Exercise plus Solution – Quick overview
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14.42
63.43
86.20
151.70
ppm
150 140 130 120 110 100 90 80 70 60 50 40 30 20 ppm
13
C
13
C{1H} NMR spectrum
measured at 62.9{250.13} MHz
C
4
H
8
O measured in CDCl
3
Deduce both constitution and
configuration, analyze each
multiplet and extract both all
homonuclear and heteronuclear
coupling constants!
13
C NMR spectrum
measured at 62.9 MHz
To see the coloured multiplets in
higher resolution, please visit the
next pages.
9442.79
9445.46
9448.09
9450.65
9451.21
9453.80
9456.41
9459.10
~
~
9623.07
9625.76
9628.38
9628.84
9630.97
9631.51
9634.10
9636.70
9639.42
Hz
151.70 ppm
To make optimal use of the
available space, parts of the
multiplets were replaced by
the symbol <--->. This area
contains baseline and noise
only about 100 Hz each.
5413.95
5418.87
5423.40
5428.32
5575.00
5584.45
~
~
~
~
5257.82
5267.28
Hz
714.16
716.8 3
719.5 2
840.66
843.37
846.07
967.1 9
969.9 0
972.6 0
~
~
~
~
~
~
1093.73
1096.42
1099.12
Hz
86.20 ppm
14.42 ppm
4122.21
4126.14
4126.79
4127.94
4128.61
4130.69
4131.34
4132.48
4133.16
4135.23
4135.83
4137.04
4137.70
4141.67
4142.24
3835.68
3840.17
3840.83
3842.04
3842.63
3844.72
3845.36
3846.54
3847.19
3849.22
3849.94
3851.07
3851.72
~
~
~
~
3978.61
3979.23
3983.15
3983.80
3984.99
3985.62
3987.70
3988.36
3989.52
3990.17
3992.25
3992.92
3994.07
3994.72
3998.63
3999.27
Hz
63.43 ppm
3.17
2.12
1.08
1.07
1.00
Inte-
gral
1596.16
1602.99
1610.52
1617.35
6.4
ppm
982.17
984.07
989.01
990.90
1027.77
1029.66
1042.12
1044.02
Hz
4.1
4.0 ppm
3.7
ppm
919.09
926.12
933.16
940.21
Hz
1.3 1.2ppm
307.47
314.53
321.56
Hz
6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5
ppm
1
H
1
H NMR spectrum
measured at 250.13 MHz
3.17
2.12
1.08
1.07
1.00
Inte-
gral
6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5
ppm
1
H
First pieces of information
Let us first try to determine the constitution and configuration from the clearly laid out proton spectrum.
Molecular formula - C
4
H
8
O
1 double bond equivalent (DBE)
𝒏
𝐃𝐁𝑬
=
𝟐𝒏
𝐂
− 𝒏
𝐇
+ 𝟐
𝟐
The protons are present as CH, one OH in principle would be possible.
There are 5 signal groups in the proton spectrum.
3.17
2.12
1.08
1.07
1.00
Inte-
gral
6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5
ppm
1
H
Integration
measured integrals of the five signal groups 1.00 1.07 1.08 2.12 3.17
sum of all integrals 8.44
number of protons according to molecular formula 8
coefficient of proportionality 0.95 (8 / 8.44)
number of protons within each signal group 0.95 1.01 1.02 2.01 3.01
1H 1H
1H
2H 3H
1596.16
1602.99
1610.52
1617.35
982.17
984.07
989.01
990.90
1027.77
1029.66
1042.12
1044.02
Hz
919.09
926.12
933.16
940.21
Hz
307.47
314.53
321.56
Hz
6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
1H 1H
1H
2H 3H
1H
1H
2H 3H1H
Multiplet structure
1596.16
1602.99
1610.52
1617.35
982.17
984.07
989.01
990.90
1027.77
1029.66
1042.12
1044.02
Hz
919.09
926.12
933.16
940.21
Hz
307.47
314.53
321.56
Hz
6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
1H
1H
2H 3H1H
Multiplet structure
The multiplet at about 6.4 ppm is a doublet of doublets (dd). For four lines with almost identical
intensity there is no other possibility if only nuclei with I = ½ are available as coupling partners.
dd
1596.16
1602.99
1610.52
1617.35
982.17
984.07
989.01
990.90
1027.77
1029.66
1042.12
1044.02
Hz
919.09
926.12
933.16
940.21
Hz
307.47
314.53
321.56
Hz
6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
1H
1H
2H 3H1H
Multiplet structure
At about 4.15 ppm and 3.95 ppm there is also a doublet of doublets (dd) in each case. The same
reasoning applies.
dd
dd
dd
1596.16
1602.99
1610.52
1617.35
982.17
984.07
989.01
990.90
1027.77
1029.66
1042.12
1044.02
Hz
919.09
926.12
933.16
940.21
Hz
307.47
314.53
321.56
Hz
6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
1H
1H
2H 3H1H
Multiplet structure
At about 3.8 ppm, there is a pure quartet (q) in the intensity ratio of the individual lines of 1 : 3 : 3
: 1. In the case of pure multiplets, the ratio of the intensity of the individual lines to each other
obeys a binomial distribution as long as coupling partners with I = ½ are involved.
