DHS H2 CHEM P3 ANSWERS Prelim

© DHS 2011 9647/03/Answers 1 Dunman High School 2011 Year 6 H2 Chemistry Preliminary Exam – Paper 3 Answers 1 Iron is the cheapest and one of the most abundant of all metals, comprising nearly 5.6% of the earth's crust and nearly the earth’s entire core. It exists in a wide range of oxidation states, from −2 to +6, although ferrous (Fe 2+) and ferric (Fe 3+) compounds are more common. (a) Acidic solutions containing ferrous ions are oxidised to ferric ions in air, with no precipitation seen. On the other hand, ferrous ions give a precipitate in alkaline solutions and the precipitate turns reddish–brown in air. With reference to the Data Booklet , explain the two reactions using relevant E values, writing equations where appropriate. [4] In acidic solution O2 + 4H+ + 4e 2H2O Eo = +1.23 V Fe3+ + e Fe2+ Eo = +0.77 V 4Fe2+ + O2 + 4H+  4Fe3+ + 2H2O Eo cell = +1.23 – (+0.77) = + 0.46 V In alkaline solution Fe2+ (aq) + OH– (aq)  Fe(OH)2 (s) OR Ferrous ion react with hydroxide ions to form the ppt, Fe(OH)2. O2 + 2H2O + 4e 4OH– Eo = +0.40 V Fe(OH)3 + e Fe(OH)2 + OH– Eo = –0.56 V Eo cell = +0.40 – (–0.56) = + 0.96 V 4Fe(OH)2 + O2 + 2H2O  4Fe(OH)3 OR The ppt will be oxidised in air to reddish–brown Fe(OH)3. (b) Ferric ions can catalyse the reaction between I–(aq) and S2O8 2–(aq). By considering relevant E values, describe and explain the role of the ferric ions in this reaction, writing equations where appropriate. Fe3+ acts as a catalyst and reacts with I–. Fe 3+ + e Fe2+ Eo = +0.77 V I2 + 2e 2I– Eo = +0.54 V 2I– (aq) + 2 Fe3+ (aq)  I2 (aq) + 2Fe2+ (aq) E cell = +0.77 – (+ 0.54) = + 0.23 V >0. Reaction is feasible. Fe2+ intermediate reacts with S2O8 2–. Fe3+ catalyst is regenerated.

© DHS 2011 9647/03/Answers 2 Fe3+ + e Fe2+ Eo = +0.77 V S2O8 2– + 2e 2SO4 – Eo = +2.01 V S2O8 2– (aq) + 2Fe2+(aq)  2SO4 2– (aq) + 2Fe3+(aq) E cell = +2.01 – (+ 0.77) = + 1.24 V >0. Reaction is feasible. [3] (c) Ferric chloride is an industrial scale commodity inorganic compound which is often used as catalyst in organic synthesis. One example is its use as a Lewis acid for catalysing the alkylation reaction of benzene by chloroethane to form phenylethane. This reaction is similar to the reaction between benzene and chlorine. (i) Write a balanced equation for the overall reaction of chloroethane and benzene. (ii) State and outline the mechanism, with equations only, for the above reaction using ferric chloride as a catalyst. Electrophilic substitution. unstable cyclic carbocation (iii) Hence, or otherwise, suggest how you w ould synthesise the following alcohol, 1–(3– nitrophenyl)propan–2–ol, starting from a ch loroalkene and benzene as your only organic reagents. Your synthesis route should be no more than three steps.

© DHS 2011 9647/03/Answers 3 1–(3–nitrophenyl)propan–2–ol Or H2O, H3PO4 catalyst at 300°C and 70 atm for the last step. (iv) Explain why 1–(3–nitrophenyl)propan–2–ol exis ts as enantiomers, and describe how pure samples of the enantiomers can be distinguished by a physical method. It has a chiral carbon centre or carbon atom with four different groups with no plane of symmetry; Pass plane–polarised light through sample and each enantiomer will rotate the light in opposite/different directions. [8] (d) Consider this iron compound, NH 4[Fe(SCN)x(NH3)y], with the following composition by mass: Fe 16.4% S 37.7% N 28.8% (i) Define the term ligand, and identify the ligands in this complex. A ligand is defined as a molecule/ion which contains at least one lone pair of electrons available for forming dative/coordinate bond with the central metal atom/ion. The ligands are thiocyanate ion (SCN –) and ammonia (NH3) (ii) Calculate the values of x and y in the formula.

