Organic Structure Analysis

1. Organic Structure Analysis Professor Marcel Jaspars

2. Aim • This course aims to extend student’s knowledge and experience with nuclear magnetic resonance (NMR) and mass spectroscopy (MS), by building on the material taught in the 3rd Year Organic Spectroscopy course, and also to develop problem solving skills in this area.

3. Learning Outcomes By the end of this course you should be able to: • Assign 1H and 13C NMR spectra of organic molecules. • Analyse complex first order multiplets. • Elucidate the structure of organic molecules using NMR and MS data. • Use data from coupling constants and NOE experiments to determine relative stereochemistry. • Understand and use data from 2D NMR experiments.

4. Synopsis • A general strategy for solving structural problems using spectroscopic methods, including dereplication methods. • Determination of molecular formulae using NMR & MS • Analysis of multiplet patterns to determine coupling constants. Single irradiation experiments. Spectral simulation. • The Karplus equation and its use in the determination of relative stereochemistry in conformationally rigid molecules. • Determination of relative stereochemistry using the nuclear Overhauser effect (nOe). • Rules to determine whether a nucleus can be studied by NMR & What other factors must be taken into consideration. • Multinuclear NMR-commonly studied heteronuclei. • Basic 2D NMR experiments and their uses in structure determination.

5. Books • Organic Structure Analysis, Crews, Rodriguez and Jaspars, OUP, 2009 • Spectroscopic Methods in Organic Chemistry, Williams and Fleming, McGraw-Hill, 2007 • Organic Structures from Spectra, Field, Sternhell and Kalman, Wiley, 2008 • Spectrometric Identification of Organic Compounds, Silverstein, Webster and Kiemle, Wiley, 2007 • Introduction to Spectroscopy, Pavia, Lampman, Kriz and Vyvyan, Brooks/Cole 2009

6. Four Types of Information from NMR

9. 1H NMR Chemical Shifts in Organic Compounds

10. 13C NMR Chemical Shifts in Organic Compounds

11. 5 Minute Problem #1 MF = C6H12O Unsaturated acyclic ether

12. Six Simple Steps for Successful Structure Solution • Get molecular formula. Use combustion analysis, mass spectrum and/or 13C NMR spectrum. Calculate double bond equivalents (DBE). • Determine functional groups from IR, 1H and 13C NMR • Compare 1H integrals to number of H’s in the MF. • Determine coupling constants (J’s) for all multiplets. • Use information from 3. and 4. to construct spin systems (substructures) • Assemble substructures in all possible ways, taking account of DBE and functional groups. Make sure the integrals and coupling patterns agree with the proposed structure.

13. Double Bond Equivalents CaHbOcNdXe [(2a+2) – (b-d+e)] DBE = 2 C2H3O2Cl =

14. Tabulate Data

15. Solution

16. Shift Prediction Prediction

17. THE PROCESS OF STRUCTURE ELUCIDATION Organic Structure Analysis, Crews, Rodriguez and Jaspars

18. DereplicationDediscovery

19. Dereplication

20. Dereplication

21. Determining the Molecular Formula Using NMR and MS Data DEPT-135 Organic Structure Analysis, Crews, Rodriguez and Jaspars

22. Determining the Molecular Formula Using NMR and MS Data

23. Determining the Molecular Formula Using NMR and MS Data

24. MS Errors • Experimental accurate mass measurement (from MS) was 136.1256 suggesting C10H16 is the correct formula. • The error between calculated and experimental mass is: • 136.1256 - 136.1248 = 0.0008 = 0.8 mmu

25. Molecular Formula Calculators James Deline MFCalc

26. Isotope Ratio Patterns:C100H200 For 12Cm13Cn 1403.6 1404.6

27. Determining molecular formulae by HR-ESI/MS ThermoFinnigan LTQ Orbitrap, Xcalibur software examples from MChem group practicals 2009 (Rainer Ebel)

29. [M+H]+

33. [M+Na]+

36. Analysis of isotope patterns experimental calculated “monoisotopic peak” (mainly 12C713C1H935Cl14N16O2)

37. 5 Minute Problem #2 • Al Kaloid, an Honours Chemistry Student at Slug State University has synthesised either A or B below. He is uncertain which one it is but he’s tabulated the 13C NMR shifts and their multiplicities. Can you help Al by determining which one it is? A B

38. Answer Base Values:

39. ChemDraw

40. The NMR effect

41. Spin-spin coupling (splitting) No coupling Coupling

42. Spin-spin coupling (splitting)

43. Origin of spin-spin coupling

44. Coupling in ethanol

45. Coupling is mutual

46. Coupling in ethanol • To see why the methyl peak is split into a triplet, let’s look at the methylene protons (CH2). • There are two of them, and each can have one of two possible orientations- aligned with, or against the applied field • This gives rise to a total of four possible states: Hence the methyl peak is split into three, with the ratio of areas 1 : 2 : 1

47. Coupling in ethanol • Similarly, the effect of the methyl protons on the methylene protons is such that there are 8 possible spin combinations for the three methyl protons: The methylene peak is split into a quartet. The areas of the peaks have the ratio of 1:3:3:1.

48. Pascal’s triangle n relative intensity multiplet 0 1 singlet 1 1 1 doublet 2 1 2 1 triplet 3 1 3 3 1 quartet 4 1 4 6 4 1 quintet 5 1 5 10 10 5 1 sextet 6 1 6 15 20 15 6 1 septet

49. Coupling patterns

50. First Order • In CH3CH2OH we could explain coupling by the n + 1 rule, this is called 1st order coupling