2019 RVHS Prelim P3 ANS

River Valley High School 9729/02/PRELIM/19 [Turn over 2019 Preliminary Examination Suggested Answers for 9729 H2 CM 2019 JC 2 Prelim Paper 3 1 (a) (i) When dissolved in a polar solvent such as water, they dissociate to form strong acids, releasing H+ and X – ions forming ion-dipole interactions with water. [1] (ii) HF has an exceptionally high boiling point compared to the other hydrogen halides as hydrogen bonds exist between HF molecules but not between the rest of the HX molecules. More energy is needed to overcome the stronger hydrogen bonds. Boiling point of the hydrogen halides increases from HCl to HBr to HI. The strength of instantaneous dipole -induced dipole interactions between the hydrogen halide molecules increases as the number of electrons in the m olecules increases. More energy is needed to overcome the stronger id-id interactions. [2] (iii) When a red-hot steel needle is introduced, HBr produces red brown bromine vapour. HI gives violet fumes of iodine . HF and H Cl do not/show little tendency to decompose. Bond Bond Energies/ kJ mol-1 H–F +562 H–Cl +431 H–Br +366 H–I +299 Less energy is required to break the weaker H–X bond. The thermal stability of hydrogen halides decreases down the Group due to decreasing H–X bond energy. [3] (b) (i) Solubility of AgBr = 0.140  10–3 / (107.9 + 79.9) = 7.45  10–7 mol dm–3 [Br–] = 4.00 10–12 / 0.100 = 4.00 10–11 mol dm–3 [Ag+] [Br–] = Ksp [Ag+] (4.00 10–11) = (7.45  10–7)2 [Ag+] = 0.0139 mol dm–3 [2] (ii) In the saturated AgCl, [Ag+] [Cl–] = Ksp (0.0139)(0.0139 + 0.001) = Ksp Ksp = 0.000206 mol2 dm–6 Solubility = (0.00206)1/2 = 0.0144 mol dm–3 [2] (iii) Mass of AgCl dissolved in 1 dm3 = (0.0139)(107.9 + 35.5) = 1.99 g Mass of residue = 5.00 – 1.99 = 3.01 g [1]

2 River Valley High School 9729/02/PRELIM/19 2019 Preliminary Examination (c) (i) The higher the number of electron donating groups attached to carbocation, the more stable the carbocation. As the 2 possible intermediate contains a primary carbocation and secondary carbocation respectively, intermediate with the secondary carbocation is more stable. [2] (ii) [2] (iii) Compound B has a 4 membered ring form in place of the 5 membered ring presented in the reaction scheme. The 4 membered ring consist of 2 carbon atoms with sp2 hybridised orbitals 120° from each other and 2 carbon atoms with sp3 hybridised orbitals at 109.5° from each other. To fit into 4 membered ring, all 4 carbons are forced into 90° bond angle which exerts high angle strains making the 4 membered rin g formation unfavourable. [2] [Total: 17]

3 River Valley High School 9729/PRELIM/19 [Turn over 2019 Preliminary Examination 2 (a) (i) [1] (ii) H2O/Water [1] (b) (i) For p K1, the H + is removed from a neutral molecule. For p K2, the removal of a H+ from the anion that already carries a negative charge is electrostatically unfavourable. OR Favourable intramolecular hydrogen bonding in the anion will be disrupted when it dissociates in pK2. [1] (ii) Phthalic acid is the limiting reagent/NaOH is in excess Amount of excess NaOH = 50 1000 × 0.1 − ( 10 1000 × 0.2 × 2) = 1.0×103 mol [OH] = 1.0×10−3 60 1000 = 1.667 × 102 mol dm3 pOH = lg(1.667 × 102) = 1.78 pH = 141.78 = 12.2 [2] (iii) [3] VNaOH/cm 3 pH = pKa = 2.89 pH = pKa = 5.51 20 (1st equiv point) 10 30 40 (2nd equiv point) 0 pH 50 Max Buffer Capacity 12.2 Max Buffer Capacity

4 River Valley High School 9729/02/PRELIM/19 2019 Preliminary Examination (iv) [1] (c) (i) Condensation [1] (ii) [1] (d) [2] (e) Reaction A: PCl5, room temperature Reaction B: limited ethanolic NH3, heat in a sealed tube Structure of C: [3] (f) (i) I: acid-base reaction (alkaline hydrolysis is not accepted) II: nucleophilic substitution III: hydrolysis [3] (ii) H2SO4(aq) and heat under reflux, followed by controlled amount of strong base. Or NaOH(aq) and heat under reflux, followed by controlled amount of strong acid [1]

