February 2020 Solution: Molecular Formula Analysis

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Explore the solution for the February 2020 Problem of the Month through molecular formula analysis and HSQC evaluation, identifying CHn fragments. Learn about the unique signals and groups present in the structure.

  • Solution
  • Molecular Formula
  • Analysis
  • HSQC Evaluation
  • CHn Fragments
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  1. Problem of the Month: February 2020 Solution

  2. 281.0 Basics Hz From the molecular formula we deduce 2 double bond equivalents A sharp proton signal at about 9.5 ppm is not very common. This might be any kind of an OH group, but with a high degree of probability this is the signal of an 288.6 273.5 aldehyde. Let us now start extracting the CHn fragments from the HSQC. ppm 9 8 7 6 5 4 3 2 1H

  3. Building blocks CHn-fragments 1.75 from 1H 9.40 6.49 1 2.38 1.12 1 2 3 3 8.98 12.78 22.28 It is very easy to evaluate an HSQC. The sensitivity, of course, is less than the sensitivity of a one dimensional proton spectrum but much higher than a one dimensional carbon spectrum. Therefore, the measurement of a HSQC is always recommended, if possible. 20 40 60 13C 80 As additional data for the proton projections we need the chemical shifts and integrals from the one dimensional proton spectrum. 100 120 138.80 140 156.21 160 HSQC 180 195.38 ppm 10 9 8 6 5 3 7 4 2 1 1H

  4. Building blocks CHn-fragments 1.75 9.40 6.49 1 2.38 1.12 1 2 3 3 8.98 12.78 22.28 Because we have one carbon signal for each of the six carbon atoms of the molecular formula, every cross peak in the HSQC is the result of a CHn group. 20 40 H 60 There is no symmetry at all. C 13C 80 12.78 HH 1.12 100 Let us start. 120 138.80 140 H 156.21 C 160 12.78 HSQC 180 HH 1.12 195.38 ppm 10 9 8 6 5 3 7 4 2 1 1H

  5. Building blocks CHn-fragments 1.75 9.40 6.49 1 2.38 1 2 3 8.98 20 There is one more methyl group. 22.28 H H 1.75 40 8.98 60 H C 13C 80 100 120 H 138.80 140 H H 1.75 156.21 C 160 8.98 12.78 H C HSQC 180 HH 1.12 195.38 ppm 10 9 8 6 5 3 7 4 2 1 1H

  6. Building blocks CHn-fragments 9.40 6.49 1 2.38 1 2 20 The next cross peak belongs to a CH2 group. 22.28 2.38 40 H H 60 13C 80 C 22.28 100 120 2.38 H 138.80 140 H H H H 1.75 156.21 C 160 8.98 12.78 H C C 22.28 HSQC 180 HH 1.12 195.38 ppm 10 9 8 6 5 3 7 4 2 1 1H

  7. Building blocks CHn-fragments 9.40 6.49 1 1 20 In the next two fragments, the carbon is sp2hybridized. The chemical shifts of both the carbon and the proton bound to the carbon are very characteristic of sp2hybridization. 40 60 13C 80 156.21 C 156.21 100 H 6.49 C 120 H 6.49 2.38 H 138.80 140 H H H H 1.75 156.21 C 160 8.98 12.78 H C C 22.28 HSQC 180 HH 1.12 195.38 ppm 10 9 8 6 5 3 7 4 2 1 1H

  8. Building blocks CHn-fragments 9.40 1 20 In the next two fragments, the carbon is sp2hybridized. The chemical shifts of both the carbon and the proton bound to the carbon are very characteristic of sp2hybridization. 40 60 13C 80 156.21 C O C 195.38 100 H 6.49 H 9.40 The aldehyde results from both proton and carbon chemical shift. 120 2.38 H 138.80 140 H H H O C H 1.75 C 160 195.38 8.98 12.78 H C H C 22.28 HSQC 180 9.40 HH 1.12 195.38 ppm 10 9 8 6 5 3 7 4 2 1 1H

  9. Building blocks CHn-fragments There is no HSQC cross peak for the carbon signal at 138.80 ppm. That s fine. molecular formula - found so far missing 20 C6H10O C5H10O C 40 - - 60 13C 80 156.21 C O C C 138.80 195.38 100 H 6.49 H 9.40 120 2.38 H 138.80 140 H H H H 1.75 C 160 8.98 12.78 H C C 22.28 HSQC 180 HH 1.12 ppm 10 9 8 6 5 3 7 4 2 1 1H

