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 f