dd
dd
dd
q
1596.16
1602.99
1610.52
1617.35
982.17
984.07
989.01
990.90
1027.77
1029.66
1042.12
1044.02
Hz
919.09
926.12
933.16
940.21
Hz
307.47
314.53
321.56
Hz
6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
1H
1H
2H 3H1H
Multiplet structure
Finally there is a triplet (t) with three individual lines in an intensity ratio of 1 : 2 : 1 at
approx. 1.3 ppm.
dd
dd
dd
q
t
6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
1H
1H
2H 3H1H
Multiplet analysis
dd
dd
dd
q
t
321.56
314.53
307.47
Hz
The highest field multiplet at approx. 1.3 ppm can only be a methyl group. OH
3
is hardly possible.
δ
1
=
321.56 Hz + 307.47 Hz
2 ∗ 250.13 MHz
= 𝟏. 𝟐𝟔 𝐩𝐩𝐦
Let us first determine the exact value of the chemical shift
and the coupling constant of this triplet.
𝐽
1
=
321.56 Hz − 307.47 Hz
2
= 𝟕. 𝟎𝟒 𝐇𝐳
1.26
7.04 Hz
CH
3
6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
1H
1H
2H 3H1H
Multiplet analysis
dd
dd
dd
q
t
1.26
7.04 Hz
CH
3
940.21
933.16
926.12
919.09
Hz
According to the n+1 rule, a triplet is the result of two equivalent
neighbouring protons.
A multiplet with integral 2 is found at about 3.7 ppm.
321.56
314.53
307.47
Hz
It can only be a CH
2
group (OH
2
is not possible).
CH
2
6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
1H
1H
2H 3H1H
Multiplet analysis
dd
dd
dd
q
t
1.26
7.04 Hz
CH
3
940.21
933.16
926.12
919.09
Hz
3.72
7.04 Hz
CH
2
δ
2
=
940.21 Hz + 919.09 Hz
2 ∗ 250.13 MHz
= 𝟑. 𝟕𝟐 𝐩𝐩𝐦
As with the triplet, let us determine the exact value of the chemical shift
and the coupling constant of this quartet.
𝐽
2
=
940.21 Hz − 919.09 Hz
3
= 𝟕. 𝟎𝟒 𝐇𝐳
6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
1H
1H
2H 3H1H
Multiplet analysis
dd
dd
dd
q
t
1.26
7.04 Hz
CH
3
940.21
933.16
926.12
919.09
Hz
7.04 Hz
3.72
CH
2
7.04 Hz
3.72
1.26
CH
2
CH
3
For a quartet, we need three equivalent neighbour protons according to the
n+1 rule.
We have already found them.
The coupling constant of 7.04 Hz appears in both the triplet
and the quartet.
6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
Multiplet analysis
1.26
7.04 Hz
3.72
1.26
CH
2
CH
3
1H
1H
1H
dd
dd
dd
1.26
7.04 Hz
3.72
1.26
CH
2
CH
3
For further analysis, we no longer need the two multipletts of the
ethyl group. In order to focus on the remaining multiplets, the signals
of the two multipletts were removed from the
1
H NMR spectrum.
6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
Multiplet analysis
1617.35
1610.52
1602.99
1596.16
Hz
1H
1H
1H
dd
dd
dd
δ
3
=
1617.35 Hz + 1596.16 Hz
2 ∗ 250.13 MHz
= 𝟔. 𝟒𝟐 𝐩𝐩𝐦
The three doublets of doublets can each be
analysed according to the same scheme. Let's start
with the exact value of the chemical shift of the
lowest field multiplet.
6.42 ppm
1.26
7.04 Hz
3.72
1.26
CH
2
CH
3
6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
Multiplet analysis
1617.35
1610.52
1602.99
1596.16
Hz
1H
1H
1H
dd
dd
dd
6.42 ppm
𝐽 =
1617.35 Hz + 1610.52 Hz
2
−
1602.99 Hz + 1596.16 Hz
2
= 𝟏𝟒. 𝟑𝟔 𝐇𝐳
The larger of the two coupling constants is
obtained by subtracting the mean value of
lines 3 and 4 from the mean value of lines 1
and 2 (counting from left to right).
1 2 3 4
J = 14.36 Hz
1.26
7.04 Hz
3.72
1.26
CH
2
CH
3
6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
Multiplet analysis
1617.35
1610.52
1602.99
1596.16
Hz
1H
1H
1H
dd
dd
dd
6.42 ppm
𝐽 =
1617.35 Hz − 1610.52 Hz + (1602.99 Hz − 1596.16 Hz)
2
= 𝟔. 𝟖𝟑 𝐇𝐳
The smaller of the two coupling constants
results from the difference of lines 3 and 4 or
lines 1 and 2, or for improved accuracy from the
mean value of both differences.
