organic structure analysis professor marcel jaspars
TRANSCRIPT
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
N
HN
HN
NH
NH
O
OO
OO
HN
O
HN
O OH
SMe
HO
1.62.43.24.04.85.66.47.2f1 (ppm)
6.9
8
0.9
21.
14
1.1
22.
03
1.1
3
3.8
33.
89
2.8
5
3.9
1
3.8
7
1.8
71.
08
0.9
9
0.9
3
0.8
90.
95
1.9
8
2.2
9
1.9
6
2.0
0
10.
00
102030405060708090100110120130140150160170
f1 (ppm)
10.
01
13.
73
18.
37
20.
22
28.
07
30.
01
37.
59
38.
70
47.
18
47.
60
52.
31
59.
30
59.
40
60.
11
99.
99
114.
95
125.
98
128.
07
128.
35
129.
02
136.
88
140.
86
155.
42
170.
56
172.
97
173.
40
173.
48
N
HN
HN
NH
NH
O
OO
OO
HN
O
HN
O OH
SMe
HO
1H NMR Chemical Shifts in Organic Compounds 12 11 10 9 8 7 6 5 4 3 2 01 -1
12 11 10 9 8 7 6 5 4 23 1 -1
RHS
R NH2
R OH
NH
H
RCONH2
CHC
ZH
H
H2C
O
HRC=CR2
H2C=CR2
R-CH2-R
M-CH3
CH3-O-
CH3-N
CH3-S-
CH3-Ph
-CO-CH2-
CH3-CO-
CH3-C-X
CH3-C=C-
Ph-CH2-
-CH2-C=C-
HO2C-
HCO-
OH
RCONHR
Z
Z
R
N H
Alcohols, Halogens
Amides
Phenol -OH
Alcohols-OHThiols-SHAmines-NH2
Methylenes (-CH2-)
Cyclopropyl
Metals (M-CH3)
ppm
ppm
Carboxylic acids-OH
AldehydesHeteroaromatics
AromaticsAlkenes
-HC-O- ó -HC-X
-CH2-O- ó -
CH2-X
Alkynes
Methyl groups (R-Me)
0
Z=O, N, S
Z=O, N, S
Amines R-NH2 -CH-N- ó -CH2-N
Relative to TMS (0 ppm)
13C NMR Chemical Shifts in Organic Compounds
220 210 200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0
KetoneAldehydeAcidEster, AmideThioketoneAzomethine
CN Nitr ile
Heteroaromatic
AlkeneAromatic
CC Alkyne
C Quat.
Halogen(C Tertiary)
Halogen(C Secondary)
Halogen(C Primary)
Halogen
230
220 210 200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0
*Relative to TMS (0 ppm)
ppm
ppm
Cyclopropanes
230
sp2
sp
sp3
4o
3o
2o
1o
Epoxides
4o Quaternary, 3o Tertiary, 2o Secondary, 1o Primary
C=OC=OC=OC=OC=SC=N-
C=N-
C=CC=CC=C
C-O-
CH-N-
C-N-
CH2-S-
C-S
CH2-X
C-XCH-CCH-O-
CH3-C
CH-S-
CH3-O-
CH-XCH2-CCH2-O-
CH3-N-
CH2-N-
CH3-S-CH3-X
Heteroaromatic
1.01.41.82.22.63.03.43.84.24.65.05.45.86.26.6
3.8
4
2.4
7
2.5
0
2.5
2
1.1
7
1.1
8
1.0
0
46
6.4
94
73.
91
48
1.2
9
69
1.1
86
98.
52
70
5.9
17
13.
43
72
1.0
38
10.
15
81
6.7
18
24.
70
83
1.6
18
38.
