DHS Prelim P3 QP

DUNMAN HIGH SCHOOL Preliminary Examination 2014 H2 CHEMISTRY (9647/03) 1 (a) Lawsone is the dye that is extracted from the henna plant, Lawsonia inermis . Although its natural colour is yellow, lawsone reacts with the proteins in hair and skin to produce the characteristic brown henna colour. Compound A is a derivative of lawsone. O O OH lawsone O O OH A (i) Name two functional groups in A, other than the phenyl group. (ii) Describe a reaction, with reagent and condition(s), to confirm one of the functional groups in A. Describe the observations you would make. (iii) Only one organic compound can be formed when lawsone is reacted with aqueous Br 2, through the loss of one water molecule. Suggest the structural formula of this compound. [4] (b) Compound B can be oxidised to lawsone by acidified K 2Cr2O7 involving two moles of electrons lost per mole of B. O O OH OH OH OH lawsoneB (i) With the use of the Data Booklet , construct a balanced half–equation for the oxidation of B and a balanced equation for the overall reaction. You are to use the molecular formulae of lawsone (C 10H6O3), B (C 10H8O3) and Cr 2O7 2– in your equations. (ii) Calculate the concentration of a solution of compound B given that 20.0 cm 3 of this solution required 12.50 cm3 of 0.050 mol dm–3 K2Cr2O7 solution to reach end–point. [4] (c) When lawsone is reacted under suitable condition, compound C is produced. Reacting C with ethanoyl chloride produces a neutral compound D, with molecular formula C 12H8O4. C12H8O4 O O O - C CH3 Cl O D (i) The production of D occurs via a two–step mechanism.

DHS Prelim Exam 2014 H2 Chemistry Paper 3 – Mark Scheme © DHS 2014 9647/03 2 Step I: B and ethanoyl chloride reacts to form an intermediate E as shown. O O O CH3 O Cl EStep II: A chloride ion is eliminated to form D. Outline the two mechanism steps by drawing relevant curly arrows, partial charges and lone pair of electrons in your answer. Label the steps clearly. (ii) Hence or otherwise, draw the structural formula of D. [3] (d) Another compound F, in addition to D, is also produced in the above reaction involving C and ethanoyl chloride. O O O CH3 O F (i) Draw the structure of the nucleophile, G, involved in the formation of F. (ii) Hence, suggest with the use of curly arrows, to show how G is formed from C. [2] (e) When added to silver nitrate solution, ethanoyl chloride forms a white precipitate, while ethanoyl bromide forms a cream precipitate. (i) Identify the precipitates. (ii) Write an equation for the formation of white precipitate from ethanoyl chloride. (iii) State and explain which of the two precipitate will be soluble in aqueous ammonia. [7] [Total: 20] 2 (a) Over 80% of all lead produced ends up in lead–acid batteries, with lead metal and lead(IV) oxide used in them. In addition to starter batteries for road vehicles, lead and lead( IV) oxide are also used for zero emission and hybrid vehicles. The extraction of lead from its ore, galena (which is lead( II) sulfide, PbS), involves several processes. Firstly, lead( II) sulfide is roasted in air to form lead( II) oxide (PbO) and sulfur dioxide. The lead( II) oxide is heated with coke (carbon) and air in a blast furnace. Some of the coke forms carbon monoxide, both carbon and carbon monoxide react with lead(II) oxide to form lead.

