SAJC H2 Chem 2012 Prelim P3 Soln

SAJC Prelim 2012 P3 solutions 1 – SAJC PRELIM 2012 P3 solutions No. Solutions 1(a)(i) Aspirin has a lower pKa hence a higher Ka thus aspirin is a stronger acid Aspirin has a RCOOH group and the conjugate base that is formed is more stable as the negative charge on the carboxylate ion is delocalised over O-C-O bond (charge delocalised over two oxygen atoms) by resonance effect. The conjugate base of phenol is less stable as the negative charge (lone pair of electrons on O) of the phenoxide ion is only delocalised into the benzene ring. (ii) H H CH3N CH3 C N C O O Na(s) -O HO or CH3CH2Br/I/ClNaOH(aq) heat H CH3N CH3CH2 C O O Phenacetin (iii) Add aqueous sodium carbonate / sodium hydrogen carbonate Aspirin gives bubbles that will form white ppt. with aqueous calcium hydroxide. Salicin does not give white ppt or bubbles. OR K2Cr2O7 (aq), aq H2SO4, heat OR Add aqueous KMnO4 (aq), aq H2SO4, heat. Salicin decolourises purple KMnO4 OR turns K2Cr2O7 from orange to green. Aspirin remains purple OR remains orange. (iv) Water is not a suitable solvent as the solute-solvent interaction (pd-pd, id-id, H-bonding) cannot overcome the H-bonding between water and id-id between aspirin. (b)(i) Ka =

2 – SAJC PRELIM 2012 P3 solutions Where HA = O C O CH3 COOH and A- = O C O CH3 COO- (ii) 10-3.5 = Hence, = 3.16 x 10-3 Percentage ionised = (3.16 x 10-3 / 1) x 100 % = 0.316 % Percentage unionised = 100 – 0.316 = 99.7 % (c)(i) Indicators are weak acids hence if added in large amount, the amount of NaOH required will be higher than expected / titration is inaccurate. (ii) [aspirin] = = 0.05 mol dm-3 Using Ka = [H+]2 = 3.98 x 10-3 pH = 2.40 (iii) Using [aspirin] = 0.05 mol dm-3 x = = 15.00 cm3 [paracetamol] = = 0.0267 mol dm-3 y = = 8.00 cm3 (iv) 2(a)(i) Cathode: O2 + 4H+ + 4e à 2H2O Anode: 13H2O + C12H22O11 à 12CO2 + 48H+ + 48e (ii) Using data booklet: Eq O2 + 4H+ + 4e ßà 2H2O + 1.23 V Using Eqcell = Eqcathode - Eqanode + 1.25 = + 1.23 – (Eqanode) Eqcathode = – 0.02 V

3 – SAJC PRELIM 2012 P3 solutions (iii) O2 + 4H+ + 4e ßà 2H2O + 1.23 V When the pH increases, the cathodic reaction shifts to the left to replenish H+. Hence, the Eqcell becomes more negative. (b)(i) A: [Cu(H2O)6]2+ B: NO2 C: Cu(OH)2 D: CuO E: [Cu(NH3)4(H2O)2]2+ F: [CuCl4]2- G/H: HNO2 and HNO3 Equations: · C: [Cu(H2O)6]2+(aq) + 2OH- (aq)à Cu(OH)2 (s) + 6H2O (l) OR Cu(NO3)2 (aq) + 2NaOH (aq) à Cu(OH)2 (s) + 2NaNO3 (aq) · D: Cu(OH)2 (s)à CuO (s) + H2O (l)/ (g) · E: [Cu(H2O)6]2+(aq) + 4NH3 (aq) à [Cu(NH3)4(H2O)2]2+ (aq) + 4H2O (l) · F: [Cu(NH3)4(H2O)2]2+ (aq) + 4HCl (aq) à [CuCl4]2- (aq) + 2H2O(l) + 4NH4+ (aq) OR [Cu(NH3)4(H2O)2]2+ (aq) + 4Cl- (aq) à [CuCl4]2- (aq) + 2H2O(l) + 4NH3 (aq) (ii) 3 bond pairs of electrons repels each other equally OR lp-lp>lp-bp>bp-bp Hence, the shape is trigonal planar. 3(a)(i) ½ N2 (g) + H2 (g) à NH3 (g) (ii) 3N2H4 (l) 4NH3 (g) + N2 (g) 3N2 (g) + 6H2 (g) + 111.8 3z y Using bond energy from the data booklet: y = [2BE (NºN) + 6BE (H-H)] – 12 BE (N-H)] y = - 76 kJ mol-1 Using energy cycle: 3z + y = 111.8 Z = +62.6 kJ mol-1 (b)(i) -3 to -2 (ii) N is oxidised, Cl- is formed or NaCl where Cl2 must be reduced. Note: HCl not accepted given that NaOH is the medium. (iii) 2OH- + 2NH3 + Cl2 à N2H4 + 2Cl- + 2H2O Note: No spectator ions.

