YIJC 2025 Prelim P3 Suggested Solutions (for exchange) H2 Chem

©YIJC 9729/03/JC2/PE/2025 [Turn over YISHUN INNOVA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME Suggested Solutions CG INDEX NO CHEMISTRY Paper 3 Free Response Candidates answer on the Question Paper. Additional Materials: Data Booklet 9729/03 16 September 2025 2 hours READ THESE INSTRUCTIONS FIRST Write your name, class and index number in the spaces at the top of this page. 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. 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 28 printed pages. For Examiner’s Use Section A 1 / 17 2 / 18 3 / 25 Section B 4 or 5 / 20 Penalty units significant figures Overall / 80

©YIJC 9729/03/JC2/PE/2025 2 Section A Answer all the questions in the spaces provided. 1 (a) Explain, with the aid of a labelled Boltzmann distribution diagram, the effect on a rate constant of increasing temperature from T1 to T2. [3] When the temperature is increased from T 1 to T 2, the average kinetic energy of the particles increases. As shown from the graph, there is an increase in the fraction of particles with energy equal to or greater than the activation energy, Ea. This result in an increase in the frequency of effective collisions, hence the reaction rate increases. A higher temperature results in an increase in reaction rate and hence a larger rate constant k. energy, E fraction of particles Ea fraction of particles with E  Ea at high T2 fraction of particles with E  Ea at low T1 T1 (lower temperature) T2 (higher temperature) 0

3 ©YIJC 9729/03/JC2/PE/2025 [Turn over (b) Hydrogen bromide, HBr, undergoes addition reaction with alkenes. With but-1-ene, 2-bromobutane is produced rather than 1-bromobutane. (i) Draw a mechanism for this reaction and use it to explain the preferential production of 2-bromobutane. [3] Electrophilic Addition Step 2: 2-bromobutane is preferentially formed over 1-bromobutane as the carbocation intermediate formed is a secondary carbocation / there are more electron-donating alkyl groups in the carbocation intermediate . This disperses the positive charge on the carbocation intermediate, making it more stable as compared to the carbocation intermediate for 1- bromobutane. (ii) Using the mechanism, write the rate equation for this reaction. [1] Rate = k [but-1-ene][HBr]

©YIJC 9729/03/JC2/PE/2025 4 (iii) Sketch a graph to show how the rate of reaction varies with the concentration of but -1-ene when hydrogen bromide is in excess. Explain your answer. [2] In excess HBr, [HBr] is relatively constant. Rate = k [but-1-ene][HBr] = k’ [but-1-ene], linear graph of rate against [but -1-ene] passing through origin with gradient of k’, i.e., rate is directly proportional to [but-1-ene] This is a psuedo first-order reaction. (c) But-2-ene is a positional isomer of but-1-ene. But-2-ene occurs in two isomeric forms, A and B. (i) Explain how A and B are stereoisomers of each other but but -1-ene does not show stereoisomerism. [2] A and B show cis-trans isomerism due to presence of C=C double bond which prevents free rotation about the C=C double bond , and there are two different groups of atoms bonded to each C atom of the C=C double bond . Whereas but-1-ene does not show cis- trans isomerism because there are 2 identical H atoms bonded to one of the C atoms of the C=C double bond. (ii) Describe a chemical test, with appropriate observations, that could distinguish between but-1-ene and but-2-ene. [2] Test: To separate test-tubes containing but-1-ene and but-2-ene, add KMnO4(aq) and dilute H2SO4, then warm. Observations: Both will decolourise purple KMnO4 but only but-1-ene will produce CO2(g) (because it is a terminal alkene). rate / mol dm−3 s−1 [but-1-ene] / mol dm−3 0 gradient = k’ = k[HBr]

5 ©YIJC 9729/03/JC2/PE/2025 [Turn over (d) But-1-ene can be converted into an ether (–C–O–C–) via the steps shown in Fig. 1.1. Fig. 1.1 In step 1, an epoxide functional group is formed when a n O atom adds across the double bond through reaction with m-CPBA. In step 2, t he epoxide formed reacts with sodium methoxide, a nucleophile , to form an ether functional group (–C–O–C–). (i) The resulting mixture of the ether contains equal quantities of two isomers. The mixture does not rotate plane-polarised light. Draw the three-dimensional structures of these two isomers. [2] (ii) Suggest the structure of the ether formed when but-2-ene is subjected to the same reaction as in Fig. 1.1. [1]

©YIJC 9729/03/JC2/PE/2025 6 (iii) A primary amine can also be used as the nucleophile in step 2, but the final product will be a secondary amine, as shown in Fig. 1.2, instead of an ether. Fig. 1.2 Suggest the structure of alkene C. [1] [Total: 17]

7 ©YIJC 9729/03/JC2/PE/2025 [Turn over 2 (a) Compound D has molecular formula CxHyOz. Its relative molecular mass is 90.0. When 2.25 g of D was burnt in excess oxygen, 4.40 g of CO2 and 2.25 g of H2O were obtained. Calculate the empirical formula of D and determine its molecular formula. [4] Method 1 (using mass) Mass of hydrogen = 2 × 1.0 18.0  2.25 = 0.250 g Mass of carbon = 12.0 44.0  4.40 = 1.20 g Mass of oxygen = 2.25 – 1.20 – 0.250 = 0.800 g C H O mass ratio 1.20 0.250 0.800 mole ratio 1.20 12.0 = 0.100 0.250 1.0 = 0.250 0.800 16.0 = 0.0500 simplest whole number ratio 0.100 0.0500 0.250 0.0500 0.0500 0.0500 2 5 1 Empirical formula of D = C2H5O Molecular formula = (C2H5O)n 90.0 = n(45.0) n = 2 Molecular formula of D = C4H10O2. Method 2 (using moles): Amount of compound D = 2.25 / 90.0 = 0.0250 mol Amount of CO2 = 4.40 / 44 = 0.100 mol Amount of H2O = 2.25 / 18 = 0.125 mol CxHyOz CO2 H2O mole 0.0250 0.100 0.125 Simplest ratio 1 4 5 CxHyO + (x + y/4 – z/2) O2 → xCO2 + y/2 H2O Coefficient of equation 1 x y/2 x = 4; y/2 = 5 so y = 10 Since the molecule formula is C4H10Oz and molecular mass is 90.0 4(12.0) + 10(1.0) + z(16.0) = 90.0 z = 2 Therefore molecular formula is C4H10O2 and empirical formula is C2H5O

©YIJC 9729/03/JC2/PE/2025 8 (b) D can undergo controlled oxidation to form E (C4H6O2). E can be further oxidised to form F (C4H6O3). No oxygen atoms are present in the carbon backbone of E and F. Four chemical tests are carried out on portions of E and F and the results are described in Table 2.1. Table 2.1 with Na2CO3(aq) with Tollens’ reagent with 2,4-DNPH with alkaline I2(aq) E no reaction silver mirror orange precipitate no reaction F effervescence no reaction orange precipitate no reaction Deduce the displayed structures of E and F. [5] Result Deductions E forms silver mirror with Tollens’ reagent Oxidation. E is an aldehyde. E and F form orange ppt with 2,4-DNPH Condensation. Both E and F are carbonyl compounds. F produces effervescence with Na2CO3(aq), but E does not produce effervescence with Na2CO3(aq). Acid-carbonate reaction. F is a carboxylic acid. E is n