TJC Prelim P3 ANS

OH OH CH2CH=CH2 OH Br Br Br Answers to H2 Chemistry Preliminary Examinations 2009 Paper 3 1 (a) • AlF3 exists as giant ionic lattice with strong ionic bonds between the ions. Large amount of energy of energy is required to overcome the strong electrostatic attraction between oppositely charged ions and hence, a high melting point. • AlCl3 exists as a simple molecular structure . Melting involves overcoming the weak Van der Waals forces of attraction. Little energy is required to overcome the intermolecular forces of attraction and thus, A lCl3 has a lower melting point as compared to NaCl. (b) (i) • High C:H ratio infers that P is an aromatic compound. • P • With anhydrous aluminium chloride P undergoes electrophilic substitution reaction to form Q. As the –OH is an activating group, the substituent will be directed to the ortho or para position. Q OR • P undergoes electrophilic substituition as it decolourises aqueous bromine to form white ppt R. R is 2,4,6-tribromophenol as -OH is a highly activating group. • Q react with aqueous bromine to form white ppt S, with the same no of bromine atoms per molecule of R. As Q has a C-C double bond, 15 OH CH2CH=CH2

CH2CH(OH)CH2Br OH Br Br OH CH2CH=CH2 electrophilic addition takes place as well. This implies that 1 Br will be attached to one of the carbons of the C-C double bonds. Thus, only 2 Br will be attached directly to the benzene. • S S is optically active because of its chiral carbon. The mirror image of S is non-superimposable. (ii) • Electrophilic Substitution • • Mechanism Step 1: Generation of electrophile CH2=CHCH2Cl + AlCl3 CH2=CHCH2 + + [AlCl4]– Step 2: Electrophilic Attack Step 3: Loss of proton + [AlCl4]– + HCl + AlCl3 Any 2 correct steps – Max 1 mark given. 3 correct steps – 2 marks awarded. 16 OH H CH2CH=CH2 + OH CH2CHCH2+ OH H CH2CH=CH2 + slow fast

(c) NCl3 (l) + 3H2O (l) → NH3 (aq) + 3HOCl (aq) PCl3 (l) + 3H2O (l) → H3PO3 (aq) + 3HCl (aq) • Nitrogen is in period 2 and it cannot expand its octet while phosphorus is in period 3 and can expand its octet structure. • P is able to use its energetically accessible 3d subshell for dative bonding with H2O in the reaction/accommodate of the lone pair of electrons from H 2O during the nucleophilic attack of H2O on PCl3. (d) (i) Ksp = [Pb2+][ CrO4 2-] Let the solubility of PbCrO4 be x mol dm-3 PbCrO4 (s) Pb2+ (aq) + CrO42- (aq) Ksp = [Pb2+][ CrO4 2-] 1.69 × 10-14 = x2 x = spK = 1.30 x 10-7 mol dm-3 The solubility of PbCrO4 is 1.30 x 10-7 mol dm-3. (ii) [Pb2+] = 2 4[ ] spK CrO − = -141.69×10 0.010 = 1.69 x 10-12 mol dm-3 2 (a) Order with respect to E = 1 Comparing experiment 1 and 3 x x 1 1 x x 3 3 x [E] [NaOH] (0.20)(0.30) 0.0150 [E] [NaOH] (0.20)(0.20) 0.0150 0.20 0.30 0.0150 0.20 0.20 0.