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