2020 HCI Prelim P3 Answers

2020 HCI C2 H2 Chemistry Preliminary Exam / Paper 3 HWA CHONG INSTITUTION 2020 C2 H2 CHEMISTRY PRELIMINARY EXAMINATION SUGGESTED SOLUTIONS Paper 3 1 (a) (i) [1] each x 3 1 (a) (ii) [1] Pick one of the products from (a)(i) and draw mirror image s using stereochemical formula with appropriate dash/wedge, showing the tetrahedral shape about the chiral carbon, and dotted line to show the mirror plane. e.g. The product has two isomers that are [1] non-superimposable mirror images, hence they are enantiomers. 1 (a) (iii) [1] The h ost cell recognises a specific conformation/chirality of the spike protein. If the amino acid in the protein is swapped with its enantiomer, it changes the conformation/3 -dimensional arrangement of the protein, and the host cell no longer recognises it and the virus cannot enter 1 (b) (i) [2] Draw one ethanol molecule and indicate correct hydrogen bond with appropriate illustrations. Students may choose any part of the phospholipid that can form hydrogen bond with the ethanol. Hydrogen bond δ+ δ‒ δ+ δ‒

1 1 (b) (ii) [1] Dispersion forces of attraction forms between the hydrocarbon chains of the soap and phospholipid ions [0.5] Water forms hydrogen bonding with the carboxylate group on the soap. [0.5] The above, together with the mechanical action of water forces the phospholipid particles to be pulled apart/ separated from each other/ washed away, hence breaking up the membrane. 1 (b) (iii) [1] Surface charge of the fiber polarises/distorts the electron cloud of the non- polar dust molecules [1] Forms ion – induced dipole attraction, OR electrostatic force of attraction between the charge on the fiber and the induced dipole on the dust molecule 1 (c) Copper metal is made up of a [1]g iant metallic lattice held together by the [1]electrostatic forces of attraction between the Cu2+ metal ions and the sea of delocalised electrons 1 (d) (i) [1] P: 1 (d) (ii) [1] Step 2: LiAlH4/ dry ether OR NaBH4 OR H2/ Ni/ high pressure [1] Step 3: PCl3 OR PCl5 OR SOCl2/ warm OR concentrated HCl/ ZnCl2/ heat [1] Step 4: ethanolic concentrated ammonia, heat in sealed tube [1] R: [1] S: 1 (d) (iii) [1/2] Lone pair of electrons on N a is delocalised into the aromatic ring (s), making the lone pair [1/2] less available for donation/ accept a proton. [1/2] Nb has 3 electron donating alkyl groups attached to it, making its [1/2] lone pair of electrons more available for donation/ accept a proton. [1] Nb more basic than Na. Awarded only if there is some attempt at explaining the difference. 1 (d) (iv) [1] Reagent/Test [1/2] Positive observation [1/2] Negative observation

2020 HCI C2 H2 Chemistry Preliminary Exam / Paper 3 K2Cr2O7(aq)/ H2SO4(aq)/ heat. Orange K2Cr2O7 turns green for hydroxychloroquine but remains orange for chloroquine or KMnO4(aq)/ H2SO4(aq)/ heat. Purple KMnO4 decolourises for hydroxychloroquine but remains purple for chloroquine or PCl5. Or SOCl2/ warm. White fumes of HCl for hydroxychloroquine but no white fumes for chloroquine or Na(s). Effervescence of (colourless odourless) H2 gas that gives pop sound with burning splint for hydroxychloroquine but no effervescence for chloroquine 2 (a) (i) Off-white precipitate [1/2] turns brown [1/2] upon reacting with dissolved oxygen. Reject: white ppt for Mn(OH)2, brown ppt of MnO2 (wrong species) 2 (a) (ii) No. of moles of iodine formed = ½ x (12.60/1000 x 0.01) = 6.30 x 10–5 mol No. of moles of Mn3+ that reacted = 2(6.30 x 10–5) = 1.26 x 10–4 mol No. of moles of dissolved oxygen = 1.26 x 10–4 /4 = 3.15 x 10–5 mol [1] [dissolved oxygen] = 3.15 x 10–5 /500 x 1000 = 6.30 x 10–5 mol dm–3 [1] = 0.002016 g dm–3 = 2.02 mg dm–3 + conclusion [1] Fish will not be able to survive in this sample as the dissolved oxygen is lower than what is needed. Ecf is allowed if no. of moles of oxygen is calculated wrongly as long as student draws the correct conclusion based on the calculated concentration. 2 (b) (i) Ksp = [Cu2+][OH–]2 [1] Ksp = (3 x 10–6)( 2x3 x 10–6)2 = 1.08 x 10–16 [1] units = mol3 dm–9 [1] 2 (b) (ii) [OH-] = 10–(14-8.73) = 5.3703 x 10-6 [1] Ksp = [Cu2+][OH–]2 [Cu2+] = Ksp ÷ [OH-]2 = 3.7448 x 10–6 mol dm–3 = 0.238 mg dm–3 + conclusion [1] Conclusion: [Cu2+] lower than MCLG hence water is possible for consumption. Ecf is allowed based on wrong Ksp from (b)(i). 2 (c) (i) Electrode with ethanol: negative Electrode with oxygen/air: positive [1] both correct polarities