DHS Prelim Exam 2014 H2 Chemistry Paper 3 – Mark Scheme © DHS 2014 9647/03 3 (i) Write an equation for the reaction that occurs during roasting. (ii) What is the compound that can be formed when sulfur dioxide is added to water? Determine the colour of universal indicator in the solution obtained when this compound is added to water. (iii) Identify the type of reaction that occurs to lead( II) oxide in the blast furnace. (iv) Deduce two equations for the reactions in which lead is formed in the blast furnace. [6] (b) Silver is an impurity present in lead obtained from the blast furnace. It is removed by a technique known as the Parkes process. In one of the steps, zinc is added to the lead and silver mixture. The mixture is then heated to a certain temperature, T, so that silver dissolves in zinc to form an alloy crust that floats. This separates silver from lead, allowing silver to be removed. The Parkes process depends on the following:  lead and zinc are almost immiscible just above their melting points  silver is much more soluble in zinc than in lead  silver/zinc alloys have higher melting points than pure zinc Use the following data to answer the following questions: metal melting point /C boiling point / C density of metal at melting point /g cm –3 lead 328 2023 10.7 silver 1233 2435 9.32 zinc 693 1180 6.57 (i) Give a reason why lead and zinc are almost immiscible above zinc’s melting point. (ii) Suggest a suitable value for temperature T. (iii) Suggest how silver can be recovered from the silver–zinc alloy crust. (iv) At a constant temperature, silver distributes itself between two immiscible solvents in such a way that the ratio of its concentration in two solvents is a constant. At 800 C, this constant is given as: 300lead molten in Agof ionconcentrat zinc molten in Agof ionconcentrat  where concentration is measured in g cm–3. Determine the mass of silver that can be extracted with zinc when 2% zinc by mass is added to 1 kg of the lead mixture containing 0.2% silver by mass as impurity. Assume that densities of zinc and lead at 800 C are the same as that given in the table. [6] (c) Lead cathodes are used with an acidic electrolyte in the organic synthesis of glyoxylic acid from oxalic acid.

DHS Prelim Exam 2014 H2 Chemistry Paper 3 – Mark Scheme © DHS 2014 9647/03 4 CC O OH O H CC O OH O OH glyoxylic acidoxalic acid + [O] Pb (i) Give the IUPAC name of oxalic acid. (ii) Glyoxylic acid has a p Ka value of 3.32. Define p Ka and hence determine the acid dissociation constant of glyoxylic acid, giving its unit. (iii) Which is a stronger acid, glyoxylic or ethanoic acid? Explain your answer. (iv) Name the type of reaction involved w hen glyoxylic acid reacts with hydrocyanic acid, HCN in the presence of trace amount of sodium cyanide. Describe the mechanism of this reaction with relevant equations, partial charges and arrows showing movement of electrons. [8] [Total: 20] 3 (a) The standard enthalpy change of formation of solid A l 2Cl6 is found to be –1364.2 kJ mol–1 using the following data: H / kJ mol–1 Al2Cl6 (s)  Al2Cl6 (aq) – 650 H2 (g) + Cl2 (g)  2HCl (g) –184 HCl (g)  HCl (aq) –72.7 2Al (s) + 6HCl (aq)  Al2Cl6 (aq) + 3H2 (g) –1026 (i) State Hess’ Law. (ii) Use the set of data and an appropriate energy cycle to verify the standard enthalpy change of formation value of solid A l2Cl6. Label the H values clearly in your cycle. [4] (b) Dimerisation of AlCl3 in the gaseous state produces Al2Cl6. (i) Draw the structure of Al2Cl6. (ii) Given that the standard enthalpy change of formation of solid A lCl3 is –705.6 kJ mol–1, calculate the standard enthalpy change for the following reaction. 2AlCl3 (s)  Al2Cl6 (s) (iii) Consider your answer in (i) and suggest with reasoning if the enthalpy value that you have determined in (ii) is as expected. (iv) The standard entropy change of reaction for the dissociation of A l 2Cl6 molecules into AlCl3 molecules is +88.0 J mol –1 K–1. Use your answer in (ii) to determine G of the dimerisation reaction. What is the significance of the sign of your answer? [5]

DHS Prelim Exam 2014 H2 Chemistry Paper 3 – Mark Scheme © DHS 2014 9647/03 5 (c) Aluminium sulfate is the active ingredient in styptic pencils, which can be used to stop bleeding from small cuts. The A l3+ ions induce coagulation in the blood, which facilita