SAJC 2022 P3 ANS

1 [TURN OVER ST ANDREW’S JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS HIGHER 2 CANDIDATE NAME CLASS 2 1 S CHEMISTRY Paper 3 Free Response Candidates answer on the Question Paper. Additional Materials: Data Booklet 9729/03 14 September 2022 2 hours READ THESE INSTRUCTIONS FIRST Write your name and class on all the work that 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. Answer all questions in the spaces provided on the Question Paper. If additional space is required, you should use the pages at the end of this booklet. The question number must be clearly shown. 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 26 printed pages (including this cover page). For Examiner’s Use Q1 21 Q2 19 Q3 20 Q4 or Q5 20 Total 80

2 [TURN OVER 1 (a) The diagram below shows how compound E may be prepared. CH3 CH3 Br Compound A O CH3 O CH3 Br Step 1 Step 2 Br O Compound B Compound C Compound D Compound E conc H2SO4 heat (i) Draw the structural formulae for C and D. [2] Compound C: OH CH3 CH3 Compound D: O OH Br (ii) Suggest the reagents and conditions for steps 1 and 2. [2] Step 1: aq NaOH, heat Step 2: aq H2SO4,aq KMnO4, heat

3 [TURN OVER or aq H2SO4 K2Cr2O7, heat (iii) Draw the structures of the products formed when E is reacted with hot aqueous sodium hydroxide. [2] CH3 CH3 OH O OH OH - (nucleophilic substitution for bromoalkane will occur) (iv) State the type of reaction when C and D react to form E. [1] Condensation / nucleophilic (acyl) substitution (v) Equal amounts of A, F and G are added separately to three test-tubes, which each contains equal concentration of ethanolic silver nitrate. The test-tubes are placed in a hot water bath. No precipitate is formed in one of the test-tubes. For the other two test-tubes, precipitates are formed at different rates. Explain these observations. CH3 CH3 Compound G CH3 CH3 I Compound F Cl [3] Compounds A and F undergo nucleophilic substitution with ethanolic AgNO3 which releases halide ion to form a ppt with Ag+. C−I bond in F is weaker than C−Br bond in A as I is larger atom than Br and has less effective orbital overlap with C. Thus Iꟷ is released faster to form ppt, followed by Brꟷ. Lone pair of electrons on the Cl atom delocalise into the π bond / C=C, resulting in C–Cl having partial double bond character, hence the bond does not break easily and no ppt is formed for compound G.

4 [TURN OVER (b) Copper is an important metal which can be used to catalyse many organic reactions. It exists naturally as an ore containing calcium and silver impurities. To obtain copper metal, the ore is purified using electrolysis. (i) Draw a labelled diagram for the purification set-up. [2] (ii) With reference to relevant data from the Data Booklet, explain what happens to the calcium and silver impurities during the purification. [3] Ag+ (aq) + e Ag(s) E = + 0.80 V Ca2+ + 2e Ca E = -2.87 V Cu2+ (aq) + 2 e Cu(s) E = + 0.34 V When an electric current is applied, copper and calcium at the anode (+), are oxidised to their respective ions. Ca is oxidised because of the negative E. Cu2+ and Ca2+ then migrate to the cathode (–). At the cathode, only Cu2+ ions are reduced to Cu due to its more positive E. Ca2+ remains as ions in the electrolyte as it is not reduce to metal easily. Ag will not be oxidised due to its positive E value and it falls off the electrode and accumulate at the bottom as anodic sludge (or words to the effect eg. drop to the bottom). (iii) A current was passed through the set up in (b)(i) for 50 minutes and the electrodes were then removed, washed, dried and weighed. It was found that the cathode had gained 0.95 g in mass. Calculate the current passing through the cell. [2] Amount of Cu = 0.95 / 63.5 = 0.01496 mol Cu2+ + 2eꟷ → Cu ( –) Cathode + – (+) Anode Impure Cu Pure Cu CuSO4 (aq) electrolyte

5 [TURN OVER Amount of e– = 2 X 0.01496 = 0.02992 mol Q = It 0.02992 X 96500 = I(50 X 60) I = 0.962 A (c) Calcium phosphate, Ca3(PO4)2, is used as a supplement for people who either do not get enough calcium from their diet or those who suffer from medical conditions like osteoporosis. (i) With the aid of relevant data from the Data Booklet, deduce whether copper(II) phosphate or calcium phosphate will decompose at a lower temperature. Explain your answer. [2] From data booklet, r+ Cu2+ = 0.073nm, r+ Ca2+ = 0.099nm Cu2+ has higher charge density and therefore, is able to distort / polarise the electron cloud of PO43ꟷ to a larger extent. P-O is weakened to a larger extent in Cu3(PO4)2 and therefore, need less energy to overcome it. Cu3(PO4)2 will decompose at lower temperature. (ii) 50 cm3 of 0.05 mol dm ꟷ3 sodium phosphate solution is mixed with 30 cm 3 of 0.05 mol dmꟷ3 calcium nitrate solution. Determine whether calcium phosphate precipitate is formed. (Ksp of calcium phosphate = 2.07 x 10ꟷ33 mol5 dmꟷ15) [2] [PO43ꟷ]mixture = (0.05 x 0.05) / 0.08 = 0.03125 mol dmꟷ3 [Ca2+]mixture = (0.03 x 0.05) / 0.08 = 0.01875 mol dmꟷ3 IP = [Ca2+]3[PO43ꟷ]2 = (0.01875)3 x (0.03125)2 = 6.4373 x 10ꟷ9 mol5 dmꟷ15 > Ksp Ca3(PO4)2 ppt will form. [Total: 21]

6 [TURN OVER 2 Chloric acid, HClO is both a strong acid and an oxidising agent. It is corrosive and will accelerate the burning of combustible materials. (a) When reacted with excess aqueous potassium hydroxide, HClO is converted into water and two chloro-containing products, one of which is a chloro-oxo anion. In an experiment, 0.5 mol of chloro-oxo anion was reacted with excess potassium iodide to form a brown solution and chloride ion. It was discovered that the brown solution required 3 moles of sodium thiosulfate for complete reaction. (i) Calculate the number of moles of electrons gained by 1 mole of chloro-oxo anion in the reaction with potassium iodide. Hence, prove that the chloro-oxo anion is ClO3ꟷ. [3] I2 + 2S2O32ꟷ → S4O62- + 2I ꟷ Amount of I2 formed = 3/2 = 1.5 mol Amount of I– reacted = 1.5 x 2 = 3 mol Chloro-oxo anion : I– 0.5 : 3 1 : 6 Given 2I – → I2 + 2e– Amount of electron gained by 1 mole of chloro-oxo anion = 6 Oxidation number of Cl in chloro-oxo anion = -1 + 6 = +5 Oxoanion = ClO3– (ii) The other chloro-containing product formed a white precipitate with silver nitrate solution. Identify this other chloro-containing product. Write an equation for the reaction between chloric acid and excess potassium hydroxide. [2] Other product = Cl– Equation: 3HClO + 3KOH → KClO3 + 2KCl + 3H2O OR 3H+ + 3ClOꟷ + 3OHꟷ → ClO3ꟷ + 2Clꟷ + 3H2O OR 3ClOꟷ → ClO3ꟷ + 2Clꟷ

7 [TURN OVER (b) Use of Data Booklet is relev