41
18
34.6
11
841.1
71
847.7
2
19
79.3
41
986.1
4
20
76.8
52
091.1
9
32
22.8
23
229.6
23
237.1
63
243.9
6
5 Minute Problem #1MF = C6H12OUnsaturated 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
DBE = [(2a+2) – (b-d+e)]
2
C2H3O2Cl =
CaHbOcNdXe
Tabulate Data
Shift (ppm)
Int. Mult (J/Hz) Inference
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
Solution
Shift Prediction
01234567PPM
Prediction
Organic Structure Analysis, Crews, Rodriguez and Jaspars
Molecularformula
Functionalgroups
Substructures
Very secure3D molecular
structure
MS, NMR
NMR, IR
UV
NMR
X-RAY
UnsaturationNumber(UN)
Working 2D
structures
List of working
2D structures
Reasonable3D molecular
structure
New 2Dmolecularstructure
NMR
ORD Totalsynthesis Molecular
modeling
Knownmolecularstructure
Dereplicate by MF
Dereplicate
Draw all isomers
NMR, MS, IR, UV
by structure
Purecompound
THE PROCESS OF STRUCTURE ELUCIDATION
DereplicationDediscovery
13C spectrum/DEPT-135
Low Resolution MS
m/z 335 [M+H]+
APT Formula C20H29
High Resolution MS
m/z 335.2222 [M+H]+
Molecular Formula C20H30O4
Crystal Unit Cell Dimensions
DATABASE
Substructures
1D NMR Spectra 2D NMR Spectra
Dereplication
1H NMR
Me
HOH
H
Me
200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10ppm
6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 ppm
13C NMR
O
OHH
Dereplication
OH
O
H
H H
TestosteroneC19H28O2
m/z 288.2089
Accurate massm/ z 288.2080 ( 3 ppm)
Molecular formulaC19H28O2
DEPT-135 FormulaC19H27
Low Resolution MSm/z 288
Su
bst
ruct
ure
s
2D NMR
1H NMR
13C NMR
Me
HOH
H
O
Me
OHH
Other Search ParametersUVmax
Crystal unit cell dimensionsMS isotope cluster
Taxonomy
Organic Structure Analysis, Crews, Rodriguez and Jaspars
Determining the Molecular Formula Using NMR and MS Data
CCH2
CH, CH3
C
DEPT-135
Determining the Molecular Formula Using NMR and MS Data
140 130 120 110 100 90 80 70 60 50 40 30 20 10ppm
DEPT-135: CH, CH3, CH2
13C NMR
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
13612565 9 10 5 96. .
.. .
ppm
Molecular Formula Calculators
James Deline MFCalc
Isotope Ratio Patterns:C100H200
For 12Cm13Cn
1403.61404.6
Determining molecular formulae by HR-ESI/MS
ThermoFinnigan LTQ Orbitrap,
Xcalibur software
examples from MChem group practicals 2009(Rainer Ebel)
O
10
HN
N-(2-hydroxy-3-methylphenyl)acetamide
OH
Chemical Formula: C9H11NO2Exact Mass: 165.0790
[M+H]+
O
H2N
OH
Chemical Formula: C9H12NO2+
Exact Mass: 166.0863
O
10
HN
N-(2-hydroxy-3-methylphenyl)acetamide
OH
Chemical Formula: C9H11NO2Exact Mass: 165.0790
[M+Na]+
O
HN
OH
Na
Chemical Formula: C9H11NNaO22+
Exact Mass: 188.0677
9
HN
2-chloro-N-(4-hydroxyphenyl)acetamide
HOO
Cl
Chemical Formula: C8H8ClNO2Exact Mass: 185.0244
Analysis of isotope patterns
H2N
HOO
Cl
Chemical Formula: C8H9ClNO2+
Exact Mass: 186.0316m/z: 186.0322 (100.0%), 188.0292 (32.0%), 187.0355 (8.7%), 189.0326 (2.8%)
experimental
calculated
“monoisotopic peak”(mainly 12C7
13C1H935Cl14N16O2)
5 Minute Problem #2
NH
MeO
NH
MeO
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?
