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Organic Structure Analysis Professor Marcel Jaspars.

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Presentation on theme: "Organic Structure Analysis Professor Marcel Jaspars."— Presentation transcript:

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 3 rd 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 1 H and 13 C 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

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9 1 H NMR Chemical Shifts in Organic Compounds

10 13 C NMR Chemical Shifts in Organic Compounds

11 5 Minute Problem #1 MF = C 6 H 12 O Unsaturated acyclic ether

12 Six Simple Steps for Successful Structure Solution Get molecular formula. Use combustion analysis, mass spectrum and/or 13 C NMR spectrum. Calculate double bond equivalents (DBE). Determine functional groups from IR, 1 H and 13 C NMR Compare 1 H 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 DBE = [(2a+2) – (b-d+e)] 2 C 2 H 3 O 2 Cl = CaHbOcNdXeCaHbOcNdXe

14 Tabulate Data Shift (ppm) Int.Mult (J/Hz)Inference 6.481Hdd, 14, Hd, Hd, Ht, Hquint, Hsext, Ht, 7

15 Solution

16 Shift Prediction Prediction

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

18 Dereplication Dediscovery

19 Dereplication

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

22 Determining the Molecular Formula Using NMR and MS Data

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24 MS Errors 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 FormuladbeAccurate mass C 10 H C 9 H 12 O C8H8O2C8H8O C7H4O3C7H4O C 9 H 14 N C 8 H 12 N

25 Molecular Formula Calculators James Deline MFCalc

26 Isotope Ratio Patterns: C 100 H 200 For 12 C m 13 C n

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

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29 [M+H] +

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33 [M+Na] +

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36 Analysis of isotope patterns experimental calculated “monoisotopic peak” (mainly 12 C 7 13 C 1 H 9 35 Cl 14 N 16 O 2 )

37 5 Minute Problem #2 A B 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?

38 Answer Base Values:

39 ChemDraw

40 The NMR effect

41 No coupling Coupling Spin-spin coupling (splitting)

42

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 (CH 2 ). –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 nrelative intensitymultiplet 01singlet 1 1 1doublet triplet quartet quintet sextet septet

49 Coupling patterns

50 First Order In CH 3 CH 2 OH we could explain coupling by the n + 1 rule, this is called 1 st order coupling

51 Second Order Like CH 3 CH 2 OH expect 7 lines but get many more.  /J < 6

52 Common Coupled Spin Systems

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54 Complex 1 st Order Spin Systems

55 Iterative application of the n + 1 rule

56 5 Minute Problem #3. Given the 1 H NMR chemical shifts and coupling constants for allyl alcohol, explain the observed spectrum (OH peak omitted)

57 A doubled quartet (dq)

58 What about this?

59 ddt

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61 Real Spectrum Spin Simulation

62 Pro-R and Pro-S 1

63 Homotopic, Enantiotopic, Diastereotopic

64 Methyl groups

65 Chemical Equivalence/Magnetic Non Equivalence

66 What is going on?

67 Result Expect:

68 Using Coupling Constants

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70 Glucose

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73 5 Minute Problem #4 Work out which of  2.1 and  2.5 is equatorial and which is axial. Also work out the 3 dihedral angles for  2.1,  2.5,  2.8,  6.8. There are also peaks at: 6.80, 1H, d, J = 0.5 Hz; 1.95, 3H, s; 0.93, 9H, s. 30 Hz

74

75 Solution

76 Removing Couplings Changing Solvents CDCl 3 C6D6C6D6 Ha, t Jab = Jac Ha, dd Jab ≠ Jac  a =  b Coupled to Hc Jab = Jac  a ≠  b Each coupled to Hc Jab ≠ Jac

77 Removing Couplings Spin decoupling CDCl 3 irradiated Irradiate at 4.11 ppm Signal due to Ha disappears Coupling due to Ha is removed See Hb, Hc at  b,  c With mutual Jbc

78 Spin Decoupling OFF A↑ X↑ A↑ X↓ A↓ X↑ A↓ X ↓

79 Spin Decoupling Two spins, A ( A ), X ( X ) with J AX Irradiate X with RF power, A loses coupling due to X ON A↑ X↑ ↓ A↓ X ↑ ↓ Average of X ↑ and X ↓

80 Nuclear Overhauser Effect

81 Size of NOE

82 Effect of NOE on 13 C NMR /5/9 8

83 13 C – 1 H NOE at equilibrium (small molecule) C↑H ↑ C↓H↓ C↓H ↑ C↑H ↓ ●●●●● ●●●● ●

