ACJC H2 Chem 2013 Prelim P3 Soln

2 © ACJC 2013 9647/03/Prelim/13 [Turn over ACJC Chemistry Prelim Paper 3 Suggested Solution 1 (a) (i) CH 2=CHCH2OCOCH3 + Br2  CH2BrCHBrCH2OCOCH3 (ii) Compare Expt No. 1 and 2, [Br 2] constant, when [propenyl ethanoate] increases by 1.5 times, rate increases by 1.5 times. Order of reaction with respect to propenyl ethanoate is 1 Compare Expt No. 1 and 3, [propenyl ethanoate] constant, when [Br 2] doubles, rate increases by 4 times. Order of reaction with respect to bromine is 2 Rate = k[Br2]2[propenyl ethanoate] (iii) 1% of ester = 1/100 x 4.0 x 10 -2 = 4.0 x 10-4 mol dm-3 Time taken = (4.0 x 10-4) / (5.16 x 10-3) = 0.0775 s (assume constant rate) (iv) Chlorine is electron-withdrawing; hence will lower the electron density on the carbon-carbon double bond, hence less susceptible to electrophiles. Rate is expected to be slower. OR Chlorine is electron-withdrawing, hence makes the carbocation intermediate less stable as positive charge on carbocation is more intensified. Thus rate is expected to be slower. (v) CH2BrCHOHCH2OCOCH3 / CH2OHCHBrCH2OCOCH3 (vi) Rate of reaction would be faster. Water is more polar than ethanoic acid. It can form ion-dipole interaction with carbocation, hence stabilize it. Activation energy would be lowered. OR Rate of reaction would be slower. Water dilutes the concentration of Br 2. Since rate depends on concentration of Br2 and concentration of Br2 is decreased, rate is expected to be slower. (b) (i) From the graph, 20 cm 3 of NaOH reacted with MgCl2. No. of moles of NaOH in 20 cm 3 = 1.00 x 20/1000 = 0.02 mol MgCl 2 + 2NaOH  2NaCl + Mg(OH)2 No. of moles of MgCl2 that reacted with NaOH = 0.02/2 = 0.01 mol Initial concentration of MgCl2 = 0.01/(50/1000) = 0.200 mol dm-3 (ii) pH = 9 [H+] = 10-9 mol dm-3 [OH-] = 10-5 mol dm-3 No. of moles of Mg 2+ = 0.01/2 = 0.005 mol [Mg2+] = 0.00500/(60/1000) = 0.0833 mol dm-3 297

3 © ACJC 2013 9647/03/Prelim/13 [Turn over Ksp of Mg(OH)2 = [Mg2+] [OH-]2 = (0.0833) (10-5)2 = 8.33 x 10-12 mol3 dm-9 (iii) Solubility decreases with increasing temperature Less OH- needed to reach Ksp. pH would be less than 9. (iv) Mg 2+ has high charge density and can polarize H2O molecules slightly, weakening the O-H bond in water. Mg(H2O)6 2+ (aq) Mg(H2O)5 (OH)+ (aq) + H+ (aq) 2 (a) (i) Two mono-brominated products: (ignore stereoisomers) (ii) Ratio of the above products = 6:4 = 3:2 (b) (i) Step I is a first order reaction. Write a balanced equation for the reaction, stating suitable reagents and conditions required. Hence, state the rate equation of the reaction. Br + NaOH  OH + NaBr aq NaOH & heat under reflux or heat rate = k[C 5H9Br] (ii) Step III: HCN in the presence of traces of NaCN or NaOH & 10C - 20C Mechanism: Nucleophilic Addition -:CN s- Os+ slow O-: CN 298

4 © ACJC 2013 9647/03/Prelim/13 [Turn over H CN   O-: CN -:CN CN OH + fast 2 (c) (i) Al2O3 (s) + 2NaOH (aq) + 3H2O (l) → 2NaAl(OH)4 (aq) Or Al2O3 (s) + 2OH- (aq) + 3H2O (l) → 2Al(OH)4 - (aq) (ii) Aluminium hydroxide Al(OH)3 2NaAl(OH)4 (aq) + CO2 (g) → 2 Al(OH)3 (s) + Na2CO3 (aq) + H2O (l) Or 2Al(OH)4 - (aq) + CO2 (g) → 2 Al(OH)3 (s) + CO3 2- (aq) + H2O (l) [Accept Al(OH)4 - (aq) + CO2 (g) → Al(OH)3 (s) + HCO3 - (aq)] (iii) Cathode: Al3+ + 3e → Al Anode: 2O2- → O2 + 4e (iv) Water is reduced preferentially over Al3+. Cryolite is added form a mixture which melts at a lower temperature. (v) Mass of aluminium oxide in bauxite = 54/100 x 1.00 x 108 = 5.40 x 107 kg Amount of aluminium = 2 x {5.40 x 1010 / [2(27) + 3(16)] } = 1.05 x 109 mol Mass of aluminium = 1.05 x 109 x 27 = 2.86 x 1010 g = 2.86 x 107 kg Amount of electrons required = 1.05 x 109 x 3 = 3.15 x 109 mol Q = 3.15 x 109 x 96500 = 3.04 x 1014 C 299

