SAJC H2 Chem 2013 Prelim P3 Soln

SAJC 2013 Prelim Paper 3 ans Page 1   Answers for SAJC 2013 Prelim Paper 3 1a 3 chiral centres [1] (b) (i) Nucleophilic addition [1] (ii)     [3] (iii) [1] (c) (i) Balancing the equation, At equilibrium, D-fructose = 1- 0.7-0.22 = 0.08 Kc = (0.7)(0.22)/(0.08) 2 = 24.1 [2] (ii) Change the temperature of the reaction. [1] (d) (i) nucleophilic substitution [1] (ii)    [1]  1194

SAJC 2013 Prelim Paper 3 ans Page 2   (iii) (I) Slower as has a larger alkyl group that increases the steric hindrance experienced by nucleophile attacking the alkyl halide. [2] (II) Slower as C-Cl bond is stronger and harder to be broken. [2] (e) (i) Hydrolysis[1] (ii) When temperature increases, the kinetic energy of invertases increases resulted in denaturation/ protein losing their native conformation. [1] The van der Waals' interactions in the tertiary structure and quaternary and the hydrogen bonds in the secondary, tertiary and quaternary structures are broken. [1] (iii) At low [substrate], active sites are not filled hence r = k[substrate]. [1] At high [substrate], all active sites are filled up hence reaction is zero order wrt [substrate] [1] 2 (a) (i) From the Data Booklet: Equation E 0 O2 + 2H+ + 2e– H2O2 +0.68 Cr2O7 2- + 14H+ + 6e 2Cr3+ + 7H2O +1.33 3H2O2 + Cr2O7 2- + 8H+ 2Cr3+ + 7H2O + 3O2 E0 cell = 1.33 – 0.68 = + 0.65 V. Therefore, reaction is feasible. [3] (ii) Hence, 3 moles of H2O2 reacts with 1 mole of Cr2O7 2-. No. of moles of Cr2O7 2- = 0.1 x 30/1000 = 3 x10-3 mol No. of moles of H2O2 in 10.0 cm3= 9 x10-3 mol [H2O2] = 0.900 mol dm-3 [1] (iii) Some hydrogen peroxide has decomposed into oxygen and water. [1] 1195

SAJC 2013 Prelim Paper 3 ans Page 3   (iv) Half-life = 13-15 min Hence, the rate order is 1 with respect to the concentration of hydrogen peroxide.[3] (v) R=k[H2O2]1 [1], k= ln2/t1/2 = ln 2/ (14) = 0.0495 min-1 [2] (vi) A more steeper curve drawn on the same graph, labeled with NH3.[1] (b) Evidence Conclusion In the presence of sulfuric acid, 2 g of MEA reacts with 4.0 g of benzoic acid to form an acidic Compound B, C 9H12NO2  2 g of MEA = 0.03279 mol 4 g of benzoic acid = 4/122 = 0.03279 mol  1 mole of MEA reacts with 1 mole of benzoic acid  condensation/ nucleophilic substitution  MEA contains an alcohol* group. 2 g of MEA require 9.2 g of benzoyl chloride to react completely to form Compound C, C 16H15NO3 and copious white fumes.  9.2 g of benzoyl chloride = 9.2/140.5 = 0.06548 mol  1 mole of MEA reacts with two mole of benzoyl chloride  condensation/ nucleophilic substitution,  producing copious white fumes of HCl  MEA contains 1 alcohol*  MEA contains 1 amine* group It is a simple aliphatic organic compound that can dissolve in water to form a weakly basic solution  MEA likely to contain an amine* MEA can also react with gaseous PCl 5 to form a cyclic Compound D, C4H10N2 as one of the products under room temperature and pressure  MEA contains an alcohol group*.  The cyclic compound is due to intra- condensation/ nucleophilic substitution. 1196

SAJC 2013 Prelim Paper 3 ans Page 4   Maximum [4] MEA: NH2CH2CH2OH [1] Compound B: [1]    Compound C: [1]    Compound D: [1]   3 (a) (i) Anode: Li  Li+ + e─ [1] Cathode: Li+ + CoO2 + e─  LiCoO2 [1] (ii) In transition elements, the difference in energy levels between 3d and 4s orbitals is small. Hence after the 4s electrons are removed, it does not require a lot more energy to remove the 3d electrons hence exhibiting variable oxidation states. [2] (b) (i) From Data Booklet Li+ + e  Li E0 = -3.04 V Overall equation: E0cell = E0 cathode – E0 anode 3.6 = E0 cathode – (-3.04) E0 cathode = + 0.56 V [3] (ii) With the increase in carbon side chain, the electron cloud size increases, results in an increase in induced dipole-induced dipole interactions in PTMA compared to TEMPO-NO radical. Hence, the pd-pd/Van der Waals/H-bonding interactions between polar solvents and PTMA will not be exothermic enough to overcome the stronger induced dipole- 1197

