2019 MI Prelim H2 Chem P3 ANS

Class Adm No Candidate Name: This question paper consists of 11 printed pages and 1 blank page. 2019 Preliminary Exams Pre-University 3 H2 CHEMISTRY 9729/03 Paper 3 Free Response 20 Sept 2019 2 hours Candidates answer on separate paper. Additional materials: Answer Paper Data Booklet Graph Paper READ THESE INSTRUCTIONS FIRST Do not turn over this question paper until you are told to do so Write your name, class and admission number 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. 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. You are reminded of the need for good English and clear presentation in your answers. 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. Question Section A Section B Total 1 2 3 4 5 Marks 19 19 22 20 20 80

2 Section A Answer all the questions in this section. 1 (a) 3–chloropropanoic acid, CH2ClCH2CO2H, is a weak Brønsted acid. A 0.100 mol dm3 solution of CH2ClCH2CO2H has a pH of 2.49. (i) Calculate the pKa of 3–chloropropanoic acid. [3] Since pH = 2.49, [H+] = 102.49 Ka = [CH2ClCH2CO2 - ][H+] [CH2ClCH2CO2H] = [10-2.49][10-2.49] [0.100] = 1.05  104 pKa = ‒lg(1.05  104) = 3.98 (ii) State and explain the differences in the relative acid strength between 3–chloropropanoic acid and propanoic acid. [2] 3–chloropropanoic acid is a stronger acid than propanoic acid. Due to the presence of the electron withdrawing group, C l, the negative charge of O on the carboxylate a nion is dispersed. ; This causes the conjugate base of 3–chloropropanoic acid to be more stable. Hence, acidity increases. ; (iii) 3-chloropropanoyl bromide is the acid derivative of 3–chloropropanoic acid . Propose a 3 -step synthetic route for the formation of 3 -chloropropanoyl bromide, using 3-hydroxypropanal as the starting reagent. State clearly the reagents and conditions used for each step. [5] 1m x 3 reagents and conditions 1m x 2 correct intermediates (b) A student investigated the rate of reaction between 3–chloropropanoic acid and aqueous sodium carbonate. The rate of the reaction may be determined by measuring how long it takes for the gas to be completely released. A series of experiments were carried out to study the order of reaction with respect to 3–chloropropanoic acid and sodium carbonate. The following results were obtained. PCl5, r.t H2SO4(aq), KMnO4/ K2Cr2O7, heat PBr3, r.t

3 [Turn over Experiment Number Volume / cm3 Time / s 3–chloropropanoic acid Na2CO3 H2O 1 20.0 40.0 40.0 78 2 20.0 30.0 50.0 100 3 5.0 20.0 25.0 75 (i) Write an equation, including state symbols, for the reaction between sodium carbonate and 3–chloropropanoic acid. [1] 2CH2ClCH2CO2H(aq) + Na2CO3(aq) → 2CH2ClCH2CO2-Na+(aq) + CO2(g) + H2O(l) (ii) State the relationship between time and rate. [1] Inverse relationship (iii) Determine the order of the reaction with respect to 3 –chloropropanoic acid and sodium carbonate. [2] Comparing Expt 1 and 2, whi le volume of 3 –chloropropanoic acid is kept constant and volume of Na2CO3 is 4/3 times, rate is 100/78 times. Order with respect to Na2CO3 is 1. Comparing Expt 1 and 3, while volume of Na 2CO3 is kept constant after multiplying the volume by two, volume of 3–chloropropanoic acid is halved, rate stays relatively constant. Order with respect to 3–chloropropanoic acid is 0. (iv) Hence, write the rate equation, stating the units of k. [2] Rate = k[Na2CO3] k = s-1

4 (v) With the aid of an appropriate diagram, explain how the addition of a catalyst can increase the rate of a reaction between 3–chloropropanoic acid and sodium carbonate. [3] A catalyst provides an alternative pathway for the reaction, lowering the activation energy required. Hence , there is a greater fraction of molecules with energy greater than activation energy. Frequency of effective collisions increases, leading to increase rate of reaction. (2m for diagram, 1m for explanation) [Total: 19] Fraction of molecules Kinetic energy Ea Fraction of molecules with kinetic energy ≥ Ea for uncatalysed reaction. E'a Fraction of molecules with kinetic energy ≥ Ea for catalysed reaction.

5 [Turn over 2 (a) Using only the elements C, H and O, draw the structural formulae of two different organic compounds, each containing a single carbon atom w ith an oxidation state of 0 and +2 respectively. [2] (b) A symmetrical organic compound A, C 2H4N2O2, upon reacting with hot d ilute NaOH(aq) produces a colourless, pungent gas that turns moist red litmus paper blue. (i) Draw the structure of compound A and state the functional group present. [2] (CONH2)2 , amide (ii) The functional group present in compound A can be reduced by lithium aluminium hydride, LiAlH4. Draw the dot-and-cross diagram for LiAlH4. [1] (iii) Other than LiAlH4, NaBH4 can also be used for the reduction of certain functional groups. LiAlH4 is able to produce hydride ion, H- more readily than NaBH4. Suggest why LiAlH4 is a more powerful reducing agent than NaBH4. [1] The electronegativity of Al is lesser than that of B. Hence, H bonded to Al can produce H- more readily than H bonded to B. Thus, reduction occurs more readily for LiAlH4, proving that it is a stronger reducing agent. OR The size of Al atom is larger compared to B, thus the orbital overlap between Al and H is less effective. Therefore the bond length of Al–H bond is longer than that of B–H bond. ; Less energy is required to break the weaker Al–H bond, resulting in a greater ease of generating H- for the reduction process. (c) Quinones are used in photography as a reducing agent. Quinone compounds are multifunctional as they exhibit properties of both ketones and alkenes.

6 The following scheme involves 1, 2-benzoquinone, where step I involves the use of LiAlH4. O O OH OH C Step I Step II HOOCCH2COOH conc H2SO4, reflux D alkaline KMnO4(aq), cold Step III C9H8O4 1,2-benzoquinone (i) Draw the structures of C and D. [2] (ii) Identify the chiral centres on B. Hence, explain why it does not display optical activity.[2] It does not display optical activity as there is a plane of symmetry in the compound. (iii) Using a simple chemical test, suggest how 1,2-benzoquinone can be distinguished from compound B. In your answer, state all reagents and conditions used and the expected observations. [2] 2,4-DNPH, warm. Orange ppt observed for 1,2-benzoquinone, no orange ppt observed for B. OR PCl5, r.t. B

7 [Turn over White f