Exercise plus Solution – Quick PDF overview
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55.53
61.71
109.64
115.43
119.41
127.48
128.46
128.94
146.13
147.69
5560657075808590100105110115120125130135140145 ppm
13
C
C
10
H
12
O
3
measured in DMSO-d
6
Coniferous wood contains significant amounts
of this compound. Can you determine the
structure of this natural compound?
13
C{
1
H} NMR spectrum
measured at 125.70{499.84} MHz
4.04.55.05.56.06.57.07.58.08.59.0 ppm
1
H
3.00
2.01
0.23
0.99
0.96
0.96
0.97
1.00
0.75
Inte-
gral
3400.45
3398.57
3392.35
3390.46
3355.45
3347.38
3216.10
3214.54
3212.91
3200.39
3198.70
3197.28
3101.52
3096.11
3090.72
3085.68
3080.25
3074.83
2384.10
2378.56
2373.14
2041.94
2040.26
2036.47
2034.82
2031.02
2029.36
3497.14
3495.28
Hz
8.99
3.78
1
H NMR spectrum using
presaturation at 4.7 ppm
recorded at 499.84 MHz
x 1.3
0.97
x 4
0.92
Please use the corrected
integrals at 4.7 and 9 ppm!
The explanation can be found on
the last page of the solution.
Don‘t worry about this correction
of two values for the moment.
Nothing is wrong.
4.04.55.05.56.06.57.07.5
1
H
ppm
60
70
80
90
100
110
120
13
C
~
126
127
128
129
6.206.40
ppm
1
H/
13
C HSQC
recorded at 499.84/125.70 MHz
There is no cross peak to the
proton signal at 9 ppm in the HSQC.
~
~
4.04.55.05.56.06.57.07.5
1
H
ppm
1
H
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
1
H TOCSY
measured at 499.84 MHz
~
4.04.55.05.56.06.57.07.58.59.09.5
1
H
ppm
110
115
120
125
130
140
145
13
C
part of
1
H/
13
C HMBC
measured at 499.84/125.70 MHz
If the cross peaks are difficult to distinguish
in some sections of the spectrum, please
check whether there is a better resolved
expansion of the same spectral region on
the next page.
4.04.55.05.56.06.57.07.5
~
61.5
62.0
13
C
4.64.85.05.25.65.86.06.26.4
1
H
ppm
128
129
13
C
~
4.04.55.06.06.57.07.5
1
H
ppm
110
115
120
125
130
140
145
13
C
further parts of the
1
H/
13
C HMBC
recorded at 499.84/125.70 MHz
4.04.55.05.56.06.57.07.58.08.59.0 ppm
1
H
3.00
2.01
0.23
0.99
0.96
0.96
0.97
1.00
0.75
Inte-
gral
8.99
3.78
Integration
Signal groups [ppm] 9,00 7,00 6,80 6,70 6,40 6,20 4,75 4,10 3,80
Integrals 0,97 1,00 0,97 0,96 0,96 0,99 0,92 2,01 3,00
Sum of integrals 11,78
Number of protons from molecular formula 12
Coefficient of proportionality 1,019
Number of protons 0,99 1,02 0,99 0,98 0,98 1,01 0,94 2,08 3,10
Proton number (rounded) 1 1 1 1 1 1 1 2 3
Total number of protons 12
x 1.3
0.97
x 4
0.92
~
~
4.04.55.05.56.06.57.07.5
1
H
ppm
1
H
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
First partial structure
Spin system taken from TOCSY
The TOCSY contains three spin systems with
- three
- four
- one
proton multiplet(s).
Let us start with the most extensive spin system.
and
~
~
4.04.55.05.56.06.57.07.5
1
H
ppm
1
H
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
Let us now determine the chemical shifts of
the proton multiplets, which belong to this
spin system.
In all subsequent 2D spectra, cross and diagonal
peaks that are no longer needed for further
evaluation after extracting the information they
contain become deleted. The same procedure
applies to the signals in the projections.
Thus signals that have already been evaluated
cannot be used a second time by mistake.
When using printed spectra on paper, simply cross
out such signals.
Spin system taken from TOCSY
First partial structure
~
~
4.04.55.05.56.06.57.07.5
1
H
ppm
1
H
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
6.42
(1H)
6.18
(1H)
6.42
(1H)
6.18
(1H)
Spin system taken from TOCSY
First partial structure
~
~
4.04.55.05.56.06.57.07.5
1
H
ppm
1
H
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
6.18
(1H)
6.42
(1H)
4.76
(1H)
4.07
(2H)
4.76
(1H)
4.07
(2H)
Spin system taken from TOCSY
First partial structure
~
~
4.04.55.05.56.06.57.07.5
1
H
ppm
1
H
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
6.18
(1H)
6.42
(1H)
4.76
(1H)
4.07
(2H)
Later we will need the TOCSY again. To simplify later use,
all cross and diagonal peaks, as well as the associated
projections of the four extracted proton multiplets, were
removed. On a sheet of paper, you can simply cross out
the signals using a pencil.
