NMR Handout - The Cook Group @ NDSU

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NMR Spectroscopy
NMR Phenomenon
‣ Nuclear Magnetic Resonance
µ
A spinning charged particle generates a magnetic
field.
A nucleus with a spin angular momentum will
generate a magnetic moment (μ).
If these tiny magnets are placed in an applied
magnetic field (Bo), they will adopt two different
states - one aligned with the field and one
aligned against the field. The energy difference
between these two states is what we are
observing with NMR.
B0
2
Nuclear Spin States
aligned against B0
Energy difference between the
states at a particular magnet
strength. In the Rf range of the
EM Spectrum.
E
aligned with B0
B0
When EM waves at this energy are directed at the nuclei - it
will absorb. Spins will flip from lower energy to higher
energy. At that energy, nuclei are “In Resonance”.
3
NMR Active Nuclei
‣ Many nuclei are “NMR Active”
‣ Spin Quantum Number I ≠ 0
H -- I = ½; 13C -- I = ½
‣
1
‣
12
C, 16O -- I = 0 -- Can’t be observed
‣ Other nuclei that are NMR active
‣
H (D), 14N, 19F, 31P
2
4
NMR Instrumentation
5
Magnetic Resonance Imaging
‣ NMR is the basis for MRI
6
To summarize
µ
A spinning charged
particle generates a
magnetic field.
A nucleus with a spin
angular momentum
will generate a
magnetic moment (m).
When placed in a
magnetic field (Bo),
they will adopt two
different states - one
aligned with the field
and one aligned
against the field.
B0
aligned against B0
Energy difference between the
states at a particular magnet
strength. In the Rf range of the
EM Spectrum.
E
aligned with B0
B0
7
Methyl Acetate - Proton NMR
O
H3C
C
O
CH3
8
Methyl Acetate - Carbon NMR
O
H3C
C
O
CH3
solvent
9
Electronic Shielding - Local Environments
Blocal
Beffective = B0 - Blocal
µ
Actual magnetic field
felt by the nucleus
B0 Beffective
10
Methyl Acetate - Proton NMR
O
H3C
C
O
CH3
11
Methyl Acetate - Carbon NMR
O
H3C
C
O
CH3
O
C
solvent
12
Chemical Shift
‣ The difference in resonance frequency of a nuclei
relative to a standard
‣ Most Shielded
‣ Relatively Inert
‣ Volatile
CH3
H3C Si CH3
‣ Resonance of standard is set to 0
CH3
TMS
TetraMethylSilane
13
NMR Scale
‣ X-Axis - frequency axis
NMR Spectrum
Low Field
Little electron shielding
(more electron withdrawing groups)
High Field
More electron shielding
(less electron withdrawing groups)
peaks measured as a shift (in Hz)
away from TMS
10 Hz
Reference
TMS
0
14
Different Spectrometer Frequencies
‣ Each specific instrument has it’s own magnetic field
strength - resonace occurs at different frequencies.
aligned against B0
100 MHz NMR
300 MHz NMR
Energy difference between the
states at a particular magnet
strength. In the Rf range of the
EM Spectrum.
E
aligned with B0
B0
Reference
TMS
300 Hz
100 Hz
0
15
Standard Scale
‣
δ = ppm = Chemical Shift from TMS (Hz)
Spectrometer Frequency (MHz)
100 MHz NMR
300 MHz NMR
100 Hz
100 MHz
=
1.0 ppm
300 Hz
300 MHz
=
1.0 ppm
Reference
TMS
1.0 ppm
0
16
NMR Scale
17
C13 NMR
‣ Difficult - Carbon 13 only 1.1% of all carbon.
