HCI Prelim_H2_CHEM_P3_ans

2016 HCI C2 H2 Chemistry Preliminary Exam / Paper 3 Paper 3 1 (a) 1-bromobutane has a higher boiling than 1 -chlorobutane. Both 1-bromobutane and 1 -chlorobutane are simple molecular. Since 1-bromobutane has a larger electron cloud size [1], there is stronger dispersion forces between its molecules as compared to 1-chlorobutane. Hence, more energy [1] is needed to break the stronger dispersion forces for 1-bromobutane. (b) (i) Standard enthalpy change of combustion is the heat evolved when one mole of the substance is completely burnt in excess oxygen at 298 K and 1 atm. [1] (ii) Heat absorbed by water q = (200)(4.18)(45.0) = 37620 J = 37.6 kJ [1] Enthalpy change of combustion = [100/80 × 37.62] / (2.35/136.9) = 2.74 × 103 kJ mol–1[1] (c) (i) Comparing Expt 1 and 2, (Total volume in Expt 2 is half that of Expt 1.) Concentration of 2-bromo-3-methylbutane is constant, when concentration of NaOH increases 4/3 times, rate increases 4/3 times. Hence, the reaction is 1st order with respect to NaOH [1] Comparing Expt 2 & 3, (Total volume in Expt 2 and 3 are the same) When concentration of 2-bromo-3-methylbutane increases 2 times and concentration of NaOH decreases 4 times, the rate decreases 2 times. Since the reaction is first order with respect to NaOH, the reaction is thus 1st order with respect to 2-bromo-3-methylbutane [1] [Calculation method also accepted] Rate = k [2-bromo-3-methylpentane][NaOH] [1] ecf (ii) SN2 Nucleophilic substitution [1]

1 2 m for mechanism (iii) Rate = k [2-bromo-3-methylbutane][NaOH] To obtain the numerical value for k, substitute the rate and the respective concentrations into the rate equation. 7.40 × 10-7 = k [ ( 10 1000)(0.0100) (10.0+30.0+60.0 1000 ) ] [ ( 30 1000)(0.0100) (10.0+30.0+60.0 1000 ) ] k = 0.247 [1] Calculation To derive the units, mol dm–3 s–1 = k (mol dm–3)2 k = mol–1 dm3 s–1 [1] Units (d) (i) Test 1: SN1 mechanism Test 2: SN2 mechanism [1] (ii) In Test 1, the rate of reaction is the fastest for the tertiary bromoalkane. The carbocation formed from the tertiary bromoalkane is stabilised by three electron-donating alkyl groups , hence it undergoes S N1 rapidly. [1] In Test 2, the rate of reaction is the fastest for the primary bromoalkane as there is least steric hindrance when the nucleophile attacks the electron deficient carbon atom of the primary bromoalkane, hence it undergoes SN2 rapidly. [1] (iii) The carbocation formation is favoured as ethanol can form ion-dipole interactions with the carbocation intermediate thereby stabilising it [1].

2016 HCI C2 H2 Chemistry Preliminary Exam / Paper 3 (e) For the second step, accept H2, Ni, high temp and high pressure. Also accept if split up last step to show reduction followed by nucleophilic substitution with heat for last step. 2 (a) (i) Ka = [CH3CO2–][H+] [CH