AQA Jun 2017 Paper 3 Q3 (with explained solutions) |

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AQA Jun 2017 Paper 3 Q3

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Benzoic acid can be prepared from ethyl benzoate. Ethyl benzoate is first hydrolysed in alkaline conditions as shown: A student used the following method. Add 5.0 cm3 of ethyl benzoate (density = 1.05 g cm3, Mr = 150) to 30.0 cm3 of aqueous 2 mol dm3 sodium hydroxide in a round-bottomed flask. Add a few anti-bumping granules and attach a condenser to the flask. Heat the mixture under reflux for half an hour. Allow the mixture to cool to room temperature. Pour 50.0 cm3 of 2 mol dm3 hydrochloric acid into the cooled mixture. Filter off the precipitate of benzoic acid under reduced pressure. 0 3 . 1 Suggest how the anti-bumping granules prevent bumping during reflux. [1 mark] 0 3 0 3 . 2 Show, by calculation, that an excess of sodium hydroxide is used in this reaction. [2 marks] *09* Question 3 continues on the next page Turn overIB/M/Jun17/7405/3<br />
100 3 . 3 Suggest why an excess of sodium hydroxide is used. 0 3 . 4 Suggest why an electric heater is used rather than a Bunsen burner in this hydrolysis. 0 3 . 5 State why reflux is used in this hydrolysis. [1 mark] [1 mark] [1 mark] 0 3 . 6 Write an equation for the reaction between sodium benzoate and hydrochloric acid. [1 mark] 0 3 . 7 Suggest why sodium benzoate is soluble in cold water but benzoic acid is insoluble in [2 marks] cold water. *10* IB/M/Jun17/7405/3<br />
[6 marks] 11 0 3 . 8 After the solid benzoic acid has been filtered off, it can be purified. Describe the method that the student should use to purify the benzoic acid. Question 3 continues on the next page *11* Turn overIB/M/Jun17/7405/3<br />
120 3 . 9 In a similar experiment, another student used 0.040 mol of ethyl benzoate and obtained 5.12 g of benzoic acid. Calculate the percentage yield of benzoic acid. Suggest why the yield is not 100%. [3 marks] Percentage yield Suggestion 18 *12* IB/M/Jun17/7405/3<br />

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Question Answers Mark Additional Comments/Guidance MARK SCHEMEA-LEVEL CHEMISTRY7405/3JUNE 2017 03.1 allows smaller bubbles to form / prevents the formation of (very) large bubbles ALLOW provides large surface area for bubbles to form on IGNORE air NOT no bubbles form / prevents bubbles forming (Mass of ester = 1.05 x 5.0 = 5.25g) amount of ester = 5.25 / 150.0 = 0.0350 mol amount of NaOH = 30 x 2 / 1000 = 0.06 mol OR (Mass of ester = 1.05 x 5.0 = 5.25g) amount of ester = 5.25 / 150.0 = 0.0350 mol Vol of 0.035 mol of NaOH = (0.035/2) x 1000 = 17.5 cm3 (so 30 cm3 used is an excess) OR amount of NaOH = 30 x 2 / 1000 = 0.06 mol 0.06 mol of ester = 9 g = 8.57 cm3 (only 5 cm3 used so NaOH in excess) To ensure that the ester is completely hydrolysed / to ensure all the ester reacts Many organic compounds / the ester / ethanol are flammable 1 1 1 1 1 1 (2 max) Mark independently ALLOW to ensure the other reagent has completely reacted ALLOW prevent ignition of any flammable vapours formed Reflux allows reactant vapours (of volatile organic compounds) to be returned to the reaction mixture / does not allow any reactant vapour to escape IGNORE reference to products 15 of 19 03.2 03.3 03.4 03.5<br />
03.6 C6H5COONa + HClC6H5COOH + NaCl Sodium benzoate soluble because it is ionic Benzoic acid insoluble because: despite the polarity of the COOH group / ability of COOH to form H-bonds, the benzene ring is non-polar. Dissolve crude product in hot solvent/water of minimum volume Filter (hot to remove insoluble impurities) Cool to recrystallise Filter under reduced pressure / with Buchner/Hirsch apparatus wash (with cold solvent) and dry 5.12 / 122 (= 0.042 mol) (0.042/0.04) x 100 = 105 % Product not dried / impurities present in product 03.7 03.8 03.9 Total MARK SCHEMEA-LEVEL CHEMISTRY7405/3JUNE 2017 Allow ionic equation. ALLOW molecular formulae (C7H5O2Na and C7H6O2) ALLOW skeletal benzene ring IGNORE polar ALLOW part of molecule or one end for COOH ALLOW ethanol If no M1 max = 4 ALLOW reference to saturated soln as alternative to min vol IGNORE use of Buchner funnel here apply list principle for each additional process in an incorrect method but IGNORE additional m.pt determination method mark ecf for M1/0.04 or calculation that 0.04 mol of benzoic = 4.88 g (M1) so % yield = (5.12/4.88) x100 = 105% Only allow M3 if M2>100% 1 1 1 1 1 1 1 1 1 1 1 18 16 of 19<br />

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