1 2 3 4
J = 14.36 Hz
𝐽 =
1617.35 Hz + 1610.52 Hz
2
−
1602.99 Hz + 1596.16 Hz
2
= 𝟏𝟒. 𝟑𝟔 𝐇𝐳
J = 6.83 Hz
1.26
7.04 Hz
3.72
1.26
CH
2
CH
3
6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
Multiplet analysis
1H
1H
1H
dd
dd
dd
6.42 ppm
J = 14.36 Hz
J = 6.83 Hz
6.42 ppm
J = 14.36 Hz
J = 6.83 Hz
4.14 ppm
3.94 ppm
J = 1.90 Hz
J = 1.90 Hz
J = 14.36 Hz
J = 6.83 Hz
We temporarily note the chemical shift and
the two coupling constants next to the
corresponding multiplet.
In the analogous analysis of the two further doublets of
doublets, one chemical shift and two coupling constants
each are also obtained.
1.26
7.04 Hz
3.72
1.26
CH
2
CH
3
6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
First structure fragment
1H
1H
1H
dd
dd
dd
6.42 ppm
J = 14.36 Hz
J = 6.83 Hz
4.14 ppm
3.94 ppm
J = 1.90 Hz
J = 1.90 Hz
J = 14.36 Hz
J = 6.83 Hz
6.42
4.14
14.36 Hz
CC
H
H
A chemical shift of 6.42 ppm fits a proton bonded to an sp
2
hybridised carbon atom.
A coupling constant of 14.36 Hz is very characteristic for two
protons in the E position on an ethene fragment.
The missing coupling partner with a coupling constant of 14.36 Hz has a
chemical shift of 4.14 ppm.
We expect the following structural fragment.
1.26
7.04 Hz
3.72
1.26
CH
2
CH
3
6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
First structure fragment
1H
1H
1H
dd
dd
dd
6.42 ppm
J = 14.36 Hz
J = 6.83 Hz
4.14 ppm
3.94 ppm
J = 1.90 Hz
J = 1.90 Hz
J = 14.36 Hz
J = 6.83 Hz
6.42
4.14
14.36 Hz
CC
H
H
But …
4.14 ppm for a proton bound to an sp2-hybridised carbon?
14.36 Hz could also correspond to a geminal coupling constant
between two diastereotopic protons on an sp
3
hybridised carbon
atom.
C
H
H
14.36 Hz
6.42
4.14
In this case, the 6.42 ppm would be very strange. Let's stay with the ethene derivative
and try to understand later. Some pieces of information are no longe necessary.
1.26
7.04 Hz
3.72
1.26
CH
2
CH
3
6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
First structure fragment
1H
1H
1H
dd
dd
dd
6.42 ppm
J = 6.83 Hz
4.14 ppm
3.94 ppm
J = 1.90 Hz
J = 1.90 Hz
J = 6.83 Hz
6.42
4.14
14.36 Hz
CC
H
H
1.26
7.04 Hz
3.72
1.26
CH
2
CH
3
What is actually missing?
molecular formula – C
4
H
8
O
known fragments – C
4
H
7
missing – OH
6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
First structure fragment
1H
1H
1H
dd
dd
dd
6.42 ppm
J = 6.83 Hz
4.14 ppm
3.94 ppm
J = 1.90 Hz
J = 1.90 Hz
J = 6.83 Hz
6.42
4.14
14.36 Hz
1.26
7.04 Hz
3.72
1.26
CH
2
CH
3
missing – OH
CC
H
H
This way?
HO
Due to the neighbouring protons of the methylene
group, a triplet structure should be visible in the
multiplet of the proton at 6.42 ppm.
OH and ethyl groups could be interchanged, then the
triplet should be visible in the multiplet of the proton
at 4.14 ppm.
A C-C linkage between ethyl group and ethene
fragment is strictly excluded!
6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
First structure fragment
1H
1H
1H
dd
dd
dd
6.42 ppm
J = 6.83 Hz
4.14 ppm
3.94 ppm
J = 1.90 Hz
J = 1.90 Hz
J = 6.83 Hz
6.42
4.14
14.36 Hz
1.26
7.04 Hz
3.72
1.26
CH
2
CH
3
missing – OH
CC
H
H
A direct C-C linkage can only be prevented with
the two atoms still available by an oxygen atom
next to the methylene group.
3.72
1.26
7.04 Hz
O CH
2
CH
3
The last remaining building block is the proton with
the chemical shift of 3.94 ppm. Two positions are
possible.
Is there another choice?
6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
First structure fragment
1H
1H
1H
dd
dd
dd
6.42 ppm
J = 6.83 Hz
4.14 ppm
3.94 ppm
J = 1.90 Hz
J = 1.90 Hz
J = 6.83 Hz
6.42
4.14
14.36 Hz
CC
H
H
3.72
1.26
7.04 Hz
O CH
2
CH
3
H
3.94
1.90 Hz
6.83 Hz
Let's just try one of the two possibilities.
A vicinal coupling constant of 1.90 Hz between the Z-
site protons with the chemical shift of 3.94 ppm and
4.14 ppm is much too small. A rough estimate for this
coupling constant is 8 Hz.
On the other hand, the typical geminal coupling
constant between the two protons of a =CH
2
group
is roughly -1 Hz. 6.83 Hz (the sign is not known) is
much too large..