84 13 C – 1 H NOE irradiation on H C↑H ↑ C↓H↓ C↓H ↑ C↑H ↓ ●●● ●● ●●● C C H sat ●●

85 13 C – 1 H NOE irradiation on H left ON C↑H ↑ C↓H↓ C↓H ↑ C↑H ↓ ●●● ●● ●●● C C H sat ●●

86 13 C – 1 H NOE result C↑H ↑ C↓H↓ C↑H ↓ ●●●● ● C C H sat ●

87 1 H- 1 H NOE example H3 H1 H4 H2OH2O

88 1 H – 1 H NOE at equilibrium (small molecule) S↑I ↑ S↓I↓ S↓I ↑ S↑I ↓ ●●●● ●● I S

89 1 H – 1 H NOE irradiation on S S↑I ↑ S↓I↓ S↓I ↑ S↑I ↓ ●●● ● W 1S (sat) W 1I I S ●

90 1 H – 1 H NOE irradiation on S left ON S↑I ↑ S↓I↓ S↑I ↓ ●●● ● W 1S (sat) W 1I ●

91 1 H – 1 H NOE result S↑I ↑ S↓I↓ S↑I ↓ ●●●½ ½ W 1S (sat) W 1I ½ I S

92 NOE 3D example

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98 Organic Structure Analysis, Crews, Rodriguez and Jaspars Events Accompanying Resonance

99 Organic Structure Analysis, Crews, Rodriguez and Jaspars ONE-PULSE SEQUENCE

100 Organic Structure Analysis, Crews, Rodriguez and Jaspars ONE-PULSE SEQUENCE 1H1H (90 o ) x PreparationDetection

101 Fourier Transformation FT

102 Relaxation and Peak Shape

103 Rotational Correlation Time  c  o = 2  o

104 Nuclear spin ExampleAtomic mass Atomic number Spin, I 13 C, 1 H, 17 O, 15 N, 3 H OddOdd or Even 1/2, 3/2, 5/2 etc 12 C, 16 OEven 0 2 H, 14 NEvenOdd1, 2, 3 etc

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107 Receptivity NucleusC Relative  ReceptivityRelative receptivity 29 Si4.7% C1.1%6.73 1H1H100%26.75

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109 Multinuclear NMR

110 15 N NMR Shifts

111 31 P NMR Shifts

112 Coupling

113 Effect of 31 P on 1 H NMR

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115 Effect of 31 P on 13 C NMR

116 The 2 nd Dimension

117 Organic Structure Analysis, Crews, Rodriguez and Jaspars BASIC LAYOUT OF A 2D NMR EXPERIMENT

118 Organic Structure Analysis, Crews, Rodriguez and Jaspars How a 2D NMR experiment works n is the number of increments Contour plot

119 TYPES OF 2D NMR EXPERIMENTS AUTOCORRELATED –Homonuclear J resolved – 1 H- 1 H COSY –TOCSY –NOESY –ROESY –INADEQUATE CROSS-CORRELATED –Heteronuclear J resolved – 1 H- 13 C COSY –HMQC –HSQC –HMBC –HSQC-TOCSY Organic Structure Analysis, Crews, Rodriguez and Jaspars

120 STRATEGY BASED ON C-H CONNECTIVITY & &

121 Organic Structure Analysis, Crews, Rodriguez and Jaspars STRATEGY BASED ON C-H CONNECTIVITY

122 Organic Structure Analysis, Crews, Rodriguez and Jaspars STRATEGY BASED ON C-H CONNECTIVITY – DEPT DATA CH CH 2 CH 3

123 Organic Structure Analysis, Crews, Rodriguez and Jaspars STRATEGY BASED ON C-H CONNECTIVITY – HSQC DATA A B C D E F  C a b c d’ d e f

124 Organic Structure Analysis, Crews, Rodriguez and Jaspars STRATEGY BASED ON C-H CONNECTIVITY – HSQC DATA ATOM  C (ppm) DEPT  H (ppm) A131CH5.5 B124CH5.2 C68CH4.0 D42CH E23CH F17CH 3 1.2

125 Organic Structure Analysis, Crews, Rodriguez and Jaspars STRATEGY BASED ON C-H CONNECTIVITY – HSQC DATA

126 Organic Structure Analysis, Crews, Rodriguez and Jaspars STRATEGY BASED ON C-H CONNECTIVITY – HSQC DATA

127 Organic Structure Analysis, Crews, Rodriguez and Jaspars STRATEGY BASED ON C-H CONNECTIVITY – HSQC DATA

128 Organic Structure Analysis, Crews, Rodriguez and Jaspars STRATEGY BASED ON C-H CONNECTIVITY – HSQC DATA diastereotopic protons