Answer
Base Values:
ChemDraw
134.6
109.9
23.9
23.4
23.820.9
NH
133
110
23
22
21
20
134.6
109.9
134.6
109.9
23.9
23.4
23.820.9
23.9
23.4
23.820.9
55.8
55.8
NH
NH
O
O
The NMR effect
I = 1/2E
Bo off Bo on
I = -1/2(, antiparallel)
I = -1/2(, parallel)
Ha Hb
Ha
HbNo coupling
Coupling
Spin-spin coupling (splitting)
Spin-spin coupling (splitting)
Origin of spin-spin coupling
Ha
C C
Hb Ha
C C
Hb
Bo
parallel antiparallel
Coupling in ethanol
H
HO
H
H
H
H
Coupling is mutualHb
HO
Hb
Hb
Ha
Ha
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 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
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. /J < 6
Common Coupled Spin Systems
Common Coupled Spin Systems
Complex 1st Order Spin Systems
Iterative application of the n + 1 rule
4.04.14.24.34.44.54.64.74.84.95.05.15.25.35.45.55.65.75.85.9
5 Minute Problem #3. Given the 1H NMR chemical shifts and coupling constants for allyl alcohol, explain
the observed spectrum (OH peak omitted)
H
H
H
OH
5.05
5.15 4.00
5.85H
H
H
OH
10.4 Hz
1.8 Hz
5.2 Hz1.8 Hz
1.8 Hz
17 Hz
5.1155.1205.1255.1305.1355.1405.1455.1505.1555.1605.1655.1705.1755.1805.185
20
48.2
1
20
49.9
3
20
51.6
4
20
53.3
3
20
65.4
3
20
67.1
5
20
68.8
6
20
70.5
5
A doubled quartet (dq)
What about this?
5.8105.8255.8405.8555.8705.8855.9005.915
1.0
0
A (ddt)5.86
23
25.0
7
23
30.2
3
23
35.4
2
23
40.6
6
23
42.2
7
23
45.8
0
23
47.4
3
23
52.6
4
23
57.8
5
23
62.9
9
ddt
5 5
5 510
17
5.8155.8255.8355.8455.8555.8655.8755.8855.8955.9055.915
4.0 3.5 3.0 2.5 2.0 1.5 1.0 ppm
4.0 3.5 3.0 2.5 2.0 1.5 ppm
Real Spectrum
Spin Simulation
Pro-R and Pro-S
Ph OH
H
H
1Ph OH
H
H
Ph
OH
HH
Homotopic, Enantiotopic, Diastereotopic
Methyl groups
Chemical Equivalence/Magnetic Non Equivalence
H
Cl
Cl
H
H
H
What is going on?
Cl
H
H
Cl
H
H
ResultH
Cl
Cl
H
H
H
8 Hz
2Hz
A
B
B
A
2Hz
8 HzA
B
J AB'J AB' J AB' J AB'
J AB JAB
Expect:
Using Coupling Constants
Glucose
4.04.24.44.64.85.05.25.45.65.86.06.26.4
O
H
HO
H
HO
H
OH
OHHH
OH
12
3
4 56
Glucose
-173.0o 9.50 Hz
O
H
HO
H
HO
H
OH
OHHH
OH 52.7o 3.51 Hz
169.4o 9.5 Hz
-170.5o 9.13 Hz
H
HO CH
CHO
12
3
45
6
C2-C3
C1-C2
H
HO COH
OH
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.