‣ Number of different carbons
‣ Functional Group Regions
13C
NMR
R
R
δ
O
C
R
200
R
O
C
O
C
C
OR
C C
NR2
C O
150
100
C N
C X
50
0 ppm
18
C13 NMR
OH
19
Symmetry
‣ Symmetry in molecules can make carbons
“Chemically Equivalent”
Cl
O
H3C
C
CH3
same
electronic
environment
Br
20
Symmetry
‣ Some molecules have more than one mirror plane
CH3
CH3
CH3
CH3
21
Symmetry
CH3
CH3
CH3
CH3
22
Symmetry
CH3
CH3
CH3
CH3
23
Substitution of Carbon
‣ The intensity of the peaks roughly correlates with
the number of hydrogens on the carbon.
24
C13 NMR Regions
13C
NMR
R
R
δ
O
C
R
200
R
O
C
O
C
C
OR
C C
NR2
C O
150
100
C N
C X
50
0 p
25
Bromooctanol
HO
Br
26
Bromooctanal
H
O
Br
27
Alanine Me-Ester HCl
O
OCH3
H3C
NH3Cl
28
Alaninol
OH
H3C
NH2
29
Alaninol - phthalimide
OH
O
H3C
N
O
30
DEPT-C13
OH
‣ A - normal
C13
‣ B - CH carbons
only
‣ C - Odd # up
(CH3 and CH)
Even # down
(CH2)
31
Example from 13.7
Cl
KOH
ethanol
or
32
A Real Example
O
O
O
NH
NH
NH
O
OR
N
O
N
O
O
O
N
O
H2, Pd/C
H2, Pd/C
O
O
O
NH
NH
NH
OR
O
N
O
In the alkane
region
there would
only be 4 peaks
due to
symmetry
O
N
O
O
N
O
In the alkane region
there would be 6
different peaks
33
The Answer Is . . .
O
NH
O
N
O
O
NH
O
N
O
34
Proton NMR
‣ Number of chemically different hydrogens
‣ Relative Ratios of protons (peak size)
‣ How many neighboring hydrogens
‣ Chemical shifts and functional groups
35
Proton Equivalency
Diastereotopic
Homotopic
H
H3C
H
H
C
H3C
CH3
H3C
C
CH3
X
H3C
H
C
C
H3C
CH3
X
H
C
C
H2
CH3
Cl
replace H's diastereomers
Enantiotopic
H
X
C
H
replace H's
- same
H
H
CH3
H3C
H
C
H
C
H
CH3
Cl
H3C
X
C
H
C
CH3
Cl
replace H's enantiomers
X
H3C
H
C
C
H2
H
CH3
H3C
X
C
C
H2
CH3
36
Proton NMR Scale
‣ Range 0-10 ppm
1H
NMR
H
OH
H
C O
C O
H
H
H
H
H
H
H
H
H
C C
H
H
δ 10
9
8
H
C
C X
7
6
5
4
3
2
1
0 ppm
37
NMR Correlation Chart
Typical NMR Chemical Shifts
1H
Chemical
Shift (ppm)
13C
Functional
Group
Type
C H
Alkane
0.7 -1.8
10 - 60
Allylic or next
to carbonyl
1.6 - 2.4
30 - 60
next to
halogen or
alcohol
2.5 - 4.0
20 - 85
next to
oxygen of an
ester
4.0 - 5.0
50 - 85
vinylic
4.5 - 6.5
110 - 150
aromatic
6.5 - 8.0
110 - 140
aldehyde
9.7 - 10.0
190 - 220
alcohol
varies widely
will exchange with D2O
N/A
carbonyl of
ester, amide, or
carboxylic acid
(X = O, N)
N/A
165 - 185
carbonyl of
ketone or
aldehyde
N/A
190 - 220
C C
H
X C H
O
C O C H
C H
C
H
O
C H
O H
O
C X
O
C
Chemical
Shift (ppm)
38
Methyl Acetate
O
H3C
C
O
CH3
Area under peak
corresponds to the
number of H’s for that
resonance
39
Triphenyl Methanol
OH
40
Ethyl Acetate
O
H3C
C
O
H2
C
CH3
41
Spin Spin Splitting
‣ Protons on adjacent carbons also have an effect
‣ Resonances will split into n+1 number of peaks
Ha Hb
C C
Hb
1
H NMR
(without
