nmr tutorial - trinity college dublin · putting a structure together three spin component -...

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