NJC Prelim P2 Solutions

1 NJC H2 Chemistry Prelim Paper 2 Suggested Answers 1 (a) Due to high temperature in the car engine, N2 and O2 from the air can react to form NO2. NO2 can be removed from the exhaust gas with the use of a catalytic converter, it can be reduced by CO to form harmless N2. (b) (i) (ii) NO2, being a radical, is very reactive and reacts with other gases in the air and gets destroyed more readily, hence a shorter atmospheric residence time (c) (i) ΔHo = +9.2 – 2(+33.2) = – 57.2 kJ mol‒1 ΔSo = +304 – 2(+240) = – 176 J mol‒1 K‒1 (ii) ΔHo has a negative sign since the dimerization is a bond formation process, hence heat is given out / reaction is exothermic. ΔSo has a negative sign since the dimerization results in fewer gas particles, hence there is a decrease in the disorderliness of the system. (iii) At equilibrium, ΔG = 0 ΔHo – TΔSo = 0 – 57.2 – T(– 176/1000) = 0 T = 325 K (iv) When N2O4 liquefies, ΔG = 0 ΔHvap – TΔSvap = 0 ΔHvap = TΔSvap = 294 x 88 = +25.9 kJ mol‒1 (iv) Using Hess’s law,  =  –  For ΔHrxn = – 57.2 – +25.9 = – 83.1 kJ mol‒1 For ΔSrxn = – 176 – 88 = – 264 J mol‒1 K‒1 2 (a) (i) H2O2 oxidises the I‒, to aqueous I2, so a brown solution / black solid would be obtained. H2O2 + 2I‒ + 2H+ → I2 + 2H2O

2 (ii) EC𝑙2/C𝑙–0 = +1.36 V EH2O2/H2O 0 = +1.77 V Eθcell = +0.41 V > 0 Since Eθcell is positive, reaction is spontaneous, H 2O2 can oxidise chloride to chlorine while it itself is reduced to H2O. Hence, the oxidation of iodide may not be complete. (iii) hexane / cyclohexane The aqueous layer will decrease in brown intensity and the colourless organic layer will turn purple / violet. (iv) Organic solvent is flammable, and can cause a fire to break out. With higher temperature, iodine may sublime and escape. (v) Excess aqueous NH3 should be added. AgCl is soluble in excess NH 3(aq) but Ag I is not. If there was significant amount of silver chloride in the precipitate obtained, most of the precipitate dissolved upon adding excess aqueous NH3 (b) (i) Amount of thioanisole used = 0.9×1.06 124.1 = 0.007687 mol Amount of NaIO4 used = 107×15.40 1000⁄ 214.0 = 0.0077 mol  NaIO4 reacts with thioanisole in a 1:1 mol ratio. (ii) C6H5SCH3 + NaIO4  C6H5SOCH3 + NaIO3 (iii) (c) As a reducing agent, H2O2 will be oxidized. Quote both eqns and Eo (In acidic) O2 + 2H+ + 2e– ⇌ H2O2 Eo = +0.68 V (In alkaline) O2 + H2O + 2e– ⇌ HO2– + OH– Eo = –0.08 V Due to a more negative E o value, HO2– is more likely to be oxidised than H 2O2. H2O2 is a better reducing agent in alkaline condition. 3 (a)

3 (b) Reagents : Dilute KMnO4, NaOH(aq) Conditions : cold (c) Structure of carbocation intermediate The carbocation ion would be formed at C 2 as it will form a more highly substituted carbocation than if the C+ is formed on C1. OR The carbocation would be formed at C 2 as it will have more electron donating alkyl group attached to C2 than if the C+ is formed at C1 (d) X Y (e) (i) (ii) 23 = 8 (f) Reagents and Conditions Step 1 KMnO4, H2SO4(aq), heat (or K2Cr2O7, H2SO4, Heat [under reflux]) Step 2 Anhydrous PBr3 (or SOBr2 or HBr(g))

