ORGANIC STRUCTURE ANALYSIS INTRODUCTION

1. ORGANIC STRUCTURE ANALYSISINTRODUCTION Prof Phil Crews, Dr Jaime Rodriguez and Dr Marcel Jaspars

3. STRUCTURAL DRAWINGS • 2D FRAMEWORK • CARBON SKELETON • FUNCTIONAL GPS • LOCATION OF FGs OK But what about: 3D FRAMEWORK 1 RELATIVE STEREO 2 ABSOLUTE STEREO 3 CONFORMATIONAL FEATURES

4. VERIFYING OR CREATING STRUCTURAL DRAWINGS 1. A COMMON TASK 2. MANY ELEMENTS 3. STEREO STRUCTURES 4. CONCISE APPROACHES

5. 4. CONCISE APPROACHES DESIRED OUTCOMES MOLECULAR FORMULA FUNCTIONAL GROUPS CH, CC, CZ, OR RINGS FUNCTIONAL GROUP POSITIONS REGIOCHEMISTRY STEREOCHEMISTRY FOUR TECHNIQUES NMR MS IR UV-VIS • OBTAIN DATA • INTERPRET • SPECTRA

6. REQUIREMENTS FOR A CONCISE APPROACH STEP-BY-STEP ANALYSIS UNDERSTANDING OF BASIC ORGANIC CHEM KNOW COMMON FUNCTIONAL GROUPS

7. REQUIREMENTS FOR A CONCISE APPROACH STEP-BY-STEP ANALYSIS KNOW STABLE VS. UNSTABLE STRUCTURES DEALING WITH ALTERNATIVES

8. 3 TYPES OF PROBLEMS • COMMERCIAL SAMPLES • - VERIFYING THE LABEL • SYNTHETIC REACTION PRODUCT • - VERIFYING THE COURSE OF A REACTION • UNKNOWN NATURAL PRODUCT • - ESTABLISHING A NEW CHEMOTYPE • MOST DIFFICULT TYPE OF APPLICATION • O.S.A. PRINCIPLES & BIOGENETIC, TAXONOMIC ASSUMPTIONS

9. SOME MARINE NATURAL PRODUCTS….

10. WHAT ABOUT PALYTOXIN? MW = 2677 MF = C129H223N3054 1ST ISOLN. 1960’s 1ST STRUCTURE 1980’s P 2

11. PALYTOXIN MW = 2677 MF = C129H223N3054 UN = 20 1ST ISOLN. 1960’s 1ST STRUCTURE 1980’s LD50 = 0.025 g/kg Rabbit LD50 = 0.45 g/kg Mouse WHAT STRUCTURAL INFO FROM NMR?

12. A solvable problem

13. RELATIVE MERITS OF TECHNIQUES DATA MS 29 PEAKS IR 12 PEAKS 1H NMR 1 PEAK 13C NMR 1 PEAK UV/VIS O PEAKS More is less??

15. USES OF MOLECULAR FORMULAOR PARTIAL FORMULA UNSATURATION NUMBER (UN) OR DOUBLE BOND EQUIVALENTS (DBE) CnH2n+2 ADD H FOR EACH X ADD CH FOR EACH N OR P EXAMPLES UN = 1 C6H12O EVEN H COUNT C6H11Cl ODD H COUNT C6H13N ODD H COUNT C6H14N2 EVEN H COUNT

16. USES OF MOLECULAR FORMULAOR PARTIAL FORMULA UNSATURATION NUMBER & FUNCTIONAL GROUP CATEGORIES SELECTED EXAMPLES UN = 0 C-C NOT A FUNCTIONAL GROUP UN = 1 RING, ALKENE, CARBONYL, IMINE, NITRO UN = 2 TWO RINGS, POLYENE, CUMULENE, ALKYNE NITRILE, ISONITRILE, ANHYDRIDE UN = 3 NON-BENZENOID AROMATICS UN = 4 ARENE, PYRIDINE