© DHS 2011 9647/03/Answers 4 Method 1 Fe (Ar = 55.8) is 16.4% of total mass.  total mass = 100 x 16.4 55.8 = 340.24 mass of S = 100 37.7 x 340.24 = 128.27 x = 128.27/Ar x = 4 mass of N = 100 28.8 x 340.24 = 98.00 y – 4 – 1 = 98.00/Ar y – 4 – 1 = 7.00 y = 2 Method 2 Fe S N mol 16.4/55.8 37.7/32.1 28.8/14 = 0.2939 = 1.174 = 2.057 ratio 1 4 7 From NH4[Fe(SCN)x(NH3)y]: Fe  x S  (1+x+y) N  x = 4 1+x+y = 7  y = 2 (iii) Calculate the oxidation number of iron in the compound. Since formula of the anion of complex is [Fe(SCN) 4(NH3)2]–, 4(–1) + 2(0) + Fe = –1 Fe = 3 Oxidation number of Fe is +3 [5] [Total: 20]

© DHS 2011 9647/03/Answers 5 2(a) The Van Slyke’s method, named after an Amercian biochemist Van Slyke Donald Dexter, refers to a method to test for primary amino groups. Amino acids can be determined by measuring the volume of nitrogen released from their reaction with nitrous acid. For example, alanine can be reacted as shown in the equation below. CH3CH(NH2)COOH + HNO2  CH3CH(OH)COOH + N2 + H2O Another method to test for primary amino groups consists of reacting amino acids with a volumetric solution of perchloric acid, for example: CH3CH(NH2)COOH + HClO4 CH3CH(N+H3)COOH + ClO4 –  The above is a weak base and strong acid neutralisation reaction. The excess perchloric acid is then determined by titration with aqueous sodium hydroxide. (i) Explain, with the aid of an equation, why alanine is a weaker base than propylamine. The basic amine (–NH2) group undergoes internal acid–base reaction with the acidic –COOH group in alanine to form a zwitterion. Extent of accepting a proton in solution decreases  alanine is a weak base. Using the above method to test for primary amino groups in lysine, 50.0 cm 3 of 0.100 mol dm–3 perchloric acid is added to a sample of lysine, in glacial acetic acid. The excess perchloric acid is determined by titration with 0.150 mol dm –3 sodium hydroxide solution. 16.0 cm3 of sodium hydroxide is needed to complete the titration. (ii) (iii) Write an equation for each of the following reaction between: (I) nitrous acid and lysine (II) perchloric acid and lysine Calculate the volume of the nitrogen released at a pressure of 103 kPa and a

© DHS 2011 9647/03/Answers 6 temperature of 20 C by the same lysine sample. n(HClO4) added = V x c = 0.0500 x 0.100 = 0.00500 mol n(NaOH) used for titration = 0.0160 x 0.150 = 0.00240 mol n(HClO4) consumed in reaction = 0.00500 – 0.00240 = 0.00260 mol Since, 2n(HClO4) = n(lysine) = 2n(N2) = 0.00260 mol p nRT)V(N2  = 103000 20)31)(2730.00260(8.  = 6.17 x 10-5 m3 or 0.0617 dm3 [6] (b) Amino acids like lysine show both acidic and basic properties. In acidic solution, lysine is completely protonated and exists as the conjugate acid. 10.0 cm 3 of completely protonated lysine is titrated with 0.10 mol dm–3 NaOH. Its titration curve is shown below. (i) Calculate the first dissociation constant Ka1 of lysine. At pt X, pH = pKa1 Ka1 = 10–2.20 = 6.31 x 10–3 mol dm–3 (ii) Identify the species present at point X. (iii) Calculate the concentration and initial pH of lysine at A. n(NaOH) used to reach 1st endpoint = 0.0010 mol n(lysine) = 0.0010 mol [lysine] = 0.0010/0.0100= 0.10 mol dm–3 HO2CCH(NH3 +)(CH2)4NH3 + –O2CCH(NH3 +)(CH2)4NH3 + + H+ Eqm 0.10 – x x