Cambridge A Level Chemistry 9701 — 2019 Oct/Nov Paper 5 · Variant 2

1 Many organic reactions performed in the laboratory require heating under reflux. 4‑nitromethylbenzene can be converted directly to 4‑aminobenzaldehyde by heating it under reflux with an excess of solution A. Solution A is an aqueous alkaline mixture of Na2S•9H2O and S8. Ethanol is used as a solvent for the reaction. 4-nitromethylbenzene 4-aminobenzaldehyde CH3 CHO solution A O2N H2N The apparatus for this reaction is shown in the diagram. reflux condenser tap funnel solution A round-bottomed flask 4-nitromethylbenzene dissolved in ethanol (a) (i) Show, using a labelled arrow, where the cooling water enters the reflux condenser. [1] (ii) After the complete addition of solution A to the solution in the round-bottomed flask, a vigorous exothermic reaction takes place. What should be done to control the vigorous exothermic reaction during the addition of solution A? ............................................................................................................................................. ....................................................................................................................................... [1] (iii) The mixture is heated to increase the rate of reaction. This is carried out under reflux. Other than to increase the rate of reaction, explain why a reflux process is used for heating the mixture. ............................................................................................................................................. ....................................................................................................................................... [1] (iv) A student suggests that placing a bung in the top of the condenser would improve the experiment. Comment on whether the student is correct. Explain your answer. ............................................................................................................................................. ............................................................................................................................................. ....................................................................................................................................... [1] (b) Once the reaction is complete, the round-bottomed flask is removed from the set‑up on page 2 and included in a new set-up for distillation. The distillation process separates the substances in the mixture in the flask. (i) Complete and label the diagram to show how the apparatus could be set up for distillation. [2] (ii) The mixture in the round-bottomed flask is heated until all of the first substance distils off. The heating is then stopped. After distillation a hot solution remains in the round-bottomed flask. The table shows some data relevant to the experiment. substance melting point / °C boiling point / °C CHO 72 138 H2N H2O 0 100 CH3CH2OH –114 78 Use the data to identify the liquid that is first distilled off. Explain your answer. ............................................................................................................................................. ....................................................................................................................................... [1] (iii) 4‑aminobenzaldehyde is soluble in hot water, but not in cold water. Starting with the hot solution in the round‑bottomed flask, suggest how pure crystals of 4‑aminobenzaldehyde can be obtained. ............................................................................................................................................. ............................................................................................................................................. ....................................................................................................................................... [2] (c) Under acidic conditions, 4‑aminobenzaldehyde (C7H7NO) forms a solid compound with molecular formula C14H12N2O. 2C7H7NO C14H12N2O + H2O A student carries out two experiments according to the method on page 2 to make C7H7NO. In Experiment 1, the student obtains 6.00 g of solid. In Experiment 2, the student uses the same method as in Experiment 1 but adds acid to neutralise the alkali in the mixture in the flask after heating under reflux. The same quantities of reagents are used in Experiment 1 and Experiment 2. Suggest how the mass of solid obtained in Experiment 2 compares to that of Experiment 1. Explain your answer. .................................................................................................................................................... .................................................................................................................................................... .............................................................................................................................................. [2] (d) 4‑aminobenzaldehyde can be reacted with a halogenoalkane to form 4-(dimethylamino)benzaldehyde, compound D. 4-aminobenzaldehyde compound D CHO CHO halogenoalkane H2N (CH3)2N Some data is given that is relevant to this reaction. carbon‑halogen boiling point halogenoalkane bond energy / °C / kJ mol–1 CH3Cl –97 340 CH3Br 4 280 CH3I 43 240 Give two reasons why CH3I is the preferred halogenoalkane in the conversion of 4‑aminobenzaldehyde to compound D. 1 ................................................................................................................................................. .................................................................................................................................................... 2 ................................................................................................................................................. .................................................................................................................................................... [2] [Total: 13]

Mark scheme: 1(a)(i) arrow pointing inwards at bottom condenser inlet 1 1(a)(ii) Solution A / alkali must be added dropwise 1 1(a)(iii) to prevent vapour loss / escape of vapour 1 1(a)(iv) (incorrect) heating causes gases to expand OR Pressure would build up (causing apparatus to pop apart) 1 1(b)(i) M1: Condenser sloping in the ‘distillation’ position able to accept distillate vapours M2: Condenser with (delivery) end open and rest of apparatus sealed 2 1(b)(ii) CH3CH2OH AND lowest boiling point. 1 1(b)(iii) M1: Cool (the solution) (until crystals form) M2: Filter AND rinse (the crystals) [with (cold) water]. 2 1(c) M1: mass obtained would be less M2: because of H2O loss / formation / production 2 1(d) M1: (only) CH3I is a liquid at room temperature M2: (C—I) bond weakest / easiest to break 2

