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ORGANIC STRUCTURE ANALYSIS INTRODUCTION Prof Phil Crews, Dr Jaime Rodriguez and Dr Marcel Jaspars.

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Presentation on theme: "ORGANIC STRUCTURE ANALYSIS INTRODUCTION Prof Phil Crews, Dr Jaime Rodriguez and Dr Marcel Jaspars."— Presentation transcript:

1 ORGANIC STRUCTURE ANALYSIS INTRODUCTION Prof Phil Crews, Dr Jaime Rodriguez and Dr Marcel Jaspars

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

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

5 FOUR TECHNIQUES NMR MS IR UV-VIS DESIRED OUTCOMES MOLECULAR FORMULA FUNCTIONAL GROUPS CH, CC, CZ, OR RINGS FUNCTIONAL GROUP POSITIONS REGIOCHEMISTRY STEREOCHEMISTRY 1)OBTAIN DATA 2)INTERPRET SPECTRA 4. CONCISE APPROACHES

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 = C 129 H 223 N ST ISOLN. 1960’s 1 ST STRUCTURE 1980’s P 2

11 PALYTOXIN MW = 2677 MF = C 129 H 223 N UN = 20 1 ST ISOLN. 1960’s 1 ST STRUCTURE 1980’s LD 50 =  g/kg Rabbit LD 50 = 0.45  g/kg Mouse WHAT STRUCTURAL INFO FROM NMR?

12 A solvable problem

13 RELATIVE MERITS OF TECHNIQUES DATA MS 29 PEAKS IR12 PEAKS 1H NMR1 PEAK 13C NMR1 PEAK UV/VISO PEAKS More is less??

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15 USES OF MOLECULAR FORMULA OR PARTIAL FORMULA UNSATURATION NUMBER (UN) OR DOUBLE BOND EQUIVALENTS (DBE) C n H 2n+2 ADD H FOR EACH X ADD CH FOR EACH N OR P EXAMPLES UN = 1 C 6 H 12 O EVEN H COUNT C 6 H 11 Cl ODD H COUNT C 6 H 13 N ODD H COUNT C 6 H 14 N 2 EVEN H COUNT

16 USES OF MOLECULAR FORMULA OR 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 = 2TWO RINGS, POLYENE, CUMULENE, ALKYNE NITRILE, ISONITRILE, ANHYDRIDE UN = 3NON-BENZENOID AROMATICS UN = 4ARENE, PYRIDINE

17 UN = [(2a+2) – (b-d+e)] 2 C 2 H 3 O 2 Cl = (4+2) = 1 CaHbOcNdXeCaHbOcNdXe Unsaturation number/Double bond equivalents

18 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 C 2 H 3 O 2 Cl ?

20 C 2 H 3 O 2 Cl ACYCLIC H

21 C 2 H 3 O 2 Cl CYCLIC

22 SUMMARY C 2 H 3 O 2 Cl H Cl

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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 (SP 3 )  10 – 100 UNFUNCTIONALIZED 

32 ALIPHATICS: CARBON SHIFTS GENERAL (SP 3 )  10 – 100 UNFUNCTIONALIZED  SHIFTS ARE ADDITIVE

33 ADDITIVE PATTERNS

34 UNDERSTANDING THE TABLE     to -3 4  0 also 

35 SHOW 13 C PREDICT!

36 FUNCTIONAL GROUP ID

37 PROBLEM SOLVING

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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) =  (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 Coupling aa bb J ab

42 FACTORS CONTROLLING J’s 1)# OF INTERVENING BONDS (1,2,3,4, etc) 2)e - DENSITY 3)ANGLE H-C-C H-C-C-H H-C-C-C H-C-C-C-H 4)BOND HYBRIDIZATION 5)ELECTRONEGATIVTY OF ATOMS ALONG PATH

43 A FOCUS ON FIRST ORDER

44 COMPLEX 1 ST ORDER

45 A FOCUS ON COMPLEX FIRST ORDER

46 TERMS 1.BIG  A/X 2.1 ST ORDER AMX 3.NON 1 ST ORDER AB 4.SPECIAL A/A’ 5.COUPLING 1 J CH 6.VALUES IN Hz NUCLEI “21 CASES”

47 NON 1 ST ORDER SPIN SYSTEMS 1.AB #2 2.ABC #3 3.A 2 B #4 4.ABCX #8 5.A 2 B 2 #9 6.AA’XX’ #10

48 P 130

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

50 MASS SPECTROMETRY

51 IONISATION – ELECTRON IMPACT (EI)

52 IONISATION – ELECTROSPRAY (ESI)

53 THE ION ANALYSER – MAGNETIC SECTOR

54 DETERMINING MOLECULAR FORMULAE FROM THE MOLECULAR ION Low resolution mass spectrum: 136 m/z

55 DETERMINING MOLECULAR FORMULAE FROM THE MOLECULAR ION Rule of 13: CH = 13 amu. Most common unit in organic compounds: Therefore (CH) 10 H 6 = C 10 H 16 Use CH 4 = H 16 = O CH 2 = H 14 = N H 12 = C

56 DETERMINING MOLECULAR FORMULAE FROM THE MOLECULAR ION Possible Formulae dbeAccurate mass C 10 H C 9 H 12 O C8H8O2C8H8O C7H4O3C7H4O C 9 H 14 N C 8 H 12 N

57 DETERMINING MOLECULAR FORMULAE FROM THE MOLECULAR ION Experimental accurate mass measurement (from MS) was suggesting C 10 H 16 is the correct formula. The error between calculated and experimental mass is: = = 0.8 mmu

58 Using DEPT 135 in conjunction with MS data to determine the MF Accurate mass (exp) ; (calc)


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