Exercise plus Solution – Quick overview
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C
10
H
12
O
2
in CDCl
3
Problem of the month
November 2022
In which fruit does this compound represent
the main odour component?
So far there is no known correlation between
spectroscopic data and odour, but the final
chemical structure should answer the question
...
13
C{
1
H} NMR spectrum
recorded at 125.83{500.36} MHz
1
H NMR spectrum
recorded at 500.13 MHz
Peak label units
< 10 - [ppm]
> 1000 - [Hz]
1
H/
1
H COSY
recorded at 500.13 MHz
1
H/
13
C HSQC
recorded at 500.13/125.77 MHz
1
H/
13
C HMBC
recorded at 500.13/125.77 MHz
Because the molecular formula is known - for example from a high-resolution
mass spectrum - 5 double bond equivalents can be calculated immediately.
Basic considerations
Double bond equivalents,
integration
C
10
H
12
O
2
in CDCl
3
Let's record this value on a sticky note.
How about the integration? All proton signal groups are well separated. Let us
start using the well known procedure.
5 DBE
Integrals [arbitrary units] 1.00 4.91 4.86 2.37 5.12 5.12 7.38 3.68
Sum of integrals [arbitrary units] 34.44
Coefficient of proportionality [a.u. / H] 2.87
Number of protons 0.35 1.71 1.69 0.83 1.78 1.78 2.57 1.28
Basic considerations
Double bond equivalents,
integration
C
10
H
12
O
2
in CDCl
3
This, of course, is nonsense. Even the most generous rounding does not result in a
reasonable distribution of protons.
What‘s wrong?
5 DBE
Please note the solvent.
Basic considerations
Double bond equivalents,
integration
C
10
H
12
O
2
in CDCl
3
The solvent contains
- the signal of the non-deuterated amount (CHCl
3
) at about 7.26 ppm and
- the water (either H
2
O or HOD) at about 1.6 … 1.7 ppm.
In the case of this set of spectra the solvent seems to be of poor quality, which means relatively large
amounts of both of these unwanted components. This is not too uncommon in routine NMR
spectroscopy.
Let us neglect these unwanted signals and repeat the integration attempt.
CHCl
3
H
2
O /
HOD
5 DBE
Integrals [arbitrary units] 4.91 4.86 2.37 5.12 5.12 7.38
Sum of integrals [arbitrary units] 29.76
Coefficient of proportionality [a.u. / H] 2.48
Number of protons 1.98 1.96 0.96 2.06 2.06 2.98
Basic considerations
Double bond equivalents,
integration
C
10
H
12
O
2
in CDCl
3
Integration quality could be slightly better, but nevertheless the 12 protons can be
easily distributed among the remaining 6 signal groups.
5 DBE
2 21 322
Basic considerations
Double bond equivalents,
integration
5 DBE
2 21 322
1376.59
1384.47
1369.65
Some more details about the chemical shifts and the estimated coupling
constants for the four multiplets are available from the enlarged parts of
the one dimensional proton spectrum.
Here is an example using the most highfield shifted multiplet.
=
1384.47 𝐻𝑧 + 1369.65 𝐻𝑧
2 ∗ 500.13 𝑀𝐻𝑧
= 𝟐. 𝟕𝟓 𝒑𝒑𝒎
𝐽 ≈
1384.47 𝐻𝑧 − 1369.65 𝐻𝑧
2
≈ 𝟕. 𝟒 𝑯𝒛
2.85
7.07
6.77
2.75
≈9Hz
≈7Hz
≈7Hz
2 21 322
2.85
7.07
6.77
2.75
≈9Hz
≈7Hz
≈7Hz
Basic considerations
Double bond equivalents,
integration
5 DBE
H
During the next steps we need the three signal groups below 3 ppm
together with their integral. Let us note the six values describing those
groups using sticky notes (chemical shifts are given in [ppm] and integrals
in number of protons).
Temporarily we don’t need the one dimensional proton spectrum. Let us
shift this spectrum to the left for possible later use.
2.852.752.16
223
To create our first building blocks additionally we need some peaks from
the one dimensional carbon spectrum.
1
H
13
C
Basic considerations
Carbon signals
5 DBE
2.852.752.16
223
45.5028.91 30.20
From the one dimensional carbon
spectrum we need the three most
highfield-shifted signals.
In the DEPT you already see, that the
signals at 45.50 ppm and 28.91 ppm
belong to CH
2
groups, whereas the signal
at 30.20 ppm belongs to a methyl group.
Although we didn’t evaluate the HSQC
until now this coincides perfectly with
the proton integrals.