2 Naphthalene, C10H8, is an aromatic organic compound that is a solid at room temperature and pressure. The enthalpy change of fusion, ΔHfusion, of naphthalene is the enthalpy change that occurs when naphthalene melts. It cannot be directly measured. C10H8(s) C10H8(l) The melting point of naphthalene depends on its purity. When other compounds dissolve in liquid naphthalene its melting point is lowered. Diphenylamine, (C6H5)2NH, dissolves in liquid naphthalene. By varying the amount of diphenylamine and measuring the melting points of the different mixtures, the enthalpy change of fusion of naphthalene can be calculated. (a) The equation that links the melting point of a naphthalene-diphenylamine mixture with its enthalpy change of fusion, ΔHfusion, is shown. ΔHfusion log Y = A – 2.30 × RTm Y = mole fraction of naphthalene R = molar gas constant, 8.31 J K–1 mol–1 Tm = melting point of naphthalene in K A is a constant (i) A series of experiments is performed using the same amount, 0.100 mol, of naphthalene each time. Calculate the mass of naphthalene, C10H8, that should be used for each of these experiments. [Ar: C, 12.0; H, 1.0] mass of naphthalene = .............................. g [1] (ii) The melting point and freezing point of a substance are the same. The melting point, Tm, of a substance can be found by recording the temperature at which the substance freezes, measured when crystals first start to appear on cooling. The results of a series of experiments using 0.100 mol of naphthalene and different masses of diphenylamine are shown. Process the results to complete the table. Record all your data to three significant figures. The mole fraction of naphthalene, Y, is calculated as shown. nN Y = nN + nD nN = amount in moles of naphthalene, C10H8, = 0.100 nD = amount in moles of diphenylamine, (C6H5)2NH temperature at amount of mole fraction which crystals (C6H5)2NH, nD log Y of C10H8, Y appear, Tm / mol / 10–3 K–1 / K 0.00 1.00 353 2.83 0.00 0.00888 349 0.0178 345 0.0266 341 0.0355 338 0.0444 334 0.0533 331 0.0621 329 0.0769 325 [3]

Mark scheme: 2(a)(i) 1 2(a)(ii) Mol Y temp 1/Tm Log Y 0.00 1.00« 1.00 1.00 353 2.83 0.00 0.00 0.00 0.00888 0.91844« 0.918 0.92 349 2.87 -0.0369 -0.0372 -0.0362 0.0178 0.84889« 0.849 0.85 345 2.90 -0.0711 -0.0711 -0.0706 0.0266 0.78988« 0.790 0.79 341 2.93 -0.102 -0.102 -0.102 0.0355 0.73800« 0.738 0.74 338 2.96 -0.132 -0.132 -0.131 0.0444 0.69252« 0.693 0.69 334 2.99 -0.160 -0.159 -0.161 0.0533 0.65231« 0.652 0.65 331 3.02 -0.186 -0.186 -0.187 0.0621 0.61690« 0.617 0.62 329 3.04 -0.210 -0.210 -0.208 0.0769 0.56529« 0.565 0.57 325 3.08 -0.248 -0.248 -0.244 M1: for column 4, ( 1 Tm) M2: for column 5 (log Y) M3: All values in columns 4 and 5 to 3 SF 3 2(b) M1: nine points plotted correctly including 2.83, 0.00. M2: best-fit straight line drawn 2 2(c)(i) M1: coordinates read & recorded correctly M2: gradient determined 2 Question Answer Marks 2(c)(ii) –∆Hfusion 2.30×R ∆Hfusion = (– 2(c)(i) × 2.30 × 8.31) / 1000 1 2(d)(i) Yes, (the data is reliable because) AND there are no anomalous points OR only 1, 2 or 3 anomalous points OR only a few points are not on/near the line OR most points on the line OR all the points are on the line OR No AND one or more anomalous points OR one or more points away from the line 1 2(d)(ii) M1: literature values: lower limit = 1.45 ÷ (10.00/128.0) = 18.56 (kJ mol–1) AND upper limit = 1.47 ÷ (10.00/128.0) = 18.82 / 18.816 (kJ mol–1) M2: experiment is not (very) accurate because experimental ∆Hfusion is outside literature range 2 2(e) M1: (on mixing) ∆H1 would be more positive M2: ∆Hmixing is positive / >0 OR ∆H1 = ∆Hfusion + ∆Hmixing 2 2(f)(i) Y is smaller AND because nD is larger (and N N D = + n Y n n ) 1 Question Answer Marks 2(f)(ii) If Y is smaller in 2(f)(i) M1: ∆Hfusion will be more positive (than actual value) M2: log Y is more negative OR Gradient is more negative If Y is greater in 2(f)(i) M1: ∆Hfusion will be less positive (than actual value) M2: log Y is less negative OR Gradient is less negative 2

More questions on Reacting masses and volumes (of solutions and gases)