Organic Structure Analysis
Professor Marcel Jaspars
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.
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.
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.
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
Four Types of Information from
NMR
1H
NMR Chemical Shifts in Organic Compounds
13C
NMR Chemical Shifts in Organic Compounds
5 Minute Problem #1
MF = C6H12O
Unsaturated acyclic ether
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.
Double Bond Equivalents
CaHbOcNdXe
[(2a+2) – (b-d+e)]
DBE =
C2H3O2Cl =
2
Tabulate Data
Shift
(ppm)
Int.
Mult (J/Hz)
6.48
1H
dd, 14, 7
4.17
1H
d, 14
3.97
1H
d, 7
3.69
2H
t, 7
1.65
2H
quint, 7
1.42
2H
sext, 7
0.95
3H
t, 7
Inference
Solution
Shift Prediction
Prediction
THE PROCESS OF STRUCTURE ELUCIDATION
Dereplicate by MF
M olecular
formula
Unsaturation
Number
(UN)
MS, NMR
NMR, IR
Pure
compound
Working
2D
structures
Functional
groups
Draw all isom ers
UV
NMR
List of
Dereplicate
working
by structure
2D structures
Substructures
X-RAY
NMR, MS, IR, UV
Very secure
Reasonable
Total
3D molecular
3D molecular
synthesis
structure
structure
Organic Structure Analysis, Crews, Rodriguez and Jaspars
NMR
ORD
Molecular
m odeling
New 2D
molecular
structure
Known
molecular
structure
Dereplication
Dediscovery
13
C spectrum/DEPT-135
APT Formula
C20H29
Low Resolution MS
High Resolution MS
m/z 335 [M+H]+
m/z 335.2222 [M+H]+
Molecular Formula C20H30O4
DATABASE
Substructures
Crystal Unit Cell Dimensions 1D NMR Spectra
2D NMR Spectra
Dereplication
13
C NMR
OH
H
O
1H
Me
NMR
Me
H
OH
H
Dereplication
Determining the Molecular Formula Using NMR and MS Data
DEPT-135
CH, CH3
C
CH2
C
Organic Structure Analysis, Crews, Rodriguez and Jaspars
Determining the Molecular Formula Using NMR and MS Data
Determining the Molecular Formula Using NMR and MS Data
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
Formula
dbe
Accurate
mass
C10H16
3
136.1248
C9H12O
4
136.0885
C8H8O2
5
136.0522
C7H4O3
6
136.0159
C9H14N
3.5
136.1123
C8H12N2
4
136.0998
1361256
.
 1361248
.
 5.9  10 6  5.9 ppm
1361256
.
Molecular Formula Calculators
James Deline MFCalc
Isotope Ratio Patterns:
C100H200
For 12Cm13Cn
1403.6
1404.6
Determining molecular formulae by HR-ESI/MS
ThermoFinnigan LTQ Orbitrap,
Xcalibur software
examples from MChem group practicals 2009
(Rainer Ebel)
[M+H]+
[M+Na]+
Analysis of isotope patterns
experimental
calculated
“monoisotopic peak”
(mainly 12C713C1H935Cl14N16O2)
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?
MeO
N
H
A
MeO
N
H
B
Answer
Base Values:
ChemDraw
The NMR effect
Bo off
Bo on
I = -1/2
(, antiparallel)
E I = 1/2
I = -1/2
(, parallel)
Spin-spin coupling (splitting)
Ha
Hb
Hb
No coupling
Ha
Coupling
Spin-spin coupling (splitting)
Origin of spin-spin coupling
Ha
Hb
Ha
Hb
C
C
C
C
Bo
parallel
antiparallel
Coupling in ethanol
H
H
HO
H
H
H
Hb
Hb
HO
Hb
Ha
Ha
Coupling is mutual
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
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.
Pascal’s triangle
n
0
1
2
3
4
5
6
relative intensity
multiplet
1
singlet
1
1
doublet
1
2
1
triplet
1
3
3
1
quartet
1
4
6
4
1
quintet
1
5
10 10
5
1
sextet
1
6
15
20
15
6
1
septet
Coupling patterns
First Order
• In CH3CH2OH we could explain coupling
by the n + 1 rule, this is called 1st order
coupling
Second Order
• Like CH3CH2OH expect 7 lines but get
many more. Dn/J < 6
Common Coupled Spin Systems
Common Coupled Spin Systems
Complex 1st Order Spin Systems
Iterative application of the n + 1 rule
5 Minute Problem #3. Given the 1H NMR chemical
shifts and coupling constants for allyl alcohol, explain
the observed spectrum (OH peak omitted)
A doubled quartet (dq)
What about this?
ddt
Spin Simulation
4.0
3.5
3.0
2.5
2.0
1.5
3.0
2.5
2.0
1.5
1.0
ppm
Real Spectrum
4.0
3.5
ppm
Pro-R and Pro-S
1
Homotopic, Enantiotopic,
Diastereotopic
Methyl groups
Chemical Equivalence/Magnetic
Non Equivalence
What is going on?