(CH, CH
3
positive, CH
2
negative)
HSQC
1
H
Building blocks
C
5 DBE
2.852.752.16
223
45.5028.91 30.20
Now we have some pieces of
information to get our first building
blocks.
Our spectrum of choice is the HSQC.
Building blocks
5 DBE
2.852.752.16
223
45.5028.91 30.20
The HSQC provides us with structural fragments. To
get their chemical shifts we need the collected data
from the one-dimensional spectra.
1
H
13
C
28.91
30.20
45.50
2.16
2.752.85
3
2
2
HSQC
Building blocks
The three cross peaks in the upper right area of the
HSQC are clearly due to CH
n
fragments with sp
3
hybridized carbon.
1
H
13
C
28.91
30.20
45.50
2.752.85
3
2
2
Let’s extract the three parts of the final molecule one
after the other.
H
3
C
30.20
2.16
H
3
C
30.20
2.16
2.16
5 DBE
HSQC
Building blocks
The three cross peaks in the upper right area of the
HSQC are clearly due to CH
n
fragments with sp
3
hybridized carbon.
1
H
13
C
28.91
30.20
45.50
2.752.85
3
2
2
Let’s extract the three parts of the final molecule one
after the other.
C
H
H
28.91
2.85
C
H
H
28.91
2.85
H
3
C
30.20
2.16
2.16
5 DBE
HSQC
Building blocks
The three cross peaks in the upper right area of the
HSQC are clearly due to CH
n
fragments with sp
3
hybridized carbon.
1
H
13
C
28.91
30.20
45.50
2.752.85
3
2
2
Let’s extract the three parts of the final molecule one
after the other.
C
H
H
45.50
2.75
C
H
H
28.91
2.85
H
3
C
30.20
2.16
C
H
H
45.50
2.75
2.16
5 DBE
HSQC
Building blocks
Connectivity
We now have the first three fragments to build our
molecule. But is there any connectivity between
these fragments? The HSQC provides us the building
blocks only.
1
H
13
C
28.91
30.20
45.50
2.752.85
3
2
2
One possibility to get connectivities is the HMBC.
Let’s try.
C
H
H
28.91
2.85
H
3
C
30.20
2.16
C
H
H
45.50
2.75
2.16
5 DBE
HSQC
HMBC
HSQC
C
H
H
28.91
2.85
H
3
C
30.20
2.16
C
H
H
45.50
2.75
1
H
13
C
HSQC
5 DBE
2.85
2.75 2.16
2.85
2.75
2.16
This subset of the HMBC is sufficient and makes the
evaluation of this HMBC rich of cross peaks easier.
Very often it is a real challenge to find the really
helpful HMBC cross peaks out of the huge number of
available signals.
We already know three of the proton signals.
HMBC
Building blocks
Connectivity
C
H
H
28.91
2.85
H
3
C
30.20
2.16
C
H
H
45.50
2.75
1
H
13
C
HSQC
5 DBE
2.85
2.75 2.16
In the lower right corner of the HMBC there are three
cross peaks showing a correlation of all three proton
signals with the same carbon atom at 208.84 ppm
(the exact value is extracted from the one
dimensional carbon spectrum).
208.84
This, of course, is a carbonyl group.
C
O
208.84
HMBC
Building blocks
Connectivity
C
H
H
28.91
2.85
H
3
C
30.20
2.16
C
H
H
45.50
2.75
1
H
13
C
HSQC
5 DBE
2.85
2.75 2.16
The carbon atom of the carbonyl group is two or three
bonds away from each of the three proton groups.
208.84
C
O
208.84
It is easy to create a substructure to explain the
neighbourhoods 208.84 ppm/2.16 ppm and 208.84
ppm/2.85 ppm (via two bonds).
HMBC
Building blocks
Connectivity
C
H
H
28.91
2.85
H
3
C
30.20
2.16
C
H
H
45.50
2.75
1
H
13
C
HSQC
5 DBE
2.85
2.75 2.16
There are no more open (free) bond at the carbonyl
group. But if we join both methylene groups, the
distance between the carbon at 208.84 ppm and the
proton at 2.75 ppm is three bonds, which is totally
fine to explain an HMBC peak.
208.84
C
O
208.84
C
H
H
C
OH
3
C
208.84
30.20
28.91
2.16
2.85
HMBC
Building blocks
Connectivity
C
H
H
45.50
2.75
1
H
13
C
HSQC
5 DBE
2.85
2.75 2.16
There are no more open (free) bond at the carbonyl
group. But if we join both methylene groups, the
distance between the carbon at 208.84 ppm and the
proton at 2.75 ppm is three bonds, which is totally
fine to explain an HMBC peak.
