HCI Prelim P3 Mark Scheme

2018 HCI C2 H2 Chemistry Preliminary Exam / Paper 3 Paper 3 1 (a) Aluminium has a giant metallic lattice structure [1] and metallic bonding between Al3+ ions and a sea of delocalised electrons. [1] [1] Number of electrons should be 3 times that of Al3+ ions (b) (i) Oxidising agent [1] (ii) n(S2O32–) = 17.5 × 0.1 / 1000 = 0.00175 mol n(I2) in 25.0 cm3 = 0.00175 / 2 = 0.000875 mol [1] n(I2) in 250.0 cm3 = 0.000875 × 10 = 0.00875 mol [1] ecf n(Cu2+) = 0.00875 × 2 = 0.0175 mol mass of Cu in alloy = 0.0175 × 63.5 = 1.11g percentage of mass of Cu in alloy = 1.11 / 1.20 × 100 = 92.6% [1] ecf (iii) The titre value would be higher. [1] Without step 3, Fe 3+ ion in the solution will undergo reaction with I–, causing more iodine to be produced. [1] (c) (i) V2 = 1830 cm3 [1] (ii) The volume will be smaller as CO 2 has significant dis persion forces between molecules. [1] = ×× =++ 11 2 2 12 281101 1500 30 273 23 273 PV PV TT V HWA CHONG INSTITUTION 2018 C2 H2 CHEMISTRY PRELIMINARY EXAM SUGGESTED SOLUTIONS

1 (d) (i) Cl Cl Cl Cl 6 x 1 = 6 2 x 3.5 = 7 3 x 1 = 3 1 x 5 = 5 [0.5] x 4 = [2] structures, [1] for correct proportions. (ii) Diagram [1] At higher temperatures (such as T 2), the proportion of particles with kinetic energy greater than or equal to the activation energy increases. [1] Thus, the frequency of effective collisions increases (rate constant increases), and reaction rate increases. [1] (iii) C–Cl bond in CFCs can be broken down by UV light to form chlorine radicals, which can react with ozone. [1] 2 (a) (i) Accept any of the following reasons: [1] Wittig reaction allows the formation of an alkene from a carbonyl compound Wittig reaction allows for number of carbon atoms to increase. (ii) [1] (b) Accept any of the following: [1] × 3 (selective) reduction, (acid) hydrolysis, nucleophilic substitution (acid) hydrolysis, nucleophilic substitution, (selective) reduction (acid) hydrolysis, (selective) reduction, nucleophilic substitution (selective) reduction, oxidation, nucleophilic substitution Note: DO NOT allow hydrogenation and DO NOT allow substitution Kinetic energy Ea No. of particles 0 T1 T2 T2 > T1 Proportion of particles with KE ≥ Ea at T1 Proportion of particles with KE ≥ Ea at T2 +

2018 HCI C2 H2 Chemistry Preliminary Exam / Paper 3 (c) [1] x 3 steps (d) [1] condensation [1] (e) Correct reagents and conditions 1: LiAlH4 in dry ether OR NaBH4 in methanol [1] Correct structure of intermediate [1] Correct reagents and conditions 2: excess concentrated H 2SO4, heat OR concentrated H3PO4, heat OR Al2O3, heat [1] (f) D is a constitutional isomer OR structural isomer OR positional isomer of olympicene. [1]

2 (g) (i) Olympicene is not a planar molecule since it contains a sp3 carbon which has a tetrahedral geometry around this carbon. [1] (ii) 18 π electrons [1] (iii) Olympicene has a larger electron cloud size which causes dispersion forces to be much stronger than those in benzene, hence requiring more energy to overcome. [2] (h) disproportionation [1] 3 (a) (i) Let the solubility of MgCO3 and Mg(OH)2 be x and y respectively in mol dm–3. For MgCO 3 Ksp = [Mg2+][CO32–] 1.0 × 10–5 = (x)(x) x = 3.16 × 10–3 mol dm–3 [1] For Mg(OH)2 Ksp = [Mg2+][OH-]2 1.1 × 10–11 = (y)(2y)2 y = 1.40 × 10–4 mol dm–3 [1] (ii) [Mg2+][OH–]2 = 1.1 × 10–11 (3.0 × 10–5)[OH−]2 = 1.1 × 10–11 [OH−] = 6.06 × 10–4 mol dm–3 [1] (iii) CO32– ions are added first to precipitate Ca2+ as CaCO3 such that the filtrate contains mainly Mg2+ and very little Ca2+. [1] The addition of CO32− must be controlled to prevent precipitation of Mg2+ as MgCO3 [1] (iv) Heat [½] solid Mg(OH)2 strongly. Mg(OH)2(s) → MgO(s) + H2O(g) [½] (b) (i) Al : 1s2 2s2 2p6 3s2 3p1 [½] Mg: 1s2 2s2 2p6 3s2 [½] The 3p subshell of A l is further away from the nucleus than the 3s subshell. There is weaker attraction between the nucleus and the outermost electron of Al. Hence less energy is needed to remove the 3p electron, resulting in lower ionisation energy. [1] (ii) MgCl2 dissolves and dissociates in aqueous solution. Mg 2+ has slightly high charge density. Slight hydrolysis occurs, [½] forming a slightly acidic solution of pH 6.5 [½] MgCl2(s) + 6H2O(l) → [Mg(H2O)6]2+(aq) + 2Cl−(aq) [Mg(H2O)6]2+(aq) + H2O(l) Ý [Mg(H2O)5(OH)]+(aq) + H3O+(aq) [½]