Result
Expect:
Using Coupling Constants
Glucose
Glucose
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
Solution
Removing Couplings
Changing Solvents
CDCl3
Ha, dd Jab ≠ Jac
a ≠ b
Each coupled to Hc
Jab ≠ Jac
C6D6
Ha, t Jab = Jac
5.5
5.0
4.5
4.0
3.5
a = b
Coupled to Hc
Jab = Jac
3.0
2.5
2.0
1.5
ppm
Removing Couplings
Spin decoupling
CDCl3
irradiated
Coupling due to Ha
is removed
See Hb, Hc at b, c
With mutual Jbc
Signal due to Ha
disappears
CDCl3
Irradiate at 4.11 ppm
Spin Decoupling
OFF
A↓ X ↓
A↓ X↑
A↑ X↓
A↑ X↑
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 ↓
Nuclear Overhauser Effect
Size of NOE
0.5
0
h
-0.5
-1
0.01
0.1
fast tumbling
1
w 0t c
10
100
slow tumbling
Effect of NOE on 13C NMR
3
2
1
6
10/5/9
8
4
7
13C
– 1H NOE at equilibrium (small molecule)
C↓H↓
C↑H ↓
●
●●●●
●●●●●
C↑H ↑
C↓H ↑
13C
– 1H NOE irradiation on H
●●
C↓H↓
C
C↑H ↓
●●●
Hsat
●●
Hsat
●●●
C
C↑H ↑
C↓H ↑
13C
– 1H NOE irradiation on H left ON
●●
C↓H↓
C
C↑H ↓
●●●
Hsat
●●
Hsat
●●●
C
C↑H ↑
C↓H ↑
13C
– 1H NOE result
●
C↓H↓
C
C↑H ↓
●●●●
Hsat
●
Hsat
●●●●
C
C↑H ↑
1H-1H
NOE example
H1
H3
H4
H2O
1H
– 1H NOE at equilibrium (small molecule)
S↓I↓
S↑I ↓
●●
●●
●●●●
S↓I ↑
S↑I ↑
nI
nS
1H
– 1H NOE irradiation on S
●
W1S (sat)
S↑I ↓
S↓I↓
W1I
●●●
●
S↓I ↑
W1S (sat)
W1I
●●●
S↑I ↑
nI
nS
1H
– 1H NOE irradiation on S left ON
●
W1S (sat)
S↑I ↓
S↓I↓
W1I
●●●
●
W1S (sat)
W1I
●●●
S↑I ↑
1H
– 1H NOE result
½
W1S (sat)
S↑I ↓
S↓I↓
W1I
●●●½
½
W1S (sat)
W1I
●●●½
S↑I ↑
nI
nS
NOE 3D example
NOE 3D example
Events Accompanying Resonance
Mo
(b) Bo on; prior to resonance
(a) No Bo
Bo
y
Mxy = 0
x
Random
orientation of
magnetic dipoles
M
(c) At resonance no = n1
z
y
x
Organic Structure Analysis, Crews, Rodriguez and Jaspars
The magnetic vectors
precess in phase with
frequency n1.
After resonance the return
to the equilibrium in (b)
occurs by the loss of Mxy via
dephasing of nuclear
dipoles by T2 and increase
in Mz by spin inversion
due to T1.
