JJC H2 CHEM P3 ANS

2009 JJC Preliminary Examination Mark Scheme 1. (a) (i) The standard enthalpy change of combustion of a substance is the enthalpy change which occurs when one mole of the substance is completely burnt in oxygen/burnt in excess oxygen under standard conditions. [1] (ii) OH (s) + 7O2 (g)  6CO2 (g) + 3H2O (l) [1] (iii) –3054 = 6 (-394) + 3 (-286) - Hf (phenol) + 0 [1] Hf (phenol) = - 168 kJ mol-1 [1] (iv) - 53.6 kJ mol-1 [1] 1. (b) (i) Three electron withdrawing Cl atoms disperses the negative charge on the anion, stabilizing the anion. Hence 2,4,6-trichlorophenol is a stronger acid than phenol. [1] 2,4,6-trichlorophenol has a larger Ka value. [1] (ii) Test: Add Br2(aq) to both samples. [1] Observation: phenol will decolourise orange brown Br 2(aq) but 2,4,6-trichlorophenol will not. [1] (c) (i) [1] (ii) Step I: dil HNO3 / HNO3(aq) [1] Step II: Sn, conc HCl, heat [1] (iii) Bulky non-polar group in 1-naphthol hinders the formation of hydrogen bonding. [1] (d) (i) Transition element complexes are coloured because of electron transition between d orbitals. In a complex ion, the presence of ligands causes the 3d orbitals to split into 2 energy levels. The difference in energies between these two sets of 3d orbitals is small such that radiation from the visible region is absorbed when an electron moves from a d orbital of lower energy to an unfilled d orbital of higher energy. The blue colour seen is the complement of the colours absorbed. [3] (ii) The prescence of energetically accessible vacant orbitals to accept lone pair of electrons from ligands. [1] NO2 OH

1. (d) (iii) [2] (iv) Deep blue solution: [Cu(NH Cu H2O O: N: :O :N : H H H H H 2O : 3)4(H2O)2]2+ [1] 2. (a) (i) A weak acid is a proton donor that dissociates partially. [1] (ii) [H+] = 5.66 x 10–3 mol dm–3 [1] pH = 2.25 [1] (iii) Amount of acid used = 5 x 1–3 mol Amount of NaOH required = 5 x 10–3 mol Volume of NaOH required = 0.0500 dm3 = 50 cm3 [1] [OH–] = solution salt of conc acid cylicacetylsali of b K = 05 . 0 05 . 0 10 5 10 2 . 3 10 00 . 1 3 4 14        = 1.25 x 10 –6 mol dm–3 pOH = 5.90 pH = 8.10 [1] (b) (i) pH = pKa =  410 2 . 3 lg  = 3.49 [ 1 ] (ii) When H+ is added: C8H7O2COO– + H+ → C8H7O2COOH [1] When OH– is added: C8H7O2COOH + OH– → C8H7O2COO– + H2O [1] (iii) Amount of HCl added = 2.00 x 10–4 mol In the resultant solution: Amount of C8H7O2COO– left = 43 10 2 10 5 . 2    3 mol 10 30 . 2  Concentration of C8H7O2COO– = 10002 25 50 10 30 . 23     = 0.0299 mol dm –3 [1] [1] Page 2 of 7

2. (b) (iii) Amount of C8H7O2COOH = 43 10 2 10 5 . 2    3 mol 10 70 . 2  Concentration of C8H7O2COOH = 10002 25 50 10 70 . 23     = 0.0351 mol dm –3 pH of the resulting solution =   0351 . 0 0299 . 0lg 10 2 . 3 lg4    = 3.43 [1] [1] (c) (i) Step I: Acidified KMnO4, heat [1] Step II: LiAlH4, dry ether [1] Step III: PBr3 [1] Structure of D: C6H5COOH [1] (ii) Br2, UV light [1] (iii) Route 1 is a better pathway as Route 2 will result in multisubstituted products being formed. [1] (iv) Add NaOH(aq) and heat, followed by excess HNO3(aq), then AgNO3(aq). C6H5CH2Br gives a cream precipitate of AgBr. [1] [1] 3. (a) (i) To increase the yield of PCl5 since reaction is exothermic. To cool the exothermic reaction for safety. [1] [1] (ii) PCl 5 will react/hydrolyse in water to give an acidic solution. [1] PCl5(s) + H2O(l) → POCl3(aq) + 2HCl(aq) OR PCl5(s) + 4H2O(l) → H3PO4(aq) + 5HCl(aq) [1] (b) A white precipitate of AgC l is obtained when sodium chloride is mixed with silver nitrate. Precipitate dissolves in aqueous ammonia to give a colourless solution of Ag(NH3)2 +. [1] Ag+(aq) + Cl–(aq) → AgCl(s) Ag+(aq) + 2NH3(aq) → Ag(NH3)2 +(aq) When NH3 is added, the formation of the complex ion decreases the concentration of Ag+, causing equilibrium position of the equilibrium AgCl  Ag+ + Cl– to shift to the right such that ionic product < Ksp(AgCl). [1] (c) NaAt(s) + H2SO4(l) → HAt(g) + NaHSO4(s) [1] 2HAt(g) + H2SO4(l) → At2(g) + SO2(g) + 2H2O(l) 8HAt(g) + H2SO4(l) → 4At2(g) + H2S(g) + 4H2O(l) [1] [1] Page 3 of 7

