2019 HCI Prelim H2 Chem P2 ANS

2019 HCI C2 H2 Chemistry Preliminary Exam / Paper 2 Paper 2 1 (a) (i) HCN with trace amount of KCN (or NaOH(aq)), cold [1] (ii) [1] for each correct structure A B C (b) (i) Mechanism: Bimolecular Nucleophilic Substitution (SN2) [1] [1] for:  OH– is the nucleophile that attacks C+ of C–I (attack from the side that’s opposite to the I)  partial charges: C+–Cl–  curly arrow going from lone pair of OH– to C+ of C–I  curly arrow going from C–I bond, to I [1] for:  transition state negatively charged (need not label “transition state”)  I– is produced at the end of the reaction  no slow or fast step −½m for every mistake HWA CHONG INSTITUTION 2019 C2 H2 CHEMISTRY PRELIMINARY EXAM SUGGESTED SOLUTIONS

(ii) [1] shape of graph [1] labelling graph  label Ea  label ∆H  label “reactants” and “products”  label axes −½m for every mistake (c) The halogens Cl2, Br2 and I2 exist as simple non-polar covalent molecules. From Cl2 to Br2 to I2, the size of the electron cloud and hence, the polarisability of the halogen molecule increases [1]. More energy is needed to overcome the stronger dispersion forces between the molecules. Hence, the volatility of the halogens decreases [1] from chlorine to iodine. (d) (i) 2HX → H2 + X2 [1] (ii) Down the group, as atomic radius increases from Cl to I, the bond length of the H–X bond increases / bond strength decreases [1]. Hence, less energy is needed to break the H –X bond . Thus, the thermal stability of the hydrogen halides decreases down the group [1]. Remarks: can also justify by quoting the bond energy data, H–Cl = +431, H–Br = +366, H–I = +299 kJ mol –1. Weaker H–X bond needs less energy to break during thermal decomposition (e) From Data Booklet, Eo/V Br2 + 2e 2Br– +1.07 I2 + 2e− 2I− +0.54 Bromine reacts with S2O32– to form S4O62–, which will be further oxidised to SO2 and subsequently SO42–. Eocell = +1.07 − (+0.09) = +0.98 V > 0 (spontaneous) Eocell = +1.07 − (+0.51) = +0.56 V > 0 (spontaneous) Eocell = +1.07 − (+0.17) = +0.90 V > 0 (spontaneous) Iodine reacts with S2O32– to form S4O62– but there is no further oxidation of S 4O62– to SO2. Eocell = +0.54 − (+0.09) = +0.45 V > 0 (spontaneous) Eocell = +0.54 − (+0.51) = +0.03 V > 0 (spontaneous but extent of reaction is too small) Energy CH3CHICH2CH3 + NaOH CH3CH(OH)CH2CH3 + NaI H < 0 Ea Progress of reaction

2019 HCI C2 H2 Chemistry Preliminary Exam / Paper 2 [1] Eocell for reaction with bromine or showing that the Eo(Br2/Br-) > all the 3 Eo [1] Eocell for reaction with iodine or showing that Eo(I2/I-) > +0.09 V but only slightly larger than +0.51 V [½] spontaneous since Eocell > 0 [½] spontaneous but extent of reaction too small 2 (a) (i) Ka = [H+][A–] / [HA] Ka = (10–5.5)(1) / 4 Ka = 7.91 × 10–7 mol dm–3 pKa = – lg(7.91 × 10–7) = 6.10 [1]