Net polarization Mz is due to
population excess in higher
energy state
The magnetic vectors
precess about Bo at
the Larmor frequency no
ONE-PULSE SEQUENCE
Bo
Bo
z
z
Mo
(90o)x
M
FT
y
y
x
x
time t2
After 90o pulse
Excess of spin
population along
the direction of
applied magnetic
field.
magnetization
is tipped into
the xy plane.
preparation
Organic Structure Analysis, Crews, Rodriguez and Jaspars
M=Magnetization which
produces the FID. It decays
as magnetization in xy
plane diminishes after
resonance
detection
frequency f2
ONE-PULSE SEQUENCE
(90o)x
1H
Preparation
Detection
Organic Structure Analysis, Crews, Rodriguez and Jaspars
Fourier Transformation
FT
Relaxation and Peak Shape
Rotational Correlation Time tc
wo = 2pno
Nuclear spin
Example
Atomic
mass
Atomic
number
13C, 1H, 17O, 15N, 3H
Odd
Odd or
Even
12C, 16O
Even
Even
2H, 14N
Even
Odd
6
1
8
6
1
7
8
7
1
Spin, I
Receptivity
Nucleus
C
Relative g
29Si
4.7%
-5.32
13C
1.1%
6.73
1H
100%
26.75
Receptivity
Relative
receptivity
Multinuclear NMR
15N
NMR Shifts
31P
NMR Shifts
Coupling
303
Effect of 31P on 1H NMR
6.18
4.00
0.54
0.51
8
7
6
5
4
3
2
ppm
Effect of 31P on 1H NMR
4.30
4.25
4.20
4.15
4.10
4.05
ppm
Effect of 31P on 13C NMR
5
4
3
1
60
50
40
30
20
ppm
The 2nd Dimension
BASIC LAYOUT OF A 2D NMR EXPERIMENT
Organic Structure Analysis, Crews, Rodriguez and Jaspars
How a 2D NMR experiment works
t2
Dt1
2Dt1
3Dt1
nDt1
t1
f2
FT
FT
FT
FT
transform
matrix
FT
FT
FT
FT
f2
n is the number of increments
Organic Structure Analysis, Crews, Rodriguez and Jaspars
f1
t1
f2
Contour plot
TYPES OF 2D NMR EXPERIMENTS
• AUTOCORRELATED
–
–
–
–
–
–
Homonuclear J resolved
1H-1H COSY
TOCSY
NOESY
ROESY
INADEQUATE
Organic Structure Analysis, Crews, Rodriguez and Jaspars
• CROSS-CORRELATED
–
–
–
–
–
–
Heteronuclear J resolved
1H-13C COSY
HMQC
HSQC
HMBC
HSQC-TOCSY
STRATEGY BASED ON C-H CONNECTIVITY
1
H
H
HSQC
HSQC
C
H-1H COSY
H
H
C
C
&
C
C
C
C
C
HMBC
H
H
HSQC
&
C
C
C
C
Organic Structure Analysis, Crews, Rodriguez and Jaspars
C
C
C
STRATEGY BASED ON C-H CONNECTIVITY
Get 1H, 13C NMR spectra
get multiplicities and integrals
Get 13C-1H correlation spectrum (eg HSQC)
Get 1H-1H correlation data (eg COSY)
Check assignment of diastereotopic protons
using COSY and HMQC
Combine substructures into
all possible working structures
Check all working structures for consistency
with 2D NMR data
Organic Structure Analysis, Crews, Rodriguez and Jaspars
Get one bond 13C-1H correlations
assign 1H resonances to 13C resonances
Assemble substructures using COSY data
Get long range 13C-1H correlation spectrum
(eg HMBC)
2D structure
STRATEGY BASED ON C-H CONNECTIVITY – DEPT DATA
CH3 CH3
CH CH
CH, CH3
C
CH2
CH
CH2
C
Organic Structure Analysis, Crews, Rodriguez and Jaspars
STRATEGY BASED ON C-H CONNECTIVITY – HSQC DATA
A B
C
f
e
d’
d
c
b
a
Organic Structure Analysis, Crews, Rodriguez and Jaspars
D
E
F
C
STRATEGY BASED ON C-H CONNECTIVITY – HSQC DATA
ATOM
C (ppm) DEPT
H (ppm)
A
131
CH
5.5
B
124
CH
5.2
C
68
CH
4.0
D
42
CH2
E
23
CH3
3.0
2.5
1.5
F
17
CH3
1.2
Organic Structure Analysis, Crews, Rodriguez and Jaspars
STRATEGY BASED ON C-H CONNECTIVITY – HSQC DATA
Organic Structure Analysis, Crews, Rodriguez and Jaspars
