OCR A Jun 2017 Paper 2 Q17 (with explained solutions) |

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OCR A Jun 2017 Paper 2 Q17

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17 This question is about the chemistry of aromatic compounds. 14 (a) Benzoic acid can be nitrated by concentrated nitric acid in the presence of concentrated sulfuric acid as a catalyst, as shown in Equation 17.1. The organic product of this reaction is 3-nitrobenzoic acid. COOH COOH + HNO3 H2SO4 + H2O Equation 17.1 benzoic acid NO2 3-nitrobenzoic acid (i) Outline the mechanism for this nitration of benzoic acid. Show how H2SO4 behaves as a catalyst. [5]<br />
15 (ii)* A chemist carries out the reaction in Equation 17.1 using 4.97 g of benzoic acid. The chemist obtains 3-nitrobenzoic acid as an impure solid. The chemist purifies the solid to obtain 4.85 g of 3-nitrobenzoic acid. Describe a method to obtain a pure sample of 3-nitrobenzoic acid from the impure solid, determine the percentage yield and check its purity.[6]<br />
(b) A student investigates the relative ease of nitration of phenol, benzene, and benzoic acid. 16 OH COOH phenol benzene benzoic acid The student finds that the conditions required for the nitration of each compound are different, as shown in Table 17.1. Compound phenol benzene benzoic acid Conditions required for nitration Dilute HNO3 Concentrated HNO3 Concentrated HNO3 20 C No catalyst 55 C 100 C H2SO4 catalyst H2SO4 catalyst Table 17.1 (i) State the trend in the relative ease of nitration of phenol, benzene, and benzoic acid.[1] (ii) Apply your knowledge of the bonding in arenes to explain the trend in part (b)(i).[3]<br />
17 (c) A student synthesises 3-bromophenylamine, shown below, starting from nitrobenzene. (i) Complete the flowchart showing the structure of the intermediate and the formulae of the reagents for each stage. NO2 bromination reagents:reduction NH2 reagents:Br 3-bromophenylamine [3] intermediate (ii) Another student attempts the same synthesis but carries out reduction before bromination. The student was surprised to find that two structural isomers of 3-bromophenylamine had been formed instead of the desired organic product. Explain this result and suggest the structures of the two isomers that formed. ExplanationStructures [3]<br />