208.84
C
H
H
C
OH
3
C
208.84
30.20
28.91
2.16
2.85
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
45.50
28.91
2.16
2.85
2.75
HMBC
Building blocks
Connectivity
1
H
13
C
HSQC
5 DBE
2.85
2.75 2.16
The carbonyl group contains one double bond
equivalent. Now there are four remaining double
bond equivalents to be assigned.
208.84
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
45.50
28.91
2.16
2.85
2.75
4 DBE
The connectivity of two methylene groups should be
visible in the COSY.
Let’s check.
HMBC
Building blocks
Connectivity
COSY
1
H
13
C
HSQC
2.85
2.75 2.16
Let’s label the pseudo projections with the three
signal groups of interest.
208.84
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
45.50
28.91
2.16
2.85
2.75
4 DBE
HMBC
HMBC
Building blocks
Connectivity
HMBC
1
H
13
C
HSQC
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
45.50
28.91
2.16
2.85
2.75
2.16
2.85
2.75
2.16
2.85
2.75
2.16
2.85
2.75
2.16
2.85
2.75
Although the chemical shifts of the two methylene
groups are similar, the connectivity between both
groups is clearly visible in the COSY.
4 DBE
Building blocks
Connectivity
HMBC
1
H
13
C
HSQC
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
45.50
28.91
2.16
2.85
2.75
2.16
2.85
2.75
2.16
2.85
2.75
But there is another really unexpected correlation
visible.
4 DBE
Building blocks
Connectivity
HMBC
1
H
13
C
HSQC
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
45.50
28.91
2.16
2.85
2.75
2.16
2.85
2.75
2.16
2.85
2.75
Using our temporary structure, this cross peak would
be the result of a coupling pathway across five single
bonds. Such coupling pathways are really very, very
rare.
Even across four single bonds coupling is not visible
very often, but apparently we have such a case in our
compound. If we swap both methylene groups, the
coupling distance reduces to four bonds.
4 DBE
Building blocks
Connectivity
HMBC
1
H
13
C
HSQC
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
2.16
2.16
2.85
2.75
2.16
2.85
2.75
Now we have four single bonds between the
coupling nuclei.
45.50
28.91
2.85
2.75
But if there is a coupling, our methyl signal should be
a triplet?
4 DBE
Building blocks
Connectivity
HMBC
13
C
HSQC
2.16
2.85
2.75
2.16
2.85
2.75
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
28.91
45.50
2.16
2.75
2.85
C
C
H
H
H
C
OH
3
C
H
200.84
30.20
28.91
45.50
2.16
2.75
2.85
COSY
1
H
Indeed! The methyl protons show a triplet with a
four bond coupling constant of 0.45 Hz.
4 DBE
Building blocks
Connectivity
2 21 322
7.07
6.77
≈9Hz
HMBC
13
C
HSQC
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
28.91
45.50
2.16
2.75
2.85
COSY
4 DBE
Indeed! The methyl protons show a triplet with a
four bond coupling constant of 0.45 Hz.
0.45 Hz
5.166.777.07
122
≈9Hz
Let us use the proton spectrum to extract the
data for the three signal groups with chemical
shifts larger than 5 ppm as done before with the
signals below 3 ppm. Additional we should
remember the coupling constant of roughly 9 Hz.
Building blocks
Connectivity
2 21 322
7.07
6.77
≈9Hz
HMBC
13
C
HSQC
Building blocks
One more substituent
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
28.91
45.50
2.16
2.75
2.85
COSY
4 DBE
For the sake of clarity let us neglect the tiny coupling
constant. We no longer need this value to get our
structure
0.45 Hz
5.166.777.07
122
≈9Hz
The signal at 5.15 ppm is certainly worth a closer
examination. A rather limited amount of NMR
signals appear in this area.
To do so, let's take another look at the HSQC.
2 21 322
7.07
6.77
≈9Hz
HMBC
13
C
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
28.91
45.50
2.16
2.75
2.85
COSY
4 DBE
5.156.777.07
122
≈9Hz
H
HSQC
As said, we are interested in the peak at 5.16 ppm.
Building blocks
One more substituent
HMBC
13
C
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
28.91
45.50
2.16
2.75
2.85
COSY
4 DBE
5.166.777.07
122
≈9Hz
1
H
5.16
1
There is no cross peak at all with the proton signal at
5.16 ppm.
