OCR A Jun 2016 Paper 4 Q2 (with explained solutions) |

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OCR A Jun 2016 Paper 4 Q2

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2 Some organic compounds contain nitrogen atoms. Examples include condensation polymers and azo dyes. (a) A section of a condensation polymer is shown below. CO(CH2)4CONH(CH 2)6NHCO(CH2)4CONH(CH 2)6NH (i) In the boxes below, draw the structures of the two monomers that form this condensation polymer. (ii) Name the type of condensation polymer and give a use for this polymer. TypeUse[1] [2]OCR 2016<br />
(b) A student plans a two-step synthesis starting with phenylamine. The steps of the synthesis are shown below. NH2 HN CH3 HN CH3 step 1 step 2 phenylamine compound A NO2 compound B (i) In step 1, phenylamine reacts with ethanoic anhydride to make compound A and one other organic product. Draw the structure of ethanoic anhydride, with the functional group displayed, and suggest the structure of the other organic product formed in step 1. ethanoic anhydride other organic product [2] (ii) Calculate the mass of compound A that can be synthesised from 3.00 g of phenylamine in step 1. The percentage yield of this reaction is 61.0%. Mr (phenylamine) = 93.0 Give your answer to three significant figures.OCR 2016 mass of compound A =g [3] Turn over<br />
The steps of the synthesis are shown again below. NH2 HN CH3 HN CH3 step 1 step 2 phenylamine compound A NO2 compound B (iii) In step 2, compound A is converted into compound B using a mixture of concentrated nitric acid and concentrated sulfuric acid. Outline, with the aid of curly arrows, the mechanism for the conversion of compound A into compound B. Use equations to explain how sulfuric acid acts as a catalyst in this reaction. [5]OCR 2016<br />
(c) An azo dye is synthesised in two steps. In step 2 the diazonium ion is reacted with compound C to form the azo dye. Complete the flowchart for this synthesis. Write your answers in the boxes. H3C H3C NH2 step 1 Reagents Conditions N2 diazonium ion step 2 Conditions H3C Structure of compound C CH3 HO azo dyeOCR 2016 [4] [Total: 17] Turn over<br />

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F324 Question (i) 2 (a) (ii) Answer H2N(CH2)6NH2 HOOC(CH2)4COOH Type of condensation polymer Polyamide AND Use of condensation polymer Fibres in clothing (b) (i) Ethanoic anhydride Other organic compound CH3COOH Mark Scheme Mark (cid:51) (cid:51) (cid:51) (cid:51) (cid:51) June 2016 Guidance ALLOW correct structural OR displayed OR skeletal formulae OR a combination of above as long as unambiguous ALLOW acid chloride, ClOC(CH2)4COCl Both answers required for one mark ALLOW nylon IGNORE numbers IGNORE polypeptide DO NOT ALLOW kevlar ALLOW any common use for nylon e.g. fibre, clothing, rope, fishing net, bristles, brushes, bags, cable ties etc. DO NOT ALLOW distinctive uses associated with kevlar or other polymers e.g. bullet-proof vests, crash helmets, bottles, cups IGNORE plastic ALLOW skeletal formula ALLOW correct structural OR displayed OR skeletal formulae OR a combination of above as long as unambiguous IGNORE names<br />
IF answer = 2.66 (g) award 3 marks IF answer = 4.36 (g) award 2 marks (% yield not used) IF answer = 7.14 (g) award 2 marks (% yield used incorrectly) n(phenylamine) (= 3.00/93.0) = 0.0323 mol n(compound A) = (0.0323 x 0.61) = 0.0197 mol Mr (compound A) = 135 AND Mass of compound A = (135)(0.0197) = 2.66 g OR n(phenylamine) (= 3.00/93.0) = 0.0323 mol Mr (compound A) = 135 AND Theoretical mass of compound A = (0.0323 x 135) = 4.36 (cid:51) Actual mass of compound A = (4.36 x 0.61) = 2.66 g F324 Mark Scheme June 2016 Question (ii) FIRST CHECK THE ANSWER ON THE ANSWER LINE Answer Mark Guidance ANNOTATE WITH TICKS AND CROSSES ETC. ALLOW 3 SF: 0.0323 up to calculator value of 0.032258064 correctly rounded ALLOW 3 SF up to calculator value Penalise rounding to 2 SF once ALLOW ECF on incorrectly rounded values Final answer must be expressed to 3 significant figures ALLOW ecf from incorrect Mr IF answer = 2.65 (g) award 2 marks unless this alternative method is used (3 marks) 93 g gives 135 g 3.00 g gives 135/93 x 3.00 = 4.35 g 4.35 x 0.61 = 2.65 g (cid:51) (cid:51) (cid:51) (cid:51) (cid:51)<br />
F324 Question (iii) M1 Answer Mark Scheme Mark (cid:51) H2SO4 + HNO3 (cid:5) HSO4 + - + H2O + NO2 M2 curly arrow from (cid:83) ring OR from within the ring to +NO2 (cid:51) M3 correct intermediate (with charge) (cid:51) M4 curly arrow from C-H to reform ring AND correct products (cid:51) June 2016 Guidance + - + H2NO3 - + H3O+ + NO2 + + + (cid:5) H2O + NO2 ANNOTATE WITH TICKS AND CROSSES ETC. Equation to show formation of the electrophile ALLOW 2H2SO4 + HNO3 (cid:5) 2HSO4 ALLOW H2SO4 + HNO3 (cid:5) HSO4 AND H2NO3 Penalise missing or incorrect NHCOCH3 on intermediate only (M3) DO NOT ALLOW intermediate with the (cid:83)-system covering less than half the ring ALLOW + charge anywhere inside the horseshoe Horseshoe must have open end towards NO2 ALLOW Kekul mechanism<br />
F324 Question Mark Scheme June 2016 Answer Mark Guidance M5 Regeneration of the catalyst: H+ + HSO4 reagents for step 1 Nitrous acid/HNO2 (and HCl) - (cid:5) H2SO4 (cid:51) conditions for step 1 (cid:100)10 oC compound C (c) (cid:51) (cid:51) (cid:51) OR - (cid:5) H2SO4 + H2O ALLOW H3O+ + HSO4 ALLOW NaNO2 + HCl IGNORE reference to concentration ALLOW OH ionised as O- ALLOW KOH(aq)/NaOH(aq)/OH-(aq)<br />
F324 Question Mark Scheme June 2016 Guidance ALLOW dilute NaOH or stated concentration IGNORE NaOH/KOH (must be aqueous) If temperature stated must be below 10oC DO NOT ALLOW heat/boil/warm Answer Mark conditions for step 2 alkaline/alkali (cid:51) Total 17 10<br />

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