17. Unsaturation number/Double bond equivalents CaHbOcNdXe [(2a+2) – (b-d+e)] UN = 2 (4+2)-3-1 C2H3O2Cl = = 1 2

18. Unsaturation number/Double bond equivalents • This is the sum of the number of multiple bonds of all kinds kinds (C=C, CC, C=O, C=N, CN, N=O, etc) and rings in a molecule, and is extremely useful in structure determination. • A double bond has a dbe = 1, a triple bond has a dbe = 2, and a benzene ring has a dbe = 4 (three double bonds plus one ring) • Halogens count as hydrogen for this purpose. • The number of oxygen atoms is immaterial. • If you can identify securely by spectroscopic or other means the number of multiply bonded functional groups, e.g. carbonyl, present, the remainder will be rings. It is therefore possible to decide if the compound is acyclic, monocyclic, bicyclic, etc. and this is extremely helpful in imagining a structure. • Once you have obtained the molecular formula, the first step in a structure determination is to work out the dbe.

19. NOW THE QUESTION HOW MANY STRUCTURES FIT C2H3O2Cl ?

20. C2H3O2Cl ACYCLIC H

21. C2H3O2Cl CYCLIC

22. C2H3O2Cl SUMMARY H Cl

23. When things go wrong.......

24. Limitations in Organic Structure Analysis Horror Stories “Even seasoned investigators sometimes experience difficulties in analyzing the structures of organic compounds”…..

25. BUT

26. Limitations in Organic Structure Analysis More Horror Stories

27. SUBSTRUCTURES – A same Z E not same & D, C, B order different JMC: A-D-C-B-Z-E JACS: A-B-C-D-Z-E

28. [3.2.0] [3.1.1] Limitations in Organic Structure Analysis More Horror Stories

29. Limitations in Organic Structure Analysis Final Horror Story

30. Chemical shift addivity Predicting carbon chemical shifts

31. ALIPHATICS: CARBON SHIFTS GENERAL (SP3) 10 – 100 UNFUNCTIONALIZED 10 - 50 21.9 31.7 29.7 11.4

32. ALIPHATICS: CARBON SHIFTS GENERAL (SP3) 10 – 100 UNFUNCTIONALIZED 10 - 50 21.9 • SHIFTS ARE ADDITIVE 31.7 29.7 11.4

33. ADDITIVE PATTERNS

34. UNDERSTANDING THE TABLE   4 2 1  3  31 • +20 • +10 • -2 to -3 •  0 +3 also 61

35. SHOW 13C PREDICT!

36. FUNCTIONAL GROUP ID

37. PROBLEM SOLVING

38. PROBLEM SOLVING

39. COUPLING

40. Coupling/Splitting • Two nuclei placed close together in the same molecule will interact and the NMR lines will be split into several components. • The splitting is also referred to as coupling and the resultant splitting or coupling patterns are also called multiplets. • The distance between the members of each multiplet in Hz is the coupling constant J, and can be calculated from the spectrum: J (Hz) = Dd (ppm) x spectrometer frequency in MHz • The proton proton interaction is transmitted through the intervening (s) electrons making up the chemical bonds, so the magnitude of J is an indication of the number and type of bonds.

41. Coupling/Splitting No coupling Jab Jab Coupling da db

42. FACTORS CONTROLLING J’s • # OF INTERVENING BONDS (1,2,3,4, etc) • e- DENSITY • ANGLE H-C-CH-C-C-HH-C-C-CH-C-C-C-H • BOND HYBRIDIZATION • ELECTRONEGATIVTY OF ATOMS ALONG PATH

43. A FOCUS ON FIRST ORDER

44. COMPLEX 1ST ORDER

45. A FOCUS ON COMPLEX FIRST ORDER

46. TERMS • BIG  A/X • 1ST ORDER AMX • NON 1ST ORDER AB • SPECIAL A/A’ • COUPLING 1JCH • VALUES IN Hz • 2-5 NUCLEI “21 CASES”

47. NON 1ST ORDER SPIN SYSTEMS • AB #2 • ABC #3 • A2B #4 • ABCX #8 • A2B2 #9 • AA’XX’ #10

48. P 130

49. Given the 1H NMR chemical shifts and coupling constants for allyl alcohol, explain the observed spectrum (OH peak omitted)

50. MASS SPECTROMETRY