NYJC 2022 Prelim P3 Answer

[Turn over NANYANG JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME CLASS TUTOR’S NAME CHEMISTRY 9729/03 Paper 3 Free Response 15 September 2022 2 hours Candidates answer on the Question Paper Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your name and 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. Circle the question you attempted in the box below. 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. For Examiner’s Use 1 /25 2 /15 3 /20 4 or 5 /20 Total /80 This document consists of 32 printed pages.

2 H2 Chemistry 9729/03 NYJC J2/22 PX [Turn Over Section A Answer all questions in this section. 1 This question is about alkene and diene. Diene is an unsaturated compound containing two double bonds between carbon atoms. (a) Table 1.1 shows the structures of hexa-1,2-diene, hexa-1,3-diene and hexa-1,5-diene. Table 1.1 hexa-1,2-diene hexa-1,3-diene hexa-1,5-diene Table 1.2 shows the enthalpy change of combustion of some organic compounds. Table 1.2 substance Hc / kJ mol−1 Hhydrogenation / kJ mol−1 Liquid hexa-1,2-diene −3867 H1 Liquid hexa-1,3-diene −3816 −225 Liquid hexa-1,5-diene −3843 −252 Liquid hexane −4163 NA Hydrogen gas −286 NA (i) Use data in Table 1.2 to draw an energy level diagram to calculate the enthalpy change of hydrogenation of hexa-1,2-diene, H1. [2] CH2=C=CHCH2CH2CH3(l) + 2H2(g) + 19 2 O2(g) −3867 −2(286) H1 C6H14(l) + 19 2 O2(g) −4163 6CO2(g) + 7H2O(l) Energy / kJ mol–1

3 H2 Chemistry 9729/03 NYJC J2/22 PX [Turn Over Applying Hess’ Law, H1 = −3867 −2(286) − (−4163) = −276 kJ mol−1 1 mark for the energy level diagram (balanced equations, correct arrow direction, arrows labelled correctly) and 1 mark for calculating H1 correctly (ii) Suggest which isomer, hexa-1,3-diene or hexa-1,5-diene, is more stable using data in Table 1.2. Explain your answer by considering the type of orbitals present in the carbon atoms of the isomer. [2] Hexa-1,3-diene is more stable as its enthalpy change of hydrogenation is the least exothermic. [1] Carbons 1 to 4 are sp2 hybridised, hence the unhybrisided p orbitals from carbons 1 to 4 are adjacent to each other and can overlap to form a delocalised  electron cloud, giving rise to resonance stability. Hence hexa-1,3-diene more stable. [1 for the underlined key phrases] (iii) Choose a suitable isomer of hexadiene from Table 1.1 and devise a three -step synthetic route to synthesise buta-1,3-diene. [5] O O OH OH CH2 CH2 OH OH step 1 step 2 step 3 step 1 KMnO4(aq), H2SO4(aq), heat under reflux step 2 LiAlH4 in dry ether step 3 excess conc. H2SO4, 170 oC or Al2O3, 350 oC 1 mark for correct choice of suitable hexadiene 2 marks for drawing the intermediates (2 of them) correctly 2 marks for stating the R&C correctly; need minimum of 2 correct R&C for 1 mark Allow for ECF for R&C ......................................................................................................................... ......................................................................................................................... ......................................................................................................................... ......................................................................................................................... ......................................................................................................................... .........................................................................................................................

4 H2 Chemistry 9729/03 NYJC J2/22 PX [Turn Over (b) When buta-1,3-diene undergoes electrophilic addition reaction with 1 mol of HBr, a mixture of two products, A and B is formed. Both A and B have molecular formula of C4H7Br. + HBr A B A exhibits cis-trans isomerism while B contains a chiral centre. (i) Draw the structures of A and B. [2] A B Br CH3 [1] CH3 CH2 Br [1] (ii) By referring to your structure in (b)(ii), explain how cis-trans isomerism arises in A. [2] Cis-trans isomerism arises in A due to the presence of restricted rotation about the C=C double bond [1] and that each C in the C=C is bonded to 2 different groups of atoms [1]. Kinetic and thermodynamic factors decide the type of addition product that is obtained. A is known as the kinetic product as it is formed faster while B is known as the thermodynamic product as it is formed more slowly and is also more thermodynamically st able than the kinetic product. (iii) In a single set of axis, sketch and label two reaction pathway diagrams for the second step of the mechanism to form A and B respectively. The carbocation intermediates used to form A and B occupy the same energy level. Use Ea1 and Ea2 to label the activation energies and H1 and H2 to label the enthalpy changes to form A and B respectively. [2] Ea2 Ea1 B A carbocation intermediate Energy / kJ mol−1 reaction pathway H1 H2 The kinetic product shld have a smaller Ea and a less exothermic H and vice versa for the thermodynamic product. 1 mark for each reaction pathway diagram; total is 2 marks

5 H2 Chemistry 9729/03 NYJC J2/22 PX [Turn Over (c) The Wittig reaction is an organic chemistry synthesis technique that involves the conversion of carbonyl compounds in to alkenes using an ylide as the reagent and via a nucleophilic addition pathway. P + Ph Ph Ph C− R2 R1 ylide O R3 R4 R2 R1 R3 R4 P O Ph Ph Ph + + where Ph is phenyl and R1, R2, R3, and R4 is either H atom or alkyl group Fig 1.1: The Wittig reaction Ylides can be synthesised from triphenyl phosphine and an alkyl halide. The first step is to react the triphenyl phosphine and alkyl halide via a nucleophilic substitution reaction. This step is an elementary reaction. The second step is to add a very strong base such a s butyl lithium, CH3CH2CH2CH2−Li+ to deprotonate the intermediate product formed in step 1. P Ph Ph Ph C H R1 R2 Br P + Ph Ph Ph C− R2 R1+ step 1 step 2 triphenyl phosphine ylide intermediate product Figure 1.2: Synthesis of ylide (i) Draw the full structural formula of the intermediate product for the reaction shown in Fig. 1.2 when R1 is hydrogen atom and R2 is methyl group. [1] P + C C H H H H H [1]

6 H2 Chemistry 9729/03 NYJC J2/22 PX [Turn Over (ii) Predict the organic products L and M of the reactions shown in Fig. 1.3. [2] P(Ph)3 O O 2 + OBr 1. PPh3 2. CH3CH2CH2CH2 −Li+ − + L M Fig. 1.3 M [1] ......................................................................................................................... .........................................