2020 ACJC Prelim Paper 3 Answers

9 Index No. Name Form Class Tutorial Class 2CH________ Subject Tutor ANGLO-CHINESE JUNIOR COLLEGE DEPARTMENT OF CHEMISTRY Preliminary Examination CHEMISTRY H2 9729/03 Paper 3 Free Response 1 September 2020 2 hours Cover page READ THESE INSTRUCTIONS FIRST Write your index number and name, form class and tutorial class on all the work you hand in. Write in dark blue or black pen. You may use an 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. The use of an approved 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. Please fill in the question number for the question attempted. Section Question No. For Examiner’s Use Marks A 1 2 3 B Presentation of answers TOTAL: 80 m This document consists of 33 printed pages and 1 blank page. 9729/03/Prelim/20 © ACJC 2020 ANGLO-CHINESE JUNIOR COLLEGE Department of Chemistry [Turn over

2 © ACJC2020 9729/03/Prelim/2020 [Turn over Section A Answer all the questions in this section. 1 Sodium borohydride, NaBH4, reduces carbonyl compounds to secondary alcohols as shown below. CH3COCH3 + NaBH4 + H2O  CH3CH(OH)CH3 + BH3 + NaOH (a) (i) The kinetics of this reaction was studied using propanone in the presence of aqueous NaBH4. Table 1.1 shows the experimental data obtained. Table 1.1 experiment initial concentration / mol dm3 initial rate of formation of CH3CH(OH)CH3 / mol dm3 s1 [NaBH4] [CH3COCH3] 1 0.0100 0.200 6.60 × 105 2 0.0080 0.200 5.28 × 105 3 0.0100 0.100 3.30 × 105 By showing approp riate working, deduce the order of reaction with respect to each compound. Hence write a rate equation for the reaction. [3] Comparing experiments 1 and 2, When [NaBH4] is increased 1.25 times, rate is increased 1.25 times. Rate is directly proportional to [NaBH4] Order of reaction with respect to NaBH4 = 1 Comparing experiments 1 and 3, When [propanone] is doubled, rate is doubled. Rate is directly proportional to [propanone] Order of reaction with respect to propanone = 1 Rate = k [NaBH4] [propanone] (ii) Calculate the value of the rate constant, stating its units. [2] Using experiment 1: k = 6.60 × 105  (0.01 x 0.200) = 0.0330 mol1 dm3 s-1 Using experiment 2: k = 5.28 × 105  (0.008 x 0.200) = 0.0330 mol1 dm3 s-1 Using experiment 3:

10 © ACJC2020 9729/03/Prelim/2020 [Turn over k = 3.33 × 105  (0.01 x 0.100) = 0.0330 mol1 dm3 s-1 (b) The structural formula of the BH4 is shown below. The reaction between propanone and sodium borohydride occurs in the following manner.  In the first step, BH4– ion transfers a hydride to the carbonyl carbon of propanone, forming the CH3CH(O–)CH3 ion.  In the next step, the CH3CH(O–)CH3 ion accepts a proton from water to form CH3CH(OH)CH3. (i) By referring to your rate equation in (a)(i) and the steps described above, name and describe the mechanism for the reaction between propanone and sodium borohydride. In your answer, you should show all charges and lone pairs and show the movement of electrons by curly arrows. [3] Nucleophilic addition/Reduction (ii) Experiment 1 was repeated under identical conditions using propanal. Explain how the initial rate of reaction will change. [2] The rate of reaction will increase with propanal. Propanal has one less electron donating alkyl group, so its carbonyl carbon is more electron deficient and more reactive towards nucleophilic attack. OR + 

11 © ACJC2020 9729/03/Prelim/2020 [Turn over The rate of reaction will increase when propanal is used. Propanal has one less alkyl group, so there is less steric hindrance around the carbonyl carbon for the nucleophilic attack. (c) (i) Propanone reacts with 2,4-dinitrophenylhydrazine to form an orange precipitate. Draw the structure of the orange precipitate. [1] (ii) Describe a simple chemical test that could distinguish between propanone and propanal. [2] Test: aqueous I2, NaOH, warm / heat Observations: Propanone – yellow precipitate Propanal – No ppt/ soln remain yellow Test: KMnO4, dilute H2SO4, heat Observations: Propanone – the solution remains purple Propanal – the solution decolourises Test: K2Cr2O7, dilute H2SO4, heat Observations: Propanone – the solution remains orange Propanal – the solution turns green Test: Tollens’ reagent, heat Observations: Propanone – no silver mirror Propanal – silver mirror is formed Test: Fehling’s solution, heat Observations: Propanone – no red-brown ppt Propanal – red-brown ppt is formed (iii) Suggest a suitable three step reaction pathway to convert propan -2-ol to 2-methylpropanoic acid. [5] CH3CH(OH)CH3  CH3CHClCH3 OR CH3CHBrCH3 OR CH3CHICH3  CH3CH(CH3)CN  CH3CH(CH3)COOH 1) I: SOC l2 / PC l5 / PCl3 and heat / NaCl, conc H 2SO4, reflux / dry HC l(g)/ concentrated HCl, anhydrous ZnCl2 OR: PBr3, heat/NaBr, conc H2SO4, reflux OR: NaI, conc H2SO4, reflux II: NaCN (ethanol), heat under reflux III: H2SO4(aq) heat under reflux

12 © ACJC2020 9729/03/Prelim/2020 [Turn over (d) In the presence of a strong base, carbonyl compounds undergo the aldol condensation to form a 3-hydroxy carbonyl compound. Fig. 1.2 shows the aldol condensation of ethanal. step 1 step 2 step 3 Fig. 1.2 (i) State the type of reaction in Step 3. [1] Step 3: Acid-base (ii) When hexanedial is reacted with a strong base, aldol condensation occurs and a product with the molecular formula C6H10O2 is formed. following compound. hexanedial Suggest the structure of this product. [1] [Total: 20]

13 © ACJC2020 9729/03/Prelim/2020 [Turn over 2 A neutral compound A, C6H8O4, is used in production of paints and adhesives. When compound A was heated with dilute hydrochloric acid, a mixture of compounds C, C4H4O4, and B was obtained. When the mixture was distilled, compound B was obtained as a distillate. When compound B was heated under reflux with acidified KMnO4, only carbon dioxide gas and water were obtained. Compound C exists as a pair of cis -trans isomers. Compound C decolourises Br2 in CCl4 and reacts with aqueous NaOH in a ratio of 1:2. (a) Deduce the structures of compounds A, B, and C and explain the reactions involved. [6] Reaction Deduction A neutral compound A, C6H8O4was heated with dilute hydrochloric acid Acid hydrolysis of ester A Degree of unsaturation in A = [2(6) 2] 8 2  = 3 Hydrolysis A is ester Compound C reacts with Br2 in CCl4 in a 1:1 ratio Electrophilic addition of C=C in C C exist as cis-trans isomers: 2 different groups a