STRATEGY BASED ON C-H CONNECTIVITY – HSQC DATA
Organic Structure Analysis, Crews, Rodriguez and Jaspars
STRATEGY BASED ON C-H CONNECTIVITY – HSQC DATA
Organic Structure Analysis, Crews, Rodriguez and Jaspars
STRATEGY BASED ON C-H CONNECTIVITY – HSQC DATA
diastereotopic protons
Organic Structure Analysis, Crews, Rodriguez and Jaspars
STRATEGY BASED ON C-H CONNECTIVITY – COSY DATA
a
b
f
e
d'
d
c
b
a
Organic Structure Analysis, Crews, Rodriguez and Jaspars
c
d
d'
e
f
STRATEGY BASED ON C-H CONNECTIVITY – COSY DATA
ATOM
C (ppm)
DEPT
H (ppm)
COSY (HH)
A
131
CH
5.5
b, c, d/d’, f
B
124
CH
5.2
a, d/d’, f
C
68
CH
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
CH3
1.2
a, b
Organic Structure Analysis, Crews, Rodriguez and Jaspars
STRATEGY BASED ON C-H CONNECTIVITY – COSY DATA
Organic Structure Analysis, Crews, Rodriguez and Jaspars
STRATEGY BASED ON C-H CONNECTIVITY – COSY DATA
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
Organic Structure Analysis, Crews, Rodriguez and Jaspars
STRATEGY BASED ON C-H CONNECTIVITY – HMBC DATA
A
B
C
D
E
F
C
f
e
d'
d
c
b
a
H
And many more…
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)
A
131
CH
5.5
b, c, d/d’, f
b, c, d, f
B
124
CH
5.2
a, d/d’, f
a, d, f
C
68
CH
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
CH3
1.2
a, b
a, b
Organic Structure Analysis, Crews, Rodriguez and Jaspars
STRATEGY BASED ON C-H CONNECTIVITY – HMBC DATA
Organic Structure Analysis, Crews, Rodriguez and Jaspars
STRATEGY BASED ON C-H CONNECTIVITY – HMBC DATA
Organic Structure Analysis, Crews, Rodriguez and Jaspars
STRATEGY BASED ON C-H CONNECTIVITY – HMBC DATA
Organic Structure Analysis, Crews, Rodriguez and Jaspars
STRATEGY BASED ON C-H CONNECTIVITY
RETROSPECTIVE CHECKING
Pieces:
H H
A C
H H
5.5
B C
131
Possibilities:
OH
A
F
D
F
E
D
B
A
5.2
OH
H 4.0
H
124
C
C 68
A
B
F
E
C
D
B
D
H 2.5
D
F
A
OH
B
F
A
E
E
C
D
A
H
H
H
E
C 42
C
B
D
E
B
A
F
C 23
F
B
E
E
C
C
A
E
OH
Organic Structure Analysis, Crews, Rodriguez and Jaspars
A
E
D
B
D
E
OH
F
A
B
E
C
OH
F
C
C 17
D
OH
C
H
OH
D
OH
E
C
E
D
B
H 1.2
H
OH
C
A
H 1.5
H
H
Combinatorial
explosion
OH
F
H
OH
C
B
H 3.0
B
E
D
A
F
C
OH
STRATEGY BASED ON C-H CONNECTIVITY
RETROSPECTIVE CHECKING
Hb
B
MeF
Hd'
Hd
D
A
Ha
MeE
C
Hc
OH
And similarly for COSY data
Organic Structure Analysis, Crews, Rodriguez and Jaspars
PROSPECTIVE CHECKING
H H
Pieces:
A C
H H
5.5
B C
131
A
5.2
H 3.0
OH
H
H 4.0
H
124
C
C 68
D
H 2.5
H 1.5
H 1.2
H
H
H
H
H
H
E
C 42
F
C 23
B
Reason: only 2 sp2 C's
F
A
B
HMBC: F-a, F-b, A-f, B-f
F
A
HMBC: D-a, D-b, A-d, B-d
B
D
F
OH
A
C
B
D
HMBC: C-d, C-a, D-c, A-c
F
OH
A
C
HMBC: E-c, E-d, C-e, C-d
Organic Structure Analysis, Crews, Rodriguez and Jaspars
E
D
H
B
C 17
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.
 13C Long range
(2 - 3 bonds)
Atom
dC (ppm)
dH (ppm)
1H
- 1H COSY
(3 bond only)
1H
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
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).
 13C Long range
(2 - 3 bonds)
Atom
dC (ppm)
dH (ppm)
1H
- 1H COSY
(3 bond only)
1H
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
38
t
3.0 d
d-b
D-b, D-c
E
25
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