120
125
130
135
140
13
C
step
by
step
136.2
136.6
137.0
137.4
137.8
13
C
Exercise plus Solution – Quick overview
It is recommended to use this version only for a quick overview of the NMR challenge. All animations of the PowerPoint
version are missing, under certain circumstances quality deficiencies may also occur.
The higher quality PowerPoint files are freely available for download at any time.
C
10
H
8
measured in CDCl
3
Deduce the structure!
13
C{
1
H} NMR spectrum
measured at 125.70{499.84} MHz
1
H NMR spectrum
measured at 499.84 MHz
1
H/
13
C HSQC
measured at 499.84/125.70 MHz
1
H/
13
C HMBC
measured at 499.84/125.7 MHz
1
H TOCSY
measured at 499.84 MHz
80ms mixing time
It is tempting to think of the solution
“naphthalene” immediately after seeing the
molecular formula C
10
H
8
and signals in the
aromatic region. And to then get stuck!
The real challenge here is to break out of this
mental cage.
C
10
H
8
in CDCl
3
7 DBE
First steps
Double bond equivalents,
Integration
From the molecular formula, 7 double bond equivalents can be derived immediately.
The high number of double bond equivalents, only proton signals between
and , further only carbon signals in the range of to , and
finally no oxygen at all quickly suggests the solution naphthalene.
However, in napththalene we would expect only 3 carbon signals and 2 proton
multiplets because of the inherent high symmetry. The compound presented here
shows 5 proton multiplets and 6 carbon signals.
One approach to avoid falling into the naphthalene trap again and again is to first
collect the data that can be easily extracted. It is possible that ideas for the further
procedure will develop from there. The proton spectrum is a good place to start.
For the moment we write down the number of double bond equivalents on a
small sticky note.
2.08
0.98
1.08
2.00
2.15
Inte
-
gral
8.4 8.3 8.2 8.1 7.98.0 7.77.8 7.6 ppm7.5 7.27.37.4 7.1
1
H
4184.71
4175.41
3963.83
3960.13
3956.38
3806.38
3796.53
3786.59
3706.41
3702.67
3596.15
3586.40
3576.56
Hz
ppm
1
H
C
10
H
8
in CDCl
3
7 DBE
The integration is simple here. The sum of the integrals rounded to the nearest
integer value is even 8.
The proton number of a single multiplet is obtained by simply rounding the
corresponding integral.
The coefficient of proportionality between the integral and the number of
protons happens to be 1.
2 21 21Number of H
First steps
Double bond equivalents,
Integration
8.4 8.3 8.2 8.1 7.98.0 7.77.8 7.6 ppm7.5 7.27.37.4 7.1
1
H
4184.71
4175.41
3963.83
3960.13
3956.38
3806.38
3796.53
3786.59
3706.41
3702.67
3596.15
3586.40
3576.56
Hz
C
10
H
8
in CDCl
3
7 DBE
=
3806.38 𝐻𝑧 + 3786.59 𝐻𝑧
2 ∗ 499.84 𝑀𝐻𝑧
=
𝐽 =
(3806.38𝐻𝑧 − 3786.59𝐻𝑧)
2
= 9.90 𝐻𝑧
2 21 21Number of H
The peak maxima are given in Hz to enable the
determination of the coupling constants. A conversion to
the ppm scale provides more common values.
The calculation of the coupling constants is based on the
assumption of pure multiplets (no pseudo triplets). The
detailed calculation is given for one of the multiplets.
First steps
Double bond equivalents,
Integration
9.30 Hz
3.73 Hz
3.74 Hz
9.90 Hz
9.79 Hz
J
8.4 8.3 8.2 8.1 7.98.0 7.77.8 7.6 ppm7.5 7.27.37.4 7.1
1
H
C
10
H
8
in CDCl
3
7 DBE
2 21 21Number of H
not yet assigned:
For further evaluation, it is helpful to keep track of
information not yet contained in substructures.
These are the nuclei not yet assigned, and the number
of double bond equivalents.
First steps
Double bond equivalents,
Integration
9.30 Hz
3.73 Hz
3.74 Hz
9.90 Hz
9.79 Hz
J
8.4 8.3 8.2 8.1 7.98.0 7.77.8 7.6 ppm7.5 7.27.37.4 7.1
1
H
7 DBE
2 21 21Number of H
not yet assigned :
Two of the proton multiplets can be immediately
assigned to structural fragments. Because of the
molecular formula, only carbon can be adjacent to the
protons. A chemical shift of about for the protons
means an sp
2
hybridisation of these carbon atoms.