The multiplet structures of all four multiplets are rather
simple, let us evaluate these multiplets as the next step.
Spin system taken from TOCSY
First partial structure
3216.10
3214.54
3212.91
3200.39
3198.70
3197.28
3101.52
3096.11
3090.72
3085.68
3080.25
3074.83
2384.10
2378.56
2373.14
2041.94
2040.26
2036.47
2034.82
2031.02
2029.36
Multiplicity and coupling constants
6.18
(1H)
6.42
(1H)
4.76
(1H)
4.07
(2H)
6.42 ppm
15.84 Hz(d)
1.58 Hz(t)
J(t) = 1.58 Hz
J(d) = 15.84 Hz
Let's start with the doublet of triplets at
lowest field (6.42 ppm).
First partial structure
3216.10
3214.54
3212.91
3200.39
3198.70
3197.28
3101.52
3096.11
3090.72
3085.68
3080.25
3074.83
2384.10
2378.56
2373.14
2041.94
2040.26
2036.47
2034.82
2031.02
2029.36
6.18
(1H)
6.42
(1H)
4.76
(1H)
4.07
(2H)
15.84 Hz(d)
1.58 Hz(t)
As an example for the second doublet of triplets (
6.18 ppm) the detailed calculation of the
coupling constants is shown here.
6.42 ppm
J(d) = 15.86 Hz
J(t) = 5.41 Hz
𝐽(d) = 3096.11 Hz − 3080.25 Hz = 𝟏𝟓. 𝟖𝟔 𝐇𝐳
𝐽 t
a
=
3101.52 Hz − 3090.72 Hz
2
= 𝟓. 𝟒 𝐇𝐳
𝐽 t
b
=
3085.68 Hz − 3074.83 Hz
2
= 𝟓. 𝟒𝟐 𝐇𝐳
𝐽 t =
𝐽 t
a
+ 𝐽 t
b
2
= 𝟓. 𝟒𝟏 𝐇𝐳
15.86 Hz(d)
5.41 Hz(t)
Multiplicity and coupling constants
First partial structure
3216.10
3214.54
3212.91
3200.39
3198.70
3197.28
3101.52
3096.11
3090.72
3085.68
3080.25
3074.83
2384.10
2378.56
2373.14
2041.94
2040.26
2036.47
2034.82
2031.02
2029.36
6.18
(1H)
6.42
(1H)
4.76
(1H)
4.07
(2H)
15.84 Hz(d)
1.58 Hz(t)
The triplet at 4.76 ppm is simple.
𝐽 t =
2384.10 Hz − 2373.14 Hz
2
= 𝟓. 𝟒𝟖 𝐇𝐳
15.86 Hz(d)
5.41 Hz(t)
4.76 ppm
5.48 Hz(t)
First partial structure
Multiplicity and coupling constants
3216.10
3214.54
3212.91
3200.39
3198.70
3197.28
3101.52
3096.11
3090.72
3085.68
3080.25
3074.83
2384.10
2378.56
2373.14
2041.94
2040.26
2036.47
2034.82
2031.02
2029.36
6.18
(1H)
6.42
(1H)
4.76
(1H)
4.07
(2H)
J(t) = 5.45 Hz
J(d) = 1.66 Hz
5.45 Hz(t)
1.66 Hz(d)
4.07 ppm
15.84 Hz(d)
1.58 Hz(t)
15.86 Hz(d)
5.41 Hz(t) 5.48 Hz(t)
There is a last doublet of triplets at 4.07 ppm.
First partial structure
Multiplicity and coupling constants
4.04.55.05.56.06.57.07.5
1
H
ppm
60
70
80
90
100
110
120
13
C
~
126
127
128
129
6.206.40
ppm
Building blocks
6.18
(1H)
6.42
(1H)
4.76
(1H)
4.07
(2H)
5.45 Hz(t)
1.66 Hz(d)15.84 Hz(d)
1.58 Hz(t)
15.86 Hz(d)
5.41 Hz(t) 5.48 Hz(t)
6.42
(1H)
6.18
(1H)
4.76
(1H)
4.07
(2H)
We have four proton multiplets. The HSQC
shows us the attached carbon atoms.
Let us label the projections with the chemical
shifts of the known proton multiplets.
6.18
(1H)
6.42
(1H)
4.76
(1H)
4.07
(2H)
Due to close chemical shifts of three carbon
signals there is an enlarged part of the HSQC
for two cross peaks only.
