2023 DHS H2 Chem Prelim Paper 2 Suggested Solutions (1)

© DHS 2023 9729/02 [Turn over Suggested Solutions DUNMAN HIGH SCHOOL Preliminary Examination Year 6 H2 CHEMISTRY Paper 2 Structured Questions Candidates answer on the Question Paper. Additional Materials: Data Booklet 9729/02 14 September 2023 2 hours READ THESE INSTRUCTIONS FIRST Write your centre number, index number, name and class 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. The use of an approved scientific calculator is expected, where appropriate. You may lose marks if you do not show your working or if you do not use appropriate units. A Data Booklet is provided. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 10 2 12 3 14 4 14 5 25 Total 75 This document consists of 25 printed pages and 1 blank page.

2 © DHS 2023 9729/02 Answer all the questions in the spaces provided. 1 (a) A fluorophore is a fluorescent chemical compound that can re -emit light upon light excitation. Fluorophores typically contain several combined aromatic groups, or planar or cyclic molecules with several π bonds. Derivatives of trans -stilbene ha ve been recently studied due to their well -known fluorophore properties and they are generally more stable than cis -stilbene. The structure of cis-stilbene is shown in Fig. 1.1. Fig. 1.1 (i) Explain why the melting point of trans -stilbene is higher than that of cis-stilbene. [1] Trans-isomer has higher symmetry and packs more closely into a crystal lattice. More energy is required to break the stronger intermolecular forces of attraction in crystal lattice in the trans–isomer than the cis–isomer. (ii) Based on the structure of stilbene, suggest a reason why trans -stilbene is more stable than cis-stilbene. [1] It has two bulky phenyl groups far from each other, making this compound more stable than cis isomers due to less steric hindrance. (b) Aluminium chloride, A lCl3, is used as a halogen carrier in order for chlorination of benzene to take place in stilbene. AlCl3 + Cl2 ⇌ AlCl4 – + Cl + (i) Name the type of bond formed when AlCl3 reacts with C l2. Explain how this bond is formed. [2] Dative bond. Al atom has a vacant low-lying orbital in its valence shell so it can accept an electron pair from Cl2. (ii) Draw the dot-and-cross diagram for the AlCl4 – ion. Use the VSEPR theory to state and explain the shape of the ion and the bond angle present. [3] x x x AlCl Cl Cl Cl x _

3 © DHS 2023 9729/02 [Turn over There are 4 bond pairs and 0 lone pairs around the central aluminium atom. As the repulsion between the 4 bond pairs are similar in strength, the shape around the central aluminium atom is thus tetrahedral to minimise repulsion between them. The bond angle is 109° . (c) Orange street-lamps contain sodium with a small amount of neon. Light is produced when the gaseous atoms are ionised in an electric field. Ne(g) → Ne+(g) + e- 1st I.E = +2080 kJ mol-1 Na(g) → Na+(g) + e- 1st I.E = +494 kJ mol-1 (i) Explain the difference in the first ionisation energies of neon and sodium. [2] Na has an additional electron shell compared to Ne , hence Na experiences stronger shielding effect . Despite having a higher nuclear charge, the electron to be lost from Na is further away and experiences weaker attraction from the nucleus . (ii) When the lamps are switched on, they first emit a red glow characteristic of neon, but after some time, the orange glow of sodium predominates. Suggest why neon is ionised first even though its first ionisation energy is much higher than that of sodium. [1] Sodium is in the solid state and energy is required to vapourise sodium first before the atoms can be ionised. [Total: 10] 2 Compound M has molecular formula, C6H8O2. M reacts with an excess of hot concentrated KMnO 4 to produce two different organic molecules L and K. (a) K has molecular formula C2H2O4 and is further oxidised to form CO2. (i) Name K. [1] ethanedioic acid (ii) Write an equation for the oxidation of K to form CO 2, using [O] to represent the oxidising agent. [1] (CO2H)2 + [O] → 2CO2 + H2O (b) L has molecular formula C4H6O3. When pure samples of L are separately added to two different reagents, the observations in Table 2.1 are recorded. Table 2.1 test reagent observation 1 alkaline aqueous iodine pale yellow precipitate 2 phosphorus(V) chloride misty acid fumes

4 © DHS 2023 9729/02 (i) Name the type of reaction occurring to L in test 1. [1] oxidation (ii) Based on the observation for test 2, name two functional groups that could be present in L. [1] alcohol or carboxylic acid reject hydroxyl / −OH groups (iii) Deduce the structure of L. [1] OO CH3 OH L test reagent observation deductions 1 alkaline aqueous iodine pale yellow precipitate Oxidation reaction. L contains either the −CH(OH)CH 3 or −COCH3 group. Since L is a product of strong oxidation of M, L contains the −COCH3 group. 2 phosphorus(V) chloride misty acid fumes Nucleophilic substitution reaction. Since L is a product of strong oxidation of M, L contains either a tertiary alcohol or a carboxylic acid group. Since L has molecular formula C 4H6O3, L has 3 oxygen atoms and contains the carboxylic acid group. (c) Suggest a possible structure of M, C6H8O2, showing the skeletal formula. [2] OO M OR OO M OR OH OH

5 © DHS 2023 9729/02 [Turn over Fig. 2.1 shows a reaction scheme of compound N. R MnO4 / H+, heat step 2 + CO2S (C8H6O4) N C2H3O C2H3O CH3 step 1 Fig. 2.1 (d) (i) N reacts with 2,4-dinitrophenylhydrazine to give an orange precipitate. Explain this observation. [1] N undergoes condensation reaction with 2,4-DNPH. N contains either a ketone or an aldehyde functional group. (ii) R is the major product of the reaction occurring in step 1. Use this information and your answer in (d)(i) to deduce the structure of N. Explain your answer. [2] CH3O N From (d)(i), the −C 2H3O group has to contain the ketone or aldehyde functional group so it can be −COCH3 or −CH2CHO. Since incoming CH3 + electrophile is directed to the 3-position, −C2H3O has to be the electron-withdrawing −COCH3 group which deactivates the benzene ring. (iii) State the reagents and conditions required for step 1. [1] CH3Cl, anhydrous AlCl3 (iv) Suggest the structure of compound S. [1] OHO O OH S [Total: 12] 3 (a) A series of kinetics experiments was carried out at a constant temperature to study

6 © DHS 2023 9729/02 the mechanism of ester hydrolysis under basic conditions. The overall chemical reaction is described below. OCH3 O CH3 + OH- O - CH3 O + CH3OH The initial rates of reaction with v arying ester concentrations were measured at [OH–] = 0.90 mol dm–3 and [OH–] = 1.50 mol dm–3. Fig. 3.1 shows the graph of the results obtained. Fig. 3.1 (i) By quoting relevant data from Fig. 3.1, determine the order of reaction with respect to each reactant. Show your reasoning clearly. [2] When [OH–] is constant, the graph of initial rate against [ester] is a straight line with a positive gradient passing through the origin . Hence rate is directly proportional to [ester] and the reaction is first order w.r.t. [ester]. When [ester] = 0.40 mol dm–3, when [OH–] from 0.90 mol dm–3 to 1.50 mol dm–3, initial