C H
3.73 Hz (t)
CH
9.90 Hz (t)
(The additionally noted
coupling constants of the
proton multiplets can be
helpful later when determining
the coupling partner.)
First steps
Double bond equivalents,
Integration
9.30 Hz
3.73 Hz
3.74 Hz
9.90 Hz
9.79 Hz
J
8.4 8.3 8.2 8.1 7.98.0 7.77.8 7.6 ppm7.5 7.27.37.4 7.1
1
H
7 DBE
2 21 21Number of H
not yet assigned :
Let us use identical colours for the protons and the
corresponding multiplets.
After assigning a total of C
2
H
2
, we still have to assign a
partial molecular formula of C
8
H
6
. Nothing changes
concerning the number of double bond equivalents.
C H
3.73 Hz (t)
CH
9.90 Hz (t)
First steps
Double bond equivalents,
Integration
9.30 Hz
3.73 Hz
3.74 Hz
9.90 Hz
9.79 Hz
J
9.30 Hz
3.73 Hz
3.74 Hz
9.90 Hz
9.79 Hz
J
8.4 8.3 8.2 8.1 7.98.0 7.77.8 7.6 ppm7.5 7.27.37.4 7.1
1
H
7 DBE
First steps
2 21 21Number of H
not yet assigned :
Three signal groups with 2 protons each remain.
There would be two possibilities.
a) Each of the signal groups corresponds to a =CH
2
group.
Then 5 quaternary carbon atoms would remain.
b) Each of the signal groups corresponds to 2 equivalent
=CH- groups (analogous to the structural elements
already found). In this case 2 quaternary carbon
atoms would remain.
Further considerations are possible, but b) is sufficiently
likely for us to continue with this version.
If it really does not work, one would have to start again
with version a).
C H
3.73 Hz (t)
CH
9.90 Hz (t)
9.30 Hz
3.73 Hz
3.74 Hz
9.90 Hz
9.79 Hz
J
8.4 8.3 8.2 8.1 7.98.0 7.77.8 7.6 ppm7.5 7.27.37.4 7.1
1
H
7 DBE
First steps
2 21 21Number of H
not yet assigned :
C
H
9.30
Hz (d)
2x
C
H
3.74
Hz (d)
2x
C
H
9.79
Hz (t)
2x
From the original molecular formula C
10
H
8
, partial
structures are now known that add up to C
8
H
8
. Only two
carbon atoms remain unknown.
C H
3.73 Hz (t)
CH
9.90 Hz (t)
8.4 8.3 8.2 8.1 7.98.0 7.77.8 7.6 ppm7.5 7.27.37.4 7.1
1
H
7 DBE
2 21 21Number of H
not yet assigned :
C
H
9.30
Hz (d)
2x
C
H
3.74
Hz (d)
2x
C
H
9.79
Hz (t)
2x
From the one-dimensional carbon spectrum, we only
need the chemical shifts.
C H
3.73 Hz (t)
CH
9.90 Hz (t)
First steps
9.30 Hz
3.73 Hz
3.74 Hz
9.90 Hz
9.79 Hz
J
8.4 8.3 8.2 8.1 7.98.0 7.77.8 7.6 ppm7.5 7.27.37.4 7.1
1
H
2 21 21Number of H
9.30 Hz
3.73 Hz
3.74 Hz
9.90 Hz
9.79 Hz
J
C
H
9.30
Hz (d)
2x
C
H
3.74
Hz (d)
2x
C
H
9.79
Hz (t)
2x
H
144 118142 140 138 136 134 126128130 120124 122
13
C
ppm
13
C
C H
3.73 Hz (t)
CH
9.90 Hz (t)
7 DBE
First steps
Carbon signals
not yet assigned :
144 118142 140 138 136 134 126128130 120124 122
13
C
ppm
(
13
C)[ppm]:
/ / / / /
1
H
7 DBE
not yet assigned :
C
H
9.30
Hz (d)
2x
C
H
3.74
Hz (d)
2x
C
H
9.79
Hz (t)
2x
C H
3.73 Hz (t)
CH
9.90 Hz (t)
The assignment of the carbon signals to the fragments is
done with the HSQC.
First, let’s reorder the fragments a little bit.
