IJC H2 CHEM P3 ANS

Answers (Prelim 2 H2 Paper 3) 1 PRELIM 2 © INNOVA 9647/03/2011 [Turn over 1 (a) (i) Manganese(IV) oxide, MnO 2 is the most common starting material for the production of compounds of manganese of other oxidation states. With reference to the Data Booklet , state which of the following ions would convert manganese(IV) oxide to manganese(II) sulfate in an acidic solution. P b 4+(aq) / Pb2+(aq) E = +1.69V C o 3+(aq) / Co2+(aq) E = +1.82V F e 3+(aq) / Fe2+(aq) E = +0.77V Co3+(aq) / Co2+(aq) E  = + 1.82V Pb4+(aq) / Pb2+(aq) E  = + 1.69V MnO2(s) / Mn2+(aq) E  = + 1.23V Fe3+(aq) / Fe2+(aq) E  = + 0.77V Only can choose Fe2+(aq) (ii) Calculate the E cell value to prove that the reaction between MnO 2 and the ion you have chosen in a(i) is feasible and write a balanced equation for the reaction. Eq(1) MnO 2 + 4H+ + 2e Mn2+ + 2H2O +1.23 Eq(2) Fe 3+ + e Fe2+ +0.77 Overall: MnO2 (s) + 4H+ (aq) + 2Fe2+ (aq)  Mn2+ (aq) + 2H2O (aq) + 2Fe3+ (aq) E θ cell = (+1.23) – (+0.77) = +0.46V Since Eθ cell > 0, reaction is feasible. (iii) What would be the colour of the final solution observed? Orange (Pale pink + Yellow) Also accept yellow [5] (b) (i) Write the full electronic configuration of Zn2+ and Mn2+. Zn2+ : 1s2 2s2 2p6 3s2 3p6 3d10 Mn2+: 1s2 2s2 2p6 3s2 3p6 3d5 (ii) Explain why manganese forms compounds that are often coloured whereas compounds of zinc are usually white. Compounds of Mn is coloured due to the presence of incompletely filled 3d– orbitals in the Mn 2+ metal ions. In the complex ion, the electrostatic field produced by the donated lone pairs on the ligands, splits the five orbitals into a group of three and a group of two . An electron in a lower d orbital energy can absorb radiation and be promoted into the higher orbital energy level by absorbing photons of light in the visible light region (d-d transitions). The colour seen is the complement of those absorbed in the visible region of the spectrum. Zn2+ is colourless because its d orbitals are completely filled. It is not possible to promote an electron to the higher energy level. [5]

Answers (Prelim 2 H2 Paper 3) 2 PRELIM 2 © INNOVA 9647/03/2011 [Turn over (c) MnO2 undergoes the following reaction as shown in the schematic diagram below: Step I: Oxidation of MnO 2 in molten KOH with oxygen from the air produces a green compound, A. Step II: Reaction of A with dilute acid produced 0.174 g of MnO 2 and a purple solution of compound B was observed. A solution of compound B is just decolourised by 40.0 cm3 of 0.500 mol dm-3 iron(II) sulphate. (i) B contains potassium, manganese and oxygen only. Suggest the chemical formula of compound B that causes the solution to appear purple. KMnO4 (ii) Calculate the number of moles of MnO2 produced in step II. No. of moles of MnO2 = 0.174 / (54.9 + 16 x 2) = 0.002 mol (iii) Calculate the number of moles of compound B produced in step II, given that compound B and iron(II) sulphate reacted in a 1:5 mole ratio. No. of moles of compound B = 40/1000 x 0.5 x 1/5 = 0.004 mol (iv) With reference to step II of the above reaction scheme, let the oxidation number of manganese in A be n, i.e. Mn n+. Using your answer to c(ii) and c(iii), write an expression in n to show (I) the no. of moles of electrons gained by Mn n+ when Mnn+ is reduced to MnO2 (II) the no. of moles of electrons lost by Mnn+ when Mnn+ is oxidised to B Let the oxidation number of A be n. [R]: Mn(n) + (n – 4) e –  Mn(IV) 0.002 0.002(n-4) 0.002 No of moles of electrons gained = 0.002 (n – 4) [O]: Mn(n)  Mn(VII) + (7-n)e– 0.004 0.004 0.004 (7 – n) No. of moles of electrons lost = 0.004 (7 – n) (v) Hence, using your answer to c(iv), calculate the oxidation number of manganese in compound A. No. of moles of electrons gained = No. of moles of electrons lost 0.002 (n – 4) = 0.004 (7 – n) n = +6 [5] (d) (i) Cl2 reacts with benzene in the presence of a suitable catalyst. Describe the mechanism and suggest the product obtained. Product

Answers (Prelim 2 H2 Paper 3) 3 PRELIM 2 © INNOVA 9647/03/2011 [Turn over Mechanism: Electrophilic Substitution Step 1: Generation of electrophile Cl2 + AlCl3 Cl + + [AlCl4] - catalyst electrophile nucleophile Step 2: Electrophilic attack on benzene ring Step 3: Loss of proton Step 4: Regeneration of catalyst H+ + [AlCl4] - AlCl3 + HCl nucleophile catalyst (ii) The reaction in d(i) can only take place under anhydrous conditions. Suggest a reason to explain why it is so. In the presence of water, A lCl3 ionises to form A l3+ and C l- OR hydrolyses in water to form acidic solution OR Hence under hydrated conditions, A lCl3 cannot act as Lewis acid OR catalyst OR halogen carrier to accept a lone pair of electrons from BrC l OR electrophile Br+ cannot be generated to brominate benzene. Thus the condition need to be anhydrous. [5] [Total:20]

Answers (Prelim 2 H2 Paper 3) 4 PRELIM 2 © INNOVA 9647/03/2011 [Turn over 2 (a) This question is related to the chemistry of carbonyl compounds to form nitrogen- containing compounds. Bupropion is an anti-depressant that was subsequently found to be useful as a smoking cessation aid. It reduces the severity of nicotine cravings and withdrawal symptoms. It can be synthesised from compound C, from which many derivatives of bupropion can also be synthesised. (i) Suggest the type of reaction occurring in step I. Step 1 is an elimination/dehydration reaction. (ii) Name the type of reaction in step III and state the reagent(s) and condition(s) involved. Nucleophilic substitution Reagents: ethanolic (CH3)3CNH2 Conditions: Heat with reflux (iii) Given that bupropion is basic, without using fractional distillation, suggest a reagent and a suitable separation technique that you will use to isolate the product bupropion from the reaction mixture. Describe what you will observe in the process. 1) Add conc HCl/ aqueous HCl to the liquid product. 2) A white ppt will be formed. 3) Obtain the crystals of bupropion hydrochloride through filtration.

Answers (Prelim 2 H2 Paper 3) 5 PRELIM 2 © INNOVA 9647/03/2011 [Turn over (iv) Suggest the reagents and conditions required in step IV and state and describe the mechanism of this step. Reagents: HCN, trace amount of NaOH(aq) OR HCN, trace amount of NaCN Conditions: 10 – 20 oC / Cold Mechanism: Nucleophilic Addition Step 1: NaOH  Na+ + OH- HCN H + + CN- OR NaCN Na+ + CN- Step 2: Step 3: (vi) Based on your answer in a(iv), suggest and explain whether the product of step IV is optically active or optically inactive. The product of step 5 is optically inactive. In the mechanism, the nucleophile/ CN - can attack the planar carbonyl carbon from the top and bottom with equal probability. So the product exists as a racemic mixture, where the optical activity of the optical isomers cancel out.