SAJC 2013 Prelim Paper 3 ans Page 5   induced dipole interactions in PTMA and pd-pd interactions between polar solvents to dissolve it. Hence, solubility is reduced. [2] (iii) Li will react with water to form LiOH 2Li + 2H2O  2 LiOH + H2 [1] (iv) Free radical substitution Initiation Propagation NO H Cl NO HCl   Termination (Any 2)     [4] 1198

SAJC 2013 Prelim Paper 3 ans Page 6   (v)   [3] [13] (c) Mass of Li to be restored = 100/87 x 1.25 = 1.4367 g No. of mole of Li = 1.4367/ 6.9 = 0.20823 mol No. of C = 0.20823 x 96500 = 20094 C Time = 20094/ 3 = 6698 sec = 1.86 hr [3] [3] [Total marks: 20] 4 (a) (i) Kp = N2O4 2 NO2 P ) (P Moles of N2O4 = 21/ 92 = 0.2283 Initial pressure: PV = nRT P = 610 x 30 273) (45 x 8.314 x 0.2283   , = 20120 kPa N 2O4 NO 2 Initial /kPa 20120 0 Equilibrium/ kPa 20120 – x 2x 20120 + x = 33400, x = 13280 kPa Kp = kPa1.03x1013280 - 20120 (2x13280) 5 2  [4] (ii) (I) At 30 0C, since vessel Y has the lowest moles of N 2O4, which shows that 1199

SAJC 2013 Prelim Paper 3 ans Page 7   equilibrium has shifted to the right. When the system is subjected to larger volume and hence lower pressure, the system will shift to increase pressure to favour the side with more moles of gases. The order of the vessel is Z < X < Y. [3] (II) From the data, when temperature is the highest, there is least number of moles of N2O4 which implies that equilibrium shift right at high temperature. When temperature is high, system favours the endothermic reaction to remove excess heat. This implies that the forward reaction is endothermic. When temperature increases, increase in rate of forward reaction is greater than increase in rate of backward reaction, Kp increases. [3] (b) (i) Initial pH: [OH-] = 0.25 x 104.2 = 3.9434 x 10-3 pH = 14 + lg 3.9434 x 10-3 pH = 11.6 [1] Maximum buffer capacity: pOH =pKb pH = 14 – 4.2 = 9.8 [1] Equivalence point: No. of moles of amphetamine = 0.25 x 1000 20 = 0.005 Volume of nitric acid needed to reach equivalence point = 0.2 0.005 = 25 cm3 [salt] = 1000 20 25 0.005 [H+] = 9 1 x 109.8 = 4.196x 10-6 pH = 5.37[1] After 30 cm3 of nitric acid was added: Excess acid = 30 – 25 = 5 1200

5 SAJC 2   (i) (ii) (iii) (a) (i) 2013 Prelim [H+] = 0.   pH= 1.70 Graph an [C6H5CH2 % compos = (12x2x % compos Diagram [ m Paper 3 a 1000 5 x 2    [1] nd labelling CH(CH3)NH sition of am 368 14)14x9   sition of am [1] ans 1000 50 = 0. g [2] H3 +]2SO4 2- [1 mphetamine 100 x ) = 73 mphetamine 02 1] in ampheta 3.91% in tablet = [1] amine sulph 73.91 / 4 = hate 18.5% [2] [Total Page 8 [10] marks: 20] ] ] 1201

SAJC 2013 Prelim Paper 3 ans Page 9   (ii) [2] By Hess Law, – 601.2 – 187.8 – 95.3 = –285.3 + x x= – 599 kJ mol -1 [1] (iii) MgO2  MgO + ½ O2 [1] (iv) H = (-601.2) – (-599) = - 2.2 kJ mol-1 [1] (v)  From the Data Booklet, the ionic radii of Group II cations: Mg2+ = 0.065nm, Ca 2+ = 0.099 nm and Ba2+ = 0.135nm.  Down the group, the size of the cation decreases while the charge remains constant. Charge density decreases in the order of Mg2+ > Ca2+ > Ba2+.  The degree of polarisation decreases in the order of Mg 2+ > Ca 2+ > Ba 2+. As a result of polarisation, the O-O is weakened decreasingly down the group.  Thus, MgO 2 is the easiest to decompose followed by CaO2 then BaO2. Hence, its rat