First steps
Carbon signals
144 118142 140 138 136 134 126128130 120124 122
13
C
ppm
(
13
C)[ppm]:
1
H
7 DBE
not yet assigned :
C
H
9.30
Hz (d)
2x
C
H
3.74
Hz (d)
2x
C
H
9.79
Hz (t)
2x
C H
3.73 Hz (t)
CH
9.90 Hz (t)
13
C
C
For three very closely adjacent carbon signals, an
enlarged section of the HSQC is necessary.
/ / / / /
Carbon signal
assignment
C
H
9.30
Hz (d)
2x
C
H
3.74
Hz (d)
2x
C
H
9.79
Hz (t)
2x
C H
3.73 Hz (t)
CH
9.90 Hz (t)
1
H
13
C
(
13
C)[ppm]:
7 DBE
not yet assigned :
C
HSQC
We can take the chemical shifts for both pseudo-
projections from the list of carbon signals and the
already known fragments.
/ / / / /
Carbon signal
assignment
136.2
136.6
136.4
137.0
137.2
137.4
137.6
137.8
13
C
8.4 8.3 8.2 8.0 7.9 7.8 7.7 7.6
1
H
C
H
9.30
Hz (d)
2x
C
H
3.74
Hz (d)
2x
C
H
9.79
Hz (t)
2x
C H
3.73 Hz (t)
CH
9.90 Hz (t)
1
H
13
C
(
13
C)[ppm]:
/
7 DBE
not yet assigned :
/ / / /
HSQC
Carbon signal
assignment
136.2
136.6
136.4
137.0
137.2
137.4
137.6
137.8
13
C
8.4 8.3 8.2 8.0 7.9 7.8 7.7 7.6
1
H
C
H
9.30
Hz (d)
2x
C
H
3.74
Hz (d)
2x
C
H
9.79
Hz (t)
2x
C H
3.73 Hz (t)
CH
9.90 Hz (t)
1
H
13
C
(
13
C)[ppm]:
/
7 DBE
not yet assigned :
Having well-separated cross peaks, a proton-carbon
assignment is easily possible without auxiliary lines.
HSQC
C
Carbon signal
assignment
136.2
136.6
136.4
137.0
137.2
137.4
137.6
137.8
13
C
8.4 8.3 8.2 8.0 7.9 7.8 7.7 7.6
1
H
C
H
3.74
Hz (d)
2x
C
H
9.79
Hz (t)
2x
C H
3.73 Hz (t)
CH
9.90 Hz (t)
1
H
13
C
(
13
C)[ppm]:
/
7 DBE
not yet assigned :
C
H
9.30
Hz (d)
2x
HSQC
C
Carbon signal
assignment
Having well-separated cross peaks, a proton-carbon
assignment is easily possible without auxiliary lines.
136.2
136.6
136.4
137.0
137.2
137.4
137.6
137.8
13
C
8.4 8.3 8.2 8.0 7.9 7.8 7.7 7.6
1
H
C
H
3.74
Hz (d)
2x
C
H
9.79
Hz (t)
2x
CH
9.90 Hz (t)
1
H
13
C
(
13
C)[ppm]:
/
7 DBE
not yet assigned :
C
H
9.30
Hz (d)
2x
C H
3.73 Hz (t)
HSQC
C
Carbon signal
assignment
Having well-separated cross peaks, a proton-carbon
assignment is easily possible without auxiliary lines.
136.2
136.6
136.4
137.0
137.2
137.4
137.6
137.8
13
C
8.4 8.3 8.2 8.0 7.9 7.8 7.7 7.6
1
H
C
H
3.74
Hz (d)
2x
C
H
9.79
Hz (t)
2x
1
H
13
C
(
13
C)[ppm]:
/
7 DBE
not yet assigned :
For the two assignments that are still missing, we need
the complete HSQC.
C
H
9.30
Hz (d)
2x
C H
3.73 Hz (t)
CH
9.90 Hz (t)
HSQC
Carbon signal
assignment
136.2
136.6
136.4
137.0
137.2
137.4
137.6
137.8
13
C
8.4 8.3 8.2 8.0 7.9 7.8 7.7 7.6
1
H
8.4 8.2 8.0 7.27.47.6
1
H
120
130
125
135
140
13
C
(
13
C)[ppm]:
/
7 DBE
not yet assigned :
1
H
13
C
The two cross peaks in the upper right corner provide
the desired information.