O
ClCl
HH H
H2.10
2.50
0.93
2.806.80
1.95
30 Hz
Solution
1.52.02.53.03.54.04.55.05.5 ppm
Removing CouplingsChanging Solvents
OH
HaHb
Hc
CDCl3
C6D6
Ha, t Jab = Jac
Ha, dd Jab ≠ Jac
a = bCoupled to HcJab = Jac
a ≠ bEach coupled to HcJab ≠ Jac
1.52.02.53.03.54.04.55.05.5 ppm
Removing CouplingsSpin decoupling
OH
HaHb
Hc
CDCl3
CDCl3irradiated
Irradiate at 4.11 ppm
Signal due to Hadisappears
Coupling due to Hais removedSee Hb, Hc at b, cWith mutual Jbc
Spin Decoupling
OFF
A↑ X↑
A↑ X↓
A↓ X↑
A↓ X ↓
Spin Decoupling• Two spins, A (A), X (X) with JAX
• Irradiate X 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
-1
-0.5
0
0.5
0.01 0.1 1 10 100w
0t
c
h
fast tumbling slow tumbling
Effect of NOE on 13C NMR(90o)x
a. Proton Decoupled
decouple
b. Distortionless Enhancement by Polarization T ransfer (DEPT, q=135o, t1= 3.5
ms)(90o)x
(90o)x (180o)x
(180o)x
decouplet1t1t1
qy
HOHO
12
3
4
5
678
9
10
pinanediol (2)
13C
1H
13C
1H
2426283032343638404244464850525456586062646668707274
1
2 6 4
7
3 10/5/9
8
13C – 1H NOE at equilibrium (small molecule)
C↑H ↑
C↓H↓
C↓H ↑
C↑H ↓
●●●●●
●●●●
●
13C – 1H NOE irradiation on H
C↑H ↑
C↓H↓
C↓H ↑
C↑H ↓
●●●
●●
●●●
C
C
Hsat
Hsat
●●
13C – 1H NOE irradiation on H left ON
C↑H ↑
C↓H↓
C↓H ↑
C↑H ↓
●●●
●●
●●●
C
C
Hsat
Hsat
●●
13C – 1H NOE result
C↑H ↑
C↓H↓
C↑H ↓
●●●●
●
●●●●
C
C
Hsat
Hsat
●
1H-1H NOE example
1.01.52.02.53.03.54.04.55.05.56.0
1
2
H3
H1H4
HO
ClZ-2-chloro-dodec-2,11-dien-1-ol
1 3 4
H H
HO
Cl
HH
H
E-2-chloro-dodec-2,11-dien-1-ol
1 3
4
nOe nOe
H2O
1H – 1H NOE at equilibrium (small molecule)
S↑I ↑
S↓I↓
S↓I ↑S↑I ↓
●●●●
●●●●
I S
1H – 1H NOE irradiation on S
S↑I ↑
S↓I↓
S↓I ↑S↑I ↓
●●●
●●●●
W1S (sat)
W1S (sat)
W1I
W1I
I S
●
1H – 1H NOE irradiation on S left ON
S↑I ↑
S↓I↓
S↑I ↓
●●●
●●●●
W1S (sat)
W1S (sat)
W1I
W1I
●
1H – 1H NOE result
S↑I ↑
S↓I↓
S↑I ↓
●●●½
●●●½ ½
W1S (sat)
W1S (sat)
W1I
W1I
½
I S
NOE 3D example
NOE 3D example
Me Me
Me
H
H
H
O
Me Me
Me
H
H
H
O
Me Me
Me
H
H
H
OH
H
H
Me Me
Me
H
H
H
OH
H
H
Organic Structure Analysis, Crews, Rodriguez and Jaspars
Random orientation of magnetic dipoles
(a) No Bo
Mo
x
yBo
Mxy = 0
(b) Bo on; prior to resonance
Net polarization Mz is due to
population excess in higher
energy state
The magnetic vectors
precess about Bo at
the Larmor frequency o
M z
y
x
(c) At resonance o = 1
The magnetic vectors
precess in phase with
frequency 1.
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.
Events Accompanying Resonance
Organic Structure Analysis, Crews, Rodriguez and Jaspars
Mo
x
y
z
Bo
Excess of spinpopulation alongthe direction ofapplied magneticfield.
(90o)x
x
y
z
Bo
After 90o pulse
magnetization
is tipped into
the xy plane.
