1. No category

NMR assignment of an unknown
By Roy Hoffman
NMR assignment of an unknown by
Roy Hoffman
1
Number of protons from integrals
16 Protons: 2x3, 4x2, 2x1
ppm
5.0
4.5
4.0
3.5
3.0
2.5
NMR assignment of an unknown by
Roy Hoffman
2.0
1.5
1.0
2
Number
of
carbon
types
11 carbon types
C11+H16O?, N?, etc.
ppm 160
140
120
100
80
NMR assignment of an unknown by
Roy Hoffman
60
40
20
3
Functionality from 1H chemical
shifts
2x1vinyl CH
1x3 methyl alkyl
4x2 CH2
Aromatic
ppm
9
8
CH, CH2
Me-alk
Me-vinyl
Vinyl CH2
Vinyl CH
CH2OH
7
6
5
4
NMR assignment of an unknown by
Roy Hoffman
3
2
1
4
Functionality from 13C chemical
2x1vinyl CH
shifts
1x3 methyl alkyl
4x2 CH2
2x1 vinyl C
1 C=O
(not CO2H because
no acidic H)
C11H16O
C=O, CO2H
ppm
175
CH, CH2
Vinyl C Vinyl CH2
CHOH,
Aromatic
CH2OH
Vinyl CH
Me
150
125
100
75
NMR assignment of an unknown by
Roy Hoffman
50
25
0
5
Connectivity from 2nd order
coupling – the roofing effect
Roofing effect
means that
J=18Hz
these two
means trans
multiplets are
coupled
No roofing Roofing effect
effect on the means that
these two
other two
multiplets multiplets are
coupled
ppm
ppm
5.40
5.35
5.30
5.25
2.9
2.8
2.7
2.6
2.5
2.4
2.3
2.2
5.20
NMR assignment of an unknown by
Roy Hoffman
6
The situation so far
•
•
•
•
•
C11H16O with C=O and 2 double bonds.
This requires one ring (Without it would be C11H18O).
There are 2 terminal methyls.
One of the vinyls has two trans H’s
Look for the following ring structures
NMR assignment of an unknown by
Roy Hoffman
7
Ring structures in the exercise
•
•
•
•
Look for the following ring structures
They may be unsaturated
They may be fused
There may not be any rings
cyclopentane cyclohexane norborane
NMR assignment of an unknown by
Roy Hoffman
8
Overlapping multiplets
• If the overlap is slight then it is still easy to
interpret.
ppm
2.24
2.22
2.20
2.18
2.16
NMR assignment of an unknown by
Roy Hoffman
2.14
9
Overlapping of similar structures
suggests almost identical
chemical environments.
These are
Similar but
no roofing
means they
are well
separated
Vinyl protons
ppm
5.45
5.40
5.35
5.30
5.25
5.20
5.15
5.10
NMR assignment of an unknown by
Roy Hoffman
5.05
10
More complex structures require
2D interpretation.
•4 pairs of protons
strongly coupled to
each other &
displays 2nd order
splitting
ppm
2.15
2.10
2.05
NMR assignment of an unknown by
Roy Hoffman
2.00
1.95
11
COSY
Two networks:
CH3CH2CH=CHCH2
1
2
and C=O
3
Strong signals are 2
& 3 bond, weak
signals are 4 bond
4
5
ppm
ppm
5
4
3
NMR assignment of an unknown by
Roy Hoffman
2
1
12
NOESY
1H
Correlation between
the two networks
NMR is assigned Me7
9
4
5 82
3
1
1
2
3
A ring is required so
close with C=O
4
O
3
10 5
1
9
6 4 2
8 7
5
ppm
NMR assignment of an unknown by
Roy Hoffman
ppm
5
4
3
2
1
13
HSQC
4
3
Me7
9
5 8 2
1
Me75 12
89
20
40
Assign carbons
attached to protons
60
80
100
120
4
3
ppm
ppm
5
4
3
NMR assignment of an unknown by
Roy Hoffman
2
1
14
HSQC to separate & correlate 1H
signals
25
Use NOESY
artifacts to
determine
proton
connectivity
30
35
40
ppm
ppm
2.1
NMR assignment of an unknown by
Roy Hoffman
2.0
15
HMBC to assign other carbons
Me7
9
5 8 2
4
3
•3-bond
correlations
strongest
1
25
1-bond
correlation
50
75
O
100
3
10 5
1
9
6 4 2
6
8 7
125
150
7
10
175
ppm
ppm
5
4
3
NMR assignment of an unknown by
Roy Hoffman
2
1
16
Use correlation table
H1
C1
O
3
10 5
1
9
6 4 2
8 7
2
C3
3
C5
H3
H4
H5
Me7
H8
3
3
H9
2
C2
C4
H2
3
2
3
3
2
3
C6
2
C7
3
3
Me7
C8
3
C9
2
2
C10
3
NMR assignment of an unknown by
Roy Hoffman
4
3
17