CH
9.90 Hz (t)
C
H
9.30
Hz (d)
2x
C
H
3.74
Hz (d)
2x
C
H
9.79
Hz (t)
2x
C H
3.73 Hz (t)
HSQC
Carbon signal
assignment
8.4 8.2 8.0 7.27.47.6
1
H
120
130
125
135
140
13
C
(
13
C)[ppm]:
/
7 DBE
not yet assigned :
1
H
13
C
C H
3.73 Hz (t)
CH
9.90 Hz (t)
The two C/H pairs ppm/ ppm and
ppm/ ppm are easy to recognise even without
auxiliary lines. We can complete the missing two
assignments.
C
H
9.30
Hz (d)
2x
C
H
9.79
Hz (t)
2x
C
H
3.74
Hz (d)
2x
HSQC
Carbon signal
assignment
8.4 8.2 8.0 7.27.47.6
1
H
120
130
125
135
140
13
C
(
13
C)[ppm]:
7 DBE
not yet assigned :
1
H
13
C
C
H
9.79
Hz (t)
2x
C
H
3.74
Hz (d)
2x
HSQC
HSQC
C
H
9.30
Hz (d)
2x
CH
9.90 Hz (t)
C H
3.73 Hz (t)
C
2x
A carbon signal remains at
ppm without a directly bonded
proton. The molecular formula tells us
about two unassigned carbon atoms,
i.e. there are two equivalent sp
2
-
hybridised quaternary C atoms.
All atoms are now assigned to
structural fragments. The puzzle of
linking these building blocks can
begin.
HSQC
Carbon signal
assignment
The two C/H pairs ppm/ ppm and
ppm/ ppm are easy to recognise even without
auxiliary lines. We can complete the missing two
assignments.
Double bond equivalents
7 DBE
not yet assigned :
1
H
13
C
C
H
9.79
Hz (t)
2x
C
H
3.74
Hz (d)
2x
HSQC
C
H
9.30
Hz (d)
2x
CH
9.90 Hz (t)
C H
3.73 Hz (t)
C
2x
Our collection of fragments contains a total of 10
carbon atoms, each with an open double bond. No
matter how the fragments are linked, a total of 5
building blocks with the structure >C=C< are
created, which means that 5 of the 7 double bond
equivalents would be assigned.
There are no more structural fragments, the two
double bond equivalents still missing can only be
two ring closures.
2 Rings
The TOCSY is a nice tool to get some pieces of
information to link at least a few fragments.
H
TOCSY
TOCSY
Linking the fragments
not yet assigned:
1
H
13
C
C
H
9.79
Hz (t)
2x
C
H
3.74
Hz (d)
2x
HSQC
C
H
9.30
Hz (d)
2x
CH
9.90 Hz (t)
C H
3.73 Hz (t)
C
2x
2 Rings
8.4
8.2
8.0
7.2
7.4
7.6
1
H
8.4 8.2 8.0 7.27.47.6
1
H
ppm
The TOCSY might be a little bit confusing at first glance.
A simple spin system is located in the middle.
TOCSY
8.4
8.2
8.0
7.2
7.4
7.6
1
H
8.4 8.2 8.0 7.27.47.6
1
H
ppm
C
H
9.79
Hz (t)
2x
C
H
3.74
Hz (d)
2x
C
H
9.30
Hz (d)
2x
CH
9.90 Hz (t)
C H
3.73 Hz (t)
C
2x
1
H
13
C
HSQC
not yet assigned :
2 Rings
Let's mark the associated structural fragments with solid
frames.
The remaining three proton multiplets form a second
spin system.
TOCSY
Linking the fragments
8.4
8.2
8.0
7.2
7.4
7.6
1
H
8.4 8.2 8.0 7.27.47.6
1
H
ppm
C
H
9.79
Hz (t)
2x
C
H
3.74
Hz (d)
2x
C
H
9.30
Hz (d)
2x
CH
9.90 Hz (t)
C H
3.73 Hz (t)
C
2x
1
H
13
C
HSQC
not yet assigned :
2 Rings
In this case we mark the associated structural fragments
with dashed frames.
