RVHS H2 Chem 2012 Prelim P3 Soln

H2 Chemistry (9647) Prelims 2012 Paper 3 1 (a) (i) C O (ii) Nickel carbonyl has a simple covalent structure with weak intermolecular forces which require little energy to overcome. (iii) M r = 2.00´8.31´(273 + 50) 1.01´105 ´314´10-6 =169 (iv) 58.7 + (12.0 +16.0)x =169 x = 3.93 = 4 (nearest whole number) Coordination number is 4 Tetrahedral (accept “square planar”) [7] (b) (i) In transition metal compounds, the presence of ligands causes the 3d orbitals to split into 2 sets of non-degenerate orbitals. The difference in energies (DE) between these 2 sets of 3d orbitals is small and radiation from the visible region of the electromagnetic spectrum is absorbed when an electron moves from a lower energy d-orbital to another unfilled/partially-filled d orbital of higher energy. The colour observed correspond to the complement of the absorbed colours. Hence, transition metal compounds are often coloured. [3] (ii) Blue solution [Cu(H2O)6]2+ Yellow solution CuCl4 2- Blue solution was formed from the oxidation of Cu (to Cu2+(aq)). Blue solution formed a white solid on heating to due to loss of water ligands. / Splitting of 3d orbitals into 2 sets of different energy no longer occurs. CuCl42- + 6H2O = [Cu(H2O)6]2+ + 4Cl - On addition of water, position of equilibrium of CuCl4 2- + 6H2O = [6][Cu(H2O)6]2+ + 4Cl - shifts to the right, forming blue [Cu(H2O)6]2+ 1

2 (c) Limonene: aqueous bromine cold alkaline KMnO4 Menthone: 2,4-DNPH Menthol: PCl 5 / SOCl2 Acidified potassium dichromate, heat Acidified KMnO 4, heat; followed by test with limewater [4] 2 (a) (i) Oxidation number = 0 – (-2) – 4(+1) = -2 (ii) Zero H C H O OR C H HO OH H +2 C HHO O OR C%O (iii) CH3CH2CH2CH2OH(l) + 6O2(g) ® 4CO2(g) + 5H2O(l) (iv) Temperature increase = 58 - 23 = 35 °C Heat evolved = 35 ´ 515 ´ 4.18 = 7.53 ´ 104 J Amount of butan-1-ol = 2.30 ¸ (4 ´ 12.0 + 10 ´ 1.0 + 16.0) = 3.11 ´ 10-2 mol DHc, = 4 2 7.53 10 3.11 10 - æö ´-ç÷ ´èø = -2.42 ´ 106 J mol-1

3 = -2.42 ´ 103 kJ mol-1 (v) The calculated value is less negative (less exothermic ) than the true value as there is heat loss to the surroundings and copper can calorimeter. [7] (b) (i) Propanedioic acid is a stronger acid than butanedioic acid due to the stabilisation of the anion by hydrogen bonding with the unionised –CO2H group. HO2C CO2H + H2O O H2C O- O O H + H3O+ A (ii) Removal of H + from an anion ( A) that already carries a negative charge is electrostatically unfavourable. OR The stabilising hydrogen bonding would be destroyed by the ionisation of the second –CO2H group. (iii) Let the acid be HA and the concentration of H+(aq) be x. 2 a 4.20 [H ][A ] [HA] 10 0.25 K +- - = = - x x Since x is small, 2 4.2010 0.25 - = x x = 3.97 ´ 10-3 mol dm-3 pH = -lg (3.97 ´ 10-3) = 2.40

4 (iv) (c) (i) Al2Cl6 (ii) AlCl3(s) + 3H2O(l) ® Al(OH)3(s) + 3HCl(g) OR 2AlCl3(s) + 3H2O(l) ® Al2O3(s) + 6HCl(g) (When a few drops of water is added, A lCl3 undergoes hydrolysis to form white fumes of HC l(g) and white insoluble solid of Al(OH)3.) AlCl3(s) + 6H2O(l) ® [Al(H2O)6]3+(aq) + 3Cl -(aq) [Al(H2O)6]3+(aq) + H2O(l) = [Al(OH)(H2O)5]2+(aq) + H3O+(aq) AlCl3 undergoes dissolves readily in water to form hydrated/aqueous Al3+ / Al3+(aq) ions which undergoes hydrolysis due to the high charge density of the polarising cation to give a weakly acidic solution. 40 20 10 30 0 2.40 pH =pK1 =4.20 pH =pK2 =5.60 Buffer Region Buffer Region VNaOH pH Al Cl Cl Cl Cl Cl Cl Al

