OCR A Jun 2013 Paper 4 Q5 (with explained solutions) |

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OCR A Jun 2013 Paper 4 Q5

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5 A chemist isolated compound L from the leaves of a common garden herb. He analysed the compound using a number of techniques. 15 (a) Compound L contains C, H and O only. Elemental analysis shows that L has the percentage composition by mass: C, 73.15%; H, 7.37%. The mass spectrum of L has a molecular ion peak at m/z = 164. Show that the molecular formula of L is C10H12O2. [2]OCR 2013 Turn over<br />
(b) The 13C NMR spectrum of compound L is shown below. 16 240 220 200 180 160 140 100 80 60 40 20 120 (cid:419)/ppm How many different carbon environments (types of carbon) are present in a molecule of compound L?[1] (c) The 1H NMR spectrum of compound L, C10H12O2, is shown below. One of the signals has been enlarged to help its analysis. expansion of multiplet centred at (cid:419) = 2.7 ppm 1H 6H (cid:419)/ppm 5HOCR 2013<br />
(i) A signal at (cid:419) = 0 ppm results from tetramethylsilane (TMS). TMS had been added to the sample of compound L before the NMR spectrum was run. 17 State why TMS was added.[1] (ii) The 1H NMR spectrum includes an integration trace. What information can be deduced about compound L from the integration trace?[1] (iii) Using the information given in parts (a), (b) and (c) of this question, determine the structure of compound L. Show all your reasoning. In your answer, you should use appropriate technical terms, spelled correctly. structure of compound LOCR 2013 Turn over<br />
18[7] [Total: 12] END OF QUESTION PAPEROCR 2013<br />

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F324 Question (a) Answer H 7.37% 7.37 6 O 19.48% 1.22 1 % mol ratio C 73.15% 6.10 5 molar ratio (C:H:O) = 6.10 : 7.37 : 1.22 OR = 5:6:1 OR empirical formula = C5H6O Mr is 164 so molecular formula = C10H12O2 (cid:57) (cid:57) (b) seven (cid:57) (c) (i) TMS is the standard (for chemical shift measurements) (cid:57) (ii) (iii) (relative) number of protons/hydrogens in each environment / peak / region OR three proton environments with protons in ratio 5:1:6 (cid:57) 13C NMR Analysis (1 mark) The peak at 185ppm suggests an ester group / AND one of the following: The peaks between 120ppm and 160ppm indicate a benzene ring OR the peaks at 18ppm AND 36ppm suggest C-C (cid:57) 12 Mark Scheme June 2013 Marks 1 1 Guidance (cid:57) } ratio = 10 OR 5 12 OR 6 } OR 1 2 } (cid:57) ALLOW alternative method 73.15%164 = 120 7.37%164 = 12.1 19.48%164 = 31.9 This mark is for some evidence of using Mr, which is twice the value that you would obtain from the empirical formula ALLOW TMS is the reference OR for calibration IGNORE unreactive / volatile / it gives a sharp peak ALLOW TMS = 0 ppm / TMS is used for comparison ALLOW (relative) number of each type of proton/hydrogen IGNORE number of protons in the compound FULL ANNOTATIONS WITH TICKS, CROSSES,CON ETC MUST BE USED Inclusion of an incorrectly assigned 13C peak CONS M1<br />
F324 Question Mark Scheme June 2013 Answer 1H ANALYSIS (4 marks) Doublet / peak at 1.2 shows R-CH AND 6 Hs / 2 CH3 (in this environment) (cid:57) Multiplet / septet / heptet / peak split into 7 / peak at 2.7ppm indicates Marks Guidance Candidates may quotevalues as ranges taken from Data Sheet, so ALLOW tolerance (ppm) eg 6.58 aromatic HC (cid:57) The doublet suggests that two CH3 groups are attached to a CH OR the multiplet / septet / heptet suggests that the CH group is attached to two CH3 groups (cid:57) (cid:4) QWC must spell one of multiplet, septet, heptet OR doublet correctly Peak at 7.3ppm indicates a benzene ring AND 5 Hs (cid:57) Compound identification (2 marks) 2.02.9 carboxyl 0.72.0 alkyl ALLOW peaks labelled on the spectrum If QWC word is not used, MAX 3 for proton NMR ALLOW C6H5 IGNORE reference to phenol IF identified as then two marks (cid:57)(cid:57) Allow has 5 Hs as C6H5 if they state that the benzene ring IF identified as then one mark (cid:57) Total 12 13<br />

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