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+] [CH3CO2H] Kb = [CH3CO2H][OH–] [CH3CO2–] [1] Ka  Kb = [𝐶𝐻3𝐶𝑂2−][𝐻+] [𝐶𝐻3𝐶𝑂2𝐻] x [𝐶𝐻3𝐶𝑂2𝐻][𝑂𝐻−] [𝐶𝐻3𝐶𝑂2−] = [H+] [OH–] = Kw (ionic product of water), which is a constant. [1] (ii) CH3CO2H is a weak acid so its conjugate base CH3CO2– is a stronger base than water. So CH3CO2– undergoes hydrolysis forming OH– ions, forming an alkaline solution. [1] CH3CO2– (aq) + H2O (l) ⇌ CH3CO2H (aq) + OH− (aq) [1] (iii) As [CH3CO2H] = [CH3CO2−], this is a maximum capacity buffer; so pH = pKa = −log(1.8 x 10–5 ) = 4.74 [1] (b) (i) A [1] (ii) B = 2,4-dinitrophenylhydrazine [1] Cl2(g), UV LiAlH4 in dry ether [1] [1] [1]

2 C [1] (iii) addition [1] (iv) Dilute H2SO4 / HNO3 / HCl, heat [1] (v) D [1] (vi) E [1] (vii) Reaction III [1] The imine is trigonal planar about the imine carbon atom . T he nucleophile can thus attack this + carbon atom in equal probability from the top as well as the botto m [1], resulting in forming equal proportions of the two optical isomers / racemic mixture. (c) (i) Functional group level of C in CH4 = 0; in CO2 = 4 [1] (ii) Functional group level of circled C at the start = 3; at the end= 1 [1] As the functional group level decreases, the reaction is reduction. [1] (iii) F G / H 1m each

2016 HCI C2 H2 Chemistry Preliminary Exam / Paper 3 3 (a) (i) [1] Americium (Am) (ii) Ts lies below astatine in Group VII, following the trend of increasing melting point down the group [1], it should be a solid [1] like iodine or astatine. (iii) [1] Group II. [0.5]: Effervescence of hydrogen gas with water => Group I or II metal or Basic oxide hydrolyses to give metal hydroxide in water => Group I or II metal [0.5]: Sulfate insoluble => Group II metal as Group I sulfates all soluble. Or Group II sulfate solubility decreases down the group. (b) (i) Titanium is able to form ions with incomplete d-subshell, Ti2+: 1s22s22p63s23p63d2 or Ti3+: 1s22s22p63s23p63d1 [1] but zinc forms only the Zn 2+ ion which has 1s22s22p63s23p63d10 configuration with a fully filled d-subshell [1]. (ii) No. of moles of titanium = no. of moles of TiO2 = 2.99  10–3 mol No. of moles of chloride = 1.72 / 143.5 = 0.0120 mol Ratio of chloride ions to titanium = 0.0120 / 2.99  10–3 = 4.01 = 4 [1] 1st mark can be awarded if student is able to provide an answer that clearly recognises from TiO 2 that the complex is formed from a Ti(IV) center [0.5] and since complex is electrically neutral, there must be 4 chloride ions [0.5] Mass of titanium and chloride in 1 g sample = 2.99  10–3  189.9 = 0.568 g Mass of coordinated THF molecules in 1 g sample = 0.432 g 0.432 / (72.1  n) = 2.99  10–3 No. of THF molecules coordinated = 0.432 / (2.99  10–3  72.1) = 2.00 = 2 [1] (working for no. of THF ligands must be shown) Coordination number = 6 [1] (this answer must be based on an actual no. of chloride and THF ligands stated earlier) Allow ecf from errors in number or chloride and THF ligands calculated, rounded up to whole numbers. (c) (i) The 3rd ionization energy of Ca ( +4940 kJ mol 1) is much higher than the 2nd (+1150 kJ mol 1) as the 3rd electron in Ca is removed from an inner quantum shell [1] which requires a lot of energy . Thus, Ca 3+ compounds do not exist.

3 For Ti, the 2nd and 3rd ionisation energies (+1310, +2720 kJ mol –1) do not differ greatly as 4s and 3d electrons are close in energy [1]. Thus, Ti3+ can be formed. [1]: some comparison/discussion of ionisation energies for Ca and Ti pertaining to the high 3rd ionisation energy for Ca compared to Ti. (ii) [2] (iii) In calcium, only the 4s electrons are donated to the sea of delocalised electrons, while in titanium, both the 4s and 3d electrons are involved (idea of more electrons contributed towards metallic bonding). OR Titanium cations have higher charge/ smaller ionic radius / greater charge density than calcium cations. [1] This leads to titanium having stronger metallic bond strength and more energy is needed to overcome the metallic bonds [1] and hence a higher melting point. (iv) Burns slowly with brick-red flame , leaving a white residue [1] 2Ca(s) + O2(g)  2CaO(s) CaO(s) + H2O(l)  Ca(OH)2(aq) [1] Ca(OH)2 dissolves / gives a colorless solution in water to give an alkaline solution of pH = 12 [1] (Accept any value above 7) 4 (a) (i) [1] (ii) z y x 3s 3px 3py 3pz 3px 3py 3pz 3s

2016 HCI C2 H2 Chemistry Preliminary Exam / Paper 3 (iii) accept Cl2 (g), UV or heat, followed NaOH (aq), heat to get phenylmethanol. (b) cis trans [1] x 2 (c) +1 [1] (d) (i) The complex is chiral