5 3 (a) (i) Determine the [ I2] at regular time intervals by quenching the sample of the solution and titrating with aqueous sodium thiosulfate. OR Measure the electrical conductivity of solution at different time intervals. OR Measure the pH of the solution at regular time intervals using a pH meter. (ii) The graph of [ I2] against time is a straight line, indicating that rate is constant despite decreasing [ I2]. Hence, the reaction is zero order with respect to iodine. From graph 1, when [CH 3COCH3] is increased from 0.10 mol dm –3 to 0.25 mol dm –3, gradient/rate of reaction is increased from 1.0 ´ 10–4 mol dm–3 min–1 to 2.5 ´ 10–4 mol dm–3 min–1. Gradient/rate is 2.5 times the original gradient/rate when [CH3COCH3} is 2.5 times the original concentration. Hence, the reaction is first order with respect to propanone. From graph 2, when [H +] is doubled from 0.20 mol dm –3 to 0.40 mol dm –3, gradient/rate of reaction is also doubled from 1.67 ´ 10–4 mol dm –3 min–1 to 3.33 ´ 10–4 mol dm –3 min–1. Hence, the reaction is first order with respect to H+. (iii) Rate = k[CH3COCH3][H+] (iv) Stage 1 is relatively slower than stage 2. / Stage 1 is the slow step. / Stage 1 is the rate determining step. Rate equation suggests that the slow step/ rate determining step involves only one molecule of propanone and a H+ ion. [8] (b) (i) NaBH4 (in aqueous ethanol) OR H2, Ni, (heat) Reduction (ii) The electron-withdrawing bromine atom in 3-bromobutanoate ion disperses the negative charge on the COO - group, making the 3- bromobutanoate ion more stable than the butanoate ion. Hence, 3- bromobutanoic acid is a stronger acid. (iii) An increase in concentration of 3- oxobutanoic acid and 3-hydroxybutanoic acid is buffered by the CO 2/HCO3- buffer system present in blood. OR Excess ketone bodies are excreted from the human body in urine (since they are water-soluble). [5] (c) P undergoes oxidation to give Q, which is further oxidised by Fehling’s solution/gives positive Fehling’s tes to form the salt of R [7]

6 Þ Q is an aldehyde and P is a primary alcohol. Both P and R undergoes acid-metal reaction with sodium to give 1 mole of hydrogen gas Þ Both P and R contain 2 O–H groups. P undergoes (intramolecular) condensation when warm ed with concentrated sulfuric acid to produce S Þ S is a (cyclic) ester. Structures: P: HO O OH Q: H O OH O R: HO O OH O S: O O 4 (a) (i) Cl2, AlCl3 OR Cl2, FeCl3 OR Cl2, Fe (ii) Electrophilic substitution Cl 2 + AlCl3 ® Cl+ + AlCl4-

7 CH3 Cl+ CH3 Cl H AlCl4 - CH3 Cl + AlCl3 + HCl (iii) Amount of methylbenzene in 118 g = 118 ¸ (7 ´ 12.0 + 8 ´ 1.0) = 1.28 mol Amount of 1-chloro-2-methylbenzene produced = 92.4 ¸ (7 ´ 12.0 + 7 ´ 1.0 + 35.5) = 0.730 mol 0.730Percentage yield 1001.28=´ = 57.0% (iv) CH3 Cl [8] (b) CH3CH2CH2Cl KCN(alc) heat under reflux CH3CH2CH2CN dilute H2SO4(aq) heat under reflux CH3CH2CH2COOH [3] (c) Compound A is optically active Þ compound A contains chiral carbon C:H ratio in Compound A is close to 1:1 Þ compound A contains benzene ring Compound A undergoes nucleophilic substitution with NaOH(aq) to form compound B Þ compound B is an alcohol Þ compound A is an alkyl bromide/alkyl halide [10]

8 Positive iodoform test with Compound B Þ compound B contains –CH(OH)CH3 Compound A undergoes elimination to form compound C Þ compound A is an alkyl bromide/alkyl halide Þ compound C contains C$C bond / alkene functional group Compound C undergoes mild oxidation to form compound D Þ compound D is a diol Compound C undergoes vigorous oxidation to form compound E and ethanoic acid Þ compound C is not a terminal alkene H3C Br A H3C OH B H3C C H3C OH D OH COOH HOOC E

9 5 (a) (i) (ii) Reactions at anode: 2H2O(l) ® O2(g) + 4H+(aq) + 4e- 2Al(s) + 3/2O2(g) ® Al2O3(s) Reaction at cathode: 2H+(aq) + 2e- ® H2(g) (iii) Volume of Al2O3 layer = 29.21 ´ 0.02 = 0.584 cm3 Mass of Al2O3 = 3.95 ´ 0.584 = 2.31 g Amount of Al2O3 = . .. 2 31 2 27 0 3 16 0´ +´ = 0.0226 m