2018 HCI C2 H2 Chemistry Preliminary Exam / Paper 3 AlCl3 dissolves in water to form aqueous ions. AlCl3(s) + 6H2O(l) → [Al(H2O)6]3+(aq) + 3Cl−(aq) Al3+ having high charge density, undergoes hydrolysis to a greater extent [½] to give an acidic solution with pH 3. [½] [Al(H2O)6]3+(aq) + H2O(l) Ý [Al(H2O)5(OH)]2+(aq) + H3O+(aq) [½] (c) (i) Q = I x t = ne x F 200 x 3 x 3600 = n e x 96500 ne = (200 x 3 x 3600)/96500 = 22.4 mol [½] Ca2+ + 2e− → Ca Mass of Ca formed = 22.4 x 40.1 2 = 449 g [½] (ii) From Data Booklet, 2H2O + 2e− = 2H2 + 2OH− E=−0.83V Ca2+ + 2e− = Ca E=−2.87 V [1] Since E(H2O/H2) is less negative than E(Ca2+/Ca), H2O will be preferentially reduced at the cathode instead of Ca2+. No Ca metal will be obtained. [1] (d) (i) H is O O [1] (ii) Electrophilic addition [1] Br2(aq) [1] (iii) J OH [1] K O O HO OH K+ [1] L OH O HO O O [1] (iv) Possible simple chemical te sts to distinguish compounds J and L where J is ethanol (C 2H5OH) and L has two functional groups (carboxylic acid and ketone): [3] KMnO 4(aq) with H2SO4(aq), heat Purple KMnO4 decolourised for J. Purple colour remains for L.

3 CH3CH2OH + 2[O] → CH3CO2H + H2O or use K2Cr2O7(aq) in place of KMnO4(aq) or aqueous iodine with aqueous NaOH, warm yellow ppt for J and no yellow ppt for L CH 3CH2OH + 4I2 + 6OH− → CHI3 + HCO2− + 5I− + 5H2O or 2,4-dinitrophenylhydrazine orange ppt for L and no orange ppt for J equation (the question wants a balanced equation) OH O HO O O + → O2N NO2NN H C HO2C HO2C + H2O or aqueous Na 2CO3 effervescence of CO 2 for L and no effervescence for J (‘CO2 evolved’ was rejected, please state what you expect to see) equation OH O HO O O + Na2CO3 ONa O NaO O O + CO2 + H2O 4 (a) For the amide –CONH, the lone pair of electrons on the nitrogen atom is delocalised over to the C=O bond [1], reducing the availability of the lone pair of electrons on nitrogen to accept a proton. [1] Hence it is a weaker base than the amine –NH2. (b) (i) Ka = 10–2.19 = 6.457 × 10–3 mol dm–3 [1] 6.457 × 10–3 = ௫మ ଴.ଵ଴଴ (assume x << 0.100) [H+] = 0.02541 mol dm–3 pH = 1.60 [1]

2018 HCI C2 H2 Chemistry Preliminary Exam / Paper 3 (ii) Correct shape [0.5] Correct axes [0.5] Label volumes [0.5] Label pH of mixture at maximum buffering capacities [0.5] (iii) 1 st equivalence point: 2nd equivalence point: [1] [1] (iv) (2.19 + 4.25) ÷ 2 = 3.22 [1] (c) (i) [1] (ii) H2A + OH− → HA− Initial / mol 0.1 × 0.1 ÷ 2 = 0.005 0.04 / 40 = 0.001 0.005 Final / mol 0.005 – 0.001 = 0.004 0 0.005 + 0.001 = 0.006 [1] final amounts of H2A and HA− After addition of small amount of NaOH, pH = pK2 + lg [HA−]/[ H2A] = 4.25 + lg [ (0.006/0.100) / (0.004/0.100)] = 4.42 [1] ecf base