  10. Linking the fragments Because protons are part of nearly each of the fragments, the best way to connect the fragments is the COSY. 1.12 The first connectivity is easily visible. 1.0 2.0 2.38 156.21 3.0 C O C C 4.0 1H 138.80 195.38 H 6.49 H 5.0 9.40 2.38 H H 6.0 2.38 H H H H 7.0 H 1.75 C 22.28 H C 8.0 8.98 12.78 C H C C 22.28 12.78 COSY 9.0 HH 1.12 HH 1.12 10 9 8 7 6 5 4 3 2 1 ppm 1H

  11. Linking the fragments The next connectivity seems to be clear. 1.12 1.0 2.0 2.38 156.21 3.0 C O C C 4.0 1H 138.80 195.38 H 6.49 H 5.0 9.40 2.38 H H 6.0 H 7.0 H 1.75 C 22.28 H 8.0 8.98 C H C 12.78 COSY 9.0 HH 1.12 10 9 8 7 6 5 4 3 2 1 ppm 1H

  12. Linking the fragments The next connectivity seems to be clear. But this would finally result in propane. That is impossible, because we would then have at least two molecules and furthermore there would have to be a proton multiplet with 6 protons. 1.0 1.75 2.0 2.38 156.21 3.0 C O C C 4.0 1H 138.80 195.38 H 6.49 H 5.0 9.40 2.38 H H 6.0 H 7.0 H 1.75 C 22.28 H 8.0 8.98 C H C 12.78 COSY 9.0 HH 1.12 10 9 8 7 6 5 4 3 2 1 ppm 1H

  13. Linking the fragments But ? There is another correlation between the protons with the chemical shifts of 2.38 ppm and 6.49 ppm. 1.0 2.0 2.38 156.21 3.0 C O C C 4.0 1H 138.80 195.38 H 6.49 H 5.0 9.40 2.38 H H 6.0 6.49 H 7.0 H 1.75 C 22.28 H 8.0 8.98 C H C 12.78 COSY 9.0 HH 1.12 10 9 8 7 6 5 4 3 2 1 ppm 1H

  14. Linking the fragments Let us connect the corresponding fragments. 1.0 2.0 2.38 156.21 3.0 C O C C 4.0 1H 138.80 195.38 H 6.49 H 5.0 9.40 2.38 H H 6.0 6.49 H 7.0 H 1.75 H C 22.28 H H 1.75 156.21 8.0 8.98 C C H C 12.78 COSY 8.98 9.0 H C H 6.49 HH 1.12 10 9 8 7 6 5 4 3 2 1 ppm 1H

  15. Linking the fragments To get the next connectivity, we don t need any COSY peak. There is only one possibility. 1.0 2.0 3.0 O C C 4.0 1H 138.80 195.38 H 5.0 9.40 2.38 2.38 H H H H 6.0 7.0 H C 22.28 C 22.28 H H H 1.75 156.21 156.21 8.0 C C C C C 138.80 12.78 12.78 COSY 8.98 9.0 H C H 6.49 6.49 H HH HH 1.12 1.12 10 9 8 7 6 5 4 3 2 1 ppm 1H

  16. Linking the fragments Now there is only one remaining possibility to connect the three pieces. 1.0 2.0 3.0 O C H H 2.38 4.0 1H 1.75 H H 195.38 C H 5.0 8.98 H 9.40 2.38 H H C C 22.28 H 6.0 156.21 C C 7.0 138.80 12.78 H O C 22.28 HH H H 195.38 1.75 156.21 H 6.49 8.0 138.80 C H C C C 1.12 12.78 COSY 8.98 9.0 H C H 6.49 HH 9.40 1.12 10 9 8 7 6 5 4 3 2 1 ppm 1H

  17. Two open questions 1. What is the reason for this cross peak? We have found one four bond correlation. Such four bond correlations are not infrequently observed as soon as electrons are part of the coupling pathway. 1.0 1.75 2.0 3.0 H 2.38 4.0 1H H 1.75 H H 5.0 8.98 H C C 22.28 H 6.0 156.21 6.49 C C C 7.0 138.80 12.78 8.0 O C H 6.49 HH 1.12 COSY 9.0 195.38 H 9.40 10 9 8 7 6 5 4 3 2 1 ppm 1H

  18. Two open questions 2. And where does this cross peak come from? Please remember, we already tried to use this cross peak in error. We have found a five bond correlation! These are not very common. The methyl group at 1.75 ppm should appear as a doublet of triplets due to both long range couplings. 1.0 1.75 2.38 2.0 3.0 H t 2.38 4.0 1H H 1.75 H H 5.0 8.98 d H C C 22.28 H 6.0 156.21 C C C 7.0 138.80 12.78 8.0 O C H 6.49 HH 1.12 COSY 9.0 195.38 H 9.40 10 9 8 7 6 5 4 3 2 1 ppm 1H