TOCSY
Linking the fragments
8.4
8.2
8.0
7.2
7.4
7.6
1
H
8.4 8.2 8.0 7.27.47.6
1
H
ppm
C
H
9.79
Hz (t)
2x
C
H
3.74
Hz (d)
2x
C
H
9.30
Hz (d)
2x
CH
9.90 Hz (t)
C H
3.73 Hz (t)
C
2x
1
H
13
C
HSQC
not yet assigned :
2 Rings
TOCSY
TOCSY
TOCSY
Linking the fragments
C
2x
1
H
13
C
HSQC
not yet assigned :
2 Rings
CH
9.90 Hz (t)
C
H
9.79
Hz (t)
2x
C
H
9.30
Hz (d)
2x
C H
3.73 Hz (t)
C
H
3.74
Hz (d)
2x
Connecting the fragments belonging to the two
independent spin systems is only possible via the two
quaternary carbon atoms.
TOCSY
Linking the fragments
1
H
13
C
HSQC
not yet assigned :
2 Rings
CH
9.90 Hz (t)
C
H
9.79
Hz (t)
2x
C
H
9.30
Hz (d)
2x
C H
3.73 Hz (t)
C
H
3.74
Hz (d)
Let's start with the fragments belonging to the smaller
spin system and the quaternary carbon atoms that serve
as spacer.
TOCSY
C H
3.73 Hz (t)
C
H
3.74
Hz (d)
2x
C
H
3.74
Hz (d)
2x
C
H
3.74
Hz (d)
C
C
2x
C
C
2x
CH
9.90 Hz (t)
C
H
9.79
Hz (t)
2x
C
H
9.30
Hz (d)
2x
Linking the fragments
C
1
H
13
C
HSQC
not yet assigned :
2 Rings
CH
9.90 Hz (t)
C
H
9.79
Hz (t)
2x
C
H
9.30
Hz (d)
2x
C H
3.73 Hz (t)
After a little cosmetic work, …
TOCSY
C
H
3.74
Hz (d)
C
C
H
H
3.74
Hz (d)
Linking the fragments
C
1
H
13
C
HSQC
not yet assigned :
2 Rings
CH
9.90 Hz (t)
C
H
9.79
Hz (t)
2x
C
H
9.30
Hz (d)
2x
C H
3.73 Hz (t)
… the first fragments can be linked together.
TOCSY
C
H
3.74
Hz (d)
C
C
H
3.74
Hz (d)
Linking the fragments
C
1
H
13
C
HSQC
not yet assigned :
2 Ringe
CH
9.90 Hz (t)
C
H
9.79
Hz (t)
2x
C
H
9.30
Hz (d)
2x
C H
3.73 Hz (t)
Due to the symmetry of the two CH groups, some
redundant information could be removed for the sake of
clarity.
TOCSY
C
C
H
3.74
Hz (d)
C
H
3.74
Hz (d)
C
C
C
H
H
H
3.74
Hz (d)
3.73 Hz (t)
(A point symmetry with the carbon marked in
blue as the centre of symmetry would also be
conceivable, but somehow we have to end
with two ring systems.)
σ
Linking the fragments
C
1
H
13
C
HSQC
not yet assigned :
2 Rings
CH
9.90 Hz (t)
C
H
9.79
Hz (t)
2x
C
H
9.30
Hz (d)
2x
The measured coupling constants and the known
multiplets are easily explained by this partial structure,
although a coupling constant of 3.73 Hz is not seen
every day.
TOCSY
C
C
C
C
H
H
H
3.74
Hz (d)
3.73 Hz (t)
σ
3.73
Hz
Linking the fragments
C
C
C C
C
CC
H
H
H
3.73
Hz
C
C
C
C
H
H
H
3.73
Hz
1
H
13
C
HSQC
CH
9.90 Hz (t)
C
H
9.79
Hz (t)
2x
C
H
9.30
Hz (d)
2x
This first substructure is terminated by the quaternary
carbon atoms, since according to TOCSY there is no
coupling to any of the remaining CH fragments.
TOCSY
not yet assigned :
2 Rings
σ
Linking the fragments
C C
C
CC
H
H
H
3.73
Hz
1
H
13
C
HSQC
CH
9.90 Hz (t)
C
H
9.79
Hz (t)
2x
C
H
9.30
Hz (d)
2x
To continue the structure beyond the quaternary carbon
atoms, but keeping the symmetry we have only two
choices. We need one of the fragments existing twice
each.
Which of the two?
TOCSY
not yet assigned :
2 Rings
σ
Linking the fragments
C C
C
CC
H
H
H
3.73
Hz
1
H
13
C
HSQC
CH
9.90 Hz (t)
C
H
9.79
Hz (t)
2x
C
H
9.30
Hz (d)
2x
Let's take a hypothetical CH fragment and attach this to
one of the the quaternary carbon atoms.