Because the molecular formula contains only H, C
and O this means, this signal belongs to an OH group.
O
H
5.16
HSQC
Building blocks
One more substituent
HSQC
HMBC
COSY
4 DBE
6.777.07
22
≈9Hz
1
H
5.16
1
There is no cross peak at all with the proton signal at
5.16 ppm.
Because the molecular formula contains only H, C
and O this means, this signal belongs to an OH group.
O
H
5.16
13
C
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
28.91
45.50
2.16
2.75
2.85
There are four signals in the one dimensional carbon spectrum, which are
not part of the already found substructures.
Let‘s create four more sticky notes with the chemical shifts of these signals.
HSQC
Building blocks
One more substituent
(CH, CH
3
positive, CH
2
negative)
HMBC
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
28.91
45.50
2.16
2.75
2.85
COSY
4 DBE
6.777.07
22
≈9Hz
1
H
O
H
5.16
HSQC
153.98 132.95 129.43 115.34
(Numbers in italics refer to the
low-intensity signals, numbers in
bold refer to the intense signals.)
Now it‘s time for an intermediate inventory.
Building blocks
One more substituent
(CH, CH
3
positive, CH
2
negative)
HMBC
COSY
4 DBE
6.777.07
22
≈9Hz
1
H
O
H
5.16
HSQC
153.98 132.95 129.43 115.34
C
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
28.91
45.50
2.16
2.75
2.85
Building blocks
One more substituent
HMBC
COSY
4 DBE
6.777.07
22
≈9Hz
1
H
O
H
5.16
HSQC
153.98 132.95 129.43 115.34
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
28.91
45.50
2.16
2.75
2.85
13
C
Our molecular formula is C
10
H
12
O
2
.
We found two fragments. Together their molecular formula is C
4
H
8
O
2
.
What else do we know?
• We have only two proton signal groups for four protons. The
chemical shift is typical for protons bound to sp
2
hybridized
carbon atoms.
• We have four carbon signals but six carbon atoms.
• The missing fragment contains 4 double bond equivalents.
Clearly there is some kind of symmetry.
Maybe there are exotic solutions to fulfill all these conditions.
But one possible structure stands out. What about benzene,
substituted in the 1 and 4 positions? Two possible substituents have
already been identified.
missing: C
6
H
4
Building blocks
The missing link
HMBC
COSY
4 DBE
6.777.07
22
≈9Hz
1
H
O
H
5.16
HSQC
153.98 132.95 129.43 115.34
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
28.91
45.50
2.16
2.75
2.85
13
C
C
C
C
C
C
C
H
HH
H
132.95
153.98
129.43
115.34
7.07
6.77
And the signal assignments fell from sky?
They didn‘t, of course. And even more, they might be wrong.
But first let‘s finalize our molecule. The explanation for the
assignment will be given afterwards.
Maybe there are exotic solutions to fulfill all these conditions.
But one possible structure stands out. What about benzene,
substituted in the 1 and 4 positions? Two possible substituents have
already been identified.
Building blocks
The missing link
HMBC
COSY
4 DBE
6.777.07
22
≈9Hz
1
H
O
H
5.16
153.98 132.95 129.43 115.34
C
C
H
H
H
C
OH
3
C
H
208.84
30.20
28.91
45.50
2.16
2.75
2.85
13
C
C
C
C
C
C
C
H
HH
H
132.95
153.98
129.43
115.34
7.07
6.77
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
129.43
115.34
2.16
2.75
2.85
7.07
6.77
5.16
Now that we know the final structure with some signal
assignment fallen from the sky, let's take another look at the
HSQC.
HSQC
Building blocks
The missing link
HMBC
COSY
4 DBE
6.777.07
22
≈9Hz
1
H
153.98 132.95 129.43 115.34
13
C
HSQC
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
129.43
115.34
2.16
2.75
2.85
7.07
6.77
5.16
For both pseudo projections we
need the chemical shifts
corresponding to these cross
peaks
Signal assignment
HMBC
COSY
4 DBE
6.777.07
22
≈9Hz
1
H
153.98 132.95 129.43 115.34
13
C
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
129.43
115.34
2.16
2.75
2.85
7.07
6.77
5.16
For both pseudo projections we
need the chemical shifts
corresponding to these cross
peaks
115.34
129.43
7.07 6.77
6.777.07
HSQC
Signal assignment
HMBC
COSY
4 DBE
6.777.07
22
≈9Hz
1
H
153.98 132.95
13
C
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
129.43
115.34
2.16
2.75
2.85
7.07
6.77
5.16
Now we can confirm a part of
the assignment that fell from
the sky before, using the two
HSQC cross peaks. The two CH
groups can of course be
interchanged.
