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 Shift (ppm) Int. Mult (J/Hz) Inference 6.48 1H dd, 14, 7
4.17
d, 14
3.97
d, 7
3.69 2H
t, 7
1.65
quint, 7
1.42
sext, 7
0.95 3H

15. Solution

16. Shift Prediction
Prediction

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

18. Dereplication Dediscovery

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 suggesting C10H16 is the correct formula.
The error between calculated and experimental mass is:
= = 0.8 mmu
Formula
dbe
Accurate mass
C10H16 3
C9H12O 4
C8H8O2 5
C7H4O3 6
C9H14N
3.5
C8H12N2

25. Molecular Formula Calculators
James Deline MFCalc

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

27. ThermoFinnigan LTQ Orbitrap,
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
doublet
triplet
quartet
quintet
sextet
septet

49. Coupling patterns

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

51. Second Order
Like CH3CH2OH expect 7 lines but get many more. Dn/J < 6

52. Common Coupled Spin Systems

53. Common Coupled Spin Systems

54. Complex 1st Order Spin Systems

55. Iterative application of the n + 1 rule

56. 5 Minute Problem #3. Given the 1H 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

61. Spin Simulation
Real Spectrum

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

70. Glucose

72. Glucose

73. 5 Minute Problem #4
Work out which of d 2.1 and d 2.5 is equatorial and which is axial. Also work out the 3 dihedral angles for d 2.1, d 2.5, d 2.8, d 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

75. Solution

76. Removing Couplings Changing Solvents
da ≠ db
Each coupled to Hc
Jab ≠ Jac
CDCl3
Ha, dd Jab ≠ Jac
C6D6
da = db
Coupled to Hc
Jab = Jac
Ha, t Jab = Jac

77. Removing Couplings Spin decoupling
Coupling due to Ha
is removed
See Hb, Hc at db, dc
With mutual Jbc
CDCl3
irradiated
Signal due to Ha
disappears
CDCl3
Irradiate at 4.11 ppm

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

79. Spin Decoupling Two spins, A (nA), X (nX) with JAX
Irradiate nX 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 13C NMR
10/5/9 3 8 2 6 4 1 7

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

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

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

86. 13C – 1H NOE result ●
C↓H↓ C
●●●●
C↑H ↓
Hsat
Hsat ● C
●●●●
C↑H ↑

87. 1H-1H NOE example H1
H2O H4 H3

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

89. 1H – 1H NOE irradiation on S●
S↓I↓
W1S (sat)
W1I
●●● ●
S↑I ↓
S↓I ↑
W1I
W1S (sat)
●●●
S↑I ↑
nI nS

90. 1H – 1H NOE irradiation on S left ON●
S↓I↓
W1S (sat)
W1I
●●● ●
S↑I ↓
W1I
W1S (sat)
●●●
S↑I ↑

91. 1H – 1H NOE result ½ S↓I↓ W1S (sat) W1I ●●●½ ½ S↑I ↓ W1I W1S (sat)
nI nS

92. NOE 3D example

93. NOE 3D example

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

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

100. ONE-PULSE SEQUENCE (90o)x 1H Preparation Detection
Organic Structure Analysis, Crews, Rodriguez and Jaspars

101. Fourier TransformationFT

102. Relaxation and Peak Shape

103. Rotational Correlation Time tc
wo = 2pno

104. Nuclear spin Example Atomic mass Atomic number Spin, I
13C, 1H, 17O, 15N, 3H
Odd
Odd or Even
1/2, 3/2, 5/2 etc
12C, 16O
Even
2H, 14N
1, 2, 3 etc 6 1 8 7 1 6 8 1 7

107. Receptivity 29Si 4.7% -5.32 13C 1.1% 6.73 1H 100% 26.75 Nucleus C
Relative g
Receptivity
Relative receptivity
29Si
4.7%
-5.32
13C
1.1%
6.73 1H
100%
26.75

109. Multinuclear NMR

110. 15N NMR Shifts

111. 31P NMR Shifts

112. Coupling

113. Effect of 31P on 1H NMR

114. Effect of 31P on 1H NMR

115. Effect of 31P on 13C NMR 5 4 3 1

116. The 2nd Dimension

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

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

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

120. STRATEGY BASED ON C-H CONNECTIVITY
&
&
Organic Structure Analysis, Crews, Rodriguez and Jaspars

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

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

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

124. STRATEGY BASED ON C-H CONNECTIVITY – HSQC DATA
ATOM
dC (ppm)
DEPT
dH (ppm) A
131 CH
5.5 B
124
5.2 C 68
4.0 D 42
CH2
3.0
2.5 E 23
CH3
1.5 F 17
1.2
Organic Structure Analysis, Crews, Rodriguez and Jaspars

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

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

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

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

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

130. STRATEGY BASED ON C-H CONNECTIVITY – COSY DATA
ATOM
dC (ppm)
DEPT
dH (ppm)
COSY (HH) A
131 CH
5.5
b, c, d/d’, f B
124
5.2
a, d/d’, f C 68
4.0
a, d/d’, e D 42
CH2
3.0
2.5
a, b, c, d, e E 23
CH3
1.5
c, d/d’ F 17
1.2
a, b
Organic Structure Analysis, Crews, Rodriguez and Jaspars

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

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

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

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

135. STRATEGY BASED ON C-H CONNECTIVITY – HMBC DATA
ATOM
dC (ppm)
DEPT
dH (ppm)
COSY (HH)
HMBC (CH) A
131 CH
5.5
b, c, d/d’, f
b, c, d, f B
124
5.2
a, d/d’, f
a, d, f C 68
4.0
a, d/d’, e
a, d, e D 42
CH2
3.0
2.5
a, b, c, d, e
a, b, c, e E 23
CH3
1.5
c, d/d’
c, d F 17
1.2
a, b
Organic Structure Analysis, Crews, Rodriguez and Jaspars

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

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

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

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

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

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

142. 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.
Atom
dC (ppm)
dH (ppm)
1H - 1H COSY
(3 bond only)
1H  13C Long range
(2 - 3 bonds) A
218 s -
A-b, A-c, A-d, A-e B
47 t
1.8 dd
b-d
B-c, B-d, B-e, B-f C
38 t
2.3 m
c-e
C-b, C-d, C-e D
32 d
1.5 m
d-b, d-e, d-f
D-b, D-c, D-e, D-f E
31 t
2.2 m
e-c, e-d
E-b, E-c, E-d, E-f F
20 q
1.1 d
f-d
F-b, F-d, F-e
Organic Structure Analysis, Crews, Rodriguez and Jaspars

143. An additional peak is present in the 1H NMR at 11.6 ppm (bs). Atom
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 1H NMR at 11.6 ppm (bs).
Atom
dC (ppm)
dH (ppm)
1H - 1H COSY
(3 bond only)
1H  13C Long range
(2 - 3 bonds) A
178 s -
A-d, A-b B
136 d
5.7 m
b-c, b-d
B-d, B-c, B-e C
118 d
5.5 m
c-b, c-e
C-b, C-d, C-e, C-f D t
3.0 d
d-b
D-b, D-c E t
2.1 m
e-c, e-f
E-b, E-c, E-f F
13 q
1.0 t
f-e
F-c, F-e
Organic Structure Analysis, Crews, Rodriguez and Jaspars