4 – SAJC PRELIM 2012 P3 solutions (c) C O CB: CA: C CHydrazone: H N N H H H H C O H H H H H H 3(d)(i) HNO3 + BF3 à NO2+ + [BF3OH]- H+ + [BF3OH]- à BF3 + H2O (ii) B has an empty orbital. Hence, B can accept a lone pair of electrons from OH- via dative bonding. (iii) Sn and concentrated HCl with heat followed by aq NaOH (iv) CH2 CH2Cl NH2 (v) Limited chlorine gas / in CCl4 + UV OR by heating. (vi) Free radical substitution Nucleophilic substitution (vii) Difficult to control the position and number of substitution by Cl during FRS. OR difficult to control NS. 4(a)(i) Using Ksp for Ag2CrO4 = [Ag+]2[CrO42-] = 9 x 10-12 [Ag+] needed to form the red ppt = [(9 x 10-12)/(0.8)]1/2 = 3.35 x 10-6 mol dm-3 Using Ksp for AgCl = [Ag+][Cl-] = 1.6 x 10-6 [Cl-] = (1.6 x 10-10)/( 3.35 x 10-6) = 4.78 x 10-5 mol dm-3 (ii) Percentage of Cl- precipitated = = 99.0 % Since almost all of the Cl- is precipitated when the first trace of red ppt appears, CrO4- acts as an indicator for the titration. (b)(i) At the cathode (negative electrode), K+ and H2O compete to be reduced. K+ + e ßà K -2.92 V 2H2O + 2e ßà H2 + 2OH- - 0.83 V Since the Eq (H2O/OH-) = - 0.83 V is more positive than Eq (K+/K) = - 2.92 V H2O is preferentially reduced. Cathode: 2H2O + 2e à H2 + 2OH- At the anode (positive electrode), Cl- and H2O compete to be oxidised. Cl2 + 2e ßà Cl- + 1.36 V

5 – SAJC PRELIM 2012 P3 solutions O2 + 4H+ + 4e ßà 2 H2O +1.23 V Although the Eq (Cl2/Cl-) = + 1.36 V is more positive than Eq (O2/H2O) = +1.23 V Cl- is preferentially oxidised due to its high concentration. Anode: 2Cl- à Cl2 + 2e (ii) 3Cl2 + 6KOH à KClO3 + 5KCl + 3H2O Note: Ionic eqn accepted. (iii) High temperature of electrolyte must be used. (iv) Cl- + 3H2O à ClO3- + 6H+ + 6e OR Thus, 1 mole of ClO3- is formed from 6F. 122.6 g of KClO3 = 1 mole of KClO3 = 579 000 C 30 g of KClO3 = 0.245 moles = 141 680 C Using Q = It t = (141 680) / 3 = 47227 seconds OR 13.1 hours. (c)(i) KClOx (s) à KCl (s) + x/2 O2 (g) (ii) Poxygen gas = Ptotal – Pwater vapour = 100525 – 2800 = 97725 Pa Using PV = nRT 97725 x 650 x 10-6 = noxygen x 8.314 x (22 + 273) noxygen = 0.0260 mol. = x = 2.99 ≈ 3 (d) KBrO3 is more easily decomposed than KClO3 OR KBrO3 is thermally less stable. BrO3- has a larger electron cloud than ClO3- hence K+ is more able to polarise the larger electron cloud more easily due to weakening of Br-O bond. 5(a)(i) Rate a 1/ time (ii) When temperature is increased, the molecules gain kinetic energy and move about faster. This increases the number of molecules having energy E ≥ E a. As a result, the frequency of effective collisions increases. Reaction rate thus increases.

6 – SAJC PRELIM 2012 P3 solutions b(i) I. Ag+ will compete with the cation of lysine to form ionic interaction with the anion of glutamic acid. OR NO3- will compete with the anion of glutamic acid to form ionic interaction with the cation of lysine. polypeptide chain II. A polar solvent such as ethanol will disrupt the ion dipole interactions between Asp and Tyr to form ion-dipole interaction with the anion of aspartic acid OR hydrogen bonding with tyrosine. polypeptide chain (ii) Ethanol in case II. Ethanol can disrupt the hydrogen bonding between peptide linkages which maintain the alpha helix structure OR by forming hydrogen bonds with the peptide linkages. Note: OR ethanol can form hydrogen bonding with the secondary structure. OR polypeptide chain

7 – SAJC PRELIM 2012 P3 solutions (c) OH C CH3 OH CH2CH3B: OH C CH3 OH CH2CH3C: Br Br OH C CH3 CHCH3D: OH C CH3 OH CHCH3E: OH OH C CH3 Cl CHCH3F: Cl ES substn at 2, 4 or 6. Information Deduction B is optically active. · B has a chiral carbon B dissolves slowly in aq. NaOH · B has phenol. · B undergoes neutralisation B reacts with HBr · B undergoes N.S · B forms alkyl bromide · B has alcohol group. B does not decolourise potassium dichromate. · B is a tertiary alcohol. B decolourises aq. Bromine to form white ppt. · B is a phenol. · B undergoes E.S. · From molecular formula of C, one of the 2,4 or 6 positions is occupied. B reacts with concentrated sulfuric acid at 170oC · B undergoes elimination. · D is an alkene. D reacts with cold alkaline KMnO4 · D undergoes mild oxidation. · D forms a diol. E reacts with PCl5 to give white fumes · E undergoes NS. END