2 Electron flow from the electrode with ethanol towards electrode with oxygen (to be drawn in the external circuit) [1] Reject: Arrow that is drawn on the electrolyte. 2 (c) (ii) CH3CH2OH + H2O → CH3CO2H + 4H+ + 4e− [1] 2 (c) (iii) CH3CH2OH + O2 → CH3CO2H + H2O [1] Ecf accepted only if they’ve combined the eqn in (c)(ii) with O2 + 4H+ + 4e → 2H2O 2 (c) (iv) ΔGor = – (–174) + (–389) + (–237) = –452 kJ mol–1 [1] ΔGor= –nFEocell –452000 = –4(96500) Eocell Eocell =+1.17 V [1] Ecf accepted if Eocell is calculated from wrong ΔGor 2 (c) (v) Cigarette smoke contains volatile compounds (e.g aldehydes or other alcohols) that may be oxidised in the cell. [1] 3 (a) (i) The standard enthalpy change of combustion is the heat evolved when one mole of the substance is completely burnt in oxygen / burnt in excess oxygen at 298 K and 1 bar. [1] 3 (a) (ii) Heat absorbed by water, q = mcT = 200 × 4.18 × (67.0 – 20.0) = 39292 J or 39.292 kJ [1] Amount of cyclobutane used = = 0.01786 mol Hc(cyclobutane) = –39.292 / (0.01786 × 0.8) = –2.75 × 103 kJ mol–1 [1] Ecf 2nd mark if student divided by correct moles and factored in 0.8 3 (a) (iii) H1 = Hc(cyclobutane) – 2Hc(CH2=CH2) = –2750 – 2(1422) = +94.0 kJ mol–1 [1] Ecf from a(ii) 3 (b) (i) Electrophilic addition [1] 8 x 1.04 x 12.0 1.00 

2020 HCI C2 H2 Chemistry Preliminary Exam / Paper 3 1m for slow step and 1m for the rest of the mechanism 3 (b) (ii) The electrons delocalise from the C=C to the C=O group, making the carbonyl carbon less electron deficient and hence less susceptible to attack by nucleophiles. This also reduces the electron density on the terminal carbon on the alkene / causes the terminal carbon on the alkene to be electron deficient and enable it to be attacked by CN nucleophile. 0.5m for each underlined point 3 (c) (i) T: [1] 3 (c) (ii) P undergoes reduction with NaBH4 to give Q  P contains an aldehyde P undergoes reduction / catalytic hydrogenation to form R  P has C=C  Since P gains 6H atoms upon reduction (2H to reduce the carbonyl group), P has two C=C  R contains 1 alcohol group (give the mark if alcohol in Q mentioned instead for reduction of P with NaBH4) P undergoes oxidation with Tollen’s reagent, followed by acidification to give S  P is an aldehyde  S contains a carboxylic acid S undergoes oxidative cleavage to form T and U.

3 P: Q: R: S: Alternative answer: P: Q: R: S: 1m each structure, 0.5m each underlined point, total: 9m 4 (a) (i) Na+ has low charge density [] and hence does not undergo hydrolysis. [] Since Al3+ has higher charge density than Na+ [], it can undergo partial hydrolysis in aq solution, resulting in pH < 7. [] [Al(H2O)6]3+ (aq) + H2O(l)  [Al(H2O)5OH]2+ (aq) +