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 point

Page 4 of 11 pH = p K2 + lg [A − ]/[HA] = 6.00 + lg (0.004/0.006) = 5.82 (b) (i) sp 2 more s character / lone pair more strongly attracted by nucleus than the lone pair in the sp 3 hybridised N and less available for protonation (ii) (c) (i) Step I: (acid-catalysed) nucleophilic addition Step II: Dehydration / elimination (ii) Any acid, heat (under reflux) (iii) R C H NH H CN R C H N+ H H CN- R C NH2 CN H (iv) Strecker’s synthesis produces racemic products and hence, do not display optical activity while naturally occurring amino acid is present as one of the enantiomers (usually L enantiomer) and will rotate plane polarised light. maximum rate of reaction substrate concentratio n0 histidine concentration

Page 5 of 11 3 (a) (i) Cu 2+ (aq) + 4NH 3(aq) ⇌ [Cu(NH 3)4]2+ (aq) (ii) The copper electrode in half-cell 1 is the negative electrode. When aqueous ammonia is added to the half-cell 1, c omplex [Cu(NH 3)4]2+ is formed and the concentration of Cu 2+ ions decreases , causing the equilibrium position of Cu 2+ + 2e − ⇌ Cu to shift to the left, favouring oxidation (loss of electrons). (iii) Kf = ( ) 2+ 3 4 2+ 4 3 [Cu NH ] [Cu ][NH ] mol − 4 dm 12 (iv) Ecell = E/ring2 cell − 2+ half-cell 1 2+ half-cell 2 [Cu ] 0.0592 lg 2 [Cu ] 0.129 = 0.00 − 2+ half-cell 1 [Cu ] 0.0592 lg 2 0.0100 [Cu 2+ ]half -cell 1 = 4.38 x 10 -7 mol dm −− −− 3 (v) Initial amount of Cu 2+ = 90/1000 × 0.0100 = 9.00 × 10 − 4 mol Initial amount of NH 3 = 10/1000 × 0.500 = 5.00 × 10 − 3 mol New volume of solution in half-cell 1 = 10 + 90 = 100 cm 3 Eqm. amount of Cu 2+ = 100/1000 × 4.384 × 10 − 7 = 4.384 × 10 − 8 mol Amount of NH 3 reacted = 4(9.00 × 10 − 4 – 4.384 × 10 − 8) = 3.60 × 10 − 3 mol Cu 2+ (aq) + 4NH 3(aq) ⇌ [Cu(NH 3)4]2+ (aq) Initial amt 9.00 × 10 − 4 5.00 × 10 − 3 – Eqm amt 4.384 × 10 − 8 5.00 × 10 − 3 – 3.60 × 10 − 3 = 1.40 x 10 -3 9.00 × 10 − 4 - 4.384 × 10 − 8 = 8.9996 × 10 − 4 ≈ 9.00 x 10 − 4 ÷ 0.100 Eqm [ ] 4.384 × 10 − 7 1.40 × 10 − 2 9.00 × 10 − 3 K f = (9.00 × 10 − 3) / (4.384 × 10 − 7) (1.40 × 10 − 2)4 = 5.34 ×× ×× 10 11 mol −− −− 4 dm 12 (b) F – Cr(OH) 3 G – Na 3Cr(OH) 6 H – Na 2CrO 4 J – Na 2Cr 2O7 (c) Fe 3+ + e− ⇌ Fe 2+ E/ring2 = +0.77 V O2 + 2H + + 2e − ⇌ H2O2 E/ring2= +0.68V H2O2 + 2H + + 2e − ⇌ 2H 2O E/ring2= +1.77 V E/ring2 cell = +0.77 – (+0.68) = +0.09 V >0 Step 1: 2Fe 3+ (aq) + H 2O2(aq) → 2Fe 2+ (aq) + O 2(g) + 2H +(aq)

Page 6 of 11 E/ring2 cell = +1.77 – (+0.77) = +1.00 V > 0 Step 2: 2Fe 2+ (aq) + H 2O2(aq) + 2H +(aq) → 2Fe 3+(aq) + 2H 2O(l) (d) (i) Two mono-brominated products: (ii) Ratio of the above products = 6:4 = 3:2 (iii) Step I : KCN(aq) in ethanol, heat (under reflux) Step II : any dilute acid, heat (under reflux) Step III : PC l5 or SOC l2 Step IV : excess conc. NH 3 in ethanol, heat in a sealed tube

Page 7 of 11 4 (a) By Hess’ law –1676 = 2(+322) + 2(+577+1820+2740) + 3/2 (+496)+ 3 (–142)+ 3(+744) + LE lattice energy of Al 2O3, LE = –15144 = –1.51 x 10 4 kJ mol -1 (b) First, add dilute HC l to the sample. If the solid is insoluble in acid, it must be the acidic oxide SiO 2 If the solid dissolves in acid, add NaOH(aq). If the solid is insoluble in NaOH(aq), it must be the basic oxide MgO. If the solid dissolves in both acid and base, it must be the amphoteric oxide Al 2O3 (c) (i) OH Al2O3 heat Compound N 2A l (s) + 3/2 O 2(g) Al2O3(s) 2A l (g) + 3/2 O 2(g) 2A l3+ (g) + 3/2 O 2(g) + 6e − 2A l3+ (g) + 3O(g) + 6e − 2A l3+ (g) + 3O − (g) + 3e − 2A l3+ (g) + 3O 2− (g) -1676 2(+322) 2(+577+1820+2740) 3/2 (+496) 3(-142) 3(+744) Lattice energy of Al 2O3, LE Energy / kJ mol -1 0

Page 8 of 11 (ii) Compound N HO OH K2Cr2O7 heat HO O O HO OH O 2-hydroxypropanoic acid cold KMnO4 NaOH(aq) NaBH4 H2SO4(aq) (d) Solution of beryllium sulfate is acidic (pH ≈ 3) thus causing skin irritations. [Be(H 2O) 4]2+ + H 2O ⇌ [Be(OH)(H 2O) 3]+ + H 3O+ The small, highly polarising beryllium cation weake ns the O–H bonds of the water molecules in its surrounding sphere of coordination and results in the release of hydrogen ions in solution. (e) (i) Step I: Sn, conc. HC l, heat (under reflux) Step II: NaOH(aq) (ii) OCH2CH2Cl NH2 + A lCl3 OCH2CH2 + NH2 + AlCl4 Al is electron deficient/ has vacant low-lying orbitals to act as electron pair acceptor, thus acting as a Lewis acid catalyst. (iii) Name: Electrophilic aromatic substitution

Page 9 of 11 Cl AlCl3 HCl AlCl3 + + fast NH2NH2 O H+ O (iv) AlCl3, being a Lewis a