CJC Prelim P3 Answers

9729/03/CJC JC2 Preliminary Examination 2018 CANDIDATE NAME CLASS 2T CHEMISTRY 9729/03 Paper 3 Free Response Friday 24 August 2018 2 hours Candidates answer on separate paper. Additional Materials: Answer Paper Data Booklet READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Section A Answer all questions. Section B Answer one question. A Data Booklet is provided. The use of an approved scientific calculator is expected, where appropriate. At the end of examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 8 printed pages. Catholic Junior College JC2 Preliminary Examinations Higher 2 WORKED SOLUTIONS

2 9729/03/CJC JC2 Preliminary Examination 2018 Section A Answer all the questions in this section. 1 Propene is a colourless gas that is produced in large quantities by the petrochemical industry. It is used in the production of synthetic rubber and as a propellant in aerosols. (a) When propene reacts with HBr, 2-bromopropane is produced as the major product. CH2=CHCH3 + HBr → CH3CHBrCH3 (i) Describe the mechanism of the above reaction, using curly arrows to show the movement of electrons. [4] Electrophilic addition Correct partial charges, lone pair of electrons, slow/fast steps Correct curly arrows Correct secondary carbocation intermediate (ii) When propene reacts with BrC l, 1-bromo-2-chloropropane is produced as the major product. CH2=CHCH3 + BrCl → CH2BrCHClCH3 With reference to your answer in (i), explain why the two reactions give major products with bromine in different positions. [2] Br is more electronegative than H but less electronegative than C l, hence Br acquires a partial positive charge in BrCl .Thus, Br adds first to form the more stable carbocation where the positive charge is on the second carbon (iii) The product of the reaction in (ii) has a chiral carbon, but it is optically inactive. Explain why this is so. [2] There is equal chance of C l‒ attacking the trigonal planar carbon in the carbocation intermediate from either side of the plane. Thus, both enantiomers are formed in equal amounts/a racemic mixture is formed.

3 9729/03/CJC JC2 Preliminary Examination 2018 [Turn over (b) 2-bromopropane reacts with hot aqueous sodium hydroxide, whereas 2-bromopropene does not. Suggest reasons for this difference in reactivity. [2] In 2-bromopropene, p orbitals of Br overlap with π bond (one of the lone pairs of electrons on Br is delocalised into the C=C π bond), hence there is a partial double bond character for the C-Br bond. This makes the C-Br bond stronger and harder to break for nucleophilic substitution to occur. The carbon of the C–Br bond also has a lower partial positive charge and is less susceptible to nucleophilic attacks. The electron rich C=C π bond will repel the negatively charged incoming nucleophile, hence the attack of the nucleophile will be less likely to occur. (c) Propene can undergo mild oxidation in the presence of cold acidified potassium manganate(VII), to give CH2(OH)CH(OH)CH3. (i) State the IUPAC name of the product of the above reaction. [1] Propane-1,2-diol (ii) Construct the half ‒equation for the oxidation of propene as described above. [1] CH 2=CHCH3 + 2H2O → CH2(OH)CH(OH)CH3 + 2H+ + 2e‒ (iii) Hence, write the equation for the overall reaction, showing clearly the stoichiometry of reaction between propene and manganate(VII) ions. [1] 5CH2=CHCH3 + 2MnO4‒ + 2H2O + 6H+ → 5CH2(OH)CH(OH)CH3 + 2Mn2+ (d) At room temperature and pressure, 28 cm 3 of propene was bubbled into 40.0 cm3 of 0.0200 mol dm‒3 acidified KMnO4(aq). The resulting solution was titrated against Fe2+(aq) of concentration 0.0750 mol dm‒3. (i) State the colour change at endpoint for this titration. [1] Pink/red-brown to colourless/yellow (ii) Given that 5 moles of Fe 2+ react with 1 mole of MnO4 ‒, determine the volume of Fe2+(aq) needed to reach endpoint. [3] No. of moles of propene = ૛ૡ ૛૝૙૙૙ = 1.167 x 10 ‒3 mol No. of moles of MnO 4‒ reacted with propene = ૚.૚૟ૠ×૚૙ష૜ ૞ ×૛ = 4.667 x 10 ‒4 mol No. of moles of MnO 4‒ originally = ૝૙.૙ ૚૙૙૙ × ૙. ૙૛૙૙ = 8.000 x 10 ‒4 mol

