2019 TJC H2 Chem Prelim P3 ANS

1 9729 / TJC Prelim / 2019 [Turn Over PRELIMINARY EXAMINATIONS HIGHER 2 CHEMISTRY 9729/03 Paper 3 Free Response 16 September 2019 2 hours Candidates answer on separate booklet. Additional Materials: 12-Page Answer Booklet Data Booklet READ THESE INSTRUCTIONS FIRST Write your Civics Group, Centre number, index number and name on all the work you hand in. Write in dark blue or black pen. You may use a HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Section A Answer all questions. Section B Answer one question. A Data Booklet is provided. The use of an approved scientific calculator is expected, where appropriate. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 14 printed pages and 2 blank pages.

2 9729 / TJC Prelim / 2019 [Turn Over Section A Answer all the questions from this section. 1 The term chelates, originates from the Greek word chele for “claw”. It refers to compounds containing ligands bonded to a central metal atom or ion at two or more points. Ligand exchange occurs when ammonia and ethylenediamine (en), H2NCH2CH2NH2, are added separately to copper sulfate solution. (a) (i) Explain all the changes observed when aqueous ammonia is added dropwise to Cu2+(aq) till in excess. Suggest the formulae of all relevant copper compounds formed and write equations for all reactions. [4] When aqueous ammonia is added dropwise to blue Cu2+(aq) solution , a blue precipitate of Cu(OH)2 is formed. Ammonia acts as a base. NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH–(aq) [Cu(H2O)6]2+(aq) + 2OH–(aq) ⇌ [Cu(OH)2(H2O)4](s) + 2H2O(l)  (1) Blue ppt OR Cu2+(aq) + 2OH–(aq) ⇌ Cu(OH)2(s) In excess ammonia, blue precipitate Cu(OH)2(s) dissolves to give a dark blue solution due to the formation of the complex ion, [Cu(NH3)4(H2O)2]2+(aq). Ammonia acts as a ligand (or ligand exchange has taken place). [Cu(H2O)6]2+(aq) + 4NH3(aq) ⇌ [Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l)  (2) Dark blue solution Formation of the complex shifts the equilibrium in equation (1) to the left, thus, the blue precipitate dissolves. (ii) Suggest why ethylenediamine can function as a chelating ligand. [1]  Ethylenediamine molecule has two lone pair of electrons on the nitrogen atoms to form two dative bonds with the Cu2+ ion. (iii) When ethylenediamine is added to Cu2+(aq), the following occurs. [Cu(H2O)6]2+ + 2en  [Cu(en)2(H2O)2]2+ + 4H2O Predict whether ethylenediamine or ammonia gives a more positive entropy change when added to Cu2+(aq). [2] Cu(H2O)6]2+ + 2en  [Cu(en)2(H2O)2]2+ + 4H2O [Cu(H2O)6]2+(aq) + 4NH3(aq)  [Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l)  Ethylenediamine (a bidentate ligand) gives a more positive change in entropy when added to Cu2+(aq) as  there is an increase in number of particles (from 3 to 5) and thus more way of arranging the particles. (iv) Suggest the coordination number and shape of [Cu(en)2(H2O)2]2+. [1]  Coordination number = 6, Shape = Octahedral

3 9729 / TJC Prelim / 2019 [Turn Over (v) The complex ion [Cu(en)2(H2O)2]2+ can exist as three stereoisomers, A, B and C. Isomers A and B rotate plane-polarised light but C does not. The structure of A is represented below. where represents Isomer A ethylenediamine (en) Draw the structure of C. [1]  Isomer C (b) Photochromic glass used for sunglasses contains AgCl crystals which are added while the glass is in molten state. The glass darkens when exposed to bright light and the following reaction is involved. Ag+ + Cl ⇌ Ag + Cl (i) Suggest why the glass darkens when exposed to bright light. [1]  In presence of UV light, colorless Ag+ cations form elemental Ag and the glass appear darker. The glass darkens significantly within about a minute of exposure to bright light but takes a longer time to clear when there is less light. Small amount of CuCl crystals are often added to the glass to speed up the reverse process. No elemental copper is involved in the process. Ag + Cl  Ag+ + Cl (ii) Explain why the copper(I) ions can be described as a homogeneous catalyst. [2] The copper(I) ions are in the ●same phase as the reactants, and is ●not consumed/used up by the reaction. (iii) State the property, typical of transition metals, which allows copper(l) ions to behave as a homogenous catalyst in the reverse reaction. Include relevant chemical equations to support your answer. [3]  The catalytic activity of transition metals in homogenous catalysis depends on their ability to exist in variable oxidation states. The catalysed pathway for the reverse reaction.  Cu+ + Cl  Cu2+ + Cl  Cu2+ + Ag  Cu+ + Ag+

4 9729 / TJC Prelim / 2019 [Turn Over (c) Microorganisms synthesise and secrete organic molecules called siderophores to increase the total concentration of available iron in the surrounding medium. Ferrichrome is a siderophore produced by fungi. Ferrichrome (i) Ferrichrome binds to iron(III) ions via its oxygen atoms. This process facilitates the transportation of iron(III) ions into the interior of a cell. Suggest what bonds are formed during this process and why iron(III) does not bind to nitrogen atoms in ferrichrome. [2]  Ferrichrome binds to iron(III) ions through the formation of dative bonds between oxygen atoms and iron(III) ions.  Lone pair of electrons on the nitrogen atoms are delocalised over the C=O group and are not available for dative bonding to iron(III) ions. (ii) Draw the structures of the products formed when ferrichrome is heated with aqueous sodium hydroxide. [3]  1 mark for each structure [Total: 20] 2 Ibuprofen is a nonsteroidal anti-inflammatory drug. It works by reducing hormones that cause inflammation and pain in the body. Ibuprofen is used to reduce fever and treat pain or inflammation. The conventional synthesis of ibuprofen from isobutylbenzene is shown below. Step 4 of the synthesis resembles the reaction between a carbonyl compound and Brady’s reagent.

5 9729 / TJC Prelim / 2019 [Turn Over (a) Name the types of reaction that are occurring during steps 4 and 5. [2] ●Step 4 : condensation reaction ●Step 5 : elimination (b) Suggest the reagents and conditions for steps 1, 4 and 6. [3] ●Step 1 : CH3COCl , anhydrous AlCl3, r.t.p ●Step 4 : NH2OH, r.t.p ●Step 6 : dilute HCl, heat (c) Draw the structures of the stereoisomers of the product formed from step 4 and state the type of stereoisomerism exhibited. [2] ●Type : cis-trans isomerism ● and

6 9729 / TJC Prelim / 2019 [Turn Over (d) Describe a simple chemical test to distinguish between the two compounds, X and Y obtained from synthesis above. [2] ●Add Tollens reagent (diammine silver(I) complex), HEAT to separate samples. ●Silver mirror observed for compound Y. No silver mirror for compound X. Alternative tests : * Alkaline aq. Iodine and warm. Yellow ppt observed for compound X and no yellow ppt for compound Y. * Fehling’s reagent, heat. Brick-red observed for compound Y and no brick-red ppt for compound X * Acidified K2Cr2O7, heat. Orange dichromate turned green when added to compound Y. No colour change for compound X. (e) Potassium chloride and potassium iodide can be distinguished by treating the compounds separately with concentrated sulfuric acid. (i) Potassium chloride and concentrated sulfuric acid reacts in an equi