First partial structure
4.04.55.05.56.06.57.07.5
1
H
ppm
60
70
80
90
100
110
120
13
C
~
126
127
128
129
6.206.40
ppm
5.45 Hz(t)
1.66 Hz(d)15.84 Hz(d)
1.58 Hz(t)
15.86 Hz(d)
5.41 Hz(t) 5.48 Hz(t)
6.42
(1H)
6.18
(1H)
4.76
(1H)
4.07
(2H)
6.18
(1H)
6.42
(1H)
4.76
(1H)
4.07
(2H)
128.94
C
H
128.94
6.42
C
H
128.94
6.42
There is another cross peak inside this enlarged
area.
First partial structure
Building blocks
4.04.55.05.56.06.57.07.5
1
H
ppm
60
70
80
90
100
110
120
13
C
~
126
127
128
129
6.206.40
ppm
5.45 Hz(t)
1.66 Hz(d)15.84 Hz(d)
1.58 Hz(t)
15.86 Hz(d)
5.41 Hz(t) 5.48 Hz(t)
6.18
(1H)
4.76
(1H)
4.07
(2H)
6.18
(1H)
6.42
(1H)
4.76
(1H)
4.07
(2H)
127.48
C
H
128.94
6.42
C
H
127.48
6.18
C
H
127.48
6.18
After removing all signals that are no longer
needed from the HSQC for sake of clarity, we can
turn to the proton signal at 4.07 ppm.
First partial structure
Building blocks
4.04.55.05.56.06.57.07.5
1
H
ppm
60
70
80
90
100
110
120
13
C
~
5.45 Hz(t)
1.66 Hz(d)15.84 Hz(d)
1.58 Hz(t)
15.86 Hz(d)
5.41 Hz(t) 5.48 Hz(t)
6.18
(1H)
6.42
(1H)
4.76
(1H)
4.07
(2H)
C
H
128.94
6.42
C
H
127.48
6.18
61.71
4.76
(1H)
4.07
(2H)
C
H
2
4.07
61.71
C
H
2
4.07
61.71
And the proton multiplet 4.76 ppm?
First partial structure
Building blocks
4.04.55.05.56.06.57.07.5
1
H
ppm
60
70
80
90
100
110
120
13
C
~
5.45 Hz(t)
1.66 Hz(d)15.84 Hz(d)
1.58 Hz(t)
15.86 Hz(d)
5.41 Hz(t) 5.48 Hz(t)
6.18
(1H)
6.42
(1H)
4.76
(1H)
4.07
(2H)
C
H
128.94
6.42
C
H
127.48
6.18
4.76
(1H)
C
H
2
4.07
61.71
There is no adjacent carbon atom. The proton
at 4.76 ppm has to be an OH group.
O
H
4.76
In the interest of clarity,
we remove all previously
used information from the
HSQC before continuing.
First partial structure
Building blocks
4.04.55.05.56.06.57.07.5
1
H
ppm
60
70
80
90
100
110
120
13
C
~
Combine the fragments
5.45 Hz(t)
1.66 Hz(d)15.84 Hz(d)
1.58 Hz(t)
15.86 Hz(d)
5.41 Hz(t) 5.48 Hz(t)
6.18
(1H)
6.42
(1H)
4.76
(1H)
4.07
(2H)
C
H
2
4.07
61.71
O
H
4.76
C
H
128.94
6.42
C
H
127.48
6.18
C
C
H
H
128.94
127.48
6.42
6.18
There is only one possible combination for the
two =CH- fragments.
First partial structure
5.45 Hz(t)
1.66 Hz(d)15.84 Hz(d)
1.58 Hz(t)
15.86 Hz(d)
5.41 Hz(t) 5.48 Hz(t)
6.18
(1H)
6.42
(1H)
4.76
(1H)
4.07
(2H)
C
H
2
4.07
61.71
O
H
4.76
C
C
H
H
128.94
127.48
6.42
6.18
15.85
Hz
Both proton multipletts have a common
coupling constant of 15.85 Hz (averaged
value). Thus the protons are trans to each
other.
We note the other two coupling constants
of 1.58 Hz and 5.41 Hz next to the protons
for future use.
1.58 Hz(t)
5.41 Hz(t)
The multiplet of the OH proton is a clean
triplet. The OH proton can only be bound to
the methylene group..
First partial structure
Combine the fragments
5.45 Hz(t)
1.66 Hz(d)
5.48 Hz(t)
4.76
(1H)
4.07
(2H)
C
C
H
H
128.94
127.48
6.42
6.18
15.85
Hz
1.58 Hz(t)
5.41 Hz(t)
C
H
2
4.07
61.71
O
H
4.76
C
H
2
O
H
4.07
61.71
4.76
5.48 Hz
And what about the two coupling constants in
the multiplet of methylene protons at 4.07
ppm?
A triplet with a coupling constant of 5.45 Hz and a
doublet with a coupling constant of 1.66 Hz?
