2020 JPJC Prelim P3 answers

 Jurong Pioneer Junior College [Turn Over NAME CLASS 19S JURONG PIONEER JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION 2020 CHEMISTRY 9729/03 Higher 2 Paper 3 Free Response 22 September 2020 2 hours Candidates answer on question paper. Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your name, class and exam index number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a HB pencil for any diagrams, graphs. Do not use staples, paper clips, glue or correction fluid. Section A Answer all questions. Section B Answer one question. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. 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 30 printed pages.

2  Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2020 Section A Answer all the questions in this section. 1 (a) In a bromine “clock” reaction , solutions of bromate(V) ions, bromide ions, bromine and acid are mixed together with a few drops of red methyl orange. There are two reactions occurring in the reaction mixture. BrO3–(aq) + 5Br–(aq) + 6H+(aq) → 3Br2(aq) + 3H2O(l) Reaction I Br2 + phenol → brominated compound Reaction II When a small but constant amount of phenol is added to the reaction mixture, the bromine being slowly produced by reaction I will immediately react with phenol in reaction II, until all the phenol has been used up. At that point, free bromine will be pres ent in the solution , which will cause a sudden decolourisation of red methyl orange. The time taken for the red methyl orange to decolourise were recorded below. Expt Volume of 0.350 mol dm3 KBrO3 / cm3 Volume of 0.650 mol dm3 KBr / cm3 Volume of 0.250 mol dm3 H2SO4 / cm3 Volume of deionised water / cm3 Time taken for methyl orange to decolourise /s 1 5.0 5.0 10.0 10.0 82 2 10.0 5.0 10.0 5.0 42 3 10.0 10.0 5.0 5.0 83 4 5.0 5.0 20.0 0 21 For Examiner’s Use (i) Explain why the amount of phenol is kept constant in each experiment. [1] As we are measuring the time taken for a fixed same amount of Br 2 to be used up for the reaction with phenol, so the amount of phenol must be kept constant. This time is related to the relative rate, (ii) Deduce the order of reaction with respect to [Br–], [BrO3–] and [H+], explaining your reasoning clearly. [4] 1Rate time

3  Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2020 [Turn Over Since total volume of reaction mixture is constant, volume of reactantconcentration of reactant in the reaction mixture. Eg. Volume of KBrO3  [KBrO3] Comparing expts 1 and 2, [KBr] and [H2SO4] are constant, when [KBrO3] doubled, rate ( 1/t) doubled . Thus the reaction is first order w.r.t [KBrO3]. Comparing expts 1 and 4, [KBrO3] and [KBr] are constant, when [H2SO4] doubled, rate (1/t) increased by 4 times . Thus the reaction is second order w.r.t [H2SO4]. Comparing expts 2 and 3, Mathematical method Let a = order of reaction w.r.t [KBr]  expt 2 3 expt 2 expt 2 expt 2 expt 3 3 expt 3 expt 3 expt 3 [Rate] [KBrO ] [KBr] [H ] [Rate] [KBrO ] [KBr] [H ] 1 a 2 1 a 2 + +  (10) (5) (10) (10) (10) (5) 1 a 2 1 a 2 1 42 1 83 a = 1 For Examiner’s Use (iii) Write a rate equation for the reaction. [1] Rate = k[BrO3- ][Br-][H+]2 (iv) Reaction of bromate( V) ions with bromide ions in acidic solution is thought to proceed via the following reaction mechanism: step 1: H+ + Br– ⇌ HBr fast step 2: H+ + BrO3– ⇌ HBrO3 fast step 3: HBr + HBrO3 → HBrO + HBrO2 slow step 4: HBrO2 + HBr → 2HBrO fast step 5: HBrO + HBr → H2O + Br2 fast Explain if this reaction mechanism is consistent with the experimentally determined rate equation. [1] 1Rate time

4  Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2020 Yes and explanation that intermediate cannot appear in rate equation and the intermediate concentration ([HBr], [HBrO3]) has to be expressed in terms of the concentrations of the reactants that form it in the previous step. For the slow step, rate = k[HBr][HBrO3] --- (*) From Step 1, K1 =  [HBr] = K1[H+][Br –] -- (1) From Step 2, K2 =  [HBrO3] = K2[H+][BrO3–] --(2) Sub (1) and (2) into (*): Thus, rate = k K1 K2 [H+][H+][Br –][BrO3–]  rate = koverall[H+]2[Br –][BrO3–] For Examiner’s Use (b) 4-nitrophenol is used to synthesise a derivative of phenacetin, a pain-relieving and fever-reducing drug. In step IV, compound C reacts with phosgene, COCl2. You may assume that –OCH2CH3 is inert. C D Derivative of phenacetin OH NO2 I II O  NO2 III CH3CH2Cl OCH2CH3 NO2 IV 4-nitrophenol Cl O Cl A B C9H10ClNO2 (i) Suggest reagents and conditions for step III. [1] Step III: Sn, concentrated HCl, heat, followed by NaOH(aq) (ii) Suggest why it is necessary to convert 4 -nitrophenol to compound A before step II is carried out [1] To convert phenol to phenoxide a stronger nucleophile to react with halogenoalkane in step II via nucleophilic substitution. (iii) In step IV, compound C and phosgene reacted in 1:1 molar ratio to form compound D, which has the molecular formula of C 9H10ClNO2. Draw the structure of D. OCH2CH3 N H O Cl [1] (iv) Name and draw the mechanism of the reaction between chloroethane and compound A in step II. [3] For Examiner’s Use  [HBr] [H][Br]  [HBrO3] [H][BrO3  ]

5  Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2020 [Turn Over Nucleophilic substitution (SN2) Penalty for: - Missing lone pair - Missing delta positive and negative charge - Wrong arrows - No 3-D/ inversion shown - Missing by-product - Wrong product (v) Compare the reactivities of chloroethane, CH3CH2Cl and ethanoyl chloride, CH3COCl with water. Explain your answer. [1] Comparing CH3COCl with CH3CH2Cl: CH3COCl, which is an acyl chloride, undergoes hydrolysis to give CH3COOH and HCl(aq).  The C atom in –COCl of CH 3COCl is highly electron – deficient or delta positive as it is bonded to two or more electronegative atoms (O and Cl)  making CH3COCl more reactive towards attack by H 2O nucleophile. CH3CH2Cl reacts with H2O only upon heating Reactivities: CH3COCl > CH3CH2Cl (c) (i) Use the data in Table 1 to calculate the enthalpy change for the reaction that forms gaseous phosgene, COCl2, H1. CO + Cl2 → COCl2 H1 Table 1 compound Hf / kJ mol−1 CO(g) -110 COCl2(g) -220 [1] H = (nHformation products) – (mHformation reactants) = -220 – (–110) = -110 kJ mol1 (ii) Use the bond energies data in the Data Booklet to calculate the enthalpy change of the above reaction, H1. [1] C Cl CH3 H H Cl + C CH3 H H O2N O: O2N O d+ d- +Cl

6  Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2020 H = +nB.E bonds broken in reactants - mB.E bonds formed in products H1 = [BE(CO) + BE(Cl─Cl)] –[1BE(C=O) + 2BE(C─Cl)] --(1) Enthalpy change of reaction = 1077 + 244–[740+2(340)] = −99.0 kJ mol1 (iii) Suggest a reason for the difference in the values obtained in (c)(i) and (c)(ii). [1] The bond energy values in Data Booklet are average values and not the true bond energies present in phosgene. (d) (i) Phosgene reacts with NaOH(aq). It is suggested that the reaction occurs in two stages. CO2 + HClCOCl2 COCl(OH) stage 1 OH