129 Organic Structure Analysis, Crews, Rodriguez and Jaspars STRATEGY BASED ON C-H CONNECTIVITY – COSY DATA

130 Organic Structure Analysis, Crews, Rodriguez and Jaspars STRATEGY BASED ON C-H CONNECTIVITY – COSY DATA ATOM  C (ppm) DEPT  H (ppm)COSY (H  H) A131CH5.5b, c, d/d’, f B124CH5.2a, d/d’, f C68CH4.0a, d/d’, e D42CH a, b, c, d, e E23CH 3 1.5c, d/d’ F17CH 3 1.2a, b

131 Organic Structure Analysis, Crews, Rodriguez and Jaspars STRATEGY BASED ON C-H CONNECTIVITY – COSY DATA

132 Organic Structure Analysis, Crews, Rodriguez and Jaspars STRATEGY BASED ON C-H CONNECTIVITY – COSY DATA

133 Organic Structure Analysis, Crews, Rodriguez and Jaspars STRATEGY BASED ON C-H CONNECTIVITY – COSY DATA HSQC suggests diastereotopic protons: 3.08/2.44 ppm 1.86/2.07 ppm

134 Organic Structure Analysis, Crews, Rodriguez and Jaspars STRATEGY BASED ON C-H CONNECTIVITY – HMBC DATA And many more…

135 Organic Structure Analysis, Crews, Rodriguez and Jaspars STRATEGY BASED ON C-H CONNECTIVITY – HMBC DATA ATOM  C (ppm) DEPT  H (ppm)COSY (H  H)HMBC (C  H) A131CH5.5b, c, d/d’, fb, c, d, f B124CH5.2a, d/d’, fa, d, f C68CH4.0a, d/d’, ea, d, e D42CH a, b, c, d, ea, b, c, e E23CH 3 1.5c, d/d’c, d F17CH 3 1.2a, b

136 Organic Structure Analysis, Crews, Rodriguez and Jaspars STRATEGY BASED ON C-H CONNECTIVITY – HMBC DATA

137 Organic Structure Analysis, Crews, Rodriguez and Jaspars STRATEGY BASED ON C-H CONNECTIVITY – HMBC DATA

138 Organic Structure Analysis, Crews, Rodriguez and Jaspars STRATEGY BASED ON C-H CONNECTIVITY – HMBC DATA

139 Organic Structure Analysis, Crews, Rodriguez and Jaspars STRATEGY BASED ON C-H CONNECTIVITY RETROSPECTIVE CHECKING Pieces: Possibilities: Combinatorial explosion

140 Organic Structure Analysis, Crews, Rodriguez and Jaspars STRATEGY BASED ON C-H CONNECTIVITY RETROSPECTIVE CHECKING And similarly for COSY data

141 Organic Structure Analysis, Crews, Rodriguez and Jaspars PROSPECTIVE CHECKING Pieces:

142 Organic Structure Analysis, Crews, Rodriguez and Jaspars 2D EXERCISE 1. For a simple organic compound the mass spectrum shows a molecular ion at m/z 98. The following data has been obtained from various 1D and 2D NMR experiments. Using this information determine the structure of the molecule in question and rationalise the 2D NMR data given. Atomd C (ppm)d H (ppm) 1 H - 1 H COSY (3 bond only) 1 H  13 C Long range (2 - 3 bonds) A218 s--A-b, A-c, A-d, A-e B47 t1.8 ddb-dB-c, B-d, B-e, B-f C38 t2.3 mc-eC-b, C-d, C-e D32 d1.5 md-b, d-e, d-fD-b, D-c, D-e, D-f E31 t2.2 me-c, e-dE-b, E-c, E-d, E-f F20 q1.1 df-dF-b, F-d, F-e

143 Organic Structure Analysis, Crews, Rodriguez and Jaspars 2D EXERCISE 2. For a simple organic compound the mass spectrum shows a molecular ion at m/z 114. The following data has been obtained from various 1D and 2D NMR experiments. Using this information determine the structure of the molecule in question and rationalise the 2D NMR data given. An additional peak is present in the 1 H NMR at 11.6 ppm (bs). Atomd C (ppm)d H (ppm) 1 H - 1 H COSY (3 bond only) 1 H  13 C Long range (2 - 3 bonds) A178 s--A-d, A-b B136 d5.7 mb-c, b-dB-d, B-c, B-e C118 d5.5 mc-b, c-eC-b, C-d, C-e, C-f D38 t3.0 dd-bD-b, D-c E25 t2.1 me-c, e-fE-b, E-c, E-f F13 q1.0 tf-eF-c, F-e


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