2021 VJC H3 Chemistry Prelims Answer

Page 1 VICTORIA JUNIOR COLLEGE PRELIMINARY EXAMINATION 2021 CHEMISTRY (Higher 3) Answers 1 (a) (i) Gaseous particles exert negligible intermolecular forces on each other. Volume of gas particles are negligible compared to the volume of the container. [2] (ii) H2(g) ⇌ H2(aq) initial amount of H2 / mol 0 6.96 × 10–4 change in amount of H2 / mol x – x final amount of H2 / mol x 6.96 × 10–4 – x Using ideal gas equation, PH2Vgas = xRT PH2 = -----(1) = 24.5 x (PH2 is in atm) Using equation 1.1, PH2 = = x = 6.96 × 10–4 – 7.82 × 10–4 × PH2 -----(2) sub eqn (2) into (1), PH2 = 24.5 × (6.96 × 10–4 – 7.82 × 10–4 × PH2) 1.019PH2 = 1.7052 × 10–2 PH2 = 1.67 × 10–2 atm [3] (iii) Partial pressure of H 2 will increase as solubility of H 2 decreases with increasing temperature . As temperature increases, kH, which is similar to Kc, decreases. As temperature increases, more energy is available to overcome the permanent dipole -induced dipole interactions between H 2 and H 2O, allowing more gaseous H 2 molecules to form. [2] (b) (i) TCE underwent reduction as the ON of C bonded to H decreases from 0 in TCE to –2 in the product. [1]

Page 2 (ii) [2] (c) (i) To allow time for equilibrium to be established between the aqueous phase and the headspace. . [1] (ii) From graph, concentration of H2 = 5.9 ppm Partial pressure of H2 = 5.9 × 10–6 × 1 atm = 5.9 × 10–6 atm Assuming temperature is constant throughout the experiment and any increase in pressure due to H 2 is negligible (assumption is valid since 5.9 × 10–6 << 1). [3] (iii) SH2 = 5.9 × 10–6 ÷ (6.87 x 104) = √√8.59 × 10–11 (no units) [H2] dissolved = no. of moles of H2 ÷ volume of H2O = 8.59 × 10–11 ÷ (1 × 18 ÷ (0.998)) × 103 = 4.76 × 10–9 mol dm–3 Since 4.76 × 10 –9 mol dm –3 > 1.0 × 10 –9 mol dm –3, it is likely that anaerobic biodegradation was taking place in this sample. [3] 2 (a) (i) [1] (ii) Ratio of M : M+2 : M+4 = ¾ x ¾ : 2 x ¾ x ¼ : ¼ x ¼ = 9 : 6 : 1 [1] (iii) Na(s) [1]

Page 3 (iv) [3] (v) E: 6.6 and 7.1 ppm, 2 doublets, 2 protons each, aromatic protons F: 7.1 ppm, complex multiplet, 4 protons, aromatic protons [1] (vi) Chemical Shift / ppm No. of Protons Multiplicity Deduction / Structural Features 1.5 6 s -C(CH3)2 6.6 4 d Two 1,4-disubstituted benzene, aromatic protons deshielded by magnetic anisotropic effect 7.0 4 d 9.1 2 s 2 phenol protons, they are labile and disappear in presence of D2O These protons deshielded by electronegative oxygen atom and magnetic anisotropic effect of benzene G is symmetrical (due to very simple 1H NMR spectrum) G: [3] (vii) [1]

Page 4 (b) H J K [3] (c) (i) OH– add to electron deficient carbon partial charges reform double bond and expel Cl– [2] (ii) Through the appearance of O-H str etch of phenol, wavenumber range 3200 – 3600 cm–1 in the organic product formed [1] δ– δ+

