DHS Prelim P3 SuggestedSolutions

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. ketone, secondary alcohol (ii) Describe a reaction, with reagent and condition(s), to confirm one of the functional groups in A. Describe the observations you would make. Reagent & conditions A Add 2,4–dintrophenylhydrazine. Orange ppt seen. Add PCl5. White fumes is seen. (iii) Only one organic compound can be formed when lawsone is reacted with aqueous Br2, through the loss of one water molecule. Suggest the structural formula of this compound. O O O Br or O O Br Br O O Br OH or [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.

DHS Prelim Exam 2014 H2 Chemistry Paper 3 – Mark Scheme © DHS 2014 9647/03 2 [O]: C10H8O3  C10H6O3 + 2e + 2H+ Overall: Cr2O7 2– + 8H+ + 3C10H8O3  2Cr3+ + 7H2O + 3C10H6O3 (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] No. of moles of B = 1000 12.50 x 0.050 x 3 = 0.001875 mol [B] = 100020 0.001875 = 0.0938 mol dm–3 (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. 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. O O O CH3 O Cl   O O O CH3 O Cl step I step II D

DHS Prelim Exam 2014 H2 Chemistry Paper 3 – Mark Scheme © DHS 2014 9647/03 3 (ii) Hence or otherwise, draw the structural formula of D. [3] O O O O CH3 (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. O O O (ii) Hence, suggest with the use of curly arrows, to show how G is formed from C. [2] O O O - [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. AgCl and AgBr (ii) Write an equation for the formation of white precipitate from ethanoyl chloride. CH 3COCl + Ag+ +H2O  CH3COOH + AgCl + H+

DHS Prelim Exam 2014 H2 Chemistry Paper 3 – Mark Scheme © DHS 2014 9647/03 4 (iii) State and explain which of the two precipitate will be soluble in aqueous ammonia. AgCl will be soluble in aqueous ammonia. AgCl(s) ⇌ Ag+ (aq) + Cl– (aq) ––––––– (1) Ag+(aq) + 2NH3(aq)  [Ag(NH3)2]+(aq) ––––––– (2) This enables aqueous NH 3 to react with free Ag + to form the soluble diammine complex /[Ag(NH3)2]+. By Le Chateliers’ Principle, equilibrium (1) shift right to increase [Ag+] hence AgCl dissolves. The Ksp value of AgBr is lower than that of AgCl. Thus there is a lower concentration of Ag +(aq) present in the solution which does not allow the formation of the soluble complex to bring about significant shift in equilibrium (1) to the right. [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. (i) Write an equation for the reaction that occurs during roasting. 2PbS + 3O 2  2PbO + 2SO2 (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. SO 2 (g) + H2O (l)  H2SO3 (aq) Red (iii) Identify the type of reaction that occurs to lead( II) oxide in the blast furnace. Reduction (iv) Deduce two equations for the reactions in which lead is formed in the blast furnace. PbO + C  Pb + CO; PbO + CO  Pb + CO2; 2PbO + C  2Pb + CO2 [6] (b) Silver is an impurity present in lead obtained from the blast furnace. It is removed

DHS Prelim Exam 2014 H2 Chemistry Paper 3 – Mark Scheme © DHS 2014 9647/03 5 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. They are of different densities/ lead is denser than zinc. (ii) Suggest a suitable value for temperature T. 700 C or any value from 693 – 1180 C. [Note: Temperature must be above melting point of zinc but below melting point of Ag so that it exists as liquid to dissolve solid silver.] (iii) Suggest how silver can be recovered from the silver–zinc alloy crust. Heat the alloy to above boiling point of zinc so that it distils off, leaving the solid silver. (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. Mass of zinc added = 2% x 1000 = 20 g Mass of silver = 0.2% x 1000 = 2 g Volume of 20 g zinc = mass/density = 20/6.57 = 3.044 cm 3 Volume of 1 kg Pb = mass/density = (97.8% x 1000)/10.7 = 91.401 cm3 Let mass of silver extracted with zinc be y g.

DHS Prelim Exam 2014 H2 Chemistry Paper 3 – Mark Scheme © DHS 2014 9647/03 6 lead molten in Agof ionconcentrat zinc molten in Agof ionconcentrat = 91.401 ) - (2 3.044