115.34
129.43
7.07 6.77
The assignment of the two
quaternary carbon atoms could
of course also be reversed.
HSQC
Signal assignment
HMBC
COSY
4 DBE
6.777.07
22
≈9Hz
1
H
153.98 132.95
13
C
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
129.43
115.34
2.16
2.75
2.85
7.07
6.77
5.16
Now we can confirm the
assignment that fell from the
sky before, using the two HSQC
cross peaks. The two CH groups
can of course be interchanged.
115.34
129.43
7.07 6.77
The assignment of the two
quaternary carbon atoms could
of course also be reversed.
HSQC
HSQC
Signal assignment
HMBC
COSY
4 DBE
6.777.07
22
≈9Hz
1
H
13
C
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
129.43
115.34
2.16
2.75
2.85
7.07
6.77
5.16
The protons of the aromatic ring form a non-trivial AA'XX'
spin system. Nevertheless, there is a dominant vicinal
coupling constant.
The protons with chemical shifts of 6.77 ppm and 7.07
ppm are in the Z-position to each other. The coupling
constant should be about 8 ... 10 Hz.
We have extracted a value of approximately this size from
the multiplets of the two protons.
And, of course, there is a cross peak in the COSY between
the two proton signals mentioned above.
There is a last argument supporting our six
membered aromatic ring.
HSQC
Signal assignment
A good possibility to check the assignments made
for the atoms of the six membered aromatic ring is
the HMBC.
COSY
1
H
13
C
HSQC
HMBC
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
129.43
115.34
2.16
2.75
2.85
7.07
6.77
5.16
Signal assignment
COSY
1
H
13
C
HSQC
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
129.43
115.34
2.16
2.75
2.85
7.07
6.77
5.16
2.85 2.75
5.16
132.95
153.98
2.75
2.85
5.16
132.95
153.98
To check the correct assigment of
the quaternary carbon atoms we
need only a few peaks.
HMBC
Signal assignment
COSY
1
H
13
C
HSQC
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
129.43
115.34
2.16
2.75
2.85
7.07
6.77
5.16
2.85 2.75
5.16
132.95
153.98
Everything is fine. The
cross peaks come from
two and three bond
correlations.
This wouldn‘t be true with
the opposite assignment
of the quaternary carbon
atoms.
But what about the carbon signals at
129.43 ppm and 115.34 ppm?
Let‘s try.
HMBC
Signal assignment
Building blocks
COSY
1
H
13
C
HSQC
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
129.43
115.34
2.16
2.75
2.85
7.07
6.77
5.16
2.85 2.75
5.16
The pseudo projection for the
protons is fine, but we need other
values for the carbon pseudo
projection.
129.43
115.34
129.43
115.34
In both cases the coupling pathway is
across four bonds.
That‘s not really impossible, but maybe
there is another possibility.
Let‘s change the CH assignment inside
the six membered ring.
HMBC
COSY
1
H
13
C
HSQC
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
2.16
2.75
2.85
5.16
2.85 2.75
5.16
129.43
115.34
129.43
115.34
7.07
6.77
HMBC
Signal assignment
COSY
1
H
13
C
HSQC
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
115.34
129.43
2.16
2.75
2.85
6.77
7.07
5.16
2.85 2.75
5.16
129.43
115.34
Now our cross peaks come from
coupling pathways across three bonds,
which is much more probable.
Fortunately, there is a last proof
supporting this assignment.
HMBC
Signal assignment
1
H
13
C
HSQC
HMBC
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
115.34
129.43
2.16
2.75
2.85
6.77
7.07
5.16
COSY
We didn‘t investigate one cross peak of
the COSY. Let's catch up.
HMBC
Signal assignment
Building blocks
1
H
13
C
HSQC
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
115.34
129.43
2.16
2.75
2.85
6.77
7.07
5.16
HMBC
2.85
7.07
2.85
7.07
We have to label two signals groups only.
And now there is a weak cross peak via four bonds. This
would be five bonds with the opposite assignment of
the CH groups.
Final solution
1
H
13
C
HSQC
C
C
C
H
H
H
C
OH
3
C
C
C
C
C
C
H
HH
H
O
H
H
208.84
30.20
28.91
45.50
132.95
153.98
115.34
129.43
2.16
2.75
2.85
6.77
7.07
5.16
HMBC
COSY
The primary question was about the smell of
the compound.
Without a dictionary, of course, you can
not answer the question.
Raspberries
