Edexcel Jan 2015 (IAL) Paper 3 Q4 (with explained solutions) |

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Edexcel Jan 2015 (IAL) Paper 3 Q4

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4 One method of preparing 1-bromobutane from butan-1-ol is given below. Procedure Step 1 10 g of sodium bromide, 10 cm3 of water and 7.5 cm3 of butan-1-ol are placed in a flask. The flask is partially immersed in a large beaker of cold water. A condenser is fitted vertically in the neck of the flask as shown in the diagram. water out tap funnel concentrated sulfuric acid water in cold water sodium bromide, butan-1-ol and water Step 2 10 cm3 of concentrated sulfuric acid is dripped slowly from the tap funnel into the reaction mixture. The flask is shaken gently. Step 3 The tap funnel is removed from the top of the condenser and the flask is taken out of the cold water bath. The flask is then heated gently for about 45 minutes. Step 4 The apparatus is then rearranged for distillation. The 1-bromobutane and water are distilled into a small beaker where they form two layers. Step 5 The 1-bromobutane layer is separated from the water. Step 6 The 1-bromobutane layer is washed with concentrated hydrochloric acid to remove unreacted butan-1-ol. Step 7 The 1-bromobutane is then washed with dilute sodium carbonate solution. 12 *P45043A01216*<br />
You will need the following data to answer the questions. Butan-1-ol, CH3CH2CH2CH2OH Mr = 74 1-bromobutane, CH3CH2CH2CH2Br Mr = 137 Liquid butan-1-ol water concentrated hydrochloric acid 1-bromobutane Density / g cm3 0.81 1.0 1.2 1.3 (a) The use of the beaker of cold water in Step 1, and the slow addition of concentrated sulfuric acid in Step 2, both prevent a reaction which gives unwanted inorganic products. Identify one of these unwanted products. State the type of reaction occurring when these products form. (2) ProductType of reaction(b) (i) Explain why the condenser is set up so that the water flows from bottom to top, as shown in the diagram. (1)(ii) Without the reflux condenser, the procedure in Step 2 would become more hazardous. Explain why. (1)*P45043A01316* 13 Turn over<br />
(c) To achieve the best possible yield of 1-bromobutane, the purification stages should involve the minimum number of transfers of the organic product from one piece of apparatus to another. (i) How could the water layer be removed from the small beaker in Step 5 without transferring the organic product? (1)(ii) Name the apparatus you would use to carry out the washing of the crude 1-bromobutane in Step 6. Describe how you would obtain the organic layer from this mixture. (2)(d) What is the purpose of Step 7? (1)(e) After Step 7, the crude 1-bromobutane is washed with pure water and separated again. Two further steps are needed to obtain a pure sample of 1-bromobutane. State what these steps are. Detailed experimental procedures are not required, but you should name any reagents which are needed. (3) Step 8Step 914 *P45043A01416*<br />
(f) (i) Calculate the mass of butan-1-ol used in Step 1. (ii) In this experiment, a student obtained 7.5 g of 1-bromobutane. Calculate the percentage yield of 1-bromobutane. Assume that each mole of butan-1-ol can produce a maximum of one mole of 1-bromobutane. Give your answer to two significant figures. (1) (3) (Total for Question 4 = 15 marks) TOTAL FOR PAPER = 50 MARKS *P45043A01516* 15<br />

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Mark Mark Acceptable Answers Reject Bromine / Br2 (1) Redox/ oxidation (1) OR sulfur dioxide / SO2 (1) Redox/ reduction (1) ALLOW Redox but no product given scores 1 mark Butanal/ butanoic acid and redox / oxidation scores 1 mark HBr and redox scores 0. Oxidation/ reduction if no product given Acceptable Answers Reject To ensure condenser is full of water / to prevent an airlock forming/ to stop air bubbles forming / to stop hot spots forming ALLOW To ensure that all of the condenser surface is covered with cold water/ So that (hot) vapour is next to the coolest water first / So the lower region (of the condenser) is colder / Makes cooling more efficient To prevent back flow of water Just So that nothing escapes Just explanation that condensation occurs Makes cooling faster Question Number 4(a) Question Number 4(b)(i)<br />
Mark Mark Mark Acceptable Answers Reject There would be escape of flammable liquid / corrosive spray / corrosive acid (spray) /poisonous gas/ toxic gas/ harmful gas IGNORE Prevents boiling over Very exothermic Any named toxic gas is only allowed if it would condense. Named substance e.g. Br2 / sulfuric acid without reference to hazard Eg bromine could escape Escape of HBr /SO2 which are toxic (because they do not condense) Risk of explosion Just escape of product Acceptable Answers Reject (teat) pipette/ syringe (to remove upper aqueous layer) ALLOW decant / description of decanting To remove lower aqueous layer Add drying agent Add dehydrating agent Just Use separating funnel Use a siphon Acceptable Answers Reject Separating funnel / tap funnel (1) Run off lower layer (1) ALLOW pipette off upper layer Run off lower aqueous layer BUT do not penalise if mark in (c)(i) lost for wrong layers. Answers showing candidate is unaware that lower layer is the product Question Number 4(b)(ii) Question Number 4(c)(i) Question Number 4(c)(ii)<br />
Mark Mark Acceptable Answers Reject To remove / neutralize (excess) acid OR to neutralize unreacted acid OR to remove / neutralize HCl ALLOW To neutralise the solution To remove all the HCl To wash out unreacted acid IGNORE To remove impurities To eliminate HCl Just to react with acid To remove/ neutralise H2SO4 (and HCl) To remove HBr Acceptable Answers Reject Dry in an oven/ evaporate to half volume scores 0 for this step. Copper sulfate Concentrated sulphuric acid Calcium hydroxide Metal carbonates Calcium sulfate recondense Step 8 Dry/ remove water from the bromobutane (1) With (anhydrous) calcium chloride / (anhydrous) magnesium sulfate / sodium sulfate/ silica gel ALLOW CaCl2 / MgSO4 / Na2SO4 If name and formula are given both must be correct (1) Step 9 (Filter / decant and then) redistil / distil (1) If only one step is given accept the answer in Step 8 or Step 9 ALLOW Description of drying carried out after redistillation max (2) Question Number 4(d) Question Number 4(e)<br />
Question Number Acceptable Answers Reject 4(f)(i) (7.5 x 0.81) = 6.075 / 6.08 (g) 6.07 Wrong units Mark Mark Ignore sf except 1sf Acceptable Answers Reject Percentages calculated from volumes with no conversion to mol or mass. 6.075/7.5 x 100 =81% scores 0 67.0 (This is 3sf) Look at final answer. 67% scores 3 marks; answers with 3sf rounding to 67 score 2 marks. If this is incorrect follow this scheme: METHOD 1 Mol butan-1-ol = (6.075/74) = 0.0820945 (1) maximum mass 1-bromobutane = (0.0820945 x 137) = 11.246959 g (1) % yield = ((7.5/11.24659)x100 = 66.85) =67% to 2 sf (1) OR METHOD 2 7.5/137 = 0.0547445 mol (bromobutane) (1) 6.075/74 = 0.0820945 mol butan-1-ol (1) % yield = ((0.05474455)x100/0.0820945) = 66.85) =67% to 2 sf (1) Also TE from one step of the calculation to the next and TE on 4f(i) unless yield > 100%. Use of 6.08 gives 0.082161 mol, 11.256216 g bromobutane, final answer 67% 11.3g bromobutane gives 66%. Total for Question 4 = 15 marks Question Number 4(f)(ii)<br />

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