5 River Valley High School 9729/PRELIM/19 [Turn over 2019 Preliminary Examination (iii) [1] (iv) Amt of D = 0.5 147 = 3.401 × 103 mol Amt of gaseous product = 𝑝𝑉 𝑅𝑇 = (101325)(42.26×10−6) 8.31×303 = 1.701× 103 mol Since mole ratio of D: Gaseous Product = 2:1, gaseous product is N2 gas. 4C8H5O2N + 33O2 → 32CO2 + 10H2O + 2N2 [3] [Total: 24] 3 (a) Amt of Fe3+ = 5 0.2 0.001001000 mol Amt of CN = 6 0.6 0.006001000 mol Since ratio of Fe3+ : CN = 1:6, H2O ligand is displaced. Complex in red solution : [Fe(CN)6]3 Formation of red solution : [Fe(H2O)6]3+ + 6CN  [Fe(CN)6]3 + 6H2O Complex in yellow solution : [Fe(CN)6]4 [2] (b) (i) 3d and 4s electrons have similar energies. More valence electrons from Fe is contributed to the sea of delocalised (mobile) electrons then Ca, thus Fe has a higher electrical conductivity. [1] (ii) Both aqueous Fe2+ and Fe3+ exist as aqua complexes with formulae [Fe(H2O)6]2+ and [Fe(H2O)6]3+ respectively. Ionic radius of Fe2+ = 0.061 nm ionic radius of Fe3+ is 0.055 nm With a smaller radius and a higher charge, Fe3+ has a higher charge density, water molecules in [Fe(H2O)6]3+ is polarised and the O–H bonds are weakened to a greater extent, [H+]/[H3O+] increases. Hence, the pH is lower. [Fe(H2O)6]3+ (aq) + H2O (l) ⇌ [Fe(H2O)5(OH)]2+ (aq) + H3O+(aq) [2]

6 River Valley High School 9729/02/PRELIM/19 2019 Preliminary Examination (c) (i) Role of Fe3+ : Homogeneous catalyst The reaction between peroxodisulfate ions S2O82− and iodide ions I− can be catalysed by either Fe2+ or Fe3+. S2O82 + 2e ⇌ 2SO42 E = +2.01 V Fe3+ + e ⇌ Fe2+ E = +0.77 V I2 + 2e ⇌ 2I  E = +0.54 V Reaction catalysed by Fe3+(aq): Step 1: Formation of an intermediate 2Fe3+(aq) + 2I (aq)  2Fe2+(aq) + I2(aq) Ecell = +0.77  (+0.54) = +0.23 V Step 2: Regeneration of catalyst S2O82(aq) + 2Fe2+(aq)  2SO42(aq) + 2Fe3+(aq) Ecell = +2.01  (+0.77) = +1.24 V [2] (ii) [3] (c) V has a molecular formula of C8H9NO2. The C:H ratio is ≈ 1:1. V contains a benzene ring V is insoluble in water and acids V contains an amide V undergoes acid-base reaction in NaOH Energy kJ mol1 [] Reaction coordinate Fe3+(aq) +I (aq) + S2O82(aq) SO42(aq) + Fe3+(aq) + I2(aq) Fe2+(aq) + S2O82(aq)+ I2(aq) Ea(2 ) Ea(1 ) H Ea(1) : activation energy for first step Ea(2) : activation energy for second step

7 River Valley High School 9729/PRELIM/19 [Turn over 2019 Preliminary Examination V contains a phenol or carboxylic acid group V undergoes electrophilic substitution with aq Br2 to form W  W has a side chain in either position 2, 4 w.r.t. OH V undergoes acidic hydrolysis to form compound X and Y  salt and carboxylic acid is formed. X is soluble in water  X is an ammonium salt Y on reduction with LiAlH4 forms Z  Z is an alcohol Z undergoes positive iodoform test/oxidation to form a yellow precipitate.  Z is –CH(CH3)(OH) group [9] [Total: 19] 4 (a) (i) ΔH vap positive due to the energy required to overcome the intermolecular forces between molecules. ΔS vap positive due to the changing from liquid to gaseous state, hence −TΔS vap negative For vaporisation to be spontaneous, |−TΔS vap| > |ΔH vap|, hence it is an entropy-driven reaction. [2] (ii) Benzene is the most volatile VOC. ΔG vap = 113.633.9 330( ) 1000 = −3.59 kJ mol−1 (For the others: ethanol +2.40 , methylbenzene +9.29 , propanone −0.05) [1] (b) (i) Excess chlorine, uv light, (limited methane) [1] (ii) Initiation U V V X OHN H C O C H3 OHN H H H C H3C OOH C H3C H2OH W Y Z OHN H C O C H3 Br Br +C l

8 River Valley High School 9729/02/PRELIM/19 2019 Preliminary Examination Cl2  2Cl Propagation CH4 + Cl  CH3 + HCl CH3 +