4 9729/03/CJC JC2 Preliminary Examination 2018 No. of moles of MnO 4‒ left = 8.000 x 10 ‒4 ‒ 4.667 x 10‒4 = 3.333 x 10 ‒4 mol No. of moles of Fe 2+ = 3.333 x 10‒4 x 5 = 1.667 x 10 ‒3 mol Volume of Fe 2+ = ૚.૟૟ૠ×૚૙ష૜ ૙.૙ૠ૞૙ ×૚ ૙ ૙ ૙ = 22.2 cm 3 (e) A mixture of propene ( Mr = 42.0) and 2-bromopropane (Mr = 122.9) kept in a vessel of volume of 3.60 dm3 maintained at 75 oC exerts a pressure of 1.66 x 105 Pa. The mole fraction of propene in the mixture is 0.28. Find the mass of the mixture of gases. [3] [Total: 20] Average Mr = 0.28 x 42.0 + (1 – 0.28) x 122.9 = 100.2 pV = ࢓ ࢘ࡹ RT m = ࢘ࡹࢂ࢖ ࢀࡾ = ૚.૟૟×૚૙૞×૜.૟૙×૚૙ష૜×૚૙૙.૛ ૡ.૜૚×(ૠ૞ା૛ૠ૜) = 20.7 g

5 9729/03/CJC JC2 Preliminary Examination 2018 [Turn over 2 Copper is a transition metal, and is one of the metals used to make coins, along with silver and gold. Most copper is used in electrical equipment such as wiring and motors. It also has uses in construction and industrial machinery. (a) One of the characteristic properties of copper is its ability to form coloured aqueous complexes shown in Fig 2.1. Fig. 2.1 (i) State the complex ion A. [1] [CuC l4]2− (ii) Draw the structure of complex ion A. [2] Cl Cu Cl Cl Cl 2- II (iii) EDTA4− is a hexadentate ligand. Deduce the formula of the complex ion B. [1] [Cu(EDTA)]2− (iv) Suggest why complex B is readily formed from complex A. [1] [Cu(EDTA)]2− is more stable as compared to [CuC l4]2+, therefore C l- ligands will be displaced by the stronger EDTA4− ligands. OR EDTA4- is a stronger ligand than Cl-. (v) Explain why aqueous Cu2+ ions are blue in colour. [3] When ligands approach/are attached/bonded to the copper ion, they will cause the incompletely/ partially-filled/3d9 degenerate d-orbitals to split into two slightly different energy levels, d and d* [OR] two groups of non-degenerate d-orbitals with small energy gap. When electrons from the lower lying d-orbitals absorbs energy (orange colour) in the visible light region, it will be excited to the higher energy d* orbital. This is known as d-d* electronic transition. The complementary colours, which is not absorbed which is blue is seen/ the colour observed is complementary to the colour that is absorbed.

6 9729/03/CJC JC2 Preliminary Examination 2018 (vi) The numerical value of the solubility product, Ksp of Cu(OH)2 is 2.20 x 10−20 at 25 oC. Write the expression for Ksp and state its units. Calculate the solubility of Cu(OH)2 at 25 oC. [3] Cu(OH)2(s) ⇌ Cu2+(aq) + 2OH-(aq) Ksp = [Cu2+][OH–]2 Units: mol3 dm−9 Let the solubility of Cu(OH)2 be x mol dm-3 Ksp = [Cu2+][OH–]2 = x(2x)2 2.20 x 10–20 = 4x3 x = 1.77 × 10–7 mol dm–3 (b) When an aqueous solution of ammonia is shaken with an organic solvent, trichloromethane, at room temperature, the following equilibrium is set up between the two immiscible liquids. NH 3(aq) ⇌ NH3(organic) equilibrium 1 The equilibrium constant, Kc, for this reaction is 0.04. Water that is contaminated with copper(II) sulfate is harmful to crops, animals and humans. Some water that has been contaminated with 0.100 mol dm−3 copper(II) sulfate was allowed to reach dynamic equilibrium at room temperature with an excess of ammonia and trichloromethane. Th