RI_H2_CHEM_P3_ANS Prelim

Page 1 of 11 RAFFLES INSTITUTION H2 CHEMISTRY 2011 YEAR 6 PRELIMINARY EXAMINATION PAPER 3 SUGGESTED SOLUTIONS 1 (a) (i) Cold: C l2(g) + 2OH – (aq)  → Cl– (aq) + C lO– (aq) + H 2O(l) Hot: 3C l2(g) + 6OH – (aq) → 5C l– (aq) + C lO3 – (aq) + 3H 2O(l) (ii) F2(g) is highly oxidizing and will react with water (violently) to give O 2(g). (b) (i) 2ClO3 − + 12H + + 10e − ⇌ Cl2 + 6H 2O ∆ G/ring2 = –(10)(1.47)(96500) = –1418.6 = – 1420 kJ mol –1 (i i) /onesans/onesans /onesans/onesans 2C lO3 – + 12H + + 12 e −  → 2C l– + 6H 2O /twosans/twosans /twosans/twosans Applying Hess Law, ∆ G /ring2 = –1420 + (–262) = – 1682 kJ mol −− −− 1 E /ring2 = – [–1682 x 10 3 / (12 x 96500)] = +1.45 V (shown) (ii i) Using Latimer diagram: E/ring2 cell = +1.45 – (+1.19) = +0.26 V Amount of electrons transferred = 6 mol ∆ G /ring2 = –(6)(96500)(+0.26) = –150 kJ mol –1 (c ) (i) A : Cl2 = 6.7/134 : 1.2/24 = 0.05 : 0.05 = 1 : 1 1 mol of A undergoes electrophilic addition with 1 mol of chlorine gas /barb4right A contains 1 alkene functional group (or C=C) Compound A undergoes nucleophilic substitution with PC l 5 to give white fumes of HC l /barb4right A is an alcohol (cannot be carboxylic acid, not enou gh O) NaBr + conc. H 2SO 4 /barb2right HBr /barb4right A undergoes both electrophilic addition and nucleophilic substitution to give C and D. Since C and D each only contains 1 chiral centre, 2ClO 3 − + 12H + + 12 e − → 2C l− + 6H 2O Cl2 + 2e− + 6H2O ∆ G/ring2 = − 262 kJ mol − 1 ∆ G/ring2 = ? (b)(i) ∆ G/ring2 = − 1420 kJ mol − 1

Page 2 of 11 A can only be H C C H CH2OH B: C C CH2OH Cl Cl HH C and D: C * C CH2Br Br H HH C C* CH2Br H Br HH major minor (ii) Compound C is the major product while D is the minor product. This is because benzylic carbocation intermediate is more stable due to resonance / positive charge dispersed into the benzene ring. (i ii) Name of mechanism: Electrophilic Addition H C C H CH2OH Cl Cl slow H C C CH2OH Cl H fast C C CH2OH Cl H Cl H Cl

Page 3 of 11 2 (a) (i ) (i i) Mark X at 50 cm 3 (iii ) COO +H3N +HN NH COO +H3N N NH + OH weak acid conjugate base + H2O (iv) Before addition of HC l: Amount of histidine present = 0.100 dm 3 x 0.1 mol dm -3 = 0.01 mol pH = p K 2 + lg [A − ]/[HA] 6 = 6.00 + lg [A − ]/[HA] lg [A − ]/[HA] = 0 [A − ]/[HA] = 1 Amount of HA = amount of A − = 0.01/2 = 0.005 mol On addition of HC l: Amount of H + added = 1 dm 3 x 0.001 mol dm − 3 = 0.001 mol HA ⇌ H+(aq) + A− (aq) Before adding HC l 0.005 0.005 H+ added +0.001 After adding HC l 0.005 + 0.001 = 0.006 0.005 – 0.001 = 0.004 12.5 25 37.5 50 62.5 75 pK1 = 1.82 pK2 = 6.00 pK3 = 9.17 pH Volume of NaOH(aq) / cm 3 X (a)(ii) isoelectric