M
time t2
M=Magnetization which produces the FID. It decays as magnetization in xyplane diminishes after resonance
FT
frequency f2
preparation detection
ONE-PULSE SEQUENCE
Organic Structure Analysis, Crews, Rodriguez and Jaspars
ONE-PULSE SEQUENCE
1H
(90o)x
Preparation Detection
Fourier Transformation
0.1 0.3 0.5 0.7 0.9 1.1 1.3 1.5 1.7t1 (sec)
1.42.02.63.23.8f1 (ppm)
FT
Relaxation and Peak Shape
Rotational Correlation Time tc
wo = 2o
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 Even 0
2H, 14N Even Odd 1, 2, 3 etc
6 1 8
6 8
1
1
7
7
Receptivity
Nucleus C Relative Receptivity Relative receptivity
29Si 4.7% -5.32
13C 1.1% 6.73
1H 100% 26.75
Multinuclear NMR
15N NMR Shifts
1000 900 800 700 600 500 400 300 100200 0
Amonium salt
Aliphatic amine PrimarySecondary
Tertiary
ppm
PrimarySecondary
TertiaryDimethyleneamine
Anilinium ion
Aniline
1000 900 800 700 600 500 400 300 100200 0ppm
Piperidine
Cyclic enamine
Aminophosphine
Guanidine
Urea, carbamate, lactame
Amide PrimarySecondary
TertiaryThiourea
Thioamide
NitramineIndole, pyrrole
ImideNitrile, isonitr ile
PyrrazoleDiazocompound
Diazonium salt
PyridineImine
Oxime
Nitrocompound
Nitrosocompound Ar-NO S-NON-NO
*Nitrometane as reference (380 ppm)
Pyrimidine
Pyridine
Pyrazine
Imidazole
InternalTerminal
Iso
N=N-N ImineAmineHidrazone, tr iazene
NO2 N
ImineAmine
31P NMR Shifts
700 600 500 400 300 200 100 0 -200-100 -300
R3P=CR2
Z3P
Z-P=P-Z
Z2N-P=NRArP=C=Z2
X =P(Y)=Z
Z6P-
ppm
ppm
Z-C P
R3-P-RR2P-PR2
Z3P=O
Z5P
700 600 500 400 300 200 100 0 -200-100 -300
ArP=C=Ar
ArP=C=O
Cl3P
(RO)3P (Ar)3P (Me)3P H3P
TMS-C P
R-C P
Cl(OR)2P=O
R2(OR)3P
R4HP
*Reference H3PO4 85%. Adapted from A Complete Introduction to Modern NMR Spectroscopy. R. S. Macomber. J. Wiley and Sons. 1998.
R2P-M+
ArP=C=Ar2
(TMS)3P
FC P
R3P=O
RO3P=O
R(OR)2P=O
Cl2RP=O
(OR)5P
Coupling
Effect of 31P on 1H NMR
0.01 0.02 0.03 0.03 0.03 0.51 0.54
4.00
6.18
8 7 6 5 4 3 2 ppm
P
O
OHO
Effect of 31P on 1H NMR
4.30 4.25 4.20 4.15 4.10 4.05 ppm
P
O
OHO
Effect of 31P on 13C NMR
60 50 40 30 20 ppm
O
P
O
O
O
1 2 34
5
4 3
1
5
The 2nd Dimension
Organic Structure Analysis, Crews, Rodriguez and Jaspars
BASIC LAYOUT OF A 2D NMR EXPERIMENT
Organic Structure Analysis, Crews, Rodriguez and Jaspars
How a 2D NMR experiment works
FT
FT
FT
FT
t2
t1
2t1
3t1
nt1
t1
f2
transformmatrix
t1
f2FT
FT
FT
FT f2
f1
n is the number of increments
Contour plot
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
Organic Structure Analysis, Crews, Rodriguez and Jaspars
STRATEGY BASED ON C-H CONNECTIVITY
H
C C
HHSQCHSQC &
H
C C
H
1H-1H COSY
C C
H
C C C
HSQC &H
C C C
HMBC
C C C
Organic Structure Analysis, Crews, Rodriguez and Jaspars
