrons) 4.08  1021 ÷ 2 = 2.04  1021 (number of atoms Fe) [1] 0.185 ÷ 55.8 = 3.31  10–3 (number of moles Fe atoms 2.04  1021 ÷ 3.31  10–3 = L = 6.153  1023 [1] 3(f)(i) S = –179 [1] 3 G = H –TS [1] –74.7 [1] 3(f)(ii) less AND G becomes more positive [1] 1

More questions on Standard electrode potentials E ⦵, standard cell potentials E ⦵ cell and the Nernst equation

4 The structure of the polydentate ligand, EDTA4–, is shown in Fig. 4.1. O O O CCH2 CH2C O NCH2CH2N O CCH2 CH2C O O O Fig. 4.1 The stability constants, at 298 K, of five octahedral complexes are given in Table 4.1. Table 4.1 complex Kstab [Cu(EDTA)]2– 6.31 × 1019 [Cr(EDTA)]2– 1.00 × 1013 [Cr(EDTA)]– 1.00 × 1024 [Fe(EDTA)]2– 2.00 × 1014 [Fe(EDTA)]– 1.26 × 1025 (a) Define stability constant. ................................................................................................................................................... ............................................................................................................................................. [1] (b) Calculate the oxidation states of Cu in [Cu(EDTA)]2– and Cr in [Cr(EDTA)]–. Cu .............................. Cr .............................. [1] (c) Deduce the number of lone pairs donated by each EDTA4– ligand in a single [Fe(EDTA)]2– complex ion. ............................................................................................................................................. [1] (d) Identify the most stable complex in Table 4.1. Explain your choice. ................................................................................................................................................... ............................................................................................................................................. [1] × 10–10 mol dm–3 and(e) In a solution at equilibrium at 298 K, [[Cu(H2O)6]2+] = 3.00 [EDTA4–] = 5.00 × 10–12 mol dm–3. Use the expression for Kstab to calculate the concentration of [Cu(EDTA)]2– in this solution. Show your working. [[Cu(EDTA)]2–] = .............................. mol dm–3 [2] (f) A solution of [Cu(EDTA)]2– ions is pale blue while a solution of [Cu(NH3)4(H2O)2]2+ ions is deep blue. Explain this difference in colour. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] [Total: 8]

Mark scheme: 4(a) equilibrium constant for the formation of a complex in a solvent / 1 from its constituents [1] 4(b) +2 AND + 3 [1] 1 4(c) Six [1] 1 4(d) [Fe(EDTA)]– AND largest Kstab [1] 1 4(e) [[CuEDTA]2–] ÷ ([[Cu(H2O)6]2+]  [EDTA2–]) = 6.311019 [1] 2 0.095 [1] 4(f) different E OR different energy gap between d-orbitals [1] 2 absorption of different wavelength OR absorption of different frequency [1]

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5 Some of the ionic compounds of Group 2 elements undergo thermal decomposition. Thermal decomposition of solid anhydrous magnesium ethanedioate, MgC2O4 , occurs above 650 °C. The products are magnesium oxide and a mixture of two different gases, one of which gives a white precipitate with saturated calcium hydroxide solution. (a) Complete the equation for the thermal decomposition of MgC2O4. MgC2O4 [1] (b) Suggest which of MgC2O4 or CaC2O4 undergoes thermal decomposition at a lower temperature. Explain your answer. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (c) The ethanedioate ion is oxidised by acidified KMnO4. 5C2O42– + 2MnO4– + 16H+ 10CO2 + 2Mn2+ + 8H2O An experiment is performed to find the solubility of MgC2O4 in water. A 40.0 cm3 sample of saturated aqueous MgC2O4 requires 27.05 cm3 of 0.00200 mol dm–3 acidified KMnO4 to oxidise all the C2O42– ions. Calculate the solubility, in mol dm–3, of MgC2O4 in water. Show your working. solubility = .............................. mol dm–3 [3] [Total: 6]