  19. Coupling constants (Although we don t know the exact value of the coupling constants just now, the two doubled sided arrows should remind us about the long range couplings.) There is only one pure multiplett: the triplet of the methyl protons at 1.12 ppm due to the two adjacent chemically equivalent methylene protons. 281.0 Hz 1.0 H 2.38 2.0 H 1.75 H H 7.55 Hz ? =288.6 Hz 273.5 Hz 3.0 = ?.?? ?? H C 2 H 288.6 2.38 H H C 273.5 4.0 1H 1.75 H H C 8.98 C C 5.0 H C C 22.28 1.12 H 6.0 O C H 6.49 156.21 HH C C C 7.0 138.80 12.78 H 8.0 O C H 6.49 HH 1.12 9.0 195.38 H 9.40 10 9 8 7 6 5 4 3 2 1 ppm ppm 1H 6 5 3 4 2 1H

  20. Coupling constants The primary structure for the proton multiplet at 6.49 ppm should be a triplet due to the two chemically equivalent methylene protons at 2.38 ppm. 281.0 Hz In the COSY we observed a long-range correlation to the protons of the methyl group at 1.75 ppm. This should result in a quartet fine splitting for each of the three lines of the triplet. H 2.38 H 1.75 H H 7.55 Hz H C Let us start analyzing the easy recognizable quartet substructure of the multiplet. 288.6 C H 273.5 C C C O C H 6.49 HH 1.12 H 1631.5 1630.2 1628.8 1627.4 1624.2 1622.9 1621.5 1620.1 1616.9 1615.5 1614.2 1612.8 6 5 3 4 2 ppm 1H

  21. Coupling constants After removing the quartet splitting due to the methyl protons at 1.75 ppm there remains (one of three) singlet. The chemical shift of this singlet is the average of the four chemical shifts of the quartet. 281.0 Hz There are some ways to get the coupling constant, here is one of them: ? =1631.5 Hz 1627.4 Hz H 2.38 = ?.?? ?? H 1.75 H H 3 7.55 Hz 1629.45 Hz H C 288.6 C H 273.5 C C C O C H 6.49 HH 1.12 1.37 Hz H 1631.5 1630.2 1628.8 1627.4 1624.2 1622.9 1621.5 1620.1 1616.9 1615.5 1614.2 1612.8 6 5 3 4 2 ppm 1H

  22. Coupling constants 281.0 After repeating the procedure twice, the three quartets collapse to three single lines in the intensity ratio of 1 : 2 : 1, which is a pure triplet. Hz H 2.38 H 1.75 H H 7.55 Hz 1629.45 Hz 1622.15 Hz 1614.85 Hz H C 288.6 C H 273.5 C C C O C H 6.49 HH 1.12 1.37 Hz H 1631.5 1630.2 1628.8 1627.4 1624.2 1622.9 1621.5 1620.1 1616.9 1615.5 1614.2 1612.8 6 5 3 4 2 ppm 1H

  23. Coupling constants ? =1629.45 Hz 1614.85 Hz = ?.?? ?? 2 281.0 After evaluating the triplet we have the coupling constant to the two equivalent methylene group protons. Hz H 2.38 H 1.75 H H 7.55 Hz 1629.45 Hz 1622.15 Hz 1614.85 Hz 7.30 Hz H C 288.6 C H 273.5 C C C O C H 6.49 HH 1.12 1.37 Hz H 1631.5 1630.2 1628.8 1627.4 1624.2 1622.9 1621.5 1620.1 1616.9 1615.5 1614.2 1612.8 6 5 3 4 2 ppm 1H

  24. Coupling constants The multiplet of the methyl group at 1.75 ppm should be a doublet (Jd = 1.37 Hz) of triplets (unknown Jtso far) due to the metyhlene group protons at 2.38 ppm. The multiplet apparently consists of 6 lines, but they are not completely resolved. Since the difference between the two outermost - well resolved - lines, however, must be 2 * Jt+ Jd, we can easily calculate the coupling constant of the triplet to Jt= ((438.5 Hz - 435.4 Hz) - 1.37 Hz) / 2 = 0.87 Hz. 281.0 Hz 0.87 Hz H 2.38 H 1.75 H H 7.55 Hz 7.30 Hz H C 288.6 C H 273.5 C C C O C H 6.49 HH 1.12 1.37 Hz H 438.5 437.6 437.1 436.3 435.4 Hz 6 5 3 4 2 ppm 1H