6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
First structure fragment
1H
1H
1H
dd
dd
dd
6.42 ppm
J = 6.83 Hz
4.14 ppm
3.94 ppm
J = 1.90 Hz
J = 1.90 Hz
J = 6.83 Hz
3.72
1.26
7.04 Hz
O CH
2
CH
3
What about the alternative position of the proton?
6.83 Hz
1.90 Hz
6.42
4.14
3.94
14.36 Hz
CC
HH
H
6.83 Hz are somewhat small for a coupling constant
between protons in Z-position to each other.
However, of the two possibilities considered for the
position of the proton with the chemical shift of
3.94 ppm, this is clearly the better choice.
Additionally, a geminal coupling constant of 1.90 Hz
is very characteristic for this structural fragment.
Final structure
3.72
1.26
7.04 Hz
O CH
2
CH
3
6.42
4.14
3.94
3.72
1.26
7.04 Hz
14.36 Hz
6.83 Hz
1.90 Hz
O CH
2
CH
3
CC
HH
H
6.83 Hz
1.90 Hz
6.42
4.14
3.94
14.36 Hz
CC
HH
H
All that remains to be done is to combine the two existing building blocks.
But what is the reasonfor these somewhat strange chemical shifts of 4.14
and 3.94 ppm?
6.42
4.14
3.94
3.72
1.26
7.04 Hz
14.36 Hz
6.83 Hz
1.90 Hz
O CH
2
CH
3
CC
HH
H
6.42
4.14
3.94
3.72
1.26
7.04 Hz
14.36 Hz
6.83 Hz
1.90 Hz
CH
2
CH
3
HH
H
O
+
CC
-
Mesomerism in ethyl vinyl ether
The oxygen has two free electron pairs.
Using one of the two electron pairs, it is easy to set up a mesomeric structure with a negative electric charge next to
the two =CH
2
protons.
This electron causes a slight additional shielding at the two protons bound to the carbon atom, resulting in a high-field
shift of their signals.
The same effect is found with the o and p protons in phenol or its derivatives.
14.42
63.43
86.20
151.70
ppm
150 140 130 120 110 100 90 80 70 60 50 40 30 20
ppm
13
C
6.42
4.14
3.94
3.72
1.26
7.04 Hz
14.36 Hz
6.83 Hz
1.90 Hz
O CH
2
CH
3
CC
HH
H
Assignment of the carbon signals
If the carbon spectrum is measured without the usual proton
broadband decoupling, the coupling pattern is dominated by
the one bond coupling constants between
13
C and
1
H (
1
J
CH
).
Because of the doublet (151.70 ppm) and quartet (14.42
ppm) structure, two of the carbon signals can be
immediately assigned without doubt.
14.42
151.70
14.42
63.43
86.20
151.70
ppm
150 140 130 120 110 100 90 80 70 60 50 40 30 20
ppm
13
C
~
~
~
~
6.42
4.14
3.94
3.72
1.26
7.04 Hz
14.36 Hz
6.83 Hz
1.90 Hz
O CH
2
CH
3
CC
HH
H
14.42
151.70
Assignment of the carbon signals
(The colour coding has been
changed to focus on carbon
assignment.)
Although of the two remaining carbon atoms, one is sp
2
and the
other sp
3
hybridised, their chemical shifts are not too different.
On closer inspection, the signal at 86.2 ppm consists of
eight lines of equal intensity, a doublet of doublets of
doublets. There are no two chemically equivalent
neighbours. The CH
2
carbon atom with two equivalent
neighbour protons is thus ruled out.
86.20
14.42
63.43
86.20
151.70
ppm
150 140 130 120 110 100 90 80 70 60 50 40 30 20
ppm
13
C
~
~
~
~
6.42
4.14
3.94
3.72
1.26
7.04 Hz
14.36 Hz
6.83 Hz
1.90 Hz
O CH
2
CH
3
CC
HH
H
14.42
151.70
Assignment of the carbon signals
86.20
After excluding the assignments already done, only the CH
2
carbon remains for the signal at 63.43 ppm. As expected,
the basic structure of this signal is a triplet.
63.43
Carbon multiplet detailed analysis
~
~
~
~
5584.45
5575.00
5423.40
5418.87
5428.32
5413.95
5267.28
5257.82
Hz
~
~
corresponds to 100 Hz
baseline each
( )
The structure of the multiplet of the carbon signal at 86.20 ppm seems
to be the easiest to solve.
It is clearly a doublet of doublets of doublets. there should be three
chemically different protons in the neighbourhood of this carbon
atom. Only the protons at 3.94, 4.14 and 6.42 ppm can be
considered for this.
Two doublets are easy to recognise, the coupling constant of these
two doublets should be observed four times in total. Let us mark
both doublets using different colours.
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
86.20
63.43
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
86.20
63.43
5423.40
5418.87
5428.32
5413.95
Carbon multiplet detailed analysis
~
~
~
~
5584.45
5575.00
5267.28
5257.82
Hz
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
86.20
63.43
1 72 3 4
5
6 98
10
1211 13
0
1 72 3 4
5
6 98
0
The other two doublets must be hidden in the middle of the composite multiplet.