4 4 (a) (i) NaOH(aq), heat (ii) Since the [OH–] decreases linearly over time with fixed gradient, rate of reaction remains the same throughout the whole experiment, the order or reaction with respect to OH– is 0. Comparing the time taken for 0.4 mol dm–3 of OH– being reacted, it takes a half the time for sample with 0.5 mol dm–3 of P (13 mins) than that of 0.25 mol dm–3 of P (26.5 mins). This means when the [P] is doubled, the rate is also doubled. The order of reaction with respect to P is 1. Rate Equation : Rate = k [Compound P] (iii) Nucleophilic Substitution (SN1) (iv) There is no optical activity present in the products. Trigonal planar carbocation intermediate allows nucleophile to approach from top and bottom of the plane at equal probability producing equimolar of enantiomers forming a racemic mixture. The effect of rotation of plane polarised light by one enantiomer is completely cancel by the other enantiomer resulting in lack of optical activity. (v) Trend: R–I, R–Br, R–Cl Rate of this nucleophilic reaction depends on the breaking of the C –X bond. C–I bond (240 kJ mol –1) is the weakest as compared to C –Br (280 kJ mol–1) and C–Cl (340 kJ mol –1). C–I bond is the easiest to be broken and hence would take the shortest time to complete the substitution (highest rate). C–Br bond is weaker than C –Cl, so C–Br would be easier to break and take shorter time than C–Cl. (b) (i) 75% yield = 20 g 100% yield = 20 x 4/3 = 26.67g Amount of diphenylethandioate = 26.67  242 = 0.11021 mol Amount of phenol = 0.11021 x 2 = 0.22042 mol Mass of phenol used = 0.22042 x 94 = 20.7 g

5 (ii) Add neutral FeC l3 (aq). Reaction is completed when there is no formation of violet complex, indicating that all phenol have been reacted. OR Add aqueous Br 2. Reaction is completed when orange aqueous Br 2 do not decolourise upon addition, indicating all phenol have been reacted. Reaction is incomplete when orange aqueous Br2 decolorised and white precipitate formed. OR Add Br2 in CCl4. Reaction is completed when orange-red Br2 do not decolourise upon addition, indicating all phenol have been reacted. Reaction is incomplete when orange-red Br2 decolorised. 5 (a) (i) Oxidation state of C1 in vitamin C : +1 Oxidation state of C1 after oxidation : +2 (ii) 2 H+ + 2e– (b) (i) 5 million red blood cells contain 150 mg of haemoglobin. Mass of haemoglobin in 10000 million red blood cells = 10000 5 × 150 = 300000 mg Mass of iron in 10000 million red blood cells = 4 100 × 300000 mg = 12000 mg = 12 g Amount of iron required = 12 55.8 = 0.215 mol (ii) Red blood cell has a lifespan of 120 days. However, even when the red blood cells dies, some of the iron still remains in the body for further use. (c) (i) dz2 dx2-y2 dxy dxz dyz

6 (ii) In the presence of ligands, the d-orbitals of the transition element ion are split into two different energy levels with a small energy gap, ∆E. Electrons in the lower energy d -orbitals can absorb light of a certain wavelength with energy corresponding to the energy gap, ∆E, and be promoted to the higher energy d-orbitals (d-d transition). The light not absorbed would be reflected and the colour of the complex is the complementary of the wavelength absorbed. (d) (i) 1s2 2s2 2p6 3s2 3p6 3d8 (ii) Tetrahedral complex Square planar complex (iii) Square planar [only contains paired electrons in the d orbitals] (e) (i) Condensation (also accept nucleophilic substitution or nucleophilic acyl substitution) (ii) Heating under reflux with conc HCl will cause the hydrolysis of the amide group in luminol. (iii) Luminol is less basic as compared to N2H4. The lone pair on all N of luminol are involved in delocalization with the neighbouring benzene and C=O groups. As such, they are less available for donation to H + and act as a base. The lone pair on N of N 2H4 do not undergo such delocalization and is more available for donation to H+. Isolated gaseous Ni2+