TJC H2 CHEM P3 ANS Prelim

2015 TJC H2 Chemistry Preliminary Exam [Turn Over 1 CHEMISTRY (H2) 9647/03 Paper 3 Free Response 16th September 2015 2 hours Candidates answer on separate paper. Additional materials: Answer paper Graph Paper Data Booklet READ THESE INSTRUCTIONS FIRST Write your name, Civics Group, Centre number and Index number in the spaces provided on the cover page and on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a HB pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, glue or correction fluid. Answer any four questions. 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. This document consists of 20 printed pages.

2015 TJC H2 Chemistry Preliminary Exam [Turn Over 2 1 (a) The halogens and their compounds are useful laboratory reagents and have many applications. (i) Describe the trend in the colour of the halogens. • Increasing intensity or darker down the group. • Cl2 (g) yellowy-green; Br2 (l) reddish-brown; I2 (s) black (ii) Explain, in terms of structure and bonding, the trend in the volatility of the halogens.  The halogens have simple molecular structure and weak van der Waals forces of attraction between molecules.  Down the group, Mr of X2 increases, giving a larger electron-cloud, which can be more easily distorted, resulting in stronger van der Waals forces hence requiring more energy to overcome. The volatility decreases from Cl2 to I2. (iii) Halogens are oxidising agents. By quoting appropriate values from the Data Booklet , explain the reactions of the halogens with sodium thiosulfate. Write balanced equations for the reactions. [6] ½Cl2 + e Cl- E = +1.36 V ½Br2 + e Br- E = +1.07 V ½I2 + e I- E = +0.54 V  (With quoting of Data) Oxidising power of the halogens decreases down the group as can be seen from the less positive E value, hence iodine can only oxidise thiosulfate to tetrathionate. [•Bonus mark] 4Cl2 + S2O3 2- + 5H2O  8Cl- + 2SO4 2- + 10H+ 4Br2 + S2O3 2- + 5H2O  8Br- + 2SO4 2- + 10H+  I2 + 2S2O3 2-  S4O6 2- + 2I- (b) Hydrocarbons react with halogens under different conditions to give a wide range of organic compounds. The Wohl-Ziegler bromination is shown below. In this reaction, a bromine atom is incorporated into a molecule at the position next to a carbon-carbon double bond. The reagent used for the reaction is N-bromosuccinimide (NBS). N-bromosuccinimide (NBS) 2015 TJ Prelim H2 Paper 3 Answers

2015 TJC H2 Chemistry Preliminary Exam [Turn Over 3 The catalytically active species is bromine, which is present in trace amount in NBS samples. Subsequent amount of bromine is generated in situ when NBS reacts with hydrogen bromide. (i) NBS reacts with hydrogen bromide in a 1:1 molar ratio to produce bromine and succinimide as the only products. Write a balanced equation for the reaction.  In one such reaction, cyclohexene is converted to 3-bromocyclohexene using NBS. (ii) Describe the mechanism for the formation of 3 -bromocyclohexene from cyclohexene and bromine.  Free radical substitution •• Mechanism Initiation Propagation Termination Br2 2Br 2Br Br2

2015 TJC H2 Chemistry Preliminary Exam [Turn Over 4 (iii) State the type(s) of stereoisomerism that could be exhibited by 3 -bromocyclohexene and draw the isomers.  Optical isomerism  In another reaction, compound A can be converted to compound B in two steps. One of the steps is the Wohl-Ziegler bromination. (iv) Suggest suitable reagents and conditions and the structural formula for the intermediate formed. •• compound A compound B * *

2015 TJC H2 Chemistry Preliminary Exam [Turn Over 5 (v) State the total number of stereoisomers of compound B and draw the isomers. [10] •• 3 (c) The mineral fluorspar, which is mainly calcium fluoride, is a major source of fluorine. The first stage in liberating the fluorine from calcium fluoride is to grind this compound up and react it with concentrated sulfuric acid. The products are hydrogen fluoride and calcium sulfate. (i) Define the standard enthalpy change of formation, ΔHf o , of a substance.  It is the enthalpy change when one mole of the substance is formed from its constituent elements in their standard states at 298 K and 1 atm. (ii) Using the following data of ΔH f and lattice energy, construct an energy cycle to calculate the enthalpy change for the reaction between calcium fluoride and concentrated sulfuric acid. Hf [HF(g)] = -271 kJ mol-1 Hf [CaSO4(s)] = -1434 kJ mol-1 Hf [Ca2+(g)] = +1918 kJ mol-1 Hf [F-(g)] = -249 kJ mol-1 L.E. [CaF2(s)] = -2640 kJ mol-1 Hf [H2SO4(l)] = -814 kJ mol-1 [4]

2015 TJC H2 Chemistry Preliminary Exam [Turn Over 6 CaF2(s) + H2SO4(l) CaSO4(s) + 2HF(g) Ca2+(g) + 2F-(g) + H2(g) + S(s) + 2O2(g) Ca(s) + F2(g) + H2(g) + S(s) + 2O2(g) By Hess’ Law, Hr – 2640 + 1918 + 2(–249) + (–814) = –1434 + 2(–271) Hr = +58 kJ mol-1 •Balanced equation for the reaction •working •Answer [Total: 20] 2 (a) In folk medicine, willow bark teas were used as headache remedies and other tonics. Its analgesic property is due to salicylic acid. Nowadays, salicylic acid is administered in the form of aspirin which is less irritating to the stomach. CO2H CO2H OCOCH3 (i) In the preparation of aspirin from salicylic acid, water is added to precipitate aspirin. Explain why aspirin is not soluble in water. Both aspirin and water have simple molecular structures. Aspirin has both van der Waals interaction and hydrogen bonding between its molecules. Water has hydrogen bonding between its molecules. The energy released from the hydrogen bonding formed between aspirin and water molecules release insufficient energy to overcome t he van der Waals interaction and hydrogen bonding between aspirin molecules and the hydrogen bonding between water molecules. Therefore, aspirin is insoluble in water. • 1m – explanation • 1m – bonding and structure Salicylic acid Aspirin Hr –1434 + 2(–271) +1918 + 2(–249) – 2640 – 814

2015 TJC H2 Chemistry Preliminary Exam [Turn Over 7 (ii) Aspirin has a pK a of 3.48 while benzoic acid has a pK a of 4.2. Explain the difference in their pKa values. • As Aspirin has a lower pKa than benzoic acid, aspirin is a stronger acid. • Aspirin has an electron-withdrawing –OCOCH3 group which further disperses the negative charge on CO2- OCOCH3 , making it more stable than benzoate. (iii) Draw the structure of the 2 organic compounds formed when aspirin is heated with NaOH(aq). •

CO2-Na+ O-Na+ • CH3CO2 -Na+ (iv) Acid Ka1 / mol dm-3 Ka2 / mol dm-3 Salicylic acid 1.07 x 10-3 1.82 x 10-14 Eth