Mark scheme: 5(a) MgC2O4 → MgO + CO2 + CO [1] 1 5(b) magnesium ethanedioate decomposes at lower T because Mg2+ has smaller radius than Ca2+ [1] 2 so anion is more polarised by Mg2+ [1] 5(c) (27.05 ÷ 1000)  0.02 = 5.41  10–5 moles MnO4– [1] 3 5.41  10–5  5/2 = 1.3525  10–4 moles C2O42– [1] 0.00338 mol dm–3 [1]

More questions on Similarities and trends in the properties of the Group 2 metals, magnesium to barium, and their compounds

6 (a) Phosphine, :PH3, and carbon monoxide, :CO, are monodentate ligands found in some transition element complexes. (i) Define monodentate ligand. ........................................................................................................................................... ..................................................................................................................................... [1] (ii) Define transition element complex. ........................................................................................................................................... ..................................................................................................................................... [1] (iii) Explain why transition elements form complexes. ........................................................................................................................................... ..................................................................................................................................... [1] (b) The formulae of six complexes are given in Table 6.1. The abbreviation en is used for 1,2-diaminoethane. The abbreviation dien is used for the tridentate ligand H2NCH2CH2NHCH2CH2NH2. The dien ligand forms three bonds to the gold ion in [Au(dien)(H2O)2Cl ]2+ and Au(dien)Cl3. These three bonds all lie in the same plane. The CO ligand coordinates through the carbon atom in [Rh(CO)2Cl 2]+. Table 6.1 formula isomerism shown geometry [Rh(en)2Cl 2]+ yes [Rh(CO)2Cl 2]+ yes [Au(dien)(H2O)2Cl ]2+ Au(dien)Cl 3 no octahedral Ni(PH3)2Cl 2 no [Ni(H2O)2(NH3)4]2+ yes (i) Complete Table 6.1 to state the geometry of the first three complexes. Each complex is either square planar, tetrahedral or octahedral. [1] (ii) Use complexes [Au(dien)(H2O)2Cl ]2+ and Au(dien)Cl 3 to write an equation showing ligand exchange. ..................................................................................................................................... [1] (iii) Draw the three-dimensional structure of Au(dien)Cl 3 in the box. The dien ligand can be drawn as N . N N Au [1] (iv) Draw the three-dimensional structure of Ni(PH3)2Cl 2 in the box. Ni [1] (v) One of the complexes, [Rh(en)2Cl 2]+ or [Rh(CO)2Cl 2]+, can exist in three isomeric forms. Identify this complex and the types of isomerism shown. ........................................................................................................................................... ..................................................................................................................................... [1] (vi) Draw the three-dimensional structures of the two isomers of [Ni(H2O)2(NH3)4]2+ in the boxes and identify the type of isomerism shown. Ni Ni type of isomerism shown .............................. [2] [Total: 10]

Mark scheme: 6(a)(i) donates one lp to metal atom or ion [1] 1 6(a)(ii) metal atom or ion bonded to one or more ligands [1] 1 6(a)(iii) has vacant d-orbitals which are energetically accessible [1] 1 6(b)(i) octahedral 1 square planar octahedral [1] 6(b(ii) [Au(dien)(H2O)2Cl]2+ + 2Cl– → [Au(dien)Cl 3] + 2H2O [1] 1 OR [Au(dien)Cl 3] + 2H2O → [Au(dien)(H2O)2Cl]2+ + 2Cl– 6(b)(iii) [1] 6(b)(iv) 1 [1] 6(b)(v) [Rh(en)2Cl2]+ AND 1 optical AND cis / trans OR geometric [1] 6(b)(vi) 2 [1] cis/trans OR geometric [1]

More questions on General characteristic chemical properties of the first set of transition elements, titanium to copper