  25. Coupling constants Let us try to simulate the multiplet with the somewhat overlapping lines using the two coupling constants. Please don t expect 100% perfection. Perfection would require to sum over 6 lorentzian lines. H 1.75 7.30 Hz 281.0 Hz Jt = 0.87 Hz 0.87 Hz H 2.38 Jd = 1.37 Hz H H 7.55 Hz H C 288.6 C H 273.5 C C C O C H 6.49 HH 1.12 1.37 Hz H 438.5 437.6 437.1 436.3 435.4 Hz 6 5 3 4 2 ppm 1H

  26. Coupling constants Everything is finished now. Nevertheless let us try to understand the structure of the multiplet at 2.38 ppm. 281.0 Hz Let us start by removing all pieces of information not necessary to understand the multiplet structure. 0.87 Hz H H 2.38 1.75 H H 7.55 Hz 7.30 Hz H C 288.6 C H 273.5 C C C O C H 6.49 HH 1.12 1.37 Hz H 6 5 3 4 2 ppm 1H

  27. Coupling constants As a first check the frequency difference between the two outmost lines should be 0.87 Hz H = 3 0.87 Hz + 7.30 Hz + 3 7.55 Hz = ??.?? ?? H 2.38 H H 7.55 Hz 7.30 Hz We have H C C H 610.5 Hz 577.9 Hz = 32.6 Hz which is perfect. 32.6 Hz, C C C O C H HH H 577.9 Hz 610.5 Hz

  28. Coupling constants A quartet with a small coupling constant is visible 5 times. This should be due to the five bond coupling pathway. Let us check. 0.87 Hz H H ? =580.5 Hz 577.9 Hz 2.38 = ?.?? ?? H H 3 7.55 Hz 7.30 Hz H C If we average five times over the four lines of each quartet there remains 5 lines in the integral ratio of about C H C C C 1 : 4 : 6 : 4 : 1. O C H HH H 577.9 Hz 580.5 Hz 610.5 Hz

  29. Coupling constants Of course that cannot be a quintet. It is a doublet (7.30 Hz) of quartets (7.55 Hz). Due to the very similar coupling constants some lines strongly overlap. H 2.38 H H H 7.55 Hz 7.30 Hz H C C H C C C O C H HH H 1 4 6 4 1

  30. Coupling constants 2.38 ppm Let us simulate the multiplet pattern of the signal at 2.38 ppm step by step. We start with the doublet splitting with a coupling constant of 7.30 Hz. H 2.38 H H H 7.55 Hz 7.30 Hz We use slightly different colous for the two lines to better see the overlapping in subsequent slides. H C C H C C C O C H HH H 1 4 6 4 1

  31. Coupling constants Now both lines should split into an quartet by a coupling constant of 7.55 Hz. Let us start with the left line. H 2.38 H H H 7.55 Hz H C C H C C C O C H HH H 1 4 6 4 1

  32. Coupling constants These four lines are the first part of our quintet . H 2.38 H H H 7.55 Hz H C C H C C C O C H HH H 1 4 6 4 1

  33. Coupling constants Let us repeat the quartet splitting with the second line of the doublet. H 2.38 H H H 7.55 Hz H C C H C C C O C H HH H 1 4 6 4 1

  34. Coupling constants If we add these four lines to the existing four lines finally we get our pseudo quintet. H 2.38 H H H 7.55 Hz H C C H C C C O C H HH H 1 4 6 4 1

  35. Coupling constants There is not a perfect proportionality between line intensities (heights) and integrals, due to the slightly different coupling constants. As a result of these tiny differences the lines of the quartet substruture in the center of the pseudo quintet are a little bit broader (and hence the intensity slightly smaller than expected) than the same substructure lines at the two outmost lines. The integrals for the pseudo quintet, however, remain in the expected 1 : 4 : 6 : 4 : 1 ratio. H 2.38 H H H 7.55 Hz H C C H C C C O C H HH H 1 4 6 4 1

  36. Complete solution 0.87 Hz H 2.38 H 1.75 H H 7.55 Hz 7.30 Hz 8.98 H C Using the available data it is not possible to determine the configuration around the double bond. C 22.28 H 156.21 C C C 138.80 12.78 O C H 6.49 HH 1.12 1.37 Hz 195.38 H 9.40

  37. Contributions Spectrometer time Measurements TU Munich Rainer Hae ner Discussions and native English language support Compilation Alan M. Kenwright Nils Schl rer Rainer Hae ner

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