DHS Prelim P3 Suggested Solutions

This question paper consists of 13 printed pages and 1 blank page. © DHS 2018 [Turn over Suggested solutions DUNMAN HIGH SCHOOL Preliminary Examination 2018 Year 6 H2 CHEMISTRY 9729/03 Paper 3 Free Response 18 September 2018 2 hours Candidates answer on separate paper. Additional Materials: Data Booklet Writing Paper Cover Sheet Graph Paper INSTRUCTIONS TO CANDIDATES 1 Write your name, index number and class on this cover page and on the Cover Sheet provided. 2 Write your answers on the separate writing papers provided. 3 Write in dark blue or black pen. 4 You may use an HB pencil for any diagrams or graphs. 5 Start each question on a fresh sheet of paper. *[Marks will be deducted if you fail to do so.] 6 At the end of the examination, fasten all your work securely together with the Cover Sheet on top. 7 Do not use staples, paper clips, glue or correction fluid. Section A 8 Answer all questions Section B 9 Answer one question. The number of marks is given in brackets [ ] at the end of each question or part question. You are advised to show all workings in calculations. You are reminded of the need for good English and clear presentation in your answers.

2 © DHS 2018 9729/03 Section A Answer all questions. 1 The Williamson ether synthesis is an organic reaction, forming an ether from a halogenoalkane and alcohol in the presence of sodium. This reaction was developed by Alexander Williamson in 1850 and still remains the simplest and most popular method of preparing ethers till today. The following equation shows the formation of dimethyl ether, a common aerosol propellant. CH3Cl CH3OH CH3OCH3 HCl+ Na CH3OH + (a) (i) State the purpose of sodium used in the Williamson ether synthesis. [1] Sodium reacts with CH3OH to form a stronger nucleophile, CH3O– (ii) Hence, name the mechanism of the reaction. [1] SN2 / bimolecular nucleophilic substitution (iii) Suggest suitable reagent(s) to synthesise each of the following ethers. • CH 3CH2OCH2CH3 • O [2] • CH 3CH2OH (in the presence of Na) and CH3CH2Cl • Cl OH (in the presence of Na) (b) Dimethyl ether is known as a symmetrical ether whereas tert-butyl ethyl ether, CH3CH2OC(CH3)3 is an example of an unsymmetrical ether. To prepare tert-butyl ethyl ether via the Williamson ether synthesis, there are two possible combinations of reagents, as shown in the table below. Combination Reagents A CH3CH2Br and (CH3)3COH B (CH3)3CBr and CH3CH2OH Identify the combination of reagents that might favour the mechanism identified in (a)(ii) and justify your choice, with reasoning. [2] Combination A CH3CH2Br is a primary alkyl halide , and hence it will be less sterically hindered for the nucleophile / alkoxide ion to attack from the back of the halogen as compared to (CH3)3CBr, a tertiary alkyl halide.

3 © DHS 2018 9729/03 [Turn Over (c) Since the middle of 1990s, dimethyl ether (DME) has been identified as a reliable diesel alternative for cars. The table bel ow compares the physical and chemical properties of DME and diesel fuel. Property Unit DME Diesel Fuel Carbon content mass % 52.2 86 Hydrogen content mass % 1 − 3 14 Oxygen content mass % 34.8 0 Liquid density kg m−3 667 831 *Auto-ignition temperature K 508 523 ^Stoichiometric air/fuel mass ratio - 9.6 14.6 Normal boiling point K 248.1 450 − 643 Enthalpy of vaporisation kJ kg−1 467.1 300 Energy released at burning MJ kg−1 27.6 42.5 *Auto-ignition temperature is the temperature at which a fuel will ignite spontaneously without an external ignition source. ^Stoichiometric air/fuel mass ratio is the mass ratio of air to fuel that completely burns the fuel with no excess air. With reference to the table, suggest one advantage and one disadvantage of using DME as compared to the conventional diesel fuel. [2] Advantage: DME has lower carbon content than diesel fuel, and thus contribute to lower carbon dioxide emission when the same mass is burnt. Disadvantage: DME releases less energy at burning than diesel fuel, and thus is a less efficient fuel for the same mass used. (d) Scientists have recently discovered a new way of synthesising DME by reacting carbon dioxide directly with hydrogen in the presence of Cu-Zn/Al2O3 catalyst. (i) Write the balanced equation for the synthesis of DME from carbon dioxide and hydrogen. [1] 2CO2 + 6H2 → CH3OCH3 + 3H2O (ii) State the type of catalysis that Cu-Zn/Al2O3 performs and explain briefly how it promotes the synthesis of DME. [3] Heterogeneous catalysis Both H2 and CO2 reactants are gases. The solid catalyst provides a surface for the gaseous molecules to be adsorbed to the surface, thus increasing the concentration of the reactants / bringing the molecules closer together and weakening the bonds of the reactants molecules , resulting in a lower Ea.