7 Benzene can be used to make benzoic acid in the two-step process shown in Fig. 7.1. benzene methylbenzene benzoic acid step 1 step 2 CH3 COOH Fig. 7.1 (a) Give the reagents and conditions for step 1 and step 2. step 1 ........................................................................................................................................ step 2 ........................................................................................................................................ [2] (b) Methylbenzene and benzoic acid each have five different peaks in the carbon (13C) NMR spectrum. Table 7.1 hybridisation environment of carbon example chemical of the carbon atom shift range atom / ppm sp3 alkyl CH3–, –CH2–, –CH<, >C< 0–50 sp3 next to alkene/arene –C–C=C, –C–Ar 25–50 sp3 next to carbonyl/carboxyl C–COR, C–O2R 30–65 sp3 next to halogen C–X 30–60 sp3 next to oxygen C–O 50–70 c c sp2 alkene or arene >C=C<, c c 110–160 c c sp2 carboxyl R–COOH, R–COOR 160–185 sp2 carbonyl R–CHO, R–CO–R 190–220 sp nitrile R–C≡N 100–125 Use Table 7.1 to complete the two sentences to suggest descriptions of these two spectra. The carbon (13C) NMR spectrum of methylbenzene: ● has ................. peak(s) in the chemical shift range of ..................... and ● has ................. peak(s) in the chemical shift range of ..................... . The carbon (13C) NMR spectrum of benzoic acid: ● has ................. peak(s) in the chemical shift range of ..................... and ● has ................. peak(s) in the chemical shift range of ..................... . [2] (c) (i) When treated with Cl 2 under suitable conditions, methylbenzene forms compound J. When treated with Cl 2 under different conditions with different reagents, methylbenzene forms compound K. Suggest and draw structures of compounds J and K in the boxes. The molecular formula of each compound is given. methylbenzene J C7H7Cl K C7H7Cl CH3 State the reagents and conditions required to form each product. to form compound J .......................................................................................................... to form compound K .......................................................................................................... [4] (ii) When treated with a chlorine-containing reagent under suitable conditions, benzoic acid forms compound L. When treated with a different chlorine-containing reagent under different conditions, benzoic acid forms compound M. Suggest and draw structures of compounds L and M in the boxes. The molecular formula of each product is given. benzoic acid L C7H5ClO2 M C7H5ClO COOH State the reagents and conditions to form compound M from benzoic acid. ..................................................................................................................................... [3] [Total: 11]

Mark scheme: 7(a) CH3Cl + AlCl3 [1] 2 hot alkaline) MnO4– [1] 7(b) 4, 110-160, 2 1, 25-50 either order [1] 4, 110-160, 1, 160-185 either order [1] 7(c)(i) Mark J and K together. 4 either or J or K [1] J or K [1] for the ring substituted product – AlCl3 [1] for the side-chain substituted product – uv [1] 7(c)(ii) 3 L [1] M [1] PCl3 + heat OR SOCl2 OR PCl5 [1]

8 Lactic acid, CH3CH(OH)COOH, is the only monomer needed to form the polymer polylactic acid, PLA. (a) (i) Draw a short length of the PLA polymer chain, including a minimum of two monomer residues. The methyl groups may be written as –CH3 but all other bonds should be shown fully displayed. Label one repeat unit of polylactic acid on your diagram. [2] (ii) Give the name of the type of polymerisation involved in the formation of PLA and the name of the functional group that forms between the monomers. type of polymerisation ....................................................................................................... functional group ................................................................................................................. [1] (iii) Predict whether PLA is readily biodegradable. Explain your answer. ........................................................................................................................................... ..................................................................................................................................... [1] (b) The proton (1H) NMR spectrum of CH3CH(OH)COOH in CDCl 3 is shown in Fig. 8.1. The proton NMR chemical shift ranges are shown in Table 8.1. Lactic acid 12 11 10 9 8 7 6 5 4 3 2 1 0 δ / ppm Fig. 8.1 Table 8.1 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 (i) Use Fig. 8.1 and Table 8.1 to complete Table 8.2. Table 8.2 proton environment chemical shift (δ) name of splitting pattern –COOH CH –OH –CH3 [3] (ii) Name the substance responsible for the peak at δ = 0.0. ..................................................................................................................................... [1] (iii) Explain why CDCl 3 is a better solvent than CHCl 3 for use in proton NMR. ........................................................................................................................................... ..................................................................................................................................... [1] (c) An impure sample of CH3CH(OH)COOH contains pentan-3-one as the only contaminant. The mixture is analysed using gas/liquid chromatography. The pentan-3-one is found to have a longer retention time than the lactic acid. (i) Explain what is meant by retention time. ........................................................................................................................................... ..................................................................................................................................... [1] (ii) Suggest suitable substances, or types of substances, that could be used as the mobile and stationary phases. mobile phase ..................................................................................................................... stationary phase ................................................................................................................ [1] (iii) Describe how the percentage composition of the mixture can be determined from the gas / liquid chromatogram. ........................................................................................................................................... ..................................................................................................................................... [1] [Total: 12]

