VJC Prelim P3 QP and Ans

1 ©VJC2018 9729/03/PRELIM/18 [Turn over CANDIDATE NAME CT GROUP VICTORIA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 ……………………………………………….………….. …………………………….. CHEMISTRY 9729/03 Paper 3 Free Response Candidates answer on separate paper. 17 September 2018 2 hours Additional Materials: Cover Page Answer Paper Data Booklet READ THESE INSTRUCTIONS FIRST Write your name and CT group on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. You must start the answer to each question on a fresh piece of writing paper. Section A Answer all questions. Section B Answer one question. 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 11 printed pages and 1 blank page.

2 ©VJC2018 9729/03/PRELIM/18 [Turn over Section A Answer all the questions in this section. 1 (a) All the elements in the third period of the Periodic Table, sodium to sulfur, form chlorides by direct combination with chlorine. Aluminium chloride may be produced by passing a stream of chlorine over heated aluminium metal in a long hard–glass tube. (i) With the aid of equations, explain t he following observations when different amounts of water were added to solid aluminium chloride. (I) When a limited amount of water was added, a white solid was formed together with steamy fumes. (II) When excess water was added, a solution of pH 3 was obtained. [2] For (I): A lCl 3 reacts with a limited amount of water to give A l(OH)3(s) and HCl(g): AlCl3(s) + 3H2O(l)→ Al(OH)3(s) + 3HCl(g) For (II): AlCl3 undergoes hydrolysis as A l3+ has high charge density / strong polarising power. Polarisation of H2O molecules favours the loss of H+, and hence, acidity of the solution increases (pH ≈ 3). AlCl3(s) + 6H2O(l) → [Al(H2O)6]3+(aq) + 3Cl–(aq) [Al(H2O)6]3+(aq) + H2O(l) ⇌ [Al(H2O)5OH]2+(aq) + H3O+(aq) (ii) Both aluminium chloride and copper( I) complex solutions are colourless whereas a solution of copper( II) sulfate appears blue. Explain these observations. [3] AlCl3 solution is colourless because the energy gap between the n=2 and n=3 electronic shells is not within the visible light region. Hence, visible light is not absorbed. Cu(I) complex is colourless because the 3d subshell is fully filled. Hence, no d- d /electronic transition can occur. CuSO4 solution appears blue because partially-filled 3d subshell is split into two different energy levels in the presence of ligands. Electron from the lower energy level absorb a wavelength of light complementary to the observed colour and get promoted to the higher energy level. Thus, d-d transition can take place. (iii) Briefly describe the process of anodisation of aluminium. Write ion-electron equations for the reactions occurring at the anode and the cathode. [2] The aluminium is made the anode in the electrolysis of dilute sulfuric acid. The oxygen released at the anode reacts with the aluminium surface to build up a thicker layer of aluminium oxide. Anode: 2H 2O(l) → O2(g) + 4H+(aq) + 4e– Cathode: 2H+(aq) + 2e– → H2(g)

3 ©VJC2018 9729/03/PRELIM/18 [Turn over (b) Compound A can be synthesised from benzene using aluminium chloride via a simple Friedel–Crafts alkylation as shown in Reaction 1 . In addition, compound B, an isomer of compound A is also formed. (i) Compound B rotates plane-polarised light. It is formed after the carbocation intermediate undergoes rearrangement through the movement of an alkyl group to an adjacent carbon atom bearing the positive charge. Draw the structure of compound B. Explain why the rearrangement of the carbocation is favoured. [2] B: A primary carbocation rearranges into a secondary carbocation that is more stable because more electron-donating alkyl groups help dispersed the positive charge and stabilised the carbocation: (ii) Explain why multi-substituted product is more favoured over mono-substituted product in Reaction 1. [1] The alkyl group that is bonded to the benzene ring exerts electron–donating inductive effect. This activates the benzene ring, making it even more susceptible towards electrophilic attack / reactive with res pect to electrophilic substitution, thus forming multi-substituted product. (c) Compound A can be formed via compound C as shown in Reaction 2 below. Step 1 involves Friedel–Crafts acylation, which have similar reaction conditions and mechanism as Friedel-Crafts alkylation. (i) Draw the mechanism for step 1. In your answer, show relevant charges, lone pairs of electrons and movement of electrons. [3] Mechanism: Electrophilic substitution Let R be (CH3)2CH– RCOCl + AlCl3 → RCO+ + AlCl4−

4 ©VJC2018 9729/03/PRELIM/18 [Turn over (ii) Hence, suggest reagents and conditions for steps I to III in the following synthesis of benzophenone from 1-hydroxyethylbenzene. Give the structural formulae of J and K. [4] step I: KMnO4, H2SO4, reflux step II: PCl5, room temperature OR PCl3, reflux OR SOCl2, reflux step III: C6H6, anhydrous AlCl3, room temperature (d) A student wants to synthesise benzophenone using the reaction pathway illustrated in (c)(ii). However, the solid sample of 1-hydroxyethylbenzene is contaminated with phenylamine. Briefly explain how you can separate 1-hydroxyethylbenzene from phenylamine via extraction. You are provided with • ethanol, hexane, HCl(aq), NaOH(aq), • separating funnel and • apparatus commonly found in a college laboratory. [3] (i) Dissolve the solid sample in hexane. (ii) Transfer the mixture to a separating funnel. (iii) Add HC l(aq) to the mixture to convert phenylamine to the salt. (iv) Shake the separating funnel and then drain off the bottom aqueous layer to get the organic layer. (v) Evaporate the organic layer to obtain 1-hydroxyethylbenzene. [Total: 20] 2 (a) Borane, BH 3, is used to synthesise alcohols from alkenes as shown in the reaction sequence below. In reaction 1, the BH2 group from BH3 is bonded to the less substituted carbon atom of the double bond. The remaining H atom from BH 3 is bonded to the other carbon atom.

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