Howe can we locate them?
Let us first mark on a ruler a doublet that has been identified without any doubt..
Using this pattern we can now try to find the doublet one more time.
1 72 3 4
5
6 98
10
1211 13
0
We label the two lines having exactly the right distance.
5423.40
5418.87
5428.32
5413.95
Carbon multiplet detailed analysis
~
~
~
~
5584.45
5575.00
5267.28
5257.82
Hz
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
86.20
63.43
1 72 3 4
5
6 98
0
1 72 3 4
5
6 8
0
The wo lines remain have to be the still missing fourth doublet that is
still missing.
Let us check.
We now have four different coloured doublets.
Carbon multiplet detailed analysis
~
~
~
~
5584.45
5575.00
5423.40
5418.87
5428.32
5413.95
5267.28
5257.82
Hz
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
86.20
63.43
𝐽 = 5584.45 Hz − 5575.00 Hz = 𝟗. 𝟒𝟓 𝐇𝐳
δ =
5584.45 Hz + 5575.00 Hz
2
= 𝟓𝟓𝟕𝟗. 𝟕𝟑 𝐇𝐳
5579.73
Analysis of one of the doublets provides a coupling constant and
the chemical shift of the signal from which the doublet comes
from.
J = 9.45 Hz
Carbon multiplet detailed analysis
~
~
~
~
5584.45
5575.00
5267.28
5257.82
Hz
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
86.20
63.43
5579.73
J = 9.45 Hz
𝐽 = 5428.32 Hz − 5418.84 Hz = 𝟗. 𝟒𝟓 𝐇𝐳
δ =
5428.32 Hz + 5418.87 Hz
2
= 𝟓𝟒𝟐𝟑. 𝟔𝟎 𝐇𝐳
The three other labelled doublets are evaluated in the same way.
5423.60
5423.40
5418.87
5428.32
5413.95
5418.68
5262.55
It is best to wait with the assignment of the coupling constant
until the analysis of this multiplet is completely finished.
Carbon multiplet detailed analysis
~
~
~
~
5584.45
5575.00
5267.28
5257.82
Hz
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
86.20
63.43
5579.73
J = 9.45 Hz
5423.60
5423.40
5418.87
5428.32
5413.95
5418.68
5262.55
𝐽 = 5579.73 Hz − 5423.60 Hz = 𝟏𝟓𝟔. 𝟏𝟑 𝐇𝐳
δ =
5579.73 Hz + 5423.60 Hz
2
= 𝟓𝟓𝟎𝟏. 𝟔𝟕 𝐇𝐳
5501.67
5340.62
The remaining doublet of doublets is easy to decompose.
J = 156.13 Hz
Carbon multiplet detailed analysis
~
~
~
~
5584.45
5575.00
5267.28
5257.82
Hz
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
86.20
63.43
5579.73
J = 9.45 Hz
5423.60
5423.40
5418.87
5428.32
5413.95
5418.68
5262.55
𝐽 = 5501.67 Hz − 5340.62 Hz = 𝟏𝟔𝟏. 𝟎𝟓 𝐇𝐳
5501.67
5340.62
J = 156.13 Hz
J = 161.05 Hz
δ =
5501.67 Hz + 5340.62 Hz
2 ∗ 62.90 MHz
= 𝟖𝟔. 𝟏𝟗 𝐩𝐩𝐦
86.19
Carbon multiplet detailed analysis
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
86.20
63.43
𝐽 = 5501.67 Hz − 5340.62 Hz = 𝟏𝟔𝟏. 𝟎𝟓 𝐇𝐳
J = 156.13 Hz
~
~
~
~
5584.45
5575.00
5267.28
5257.82
Hz
5579.73
5423.60
5423.40
5418.87
5428.32
5413.95
5418.68
5262.55
5501.67
5340.62
J = 161.05 Hz
δ =
5501.67 Hz + 5340.62 Hz
2 ∗ 62.90 MHz
= 𝟖𝟔. 𝟏𝟗 𝐩𝐩𝐦
86.19
J = 9.45 Hz
J = 156.13 Hz
J = 161.05 Hz
J = 9.45 Hz
Aber …
Es geht zwar nur um 0.01 ppm, aber wieso
unterscheiden sich die beiden Werte?
We still have to assign the coupling constants.
Carbon multiplet detailed analysis
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
86.20
63.43
86.19
J = 9.45 Hz
J = 156.13 Hz
J = 161.05 Hz
It is not necessary for the evaluation of this
spectroscopic challenge, but at this point a more
detailed explanation might be a reasonale idea.
ν
measurement
(62.902 MHz)
ν
signal
ν
reference
(62.896 MHz)
spectral width
δ =
ν
Signal
− ν
Referenz
ν
Referenz
Measurements are made with a fixed frequency and a spectrum width
determined by the dwell time.
Within the spectrum width are the observed signal and a reference signal,
from which according to the known formula
the chemical shift is determined.
The frequency of the reference signal has nothing to do with the
measuring frequency. In this example the two frequencies differ by
about 6 kHz.