5 © ACJC 2013 9647/03/Prelim/13 [Turn over 3 (a) Bonds broken = 740 + 410 +350 = +1500 kJ mol-1 Bonds formed = 460 + 610 + 360 = +1430 kJ mol-1 ∆Hr = 1500 – 1430 = +70 kJ mol-1 (b) (i) K a (H3O+) = [H+] [H2O] / [H3O+] and Ka ((CH3)2C=OH+) = [H+] [(CH3)2C=O] / [(CH3)2C=OH+] (ii)  Kc = (iii) Kc = (101.7 / 107.2) = 10-5.5 = 3.16 x 10-6 (iv) ∆Go = - 8.31 x 298 x ln (10-5.5) = + 31361 J mol-1 = + 31.4 kJ mol-1 (v) Kc value is very small indicating that the keto form is predominant in the overall equilibrium Also ∆Go is very positive indicating that the enol form is not energetically favourable. (c) (i) Enol form is stabilised by intra-molecular hydrogen bonding OR Conjugation of the C=O bond with the C=C bond also contributes to the stability. (ii) No / less likely No conjugation between the C=O bond with the C=C bond. (iii) Keto form participates in dipole-dipole interaction or hydrogen bond formation with polar solvents which suppresses intra-molecular hydrogen bonding In a non polar solvent, the enol form becomes more stable through intramolecular hydrogen bonding (d)  MF = C 10H12  P has no alkene functional group or C=C  Q has 2 –COOH groups  R is an acid derivative since it is neutral 300

6 © ACJC 2013 9647/03/Prelim/13 [Turn over O OH O OH +1 1 [ O ] +2 C O2 +3 H2O O OH O OH O O O +H 2O 301

7 © ACJC 2013 9647/03/Prelim/13 [Turn over 4 (a) (i) Cu+ 1s2 2s2 2p6 3s2 3p6 3d10 Cu2+ 1s2 2s2 2p6 3s2 3p6 3d9 (ii) For Cu 2+(aq), the d-orbitals are not completely filled with electrons . An electron from a lower energy level can be excited to higher energy level when light is absorbed (d-d* electron transition). Light not absorbed is seen as colour of complex ion. For Cu +(aq), the d-orbitals are completely filled with electrons . Hence d-d* electron transition is not possible. Complex ion appears colourless. (iii) NH3(aq) + H2O(l)  NH4 +(aq) + OH(aq) [Cu(H2O)6]2+(aq) + 2OH(aq)  Cu(OH)2(s) + 6H2O(l) When NH 3(aq) is added slowly, blue ppt of Cu(OH)2 is formed. Cu(OH)2(s) + 4NH3(aq) + 2H2O(l) [Cu(NH3)4(H2O)2]2+(aq) + 2OH(aq) or Cu(OH)2(s) + 4NH3(aq) [Cu(NH3)4]2+(aq) + 2OH(aq) or Cu2+(s) + 4NH3(aq) [Cu(NH3)4]2+(aq) When NH3(aq) is added in excess, blue ppt dissolves and a deep blue solution is formed. (iv) [Cu(H2O)6]2+(aq) + 4Cl(aq) [CuCl4]2(aq) + 6H2O(l) or Cu2+(aq) + 4Cl(aq) [CuCl4]2(aq) When HCl(aq) is added, an intermediate green colour is observed because blue [Cu(H 2O)6]2+(aq) is mixed with yellow [CuCl4]2(aq). Final colour of solution is greenish-yellow. (v) 2Cu2+(aq) + 4I(aq)  2CuI(s) + I2(s) . When KI(aq) is added, CuI white ppt is formed in I2 brown solution. (vi) Cu+ + e Cu +0.52V Cu2+ + e Cu+ +0.15V Overall: 2Cu+  Cu2+ + Cu Ecell = +0.52 – (+0.15) = +0.37V Since Ecell > 0 Volt, the reaction is thermodynamically feasible. Equation: Cu2O(s) + H2SO4(aq)  Cu(s) + CuSO4(aq) + H2O(l) Acid-Base and Disproportionation (b) V2+ + Ce4+  V3+ + Ce3+ Ecell = +1.61 + (+0.26) = +1.87V Since Ecell > 0 Volt, the reaction is thermodynamically feasi