Mark scheme: 8(a)(i) 2 diagram has trailing bonds, two or more monomer residues AND either repeat unit correct OR ester linkage correct [1] diagram fully correct [1] 8(a)(ii) condensation AND ester [1] 1 8(a)(iii) it is biodegradable because it can hydrolyse [1] 1 8(b)(i) d values: 3 12, 4.4 [1] 5.0, 1.6 [1] splitting patterns: singlet, quartet / quadruplet, singlet, doublet [1] 8(b)(ii) TMS / tetramethylsilane [1] 1 8(b)(iii) CDCl3 will not give an absorption / peak OR 1 CHCl3 will give an absorption / peak [1] 8(c)(i) time between injection and detection [1] 1 8(c)(ii) an unreactive gas AND a non-polar liquid [1] 1 8(c)(iii) area of peak divided by total area of all peaks  100% [1] 1

9 (a) State the reactants and conditions for two different types of reactions that both produce diethylamine, CH3CH2NHCH2CH3. reaction one .............................................................................................................................. ................................................................................................................................................... reaction two ............................................................................................................................... ................................................................................................................................................... [4] (b) Describe the relative basicities of diethylamine, phenylamine and ammonia in aqueous solution. Explain your answer in terms of structure. ............................................... ............................................... ............................................... least basic most basic explanation ............................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... [3] (c) Phenylamine reacts with HNO2(aq) at 4° C to form compound P. Compound P reacts with phenol under alkaline conditions at 4° C. The product of this reaction is acidified, forming azo compound Q. Draw the structure of compound Q. Circle the azo group on your structure. State one use of an azo compound such as Q. compound Q: An azo compound can be used .............................................................................................. . [2] (d) CH3CH2NHCH2CH3 reacts with ethanoyl chloride, CH3COCl, to give the amide N,N-diethylethanamide, CH3CON(C2H5)2. An incomplete description of the mechanism of this reaction is shown in Fig. 9.1. box one box two box three O Cl Cl H3C C N C2H5 H3C C O H3C C O C2H5 N C2H5 N C2H5 H H C2H5 C2H5 and ......................... reactants intermediate products Fig. 9.1 (i) Complete the mechanism in Fig. 9.1. You should include: ● all relevant dipoles (δ+ and δ–) and full electric charges (+ and –) on the species in box one and in box two ● all relevant lone pairs on the species in box one and in box two ● all relevant curly arrows to show the movement of electron pairs in box one and in box two ● the formula of the second product in box three. [4] (ii) Name this mechanism. ..................................................................................................................................... [1] [Total: 14]

Mark scheme: 9(a) chloroethane and ethylamine [1] 4 heat, pressure, ethanol [1] N-ethyl ethanamide CH3CONHC2H5 [1] LiAlH4 [1] Other syntheses that work also got credit / 9(b) phenylamine, ammonia, diethylamine [1] 3 • Availability of lone pair on N to receive H+ • phenylamine – LP, delocalised / overlaps into ring [1] • diethylamine – ethyl are electron donating [1] Two bullet points for one mark, three bullet points for two marks. 9(c) 2 Q or [1] N2 or N=N circled AND dye / colouring[1] 9(d)(i) box one box two box three 11 HCl box one: 4 1 dipole on C=O 2 LP on N of amine [1] 3 CA from N of amine to C of acyl chloride 4 CA from = to O of acyl chloride [1] box two: 5 + on what was N of amine and nowhere else 6 – on what was O of acyl chloride and nowhere else 7 LP on what was O of acyl chloride [1] 8 CA from N-H bond to what was N of amine 9 CA from C-Cl bond to what was Cl of acyl chloride 10 CA from (LP on) O to reform C=O [1] box three: 11 HCl 9(d)(ii) Addition-elimination 1

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