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H432/02 Question Answer 17 (a) (i) Generation of electrophile Mark Scheme June 2017 Marks Guidance HNO3 + H2SO4H2O + HSO4 Electrophilic substitution+ NO2 +Curly arrow from -bond to NO2 +--- ------------------------------------------------------------------ Correct intermediateCurly arrow back from C-H bond to reform -ring AND H+ as product+ + HSO4+ then H2NO3 ANNOTATE ANSWER WITH TICKS AND CROSSES ALLOW HNO3 + 2H2SO4H3O+ + 2HSO4 NO2 ALLOW HNO3 + H2SO4H2NO3 ALLOW +NO2 OR NO2 First curly arrow must come from the ring to NO2 DO NOT ALLOW the following intermediate: +H2O + NO2 Regeneration of catalyst H+ + HSO4H2SO4 -ring should cover approximately 4 of the 6 sides of the benzene ring structure AND the correct orientation, i.e. gap towards C with NO2 ALLOW + sign anywhere inside the hexagon of intermediate 13<br />
H432/02 Mark Scheme June 2017 Question Answer (ii)* Please refer to the marking instructions on page 5 of this mark scheme for guidance on how to mark this question. Level 3 (56 marks) Outlines the main steps of recrystallisation to produce a pure sample of 3-nitrobenzoic acid from the impure solid. AND Calculates correct percentage yield of 3-nitrobenzoic acid. AND Method of checking purity to include comparison to relevant data. A well-structured response with the steps for recrystallisation and the determination of purity being given in the correct order. Correct use of terminology throughout. Level 2 (34 marks) Attempts all three scientific points but explanations may be incomplete. OR Explains two scientific points thoroughly with very few omissions. The description of checking for purity or recrystallisation is clear and any calculations structured. Key terminology used appropriately. Level 1 (12 marks) A simple explanation based on at least two of the main scientific points. OR Marks Guidance Indicative scientific points, with bulleted elements, may include: 1. PurificationRecrystallisationDissolve impure solid in minimum volume of hot water/solventCool solution and filter solidWash with cold water/solvent and dry 2. Percentage yieldn(benzoic acid) used = = 0.0407 (mol) 4.97 122n(3-nitrobenzoic acid) made = = 0.0290 (mol) 4.85 167percentage yield =100 = 71.3 (%) 0.0290 0.0407 ALLOW 71 to calculator value of 71.29001554 correctly rounded. CHECK for extent of errors by ECF Alternative correct calculation may calculate theoretical mass of 3-nitrobenzoic acid that can be produced as 0.0407167 = 6.80 (g) followed by: percentage yield =100 = 71.3 (%) 4.85 6.80 Calculation must attempt to calculate n(benzoic acid) in mol. 3. Checking purity 14<br />
H432/02 Question Answer Mark Scheme Marks Explains one scientific point thoroughly with few omissions. There is an attempt at a logical structure. The description of the practical techniques provides some detail but may not be in the correct order.Purification step is unclear with few scientific terms and little detail, e.g. just recrystallise.Calculation is difficult to follow, may just include a calculation of moles of reactants and/or products.Purity check specifies a method but this is unclear with little detail, e.g. take melting point. 0 marks No response or no response worthy of credit. (b) (i) Phenol is the most easily nitrated/ most reactive AND Benzoic acid is the least easily nitrated /least reactive(ii) Reactivity of phenol a (lone) pair of electrons on O is (partially) delocalised/donated into the -system / ring15 June 2017 GuidanceObtain melting pointCompare to known valuesPure sample will have a (sharp) melting point very close to data book value ALLOW alternative approach based on spectroscopy or TLC SpectroscopyRun an NMR/IR spectrumCompare to (spectral) databaseSpectrum of pure sample will contain same peaks and not others TLCRun a TLCCompare (Rf value) to known dataPure sample will have a very similar Rf Response must give rank order of reactivity e.g. nitration becomes more difficult from phenol (to benzene) to benzoic acid OR nitration becomes easier from right to left in the table ANNOTATE ANSWER WITH TICKS AND CROSSES ALLOW the electron pair in the p orbitals of the O atom becomes part of the -system / ring ALLOW diagram to show movement of lone pair into ring<br />
H432/02 Question Answer Mark Scheme Marks Guidance June 2017 Reactivity of benzoic acid The COOH group on benzoic acid is an electron withdrawing groupLinks electron density in -bond to reactivity In phenol electron density is higher AND The ring is more susceptible to attack OR In benzoic acid electron density is lower AND The ring is less susceptible to attack(c) (i) Bromination: Br2 AND AlBr3/FeBr3/Fe Intermediate 16 ALLOW lone pair of electrons on O is (partially) drawn/attracted/pulled into -system / ring IGNORE activating and deactivating. ALLOW the following alternatives for susceptibility to attack:phenol attracts electrophiles / NO2 + morephenol polarises electrophiles / NO2 + morebenzoic acid attracts electrophiles / NO2 + lessbenzoic acid polarises electrophiles / NO2 + less ALLOW any combination of skeletal OR structural OR displayed formula as long as unambiguous ALLOW any suitable halogen carrier catalyst<br />
H432/02 Question Answer Mark Scheme Marks Guidance June 2017 ALLOW Kekul structure IGNORE names (question asks for formulae) IGNORE reaction conditions even if incorrect IGNORE dilute for HCl IGNORE H2 IGNORE NaOH if seen as a reagent to convert nitro group into amine e.g Sn/(concentrated) HCl then NaOH scores the mark IGNORE references to electron donating/withdrawing groups ALLOW NH2 activates the ring causing the new group to join at positions 2 and 4. ALLOW ortho and para directing for 2,4 directing IGNORE 6-directing ALLOW Kekul structure IGNORE names Reduction: Sn AND (concentrated) HCl(ii) NH2 is 2,4 directingProducts (1 mark for each):Total 21 17<br />

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