coupling)
Ha
Hb
J
J
1
H NMR
(with
coupling)
42
Spin Spin Splitting
‣ Two hydrogens split neighbors into a triplet
Ha Hb
C
C Hb
1H
NMR
(without
coupling)
Ha
Hb
Hb
1H
NMR
(with
coupling)
J
1
1
:
J
2
:
J
1
1
43
Spin Spin Splitting
‣ Every splitting can be broken down into a series of
doublets
Ha Hb
C
1H
NMR
(without
coupling)
Ha
C Hb
J
Hb
1
1
1
H NMR
(with
coupling)
J
1
1
:
J
2
:
1
1
44
Spin Spin Splitting
‣ Three neighbors - Quartet
Ha Hb
C
C Hb
Hb
1
Ha
H NMR
(without
coupling)
Hb
Hb
1
H NMR
(with
coupling)
J
1
1
:
J
3
:
J
J
1
3
:
1
1
45
Higher Spin Spin Splitting
Pascal’s Triangle
Hb Ha Hb
Hb C C
Hb
C Hb
Hb
singlet
doublet
Ha will split into 7 peaks
64 different
combinations
of 6 spins
triplet
quartet 1
1
1
1
1
2
3
1
3
1
quintet 1 4 6 4 1
sextet
1 5 10 10 5 1
septet
1 6 15 20 15 6 1
46
Summary of Spin Spin Splitting
‣ Proton resonance split into n+1 number of peaks
‣ Relative ratio of peaks depends on number of spin
states of the neighbors.
‣ Adjacent protons will couple with the same
coupling constant.
‣ Protons farther away usually do not couple.
‣ Chemically equivalent protons cannot couple (eg.
ClCH2CH2Cl).
47
Doublet Splitting
O
H3C
C
C
H
O
CH3
NH3Cl
Methyl sees 1
neighbor,
Methine sees 3
48
Ethanol
‣ Note that the OH (and NH) usually don’t couple.
49
1,1,2-Trichloroethane
50
2-Bromopropane
51
Butanone
O
52
para-Methoxypropiophenone
53
Toluene
‣ Sometimes peaks overlap
54
Cinnamaldehyde
‣ Multiple Coupling
55
Spin Spin Splitting
‣ Every splitting can be broken down into a series of
doublets
Ha Hb
C
1H
NMR
(without
coupling)
Ha
C Hb
J
Hb
1
1
1
H NMR
(with
coupling)
J
1
1
:
J
2
:
1
1
56
Coupling with the same J
Ha
Ja-b = 5 Hz
Ja-c = 5 Hz
Hb Ha Hc
5
coupling with Hb
C C C
5
coupling with Hc
1
5
2
1
57
Coupling with different J values
Ha
Ja-b = 5 Hz
Ja-c = 10 Hz
Hb Ha Hc
5
coupling with Hb
10
coupling with Hc
1
10
1
C C C
1
1
Ha
10
coupling with Hc
coupling with Hb
5
1
5
1
1
1
58
Cinnamaldehyde
‣ Multiple Coupling
‣ J H1-H2 = 6 Hz, H2-H3 = 12 Hz
59
Cinnemaldehyde
60
Multiple Coupling - Identical J
Ha
Ja-b = 5 Hz
Ja-c = 5 Hz
coupling with Hb
Hb Ha Hc
C
C
C
5
Hc
5
coupling with Hc
5
1
5
coupling with Hc
1
1
2
5
3
5
3
1
61
Multiple Coupling - Different J
Ja-b = 10 Hz
Ha
Ja-c = 5 Hz
Hb Ha Hc
C
C
C
Hc
5
coupling with Hc
5
coupling with Hc
1
10
coupling with Hb
1
5
2
1
10
10
2
2
2
1
62
Nitropropane
63
Strategies for Determining Unknows
‣ Given the Molecular Formula - calculate degrees of
unsaturation.
‣ Identify functional groups
‣ Identify pieces of the structure
‣ Put the pieces together in a reasonable way
‣ Double check that your structure matches all the
data given.
64
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