OCR A Jun 2011 Paper 2 Q7 (with explained solutions) |

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OCR A Jun 2011 Paper 2 Q7

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7 Compound X and compound Y react together to make an ester Z. Samples of X and Y were analysed by a research chemist. A summary of the chemists results are shown below. 20 Analysis of compound X type of analysis infrared spectroscopy percentage composition by mass mass spectrometry evidence absorption at 1720 cm1 and a very broad absorption between 2500 and 3300 cm1 C, 48.65%; H, 8.11%; O, 43.24% molecular ion peak at m/z = 74.0 Analysis of compound Y infrared spectrum of Y transmittance (%) 100 80 60 40 20 4000 3500 3000 mass spectrum of Y 2000 2500 wavenumber / cm1 1500 1000 500 relative abundance (%) 100 80 60 40 20 0.0 10 20 30 m / z 40 50OCR 2011<br />
Use this information to suggest the identity of compound X, compound Y and ester Z. In your answer you should make clear how your explanation is linked to the evidence. 21[10] END OF QUESTION PAPER [Total:10]OCR 2011<br />

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F322 Question Mark Scheme June 2011 Answer ANY SEVEN FROM: Compound X QWC: X contains C=O because of absorption at 1720 cm1 AND contains OH because of (broad) absorption between 2500 to 3300 cm1 (cid:57) So X is a carboxylic acid (cid:57) Molar ratio (C:H:O) of X is 4.05 : 8.1 : 2.7 OR (Empirical formula) is C3H6O2 (cid:57) Mr is 74.0 so X is C3H6O2 (cid:57) 65.48 0.12 24.43 0.16 11.8 0.1 (cid:57) 17 Mark Guidance ANNOTATE ANSWER WITH TICKS AND CROSSES PLEASE ENSURE YOU LOOK AT THE DATA AND SPECTRA ON PAGE 20 IN CASE THEY INCLUDE COMMENTS THAT ARE WORTHY OF CREDIT. MARK THIS PAGE WITH AN OMISSION MARK, ^ , IF BLANK QWC: mark is integrated into the chemistry marks. These marks need to link evidence with an explanation ALLOW X contains C=O and OH because of absorptions at 1720 cm1 and 2500 to 3300 cm1 ALLOW X contains carboxylic acid/COOH because of absorption at 1720 cm1 and (broad) absorption between 2500 to 3300 cm1 (cid:57)(cid:57) ALLOW alternative approach to molecular formula Mr is 74.0 (cid:57) 65.48 74 x 100 C3H6O2 (cid:57) This mark is for some evidence of using Mr to deduce the molecular or structural formula ALLOW Mr is 74.0 so X is CH3CH2COOH (cid:57) DO NOT ALLOW ECF from the empirical formula with the wrong molar ratio = 36 : 6 : 32 (cid:57) 24.43 100 : 74 x 11.8 100 : 74 x<br />
F322 Question Mark Scheme June 2011 Mark Guidance ANNOTATE ANSWER WITH TICKS AND CROSSES ALLOW Y is an alcohol (or phenol) because of absorption between 3200 and 3550 cm1 ALLOW Y contains CO, CH and OH bonds because of absorptions at approximately 1030, 2950 and 3350 cm-1 ALLOW m/z = 46 so Mr is 46 OR mass spectrum has a peak at 46 which is the Mr OR Mr is 46 because of m/z peak shown on the actual spectra Mr = 46 on its own is not sufficient m/z = 46 on its own is not sufficient ALLOW m/z = 31 shows CH2OH (fragment); m/z = 29 shows C2H5 m/z = 15 is CH3 (fragment) (fragment); Answer Compound Y QWC Y contains OH because of absorption between 3100 and 3500 cm1 (cid:57) QWC Mass spec of Y has molecular ion, m/z = 46 so Mr is 46 (cid:57) Correct identification of one fragment from a m/z value e.g. m/z = 31 is CH2OH+; m/z = 29 is C2H5 (cid:57) Identification of compounds So X must be CH3CH2COOH OR propanoic acid (cid:57) So Y is ethanol OR C2H5OH OR CH3CH2OH (cid:57) Z is CH3CH2COOC2H5 OR ethyl propanoate (cid:57) +; m/z = 15 is CH3 Note: an incorrect name CONs a correct structure ALLOW skeletal OR displayed formula throughout DO NOT ALLOW propanoic acid with wrong structure or incorrect molecular formula DO NOT ALLOW ethanol with wrong structure or incorrect molecular formula DO NOT ALLOW ethyl propanoate with wrong structure or incorrect molecular formula ALLOW ECF for identification of Z from incorrect X and Y. DO NOT ALLOW this ECF if name and structures of X or Y do not match Total 10 18<br />

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