Carbon multiplet detailed analysis
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
86.20
63.43
J = 9.45 Hz
J = 156.13 Hz
J = 161.05 Hz
ν
measurement
(62.902 MHz)
spectral width
ν
signal
ν
reference
(62.896 MHz)
It is not necessary for the evaluation of this
spectroscopic challenge, but at this point a more
detailed explanation might be a reasonale idea.
86.19
The difference of about 0.01 ppm here is due to the fact that the processing
software correctly converts from Hz to ppm with ν
reference
, but when
publishing spectra (as done here), the measurement frequency is usually
specified.
For nuclei with a very wide range of chemical shift (e.g. platinum), this can
result in significantly larger errors than 0.01 ppm.
Carbon multiplet detailed analysis
86.20
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
63.43
J = 9.45 Hz
J = 156.13 Hz
J = 161.05 Hz
161.05
Hz
156.13
Hz
9.45 Hz
There remains some ambiguity assigning the coupling constant.
A typical value for
1
J
CH
(with sp
2
hybridised carbon) is about 165 Hz.
A clear assignment of the two values would only be possible by
selectively decoupling one of the two protons. Technically, this is
not quite trivial because of the multiplet structure and the slight
difference in the chemical shift between the two protons.
The last copulation constant is easy to assign.
86.20
~
~
~
~
~
~
Carbon multiplet detailed analysis
1099.12
1096.42
1093.73
972.60
969.90
967.19
846.07
843.37
840.66
719.52
716.83
714.16
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
63.43
161.05
Hz
156.13
Hz
9.45 Hz
86.20
The multiplet of the carbon atom at 14.42 ppm is much easier to
decompose.
A triplet structure occurs a total of four times, the intensity distribution
of the four blocks with triplet structure is about 1 : 3 : 3 : 1.
~
~
~
~
~
~
Carbon multiplet detailed analysis
1099.12
1096.42
1093.73
972.60
969.90
967.19
846.07
843.37
840.66
719.52
716.83
714.16
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
63.43
161.05
Hz
156.13
Hz
9.45 Hz
86.20
𝐽 =
1099.12 Hz − 1093.73 Hz
2
= 𝟐. 𝟔𝟗 𝐇𝐳
J = 2.69 Hz
1096.42
969.90
843.37
716.83
For the detailed analysis, the coupling constants of the triplets are extracted first.
The chemical shift for the analysis of the parent multiplet is simply taken from the middle
line of each triplet.
~
~
~
~
~
~
Carbon multiplet detailed analysis
1099.12
1096.42
1093.73
972.60
969.90
967.19
846.07
843.37
840.66
719.52
716.83
714.16
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
63.43
161.05
Hz
156.13
Hz
9.45 Hz
86.20
𝐽 =
1096.42 Hz − 716.83 Hz
3
= 𝟏𝟐𝟔. 𝟓𝟑 𝐇𝐳
J = 2.69 Hz
1096.42
969.90
843.37
716.83
What remains is the simple analysis of the quartet.
J = 126.53 Hz
δ =
1096.42 Hz + 716.83 Hz
2 ∗ 62.90 MHz
= 𝟏𝟒. 𝟒𝟏 𝐩𝐩𝐦
14.41 ppm
The discrepancy of 0.01 ppm has already been explained.
Carbon multiplet detailed analysis
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
63.43
161.05
Hz
156.13
Hz
9.45 Hz
86.20
J = 2.69 Hz
J = 126.53 Hz
~
~
~
~
~
~
1099.12
1096.42
1093.73
972.60
969.90
967.19
846.07
843.37
840.66
719.52
716.83
714.16
1096.42
969.90
843.37
716.83
14.41 ppm
126.53 Hz
2.69 Hz
The assignment of the coupling constants is
very simple thanks to the multiplicities.
Carbon multiplet detailed analysis
~
~
~
~
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
63.43
86.20
161.05
Hz
156.13
Hz
9.45 Hz
126.53 Hz
2.69 Hz
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
63.43
86.20
The underlying structure of the carbon signal at 63.43 ppm must be a
triplet with a coupling constant of about 130 Hz because there are two
chemically equivalent methylene protons separated by one bond.
To determine the coupling constant of this
triplet we need the centre of the left and
the right submultiplet.
Carbon multiplet detailed analysis
~
~
~
~
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
63.43
86.20
4137.70
4126.14
3851.72
3840.17
4131.92 Hz 3845.94 Hz
𝐽 =
4131.92 Hz − 3845.94 Hz
2
= 𝟏𝟒𝟐. 𝟗𝟗 𝐇𝐳
We only need the frequencies of a few selected peaks.
142.99 Hz
Carbon multiplet detailed analysis
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
63.43
86.20
142.99 Hz
The three submultiplets each have an identical fine structure.
For further analysis, one of the three partial multiplets is
sufficient; of course, the one with the best signal-to-noise
ratio is used..
3999.27
3998.63
3994.72
3994.07
3992.92
3992.25
3990.17
3989.52
3988.36
3987.70
3985.62
3984.99
3983.80
3983.15
3979.23
3978.61
Hz
Carbon multiplet detailed analysiss
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
63.43
86.20
142.99 Hz
3999.27
3998.63
3994.72
3994.07
3992.92
3992.25
3990.17
3989.52
3988.36
3987.70
3985.62
3984.99
3983.80
3983.15
3979.23
3978.61
Hz
𝐽 = 3990.17 Hz − 3989.52 Hz = 𝟎. 𝟔𝟓 𝐇𝐳
In the very confusing multiplet, one immediately notices that all lines
appear twice.
