JPJC Prelim P2 Ans

¤ Jurong Pioneer Junior College [Turn Over Mark Scheme for JC2 H2 Chemistry (9729) Preliminary Examination Paper 2 1 (a) (i) Li+: 1s2 H–: 1s2 [1m] for both (ii) Both have same number of quantum shells and no shielding by inner shell electrons. Li+ has a larger nuclear charge/more protons and hence smaller radius. [1m] (iii) LiAlH4 o LiAl + 2H2 [1m] (iv) LiAl has giant metallic structure. [1m] Large amount of heat energy is required to overcome the strong electrostatic attraction between a lattice of cations and delocalised electrons. [1m] (v) The effect of a smaller Ar is more significant than that of a smaller volume of Li atom due to its smaller (atomic) radius. [1m] (b) (i) (ii) H is 2-hydroxybutanoic acid. [1m] NaBH4 reduces ketone only while LiAlH4 reduce both ketone and carboxylic acid, forming back butane-1,2-diol [1m] (c) (d) J CO2 H2O Amount/mol 4.32 192 = 0.0225 5.94 44.0 = 0.135 1.62 18.0 = 0.0900 Mol ratio 0.0225 0.0225 = 1 0.135 0.0225 = 6 0.0900 0.0225 = 4 Mole ratio of J : C : H = 1 : 6 : 8 [1m] DFFHSW´µ Let molecular formula of J be C6H8Ox. 6(12.0) + 8(1.0) + 16.0x = 192.0 Ÿ x = 7 ? molecular formula is C6H8O7. [1m] Since J : NaOH = 1 : 3, J has 3 ‒COOH groups. [1m] for any achiral C6H8O7 e.g. [1m] dotted lines between O and H for two hydrogen bonds [1m] dipoles on 2 O-H, lone pair on O + label ´K\GURJHQERQGµIRU one hydrogen bond hydrogen bond [1m]

2 ¤ Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2021 2 (a) (i) x reaction of solid NaCl with conc. H2SO4 role of conc. H2SO4: Bronsted acid (9) explanation: H2SO4 donates a H+ to form HSO4‒ OR Cl‒ accepts a H+ to form HCl. [1m] x reaction of solid NaI with conc. H2SO4 role of conc. H2SO4: Oxidising agent (9) explanation: O.S. of I increases from ‒1 in NaI to 0 in I2 OR O.S. of S decreases from +6 in H2SO4 to ‒2 in H2S. [1m] 2(9): [1m] (ii) HI/I‒/NaI is a stronger reducing agent than HCl/Cl‒/NaCl. [1m] (iii) 2NaBr(s) + 3H2SO4(l) o 2NaHSO4(s) + Br2(g) + SO2(g) + 2H2O(l) [1m] for Br2 + SO2; [1m] for balanced eqn (b) (i) There is p-p orbital overlap between C of C=O bond and the adjacent C of C=C. [1m] (ii) (iii) Electron-withdrawing C=O in the secondary carbocation that forms Q intensifies the positive charge, making it less stable and less readily formed than the primary carbocation that forms P. [1m] (c) (i) step 1: hydrolysis/ nucleophilic substitution [1m] step 2: condensation [1m] (91) (92) (93) (94) (95) 5(9): [2m] 2-4(9): [1m]

3 ¤ Jurong Pioneer Junior College 9729/02/J2 PRELIMINARY EXAM/2021 (ii) T U V (iii) When the reactants are adsorbed on the catalyst surface (9), the bonds in the reactants are weakened (9) which lowers Ea. Surface concentration of reactants also increases. Thus, the rate of the reaction increases (9). The products are desorbed (9) from the catalyst surface, ma