STRATEGY BASED ON C-H CONNECTIVITY
Get 1H, 13C NMR spectra
get multiplicities and integrals
Get one bond 13C-1H correlations
assign 1H resonances to 13C resonancesGet 1H-1H correlation data (eg COSY)
Check assignment of diastereotopic protonsusing COSY and HMQC
Assemble substructures using COSY data
Get long range 13C-1H correlation spectrum
(eg HMBC)Combine substructures into
all possible working structures
Check all working structures for consistencywith 2D NMR data
2D structure
Get 13C-1H correlation spectrum (eg HSQC)
Organic Structure Analysis, Crews, Rodriguez and Jaspars
STRATEGY BASED ON C-H CONNECTIVITY – DEPT DATA
CCH2
CH, CH3
CHCH CH
CH2
CH3 CH3
C
Organic Structure Analysis, Crews, Rodriguez and Jaspars
STRATEGY BASED ON C-H CONNECTIVITY – HSQC DATA
A B C D E F C
ab
c
d’d
ef
Organic Structure Analysis, Crews, Rodriguez and Jaspars
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 3.0
2.5
E 23 CH3 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
ab
c
d
d'
ef
a b c d d' e f
Organic Structure Analysis, Crews, Rodriguez and Jaspars
STRATEGY BASED ON C-H CONNECTIVITY – COSY DATA
ATOM C (ppm) DEPT H (ppm) COSY (HH)
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 DATAHSQC suggests diastereotopic protons:3.08/2.44 ppm1.86/2.07 ppm
Organic Structure Analysis, Crews, Rodriguez and Jaspars
STRATEGY BASED ON C-H CONNECTIVITY – HMBC DATA
A B C D E F
ab
c
d
d'
ef
C
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 (HH) HMBC (CH)
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 CONNECTIVITYRETROSPECTIVE CHECKING
Pieces:
A
H
B
H
C
OH
H
D
HH
E
H H
H
F
H H
H
C 131 C 124 C 68 C 42 C 23 C 17
H 5.5 H 5.2H 4.0
H 3.0H 2.5
H 1.5 H 1.2
Possibilities:
F
A
B
C
D
E
OH
F
B
A
C
D
E
OH
E
B
A
C
D
F
OH
E
A
B
C
D
F
OH
F
D
A
B
C
E
OH
E
D
B
A
C
E
OH
E
D
A
B
C
E
OH
F
D
B
A
C
E
OH
F
A
B
D
C
E
OH
E
B
A
D
C
F
OH
E
A
B
D
C
F
OH
F
B
A
D
C
E
OH
Combinatorialexplosion
Organic Structure Analysis, Crews, Rodriguez and Jaspars
STRATEGY BASED ON C-H CONNECTIVITYRETROSPECTIVE CHECKING
MeF
B
A
D
C
MeE
OH
Hb
Ha
Hd Hd'
Hc
And similarly for COSY data
Organic Structure Analysis, Crews, Rodriguez and Jaspars
PROSPECTIVE CHECKING
Pieces: A
H
B
H
C
OH
H
D
HH
E
H H
H
F
H H
H
C 131 C 124 C 68 C 42 C 23 C 17
H 5.5 H 5.2H 4.0
H 3.0H 2.5
H 1.5 H 1.2
A B
Reason: only 2 sp2 C's
A B
F
HMBC: F-a, F-b, A-f, B-f
A B
F
DHMBC: D-a, D-b, A-d, B-d
A B
F
DC
OH
HMBC: C-d, C-a, D-c, A-cA B
F
DC
E
OH
HMBC: E-c, E-d, C-e, C-d
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.
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
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 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