After reduction of all 8 doublets to one line each (the coupling constant
determined as an example for one pair of lines can also be averaged over
all 8 doublets), a no less clear multiplet of 8 lines in the intensity ratio 1 :
3 : 1 : 3 : 3 : 1 : 3 : 1 remains.
J = 0.65 Hz
Carbon multiplet detailed analysis
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
63.43
86.20
142.99 Hz
3999.27
3998.63
3994.72
3994.07
3992.92
3992.25
3990.17
3989.52
3988.36
3987.70
3985.62
3984.99
3983.80
3983.15
3979.23
3978.61
Hz
With the "ruler method" already used before, one finds four
equidistant lines.
It is a quartet. A second quartet is now quite easy to recognise.
J = 0.65 Hz
Carbon multiplet detailed analysis
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
63.43
86.20
142.99 Hz
3999.27
3998.63
3994.72
3994.07
3992.92
3992.25
3990.17
3989.52
3988.36
3987.70
3985.62
3984.99
3983.80
3983.15
3979.23
3978.61
Hz
The base of the quartets are two lines of equal intensity.
J = 0.65 Hz
Carbon multiplet detailed analysis
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
63.43
86.20
142.99 Hz
3999.27
3998.63
3994.72
3994.07
3992.92
3992.25
3990.17
3989.52
3988.36
3987.70
3985.62
3984.99
3983.80
3983.15
3979.23
3978.61
Hz
The coupling constant of the quartet is three times the distance
between these two lines.
We can read off the numeric values here, for example.
𝐽 = 3999.27 Hz − 3985.62 Hz /3 = 𝟒. 𝟓𝟓 𝐇𝐳
J = 0.65 Hz
J = 4.55 Hz
Carbon multiplet detailed analysis
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
63.43
86.20
142.99 Hz
3999.27
3998.63
3994.72
3994.07
3992.92
3992.25
3990.17
3989.52
3988.36
3987.70
3985.62
3984.99
3983.80
3983.15
3979.23
3978.61
Hz
The remaining doublet is of course
very easy to analyse.
Here is one way to
extract the coupling
constant.
𝐽 = 3994.72 Hz − 3988.36 Hz = 𝟔. 𝟑𝟔 𝐇𝐳
J = 0.65 Hz
J = 4.55 Hz
J = 6.36 Hz
Carbon multiplet detailed analysis
63.43
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
86.20
142.99 Hz
J = 0.65 Hz
J = 4.55 Hz
J = 6.36 Hz
63.43
The assignment of the three coupling constants is not too difficult.
There is only one option for the quartet.
4.55 Hz
The source of the doublet with the coupling constant of 6.36 Hz is one of three protons. Two of them
are four bonds away from the carbon atom. 6.36 Hz are extremely unlikely for such a 4-bond coupling
constant.
6.36 Hz
The value of 0.65 Hz must be a 4-bond coupling constant to one of the remaining two protons.
Which of the two protons causes the splitting cannot be determined here.
0.65 Hz
~
~
9639.42
9636.70
9634.10
9631.51
9630.97
9628.84
9628.38
9625.76
9623.07
9459.10
9456.41
9453.80
9451.21
9450.65
9448.09
9445.46
9442.79
Hz
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
86.20
63.43
Carbon multiplet detailed analysis
𝐽 =
9639.42 Hz + 9623.07 Hz
2
−
9459.10 Hz − 9442.79 Hz
2
= 𝟏𝟖𝟎. 𝟑𝟎 𝐇𝐳
The basic structure of the multiplet of the carbon atom at 151.70
ppm is a doublet, caused by the proton at 6.42 ppm.
180.30 Hz
To extract the other coupling
constants, one of the two partial
multiples is sufficient..
The coupling constant is determined from the difference of the mean
value of the two partial multiplets with currently unknown fine structure.
9459.10
9456.41
9453.80
9451.21
9450.65
9448.09
9445.46
9442.79
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
86.20
63.43
Carbon multiplet detailed analysis
Hz
180.30 Hz
This partial multiplet should be an
doublet
of doublets
of triplets
with alltogether 12 lines.
Eight of these lines are visible (ten, if you include two shoulders).
Translation in progress beyond this slide.
9459.10
9456.41
9453.80
9451.21
9450.65
9448.09
9445.46
9442.79
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
86.20
63.43
Analyse der Kohlenstoffmultipletts
Hz
180.30 Hz
1 72 3 4
5
6 98
10
1211 13 14
150
Am leichtesten sollte man eine Triplettstruktur erkennen können,
vorausgesetzt, mindestens zwei der drei Linien des Tripletts
zeigen keine Überlagerung mit anderen Linien.
Zwei der Linien des Multipletts scheinen im Intensitätsverhältnis
1 : 2 vorzuliegen.
