Junior Sophister
NMR tutorial
Typical NMR course questions
• Theory
• Theory and an associated problem
• Problem – Identify a structure (s)– You may have to comment on the chemistry
– Do not forget the mass spec
The spectroscopy question
• Look at the question !
• Look at the information provided– All the relevant details are provided
• Peak positions, spin couplings• Homonuclear (HH), Heteronuclear (CH) COSY• NOE, addition of D2 O
2008
• NMR spectra for an unknown compound (C7 H7 N) are shown below.
• Identify the structure of the unknown compound based on as much evidence as you can gather from each of the spectra as possible.
Proton NMR 400MHz0.
9779
1.01
14
0.99
54
0.99
88
1.00
00
1.03
58
1.04
33
Inte
gral
(ppm)5.45.65.86.06.26.46.66.87.07.27.47.67.88.08.28.48.6
8.58 ppm
solvent
5.48 ppm6.206.827.157.347.64
J 11HzJ 17HzJ 11, 17Hz
J 4.5Hz J 4.5, 8HzJ 8, 8Hz
J 8Hz
C7 H7 NPeaks - ppm
Spin coupling (J, HZ)
Integrals
Carbon-13 and DEPT 135°
116118120122124126128130132134136138140142144146148150152154156
116118120122124126128130132134136138140142144146148150152154156(ppm)
155.8 ppm 149.5 137136.4
122.4 121.2
118.3 ppm
Summary : carbon 7 peaks - 1 C , 5 CH (aromatic) , 1 CH2 (terminal double bond ?)proton 7 peaks – all in the aromatic or sp2 region
CH COSY
(ppm) 8.0 7.2 6.4 5.6
(pp
144
136
128
120
CH2
149.5, 8.58
136.4, 7.64
137, 6.82
122.4, 7.15
121.2, 7.34
118.3, 6.20, 5.48
HH COSY
(ppm) 8.0 7.2 6.4 5.6
(pp
8.0
7.2
6.4
5.6
Three spins Linked
Four spins Linked
5.48, 6.20, 6.82 ppm
8.58, 7.15, 7.64, 7.34 ppm
Putting a structure together
Three spin component - contains a double bondthe spin-spin couplings determine the structure
trans J coupling > cis J coupling across a double bond
H
H
H
6.82 J= 11,17Hz
6.20 J=11Hz
5.48 J=17Hz
137118.3
More pieces of the structural jigsaw
The four spin component are all linked in a contiguous abcd pattern in an aromatic system i.e. 8.58 (d) – 7.15 (dd) – 7.34 (t) – 7.34 (d) ppm
The ring system has to contain the N atom - downfield proton at 8.58ppm and a small J coupling - 4.5 Hz- downfield quaternary C at 155.8ppm
The obvious structure is a pyridine ring
N RH
H
H
H
8.58
7.15
7.64
7.34136.4122.4
155.8
121.2
149.5
The complete structure
NH
H
H
H
8.58
7.15
7.64
7.34136.4122.4
155.8
121.2
149.5H
HH
6.82 J= 11,17Hz
6.20 J=11Hz
5.48 J=17Hz
137
118.3
J=8Hz
J=8m,8Hz
J=4.5, 8Hz
J=4.5Hz
2006
• NMR spectra for an unknown compound (C6 H5 NO) are shown below.
• Identify the structure of the unknown compound based on as much evidence as you can gather from each of the spectra as possible
Proton NMR 400MHzC6 H5 NO
Peaks - ppm
Spin coupling (J, HZ)
Integrals
0.98
43
0.96
53
0.95
94
0.98
16
1.01
11
Inte
gral
(ppm)7.27.47.67.88.08.28.48.68.89.09.29.49.69.810.010.210.4
10.1 9.1 8.8 8.2 7.5ppm
J=5, 8 Hz
J= 2Hz
J=2,2, 8 HzJ=2,5 Hz
Carbon-13 and DEPT 135°
Summary : carbon 6 peaks - 1 C , 4 CH (aromatic) , 1 CHO (aldehyde)proton 5 peaks – 4 aromatic CH , 1 downfield CH
110115120125130135140145150155160165170175180185190195
110115120125130135140145150155160165170175180185190195(ppm)
190.7154.7
152.0
135.7
131.3
124.0
CH COSY
(ppm) 9.6 8.8 8.0 7.2
(pp
180
160
140
120
135.7, 8.2
C
O
H
190.7
10.1
190, 10.1
124.0, 7.5
152.0, 9.1
154.7, 8.8
Aldehyde
C6 H5 NO
Possible two signals beside a N152.0, 154.7 ppm
HH COSY
(ppm) 9.6 8.8 8.0 7.2
(pp
9.6
8.8
8.0
7.2
7.59.110.1
8.8 ppm8.2
dd J= 5, 8 Hz
ddd J= 2, 2, 8 Hz
dd J= 2, 5 Hz
d J= 2 Hz
Clearcut 3 spin system8.8 – 7.5 - 8.2 ppm
One remote spin at 9.1ppmwith small J coupling of 2Hz
NMR structure
C
O
H
190.7
10.1
N
H
H
H H
8.2
8.8
7.5
9.1J= 5, 2 Hz J= 2 Hz
J=5, 8 Hz
J= 8,2, 2 Hz
152.0 154.7
124.0
135.7 131.3
Note the smaller J coupling for protons adjacent to a nitrogen
2005• NMR spectra for compound
which is suspected to have the structure A
• Is this the correct structure ?