Page 5 (d) (i) Number of Au gold atoms in unit cell = 8 x 1/8 + 6 x ½ = 4 [1] (ii) Volume = (0.408 x 10–9)3 = 6.79 ×10–29 m3 Mass = 1.93 ×104 x 6.79 ×10–29 = 1.31 ×10–24 kg [1] (iii) Mass of one Au atom = 1.31 ×10–24 / 4 = 3.28 x 10–25 kg NA = 196.97 / (3.28 ×10–22) = 6.01 x 1023 [2] (iv) ln (nh / nh0) = – (Eh – Eh0) / RT = – mg (h – h0) / RT = – m*NAg (h – h0) / RT Gradient = Numerical gradient = –0.0235 x 106 m NA = (gradient x RT) / (–m*g) = (–0.0235 x 106 x 8.31 x (15 + 273)) / (–8.3 x 10–18 x 9.81) = 6.91 x 1023 [4] 3 (a) (i) When a photon is absorbed by a molecule, an electron gains the photon’s energy and is promoted to a higher energy orbital. The electrons are normally resident in the bonding orbitals , or in the non -bonding orbitals if they are lone pairs on atoms such as nitrogen or oxygen. The higher energy orbital to which the electrons are promoted into is invariably an anti bonding orbital . Only π → π*, n → π* and n → σ* symmetry allowed normally produce absorption in the uv/visible region and give the characteristic uv/visible spectra. A system containing the electrons responsible for the absorptio n of radiation is called a chromophore. This system is usually a conjugated system of double bonds.

Page 6 Different chromophores absorb at different wavelengths which allows some analysis of structure. [2] (ii) The π electrons in benzene can delo calise into -NO2 or –SO2Cl which increases the extent of conjugation . Increased conjugation will decrease the energy gap between π and the π* orbitals (or HOMO and LUMO) . Since E = hc/ , the decrease in the energy gap will result in a longer wavelength of absorption for the reagent. [1] (iii) Chiral carbon at 5-membered ring: S Chiral carbon joined to –CH2C6H5: S Chiral carbon joined to –OH: R [2] (iv) Enantiomer: Maximum optical purity = 100 (+2) / (+8) = 25% [1] (v) [1] (b) (i) conformation showing anti periplanar geometry arrows for E2 mechanism For E2 mechanism to occur, the Br and H removed must be in the anti periplanar arrangement. To form the Zaitsev product , the Br and H cannot be in the required arrangement to form double bond across to the carbon containing CH3 group. [2]

Page 7 (ii) Structural isomer: Thermodynamic product (Zaitsev product) will be f ormed when the reactions operate at higher temperatures. Although the reaction rate leading to the more stable product may be slower, because of higher temperature and equilibriation, eventually the lower energy product will be favoured to be formed. Kinetic product (Hoffman product) arises (usually) from steric hindrance due to the size of the attacking bulky base (eg (CH3)3CONa) which would result in preferentially removal of H from the less substituted carbon. Another instance is when temperature of reaction is lowered, the activation energy leading to formation of less substituted alkene is lower and hence more favoured. [2] (c) (i) Multiplying the Kc expressions for the four given reactions produces = (1 x 10–7)(1 × 10–6)4 / [MoS42–] = 1.352 x 10–20 [MoS42–] = 7.40 x 10–12 mol dm–3 [1] (ii) 0.365 = 11870 x 10.0 x [MoS42–] (use of Beer’s law) [MoS42–] = 3.07 x 10–6 mol dm–3 Using [MoO2S22–] + [MoOS32–] + [MoS42–] = 6.0 x10–6 [MoO2S22–] + [MoOS32–] = 2.93 x10–6 [MoO2S22–] = 2.93 10–6 – [MoOS32–] 0.213 = 120 x 10.0 x 3.07 x 10 –6 + 9030 x 10.0 x [MoOS32–] + 3230 x 10.0 x [MoO2S22–] Solving above two equations gives [MoOS32–] = 2.00 x 10–6 mol dm–3 [MoO2S22–] = 9.30 x 10–7 mol dm–3 [2] (iii) Subtracting two equations given in the question gives [MoO42–] = [H2S] – 6.0 x 10–7 [MoO3S2–] = 8.0 x 10–7 – [H2S] Using Kc expression for K4:

Page 8 6.5 x 10–6 = = ([H2S] – 6.0 x 10–7)[H2S] / (8.0 x 10–7 – [H2S]) Solving gives [H2S] = 7.8 x 10–7 mol dm–3 [MoO3S2–] = 2.0 x 10–8 mol dm–3 [MoO42–] = 1.8 x 10–7 mol dm–3 Using Kc expression for the other equilibria gives [MoO2S22–] = 9.8 x 10–10 mol dm–3 [MoOS32–] = 7.6 x 10–11 mol dm–3 [MoS42–] = 4.6 x 10–12 mol dm–3 relevant working [4] 4 (a) (i) correct structure of M nucleophilic attack on C by N Deprotonation by Cl – [2] (ii) [1] (iii) Type of stereoisomerism: E, Z isomerism L M