Markieren wir versuchsweise diese beiden Linien und markieren
gleichzeitig die Differenz auf einem Lineal.
Finden wir durch Verschieben des Lineals die dritte Linie des
Tripletts? Versuchen wir es einfach.
9459.10
9456.41
9453.80
9451.21
9450.65
9448.09
9445.46
9442.79
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
86.20
63.43
Analyse der Kohlenstoffmultipletts
Hz
180.30 Hz
1 72 3 4
5
6 98
10
1211 13 14
150
Am leichtesten sollte man eine Triplettstruktur erkennen können,
vorausgesetzt, mindestens zwei der drei Linien des Tripletts
zeigen keine Überlagerung mit anderen Linien.
Zwei der Linien des Multipletts scheinen im Intensitätsverhältnis
1 : 2 vorzuliegen.
Markieren wir versuchsweise diese beiden Linien und markieren
gleichzeitig die Differenz auf einem Lineal.
Finden wir durch Verschieben des Lineals die dritte Linie des
Tripletts? Versuchen wir es einfach.
1 72 3 4
5
6 98
10
1211 13 14
150
9459.10
9456.41
9453.80
9451.21
9450.65
9448.09
9445.46
9442.79
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
86.20
63.43
Analyse der Kohlenstoffmultipletts
Hz
180.30 Hz
1 72 3 4
5
6 98
10
1211 13 14
150
Die dritte Linie unseres Tripletts könnte sich in diesem Bereich
überlagerter Linien verbergen.
Setzen wir eine Markierung und markieren anschließend das
komplette Triplett auf dem Lineal.
1 72 3 4
5
6 98
10
1211 13 14
150
Die gleiche Struktur finden wir auf der rechten Seite des
Multipletts noch einmal. Wir können jetzt aus beiden Tripletts
die Linie rekonstruieren, aus der durch Koplung mit den
Methylenprotonen die Triplettstruktur entsteht. Das Lineal mit
den drei Markierungen heben wir für eine spätere Verwendung
auf.
9459.10
9456.41
9453.80
9451.21
9450.65
9448.09
9445.46
9442.79
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
86.20
63.43
Analyse der Kohlenstoffmultipletts
Hz
180.30 Hz
Um das theoretisch vorausgesagte Dublett von Dubletts von Tripletts zu
erhalten, müssten sich in dem Multiplett noch zwei weitere Tripletts
verstecken.
Hierfür kommt nur der markierte Bereich des Multipletts in Frage.
Die Linien außerhalb des Bereiches werden mit sehr hoher Sicherheit
durch die bereits gefundenen Tripletts vollständig erklärt.
J = 9459.10 Hz – 9456.41 Hz = 2.69 Hz
2.69 Hz
9459.10
9456.41
9453.80
9451.21
9450.65
9448.09
9445.46
9442.79
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
86.20
63.43
Analyse der Kohlenstoffmultipletts
Hz
180.30 Hz
1 72 3 4
5
6 98
10
1211 13 14
150
1 72 3 4
5
6 98
10
1211 13 14
0
1 72 3 4
5
6 98
10
1211 13
0
Probieren wir es wieder mit dem Lineal, auf dem bereits die drei
Linien des Tripletts markiert sind.
Hier scheint sich ein Triplett zu befinden.
Und hier ein weiteres Triplett.
2.69 Hz
9459.10
9456.41
9453.80
9451.21
9450.65
9448.09
9445.46
9442.79
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
86.20
63.43
Analyse der Kohlenstoffmultipletts
Hz
180.30 Hz
1 72 3 4
5
6 98
10
1211 13
0
Nur für die mittlere Linie der beiden Tripletts beobachtet
man keine Überlagerung. Eine Kopplungskonstante kann
hier nicht gemessen werden, die ist allerdngs auch bereits
bekannt.
2.69 Hz
9459.10
9456.41
9453.80
9451.21
9450.65
9448.09
9445.46
9442.79
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
86.20
63.43
Analyse der Kohlenstoffmultipletts
Hz
180.30 Hz
Die Analyse des verbleibenden Dubletts von
Dubletts ist jetzt einfach.
Die nötigen chemischen Verschiebungen
können ohne Rechnung direkt dem
Multiplett entnommen werden.
2.69 Hz
9456.41
9451.21
9450.65
9445.46
J = 5.20 Hz
J = 5.75 Hz
5.20 Hz
5.75 Hz
Carbon chemical shifts and carbon proton coupling constants
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
86.20
63.43
180.30 Hz
2.69 Hz
5.20 Hz
5.75 Hz
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
86.20
142.99 Hz
63.43
4.55 Hz
6.36 Hz
0.65 Hz
6.42
4.14
3.94
O CH
2
CH
3
CC
HH
H
14.42
151.70
63.43
161.05
Hz
156.13
Hz
9.45 Hz
86.20
126.53 Hz
2.69 Hz
(Bei den gestrichelten Linien ist die
Zuordnung nicht eindeutig.)
6.42
4.14
3.94
3.72
1.26
7.04 Hz
14.36 Hz
6.83 Hz
1.90 Hz
O CH
2
CH
3
CC
HH
H
Proton chemical shifts and proton proton coupling constants