• Support your answer with as much evidence from each of the spectra as possible
O
Br
Proton NMR 400MHzC10 H9 BrO
Peaks - ppm
Spin coupling (J, HZ)
Integrals
1.00
00
1.01
82
1.02
72
0.95
32
1.00
61
2.20
20
2.21
27
Inte
gral
(ppm)2.83.23.64.04.44.85.25.66.06.46.87.27.68.0
2.59
4.30
7.02 7.85 7.14
7.25 6.77
(ppm)6.86.97.07.17.27.37.47.57.67.77.87.9
5.3 HzJ (Hz)
8 Hz 8 Hz
8 Hz 8 Hz
(ppm)4.244.284.32
(ppm)2.55
5.8 Hz ? Hz
Carbon-13 and DEPT 135°
Summary : carbon 10 peaks - 3 C , 5 CH (aromatic, sp2) , 2 CH2proton 7 peaks – 4 aromatic CH , 1 double bond CH, 2 CH2
35404550556065707580859095100105110115120125130135140145150155160
35404550556065707580859095100105110115120125130135140145150155160(ppm)
157.8
72.6 33.3
120.9
120.5128.3
133.5
133.1 123.3
129.5
CH COSY
(ppm) 8.00 7.00 6.00 5.00 4.00 3.00
(pp
120
100
80
60
40
C10 H9 BrO
Aromatic , sp2 CHs
72.6, 4.30
33.3, 2.59
CH2
CH2O
Br
72.4, 4.30
33.3, 2.59
CH COSY
(ppm) 7.80 7.60 7.40 7.20 7.00 6.80
(pp
134
132
130
128
126
124
122
120
C10 H9 BrO
120.9, 7.02
CH2
CH2O
Br 133.5, 6.77
72.4, 4.30
33.3, 2.59
H
133.1, 7.85
129.5, 7.25
123.3, 7.14
HH COSY
(ppm)8.0 7.2 6.4 5.6 4.8 4.0 3.2 2.4
(pp
8 0
7.2
6.4
5.6
4.8
4.0
3.2
2.4
Three spin system
=CH- CH2-CH2-O-
6.77- 2.59- 4.30 ppm
HH COSYFour spin system abcdorder on the ring has to be guessed
(ppm) 7.80 7.60 7.40 7.20 7.00
(pp
7.80
7.60
7.40
7.20
7.00
7.85, 7.14, 7.25, 7.02 ppm
The Problem !
O
Br
H
H
H
H
7.85
7.14
7.25
7.02
O
Br
H
H
H
H
7.02
7.25
7.14
7.85
Which is the correct structure ?
The NMR spectroscopists fix - a long range CH COSY
(ppm)8.0 7.2 6.4 5.6 4.8 4.0 3.2 2.4
(pp
160
140
120
100
80
60
40
HC
HCCH
HC
CH
CH2
CH2O
Br
2.59
157.8
120.5128.3
2.59ppm
More long range CH COSY conformation
(ppm) 7.80 7.60 7.40 7.20 7.00 6.80 6.60
(pp
132
128
124
120
CH
CH2
CH2O
Br
H
H
H
H
7.02
7.25
7.14
7.85120.5
128.3
157.8
120.9129.5
133.1
72.4, 4.30
33.3, 2.59
133.5, 6.77
123.3
7.85
120.5
128.3
7.14 6.77
3JCH links prove the structure
2004
• A set of 7 NMR spectra of an alcohol with the molecular formula C9 H10 O are shown below
• What structural information which can be extracted from each spectrum ?
• Use this information to assign the structure of the molecule
Proton NMR 400MHzC9 H10 O
Peaks - ppm
Spin coupling (J, HZ)
Integrals
5 1 1 2 1 l
(ppm)1.52.02.53.03.54.04.55.05.56.06.5 7.07.5
(ppm)6.46.56.66.7
(ppm)4.304.40
(ppm) 7.307.40
6.64
4.35
1.64 6.40
7.41 7.35
7.28
7.5 Hz 7.5 Hz
7.5 Hz
16.4 Hz 16.4, 6.0 Hz 6.0 Hz
CHCl 3
OH
Possible double bond (large coupling)
Carbon-13 NMR
Summary : carbon 7 peaks proton 7 peaks – 3 aromatic CH , 2 double bond CH, 1 CH2 1 OH
(ppm)60657075808590 95100105110115120125130 135 140
(ppm)127.0128.0129.0
128.6 128.4
127.6
63.7
136.6
131.1
126.4
CDCl3
DEPT 135° and 90°
Summary : carbon 7 peaks - 1 C , 5 CH (aromatic, sp2) , 1 CH2proton 6 peaks – 3 phenyl CHs , 2 double bond CHs, 1 OCH2
65707580859095100105110115120125 130135
65707580859095100105110115120125 130135
(p pm )
DEPT 90
D EPT 135
5 s ig n als
CH COSY
ppm 7.00 6.00 5.00 4.00 3.00 2.00
140
120
100
80
60
(ppm) 7.40 7.20 7.00 6.80 6.60 6.40
131
130
129
128
127
OH-OCH2 -
63.7 4.357.41, 126.6
7.28, 127.6
6.64, 131.1
6.40, 128.4
7.35, 128.6
Mono substituted Phenyl group (5H)
o m p
1D NOE NMR - Through space connection
6 .6 06 .7 06 .8 06 .907 .00 7 .1 07 .2 07 .3 07 .4 07 .50
6 .6 06 .7 06 .8 06 .907 .00 7 .1 07 .2 07 .3 07 .4 07 .50 (pp m)
Irrad ia ted p eak
Peak at 6.6pp is irradiated
positive NOE observed to ortho proton on the phenyl ring at 7.41ppm
structure
C C
H CH2
H
OH
1.64
4.35
6.64
6.40
NOE
7.41
7.25, 127.6
d, J= 6 Hz
dt, J=16.4, 6 Hz
d, J= 16.4 Hz
7.35
126.4
131.1
136.6
128.6
63.7128.4