4 © DHS 2018 9729/03 (iii) DME can also be synthesised from carbon dioxide via a two-step reaction in the laboratory. CO2(g) + 3H2(g) → CH3OH(l) + H2O(l) ΔH = −13.1 kJ mol–1 2CH3OH(l) → CH3OCH3(l) + H2O(l) ΔH = –459.1 kJ mol–1 Using the thermochemical equations given and any relevant data in part (c), draw an energy cycle to determine the enthalpy change of reaction for the synthesis of DME from carbon dioxide and hydrogen at room temperature and pressure. [5] ΔH vap = 467.1/1000 × 46 = +21.49 kJ mol−1 By Hess’ Law, ΔH rxn = −26.2 − 459.1 + 21.49 = −464 kJ mol−1 (e) Both dimethyl ether and dimethyl amine have similar hybridisation around the heteroatoms, O and N, respectively. (i) State the type of hybridisation of the O and N atoms in dimethyl ether and dimethyl amine respectively. [1] sp 3 (ii) Dimethyl ether has a solubility of 7.1 g per litre of water but dimethyl amine has a solubility of 3.54 kg per litre of water instead. Using suitable equation(s), explain briefly the difference in the solubilities between the two compounds. [3] DME can form hydrogen bonds with water molecules whereas dimethyl amine is a weak base and can partially dissociate in water in to form ions which can form ion-dipole interactions with water molecules , hence dimethyl amine has a higher solubility in water. (CH3)2NH + H2O ⇌ (CH3)2NH2+ + OH−

5 © DHS 2018 9729/03 [Turn Over (iii) A 0.1 mol dm –3 solution of dimethyl amine containing an unknown concentration of dimethylamine hydrochloride has a pH of 10.57. Given that the numerical value of Kb of dimethyl amine is 7.4 × 10 –4, determine the concentration of dimethylamine hydrochloride in the solution. [3] A buffer system is set up. (CH3)2NH + H2O ⇌ (CH3)2NH2+ + OH− Kb = 7.4 × 10–4 mol dm−3 pH = 10.57 pOH = 14 – 10.57 = 3.43 [OH −] = 10−3.43 mol dm−3 Kb = ൣሺCH3ሻ2NH2 +൧ሾOH-ሿ ሾሺCH3ሻ2NHሿ 7.4 x 10–4 = ൣሺCH3ሻ2NH2 +൧ቂ10-3.43ቃ 0.1 [(CH3)2NH2+] = 7.4×10-4×0.1 10-3.43 = 0.199 mol dm −3 OR pOH = pK b + log10൬ [(CH3)2NH2 +] [(CH3)2NH] ൰ 14 – 10.57 = −log10(7.4 × 10–4) + log10൬ [(CH3)2NH2 +] 0.1 ൰ [(CH3)2NH2+] = 0.199 mol dm−3 (iv) Explain why trimethyl amine has a higher pKb than dimethyl amine. [1] The lone pair of electrons on N in trimet