TOCSY
not yet assigned :
2 Ringe
σ
C
H
C
H
C
H
C
H
Only one more neighbour is possible, i.e. we have to
observe a doublet.
Doublet
The proton at ppm appears as a doublet. The CH group
containing this proton follows the quaternary C atoms.
Linking the fragments
C C
C
CC
H
H
H
3.73
Hz
1
H
13
C
HSQC
CH
9.90 Hz (t)
C
H
9.79
Hz (t)
2x
C
H
9.30
Hz (d)
2x
Let's add two identical CH groups to our previous
fragment.
TOCSY
not yet assigned :
2 Rings
σ
C
H
9.30
Hz (d)
C
H
Linking the fragments
C C
C
CC
H
H
H
3.73
Hz
1
H
13
C
HSQC
CH
9.90 Hz (t)
C
H
9.79
Hz (t)
2x
TOCSY
not yet assigned :
2 Rings
σ
C
H
9.30
Hz (d)
C
H
C
H
9.79
Hz (t)
C
H
To continue, we need two more identical fragments.
There is only one possibility.
Linking the fragments
C C
C
CC
H
H
H
3.73
Hz
1
H
13
C
HSQC
CH
9.90 Hz (t)
TOCSY
not yet assigned :
2 Rings
σ
C
H
9.30
Hz (d)
C
H
C
H
9.79
Hz (t)
C
H
CH
9.90 Hz (t)
There is only one possible position for the last CH group.
Of course, we create a bond using the two free valences.
Now the molecule is complete.
Linking the fragments
C
H
C
H
CH
C
C
C
H
H
H
C
C
C
H
C
H
C
C
C
C
H
H
H
C
HC
C
H
H
C
HC
C
H
9.30
Hz (d)
9.79
Hz (t)
9.90 Hz (t)
3.73
Hz
9.30
Hz (d)
9.79
Hz (t)
HSQC
not yet assigned :
2 Rings
The molecule looks a little bit strange, but that's just a
matter of cosmetics.
1
H
13
C
TOCSY
Linking the fragments
C
C
C
H
H
H
C
C
C
C
H
H
H
C
HC
C
H
9.90 Hz (t)
3.73
Hz
9.30
Hz (d)
9.79
Hz (t)
HSQC
not yet assigned :
2 Rings
The postulated structure (azulene) also contains the two
missing ring systems.
1
H
13
C
TOCSY
For the sake of completeness, one can add the two
missing homonuclear coupling constants in the 7-ring
system.
9.90
Hz
9.30
Hz
The multiplet of the proton at ppm then, of course,
is not a true triplet but, rather, a pseudo triplet.
Linking the fragments
C
C
C
H
H
H
C
C
C
C
H
H
H
C
HC
C
H
3.73
Hz
9.90
Hz
9.30
Hz
Final check
HSQC
1
H
13
C
TOCSY
Even without using the HMBC, it was possible to solve
this challenge.
But, of course, we can use the HMBC to independently
verify the result.
C
C
C
H
H
H
C
C
C
C
H
H
H
C
HC
C
H
120
130
125
135
140
C
C
C
C
H
H
H
C
C
C
C
H
H
H
C
HC
C
H
HSQC
1
H
13
C
TOCSY
HMBC
As an example let's follow the coupling path that
belongs to one of the cross peaks.
Final check
8.4 8.2 8.0 7.27.47.6
1
H
120
130
125
135
140
13
C
C
C
C
H
H
H
C
C
C
C
H
H
H
C
HC
C
H
HSQC
1
H
13
C
TOCSY
As an example let's follow the coupling path that
belongs to one of the cross peaks.
On the other hand, we do not see a cross-peak for a
four-bond coupling pathway.
HMBC
Final check
8.4 8.3 8.2 8.1 8.0 7.9 7.8 7.7 7.6 7.5 7.4 7.3 7.2 7.1 ppm
1
H
144 142 140 138 136 134 132 130 126 124 122 120 118 ppm
13
C
H
H
H
H
H
H
H
H
C
C
C
C
C
C
C
C
C
C
Solution at a glance
Contributions
Measurements
Maria Nesterova,
Lawrence J. Clos,
Christopher Stancic,
Mark E. Anderson,
John L. Markley
Spectrometer time
University of Wisconsin-Madison
(BioMagResBank)
Discussions and
native English
language support
Alan Kenwright
Compilation
Rainer Haeßner
More exercises …