Exercise plus Solution – Quick overview
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2.39
0.81.01.21.41.61.82.02.22.42.62.83.03.23.4
ppm
1
H
185.20
190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20
ppm
13
C
36.92
13
C{
1
H} NMR spectrum
recorded at 62.90{250.13} MHz
1
H NMR spectrum
recorded at 250.13 MHz
C
4
H
6
O
4
measured in D
2
O
Deduce the structure.
Solution
𝒏
𝐃𝐁Ä
=
𝟐𝒏
𝐂
− 𝒏
𝐇
+ 𝟐
𝟐
C
4
H
6
O
4
First results:
- two double bond equivalents
- four carbon atoms, but only two carbon signals in the NMR spectrum
- six protons, but one singlet only
Probable explanation:
- a symmetrical molecule R – R, where – R corresponds to the molecular
formula C
2
H
3
O
2
, contains an open binding site and a double bond
equivalent
𝒏
𝐃𝐁𝐄
=
𝟐𝒏
𝐂
− 𝒏
𝐇
+ 𝟐
𝟐
C
4
H
6
O
4
Let's write down the half-molecule fragment we are looking for
and the little known information on sticky notes.
C : 36.92 ppm C : 185.20 ppmH : 2.39 ppmwanted: -C
2
H
3
O
2
with 1 DBE
Solution
𝒏
𝐃𝐁𝐄
=
𝟐𝒏
𝐂
− 𝒏
𝐇
+ 𝟐
𝟐
C
4
H
6
O
4
Let's write down the half-molecule fragment we are looking for
and the little known information on sticky notes.
C : 36.92 ppm C : 185.20 ppmH : 2.39 ppmwanted: -C
2
H
3
O
2
with 1 DBE
Solution
We no longer need the information-poor spectra.
Our first view is the carbon signal at 185.20 ppm.
C : 36.92 ppm C : 185.20 ppmH : 2.39 ppmwanted: -C