First partial structure
Combine the fragments
5.45 Hz(t)
1.66 Hz(d)
4.07
(2H)
C
C
H
H
128.94
127.48
6.42
6.18
15.85
Hz
1.58 Hz(t)
5.41 Hz(t)
C
H
2
O
H
4.07
61.71
4.76
5.48 Hz
5.45 Hz(t)
1.66 Hz(d)
The multiplet of the methylene protons cannot
possibly be a triplet. There are no two
neighbouring protons that are equivalent.
However, a doublet of doublets with the very
similar coupling constants of 5.48 Hz (OH group)
and 5.41 Hz (=CH- group) would be a good
explanation.
This results in a pseudo triplet with an
apparent coupling constant that corresponds to
the averaged value of the two values
mentioned above.
5.41 Hz is a typical vicinal coupling constant, the -
CH
2
-OH group is bound to the =CH- group with
the proton signal at 6.18 ppm.
We keep the doublet splitting of 1.66 Hz for future use.
First partial structure
Combine the fragments
C
C
H
H
128.94
127.48
6.42
6.18
15.85
Hz
1.58 Hz(t)
5.41 Hz(t)
C
H
2
O
H
4.07
61.71
4.76
5.48 Hz
5.45 Hz(t)
1.66 Hz(d)
The coupling constants unassigned so far can
now be easily explained.
C
C
H
H
C
H
2
O
H
4.07
61.71
128.94
127.48
6.42
6.18
4.76
15.85
Hz
5.48 Hz
5.45 Hz(t)
1.66 Hz(d)
1.58 Hz(t)
5.41 Hz(t)
First partial structure
Combine the fragments
C
C
H
H
C
H
2
O
H
4.07
61.71
128.94
127.48
6.42
6.18
4.76
15.85
Hz
5.48 Hz
5.45 Hz(t)
1.66 Hz(d)
1.58 Hz(t)
5.41 Hz(t)
5.41 Hz
1.62 Hz
At this point, it makes sense to temporarily “hide" the
structural fragment found so far with the molecular
formula C
3
H
5
O and to continue with the extraction of
the remaining pieces of information from the HSQC.
If we subtract C
3
H
5
O from the molecular formula, we
still have to assign
- C
7
H
7
O
2
and
- four double bond equivalents.
First partial structure
Combine the fragments
C
C
H
H
C
H
2
O
H
4.07
61.71
128.94
127.48
6.42
6.18
4.76
15.85
Hz
5.48 Hz
5.41 Hz
1.62 Hz
4.04.55.05.56.06.57.07.5
1
H
ppm
60
70
80
90
100
110
120
13
C
~
More building blocks
HSQC
First we take the chemical shifts from the proton and
carbon one dimensional spectra, as well as the
integrals from the proton spectrum and add those
values to the HSQC projections.
4.04.55.05.56.06.57.07.5
1
H
ppm
60
70
80
90
100
110
120
13
C
~
HSQC
6.99
(1H)
6.79
(1H)
6.70
(1H)
3.78
(3H)
55.53
109.64
115.43
119.41
128.46
Three cross peaks clearly belong to three different
=CH- groups.
More building blocks
4.04.55.05.56.06.57.07.5
1
H
ppm
60
70
80
90
100
110
120
13
C
~
HSQC
6.99
(1H)
6.79
(1H)
6.70
(1H)
3.78
(3H)
55.53
109.64
115.43
119.41
128.46
C
H
6.99
109.64
C
H
6.99
109.64
More building blocks
4.04.55.05.56.06.57.07.5
1
H
ppm
60
70
80
90
100
110
120
13
C
~
HSQC
6.79
(1H)
6.70
(1H)
3.78
(3H)
55.53
115.43
119.41
128.46
C
H
6.99
109.64
C
H
6.70
115.43
C
H
6.70
115.43
More building blocks
4.04.55.05.56.06.57.07.5
1
H
ppm
60
70
80
90
100
110
120
13
C
~
HSQC
6.79
(1H)
3.78
(3H)
55.53
119.41
128.46
C
H
6.99
109.64
C
H
6.70
115.43
C
H
6.79
119.41
C
H
6.79
119.41
Finally there is a methyl group.
More building blocks
4.04.55.05.56.06.57.07.5
1
H
ppm
60
70
80
90
100
110
120
13
C
~
HSQC
3.78
(3H)
55.53
128.46
C
H
6.99
109.64
C
H
6.70
115.43
C
H
6.79
119.41
H
3
C
3.78
55.53
H
3
C
3.78
55.53
And the carbon signal with the chemical shift
of 128.46 ppm?
More building blocks
4.04.55.05.56.06.57.07.5
1
H
ppm
60
70
80
90
100
110
120
13
C
~
HSQC
128.46
C
H
6.99
109.64
C
H
6.70
115.43
C
H
6.79
119.41
H
3
C
3.78
55.53
C
128.46
C
128.46
And the carbon signal with the chemical shift
of 128.46 ppm?
There is no cross peak in the HSQC, that‘s a quaternary
carbon atom.
What else is missing?
More building blocks
9.0
ppm1
H
0.75
Inte-
gral
8.99
146.13
147.69
140145 ppm13
C
C
C
H
H
C
H
2
O
H
4.07
61.71
128.94
127.48
6.42
6.18
4.76
15.85
Hz
5.48 Hz
5.41 Hz
1.62 Hz
4.04.55.05.56.06.57.07.5
1
H
ppm
60
70
80
90
100
110
120
13
C
~
C
H
6.99
109.64
C
H
6.70
115.43
C
H
6.79
119.41
H
3
C
3.78
55.53
C
128.46
1D signals without
corresponding HSQC cross peak
Three signals in the
1
H and
13
C spectra do not show any cross peak
in the HSQC. This immediately results in three structural fragments
(For the proton, the binding partner has to be oxygen once we
exclude the possibility of a CH fragment since we only have C, O
and H in the molecular formula).
O
H
8.99
C
147.69
C
146.13
Last building blocks
C
C
H
H
C
H
2
O
H
4.07
61.71
128.94
127.48
6.42
6.18
4.76
15.85
Hz
5.48 Hz
5.41 Hz
1.62 Hz
C
H
6.99
109.64
C
H
6.70
115.43
C
H
6.79
119.41
H
3
C
3.78
55.53
C
128.46
Final building blocks
A short inventory
O
H
8.99
C
147.69
C
146.13
atoms found so far C
10
H
12
O
2
double bond equivalents 4
molecular formula C
10
H
12
O
3
resulting double bond
equivalents from 5
molecular formula
still missing
- one oxygen atom
- one double bond equivalent
C
C
H
H
C
H
2
O
H
4.07
61.71
128.94
127.48
6.42
6.18
4.76
15.85
Hz
5.48 Hz
5.41 Hz
1.62 Hz
C
H
6.99
109.64
C
H
6.70
115.43
C
H
6.79
119.41
H
3
C
3.78
55.53
C
128.46
O
H
8.99
C
147.69
C
146.13
The remaining oxygen fits best next to the methyl group.
A carbon signal with a chemical shift of about 60 ppm and a
proton signal at about 4 ppm are very characteristic for a
methoxy group. With the methyl group next to an sp
2
carbon
atom, we would expect chemical shifts of about 2.5 … 3 ppm in
the proton spectrum and 25 ppm in the carbon spectrum.
O
H
3
C
3.78
55.53
And these six fragments, when put together, would make a nice phenyl ring with
three free valences - matching the three remaining structural fragments - and the
still missing double bond equivalent.
Final building blocks
A short inventory
C
C
H
H
C
H
2
O
H
4.07
61.71
128.94
127.48
6.42
6.18
4.76
15.85
Hz
5.48 Hz
5.41 Hz
1.62 Hz
C
H
6.99
109.64
C
H
6.70
115.43
C
H
6.79
119.41
H
3
C
3.78
55.53
C
128.46
O
H
8.99
C
147.69
C
146.13
O
H
3
C
3.78
55.53
C
H
6.99
109.64
C
H
6.70
115.43
C
H
6.79
119.41
H
3
C
3.78
55.53
C
128.46
O
H
8.99
C
147.69
C
146.13
O
H
3
C
3.78
55.53
Which patterns are possible, if we assume a
triple substituted phenyl ring?
Final building blocks
A short inventory
Phenyl ring
Three base structures
C
C
H
H
C
H
2
O
H
4.07
61.71
128.94
127.48
6.42
6.18
4.76
15.85
Hz
5.48 Hz
5.41 Hz
1.62 Hz
C
H
6.99
109.64
C
H
6.70
115.43
C
H
6.79
119.41
C
128.46
O
H
8.99
C
147.69
C
146.13
O
H
3
C
3.78
55.53
We expect the following coupling constants [Hz] and multiplets.
The dashed coupling path and the resulting multiplet structure might not be
visible because of the very small coupling constant.
dd
H
H
H
2
8
8
dd
dd
H
H
H
8
2
0.5
d(d)
dd
d(d)
H
H
H
2
dd
dd
dd
2
2
C
C
H
H
C
H
2
O
H
4.07
61.71
128.94
127.48
6.42
6.18
4.76
15.85
Hz
5.48 Hz
5.41 Hz
1.62 Hz
C
H
6.99
109.64
C
H
6.70
115.43
C
H
6.79
119.41
C
128.46
O
H
8.99
C
147.69
C
146.13
O
H
3
C
3.78
55.53
H
H
H
H
H
H
H
H
H
2
8
8
8
2
2
0.5
dd
dd
dd
d(d)
dd
d(d)
dd
dd
dd
2
2
If we have a look into the TOCSY,
where we removed the signals of
the already known substructure,
there are still three aromatic
protons at 6.70 ppm, 6.79 ppm
and 6.99 ppm which might be
part of a phenyl ring.
6.70
6.79
6.99
3.78
The last remaining signal at
3.78 ppm belongs to the
methoxy group, already
assigned.
First lets have to look for three aromatic protons.
Phenyl ring
Three base structures
C
C
H
H
C
H
2
O
H
4.07
61.71
128.94
127.48
6.42
6.18
4.76
15.85
Hz
5.48 Hz
5.41 Hz
1.62 Hz
C
H
6.99
109.64
C
H
6.70
115.43
C
H
6.79
119.41
C
128.46
O
H
8.99
C
147.69
C
146.13
O
H
3
C
3.78
55.53
Multiplet structure of three
aromatic protons
H
H
H
H
H
H
H
H
H
2
8
8
8
2
2
0.5
dd
dd
dd
d(d)
dd
d(d)
dd
dd
dd
2
2
1.86 Hz
8.10 Hz
1.88 Hz
8.07 Hz
We no longer need the TOCSY.
We need the multiplet structure and the coupling
constants of the three aromatic protons.
dd dd
It is easy to analyze the multiplets.
Everything fits to the upper right structure.
6.99 6.79 6.70
6.99
6.79
6.70
Phenyl ring
C
C
H
H
C
H
2
O
H
4.07
61.71
128.94
127.48
6.42
6.18
4.76
15.85
Hz
5.48 Hz
5.41 Hz
1.62 Hz
O
H
8.99
O
H
3
C
3.78
55.53
C
H
6.99
109.64
C
H
6.70
115.43
C
H
6.79
119.41
C
C
C
C
C
C
C
C
C
H
H
H
6.99
6.79
6.70
109.64
115.43
119.41
8.09 Hz
1.88
Hz
H
H
H
8
2
0.5
d(d)
dd
d(d)
From our pile of structural fragments, we can
now pick the 6 fragments with sp
2
hybridised
C atoms and combine them according to the
pattern.
6.99
6.79
6.70
Multiplet structure of three
aromatic protons
Phenyl ring
O
H
8.99
O
H
3
C
3.78
55.53
Combine two
substructures
C
C
C
C
C
C
H
H
H
6.99
6.79
6.70
109.64
115.43
119.41
8.09 Hz
1.88
Hz
C
C
H
H
C
H
2
O
H
4.07
61.71
128.94
127.48
6.42
6.18
4.76
15.85
Hz
5.48 Hz
5.41 Hz
1.62 Hz
6.79
6.99
128.94
The HMBC contains two cross peaks that
provide the information how to link the
two large partial fragments.
O
H
8.99
O
H
3
C
3.78
55.53
C
C
C
C
C
C
H
H
H
6.99
6.79
6.70
109.64
115.43
119.41
8.09 Hz
1.88
Hz
6.79
6.99
128.94
C
C
H
H
C
H
2
O
H
4.07
61.71
128.94
127.48
6.42
6.18
4.76
15.85
Hz
5.48 Hz
5.41 Hz
1.62 Hz
C
C
H
H
C
H
2
O
H
4.07
61.71
128.94
127.48
6.42
6.18
4.76
15.85
Hz
5.48 Hz
5.41 Hz
1.62 Hz
The HMBC contains two cross peaks that
provide the information how to link the
two large partial fragments.
Altogether there are three possibility to link the fragments.
Why should we prefer the connectivity shown here and exclude the
two others?
Combine two
substructures
O
H
8.99
O
H
3
C
3.78
55.53
6.79
6.99
128.94
C
C
C
C
C
C
H
C
H
H
C
H
H
C
H
2
O
H
4.07
61.71
128.94
127.48
6.42
6.18
6.99
6.79
6.70
109.64
115.43
119.41
4.76
15.85
Hz
5.41 Hz
5.48 Hz
1.62 Hz
8.09 Hz
1.88
Hz
Starting from the C atom with the chemical shift of 128.94 ppm, the
protons with the chemical shifts of 6.79 ppm and 6.99 ppm are
comparable. Each of the protons is three bonds away from the
carbon; moreover, the geometry of the coupling path is identical in
both cases. Which means, at least in theory, we should expect two
cross peaks with very similar intensity, as visible in the HMBC.
Combine two
substructures
O
H
8.99
O
H
3
C
3.78
55.53
6.79
6.99
128.94
C
C
C
C
C
C
H
C
H
H
C
H
H
C
H
2
O
H
4.07
61.71
128.94
127.48
6.42
6.18
6.99
6.79
6.70
109.64
115.43
119.41
4.76
15.85
Hz
5.41 Hz
5.48 Hz
1.62 Hz
8.09 Hz
1.88
Hz
If we choose this C atom to combine the two fragments, there
would be five bonds from the C atom with the chemical shift of
128.94 ppm to the proton with the chemical shift of 6.79 ppm.
Such five bond correlations are not totally impossible, but really
very rare.
1
2
3
4
5
Exclude two other options
Combine two
substructures
O
H
8.99
O
H
3
C
3.78
55.53
6.79
6.99
128.94
C
C
C
C
C
C
H
C
H
H
C
H
H
C
H
2
O
H
4.07
61.71
128.94
127.48
6.42
6.18
6.99
6.79
6.70
109.64
115.43
119.41
4.76
15.85
Hz
5.41 Hz
5.48 Hz
1.62 Hz
8.09 Hz
1.88
Hz
If we link our fragment to this C atom, there would be four bonds between the C
atom with the chemical shift of 128.94 ppm and the proton with the chemical
shift of 6.99 ppm. Such a 4-bond correlation is possible.
On the other hand, between the mentioned C atom (128.94 ppm) and the proton
with the chemical shift of 6.70 ppm there are only three bonds. However, a cross
peak linking those two is not visible even though there should be one with a high
degree of probability. The absence of a cross peak in the HMBC is not definitive
proof, but both arguments mentioned above taken together exclude this
connection with a very high probability.
1
2
3
X
Combine two
substructures
Exclude two other options
O
H
8.99
O
H
3
C
3.78
55.53
C
C
C
C
C
C
H
C
H
H
C
H
H
C
H
2
O
H
4.07
61.71
128.94
127.48
6.42
6.18
6.99
6.79
6.70
109.64
115.43
119.41
4.76
15.85
Hz
5.41 Hz
5.48 Hz
1.62 Hz
8.09 Hz
1.88
Hz
Assign the first quaternary
carbon atom
128.46
6.70
6.42
6.18
128.46
Combine two
substructures
Our pile of building blocks includes three quaternary =C-
fragments with chemical shifts of 128.46, 146.13 and 147.69
ppm. Let’s inspect the signal at 128.46 ppm in the HMBC.
Two intense and one weak cross peaks belong to this carbon
atom.
For assignment of the quaternary C atom at 128.46 ppm, only
one position is possible..
O
H
8.99
O
H
3
C
3.78
55.53
C
C
C
C
C
C
H
C
H
H
C
H
H
C
H
2
O
H
4.07
61.71
128.94
127.48
6.42
6.18
6.99
6.79
6.70
109.64
115.43
119.41
4.76
15.85
Hz
5.41 Hz
5.48 Hz
1.62 Hz
8.09 Hz
1.88
Hz
128.46
6.70
6.42
6.18
128.46
But why is the intensity of the cross-peak to the proton at 6.42 ppm
that weak and why are there no cross-peaks at all to the protons at
6.79 ppm and 6.99 ppm?
The size of coupling constants only tends to depend on the number of
bonds between the coupling partners.
Specifically, geminal C-H coupling constants - and thus indirectly the
intensity of cross peaks - are often very small if the bond angle
between the coupling partners is 120°.
Combine two
substructures
Assign the first quaternary
carbon atom
O
H
3
C
3.78
55.53
OH group
C
C
C
C
C
C
H
C
H
H
C
H
H
C
H
2
O
H
4.07
61.71
128.94
127.48
6.42
6.18
6.99
6.79
6.70
109.64
115.43
119.41
4.76
15.85
Hz
5.41 Hz
5.48 Hz
1.62 Hz
8.09 Hz
1.88
Hz
128.46
O
H
8.99
O
H
8.99
C
C
C
C
C
C
H
C
H
H
C
H
H
C
H
2
O
H
4.07
61.71
128.94
127.48
6.42
6.18
6.99
6.79
6.70
109.64
115.43
119.41
4.76
15.85
Hz
5.41 Hz
5.48 Hz
1.62 Hz
8.09 Hz
1.88
Hz
128.46
There are only two possibilities to attach the OH group.
Let's just try one of them.
Which cross peaks are visible in the HMBC?
(The assignment of the two remaining quaternary =C-
atoms is still uncertain. There are only two possibilities
and maybe the uncertain assignment doesn’t matter).
O
H
3
C
3.78
55.53
128.46
C
C
C
C
C
CO
H
C
H
H
C
H
H
H
C
H
2
O
H
4.07
61.71
128.94
127.48
6.42
6.18
8.99
6.99
6.79
6.70
109.64
115.43
119.41
146.13/
147.69
4.76
15.85
Hz
5.41 Hz
5.48 Hz
1.62 Hz
8.09 Hz
1.88
Hz
8.99
Starting from the signal of the OH group at 8.99 ppm,
there are three cross peaks in the HMBC.
115.43
146.13
147.69
There are a maximum of three bonds between the
proton and each of the carbon atoms, no matter how
we assign the signals at 146.13/147.69 ppm.
OH group
O
H
3
C
3.78
55.53
128.46
C
C
C
C
C
C
O
H
C
H
H
C
H
HH
C
H
2
O
H
4.07
61.71
128.94
127.48
6.42
6.18
8.99
6.99
6.79
6.70
109.64
115.43
119.41
146.13/
147.69
4.76
15.85
Hz
5.41 Hz
5.48 Hz
1.62 Hz
8.09 Hz
1.88
Hz
x
x
The second possible position of the OH group would
require a four-bond correlation, while on the other
hand an expected three bond correlation does not
appear in the HMBC.
OH group
128.46
C
C
C
C
C
CO
H
C
H
H
C
H
H
H
C
H
2
O
H
4.07
61.71
128.94
127.48
6.42
6.18
8.99
6.99
6.79
6.70
109.64
115.43
119.41
146.13/
147.69
4.76
15.85
Hz
5.41 Hz
5.48 Hz
1.62 Hz
8.09 Hz
1.88
Hz
128.46
C
C
C
C
C
CO
H
C
H
H
C
H
H
H
C
H
2
O
H
4.07
61.71
128.94
127.48
6.42
6.18
8.99
6.99
6.79
6.70
109.64
115.43
119.41
146.13/
147.69
4.76
15.85
Hz
5.41 Hz
5.48 Hz
1.62 Hz
8.09 Hz
1.88
Hz
O
H
3
C
3.78
55.53
O
H
3
C
3.78
55.53
Final structure
Let's return to the first way of inserting the OH group.
There is only one remaining possibility to attach the
methoxy group.
Now we have the final constitution. We might try
to assign the two quaternary carbon atoms
unambiguously.
~
4.04.55.06.06.57.07.5
1
H
ppm
145
13
C
147.69
146.13
C
C
C
C
C
CO
O
H
C
H
H
C
H
H
H
3
C
H
C
H
2
O
H
3.78
55.53
4.07
61.71
128.94
127.48
6.42
6.18
8.99
6.99
6.79
6.70
109.64
115.43
119.41
128.46
4.76
15.85
Hz
5.41 Hz
5.48 Hz
1.62 Hz
8.09 Hz
1.88
Hz
O
H
3
C
3.78
55.53
3.78
147.69
146.13/
147.69
Assign the two remaining
quaternary carbon atoms
The cross peak is easily explained if the carbon
atom with the chemical shift of 147.69 ppm is
next to the methoxy group.
If the carbon with the chemical shift of 147.69
ppm were next to the OH group, the above
cross-peak would correspond to a four-bond
correlation. On the other hand, there would be
three bonds between C(146.13ppm) and H(3.78
ppm). However, a corresponding cross-peak for
that is not observed.
Final structure
147.69
146.13
C
C
C
C
C
CO
O
H
C
H
H
C
H
H
H
3
C
H
C
H
2
O
H
3.78
55.53
4.07
61.71
128.94
127.48
6.42
6.18
8.99
6.99
6.79
6.70
109.64
115.43
119.41
128.46
4.76
15.85
Hz
5.41 Hz
5.48 Hz
1.62 Hz
8.09 Hz
1.88
Hz
O
H
3
C
3.78
55.53
Some remarks:
From the given data, the structure can be
deduced with a very high probability (> 99.99%),
but not with 100% certainty.
To be absolutely sure, additional data would be
necessary. In this example a NOESY might be
very helpful.
The assignment of the quaternary carbon atoms
at 146.13 ppm and 147.69 ppm is very close to
perfection, but the ultimate proof would require
the evaluation of
13
C-
13
C coupling constants.
Alternatively, “external” pieces of information
are possible, for example knowledge of the
synthesis pathway.
But, there is still one open question ...
Final structure
Wrong integrals
The integral of the proton signal at 4.76 ppm was too small by a factor of 4 and the integral of the OH group at 8.99 ppm
was too small by about 25%.
The solution to the riddle is hidden on page 2 of the exercise: " ... using presaturation ...".
Presaturation is used to suppress the water signal at 4.75 ppm and as a side effect also almost extinguishes a
closely neighbouring signal at 4.76 ppm. Through chemical exchange, the second OH group at 8.99 ppm is also
indirectly influenced by the presaturation.
But why to use water suppression at 4.75 ppm for a sample measured in DMSO-d
6
? We expect the water signal
in DMSO at about 3.3 ppm. Even if we assume a strong contamination with water, the presaturation would have
been done at the wrong chemical shift.
This nice measurement is a gift from a laboratory mainly dealing with aqueous samples. To see the spectrum of
a substance with a concentration of about 2 mMol measured in an environment containing more than 100 Mol
of protons coming from the solvent requires special techniques, such as presaturation.
Apparently accidentally one of such special technique was used (inappropriately) here. If there were no signals
close to 4.75 ppm that would not matter. Unfortunately, correction of unfavourable measuring conditions post-
acquisition is not possible.