3. (d) P undergoes alkaline hydrolysis to form Q. => P is an ester. => Q is an alcohol. Q undergoes substitution to form R => R is an alkyl halide. R undergoes nucleophilic substitution to from S. => S contains a CN group and 1 alcohol group. S undergoes acid hydrolysis to from T. => T contains one COOH group and 1 alcohol group T undergoes dehydration to form U and since U undergoes electrophilic addition with Br2 => U contains an alkene and a COOH group [5 – for explanation] CH2 CH2 CO H O O H T C H2 CH C OH O U CH2 CH2 CNO H S CH2 CH2O H Cl R CH2 CH2O H OH Q CH2 CH2OCC H 3 O O C O C H3 P [1] [1] [1] [1] [1] [1] 4. (a) (i) From experiments 2 and 3, [H2O2] and [H+] = constant. When [I] increases by 1.2 times, rate increases by 1.2 times. Therefore, Rate  [I ]. Reaction is first order with respect to I . [1] From experiments 1 and 3, [H2O2] and [I] = constant. When [H+] increases by 1.2 times, rate remains the same. Reaction is zero order with respect to H+. [1] From experiments 2 and 4, [H+] and [I] = constant. When [H2O2] increases by 1.25 times, rate increases by 1.25 times. Therefore, Rate  [ 2 2 Reaction is H O ]. first order with respect to H2O2. [1] rate = k [H2O2] [I ] [1] (ii) Meachanism A because it has a slow step (Step 1) involving one molecule of H2O2 and one I− ion. [1] 4. (b) (i) Transition State Energy Page 4 of 7 79 kJ mol-1

[2] [2] (ii) (ii) Activation energy (Ea) is the minimum amount of energy that Activation energy (E ) is + :CN CO  a the minimum amount of energy that reactants must possess before a reaction can occur. [1] Consider the rate equation, Rate = k [A][B] Rate constant is the proportionality constant, k, in the experimentally determined rate equation. [1] (iii) When temperature increases, activation energy remains constant/unaffected, rate constant increases. Explanation: Activation energy of a reaction depends on the reaction pathway[1] When temperature increases, the number of molecules with energy >/ activation energy increases, Therefore, frequency of effective collisions between molecules with energy >/ activation energy increases and hence, rate constant increases. [1] [1 for properly labelled Boltzmann Distribution Curve] (c) (i) [2m for mechanism] 4. (c) (ii) The first step [1] is the rate determining (slow) step. Ea energy T No. of molecules with energy ≥ Ea at T2 No. of molecules with energy ≥ Ea at T1 2 > T1 Fraction of molecules T2 0 R R C ═ O +  R R CN + HCN COH + CN R - R CN Page 5 of 7

Step 1 involves breaking the C=O π bond. [1] (iii) HCN is a very weak acid, almost completely un-ionised OR −CN is a stronger nucleophile than HCN. [1] (iv) Nucleophilic addition is dependent on δ+ on the carbon of C=O, but this is neutralised by the oxygen, with delocalisation . [1] (v) CO H C CN OH H CH3CH2 HCN traces of NaCN or NaOH CH3CH2 C CH2NH2 OH H CH3CH2LiAlH4, dry ether or H2, Ni, heat C O OC H 3 [1] [1] 5. (a) (i) (ii) E cell = +1.77 – (+0.77) = +1.00 V (energetically feasible) [1] H2O2 + 2H+ + 2Fe2+ → 2Fe3+ + 2H2O [1] (b) (i) [Fe2+] = 1 mol dm−3 [1] (ii) Ksp = [Fe2+] [OH−]2 [1] (iii) Ksp = [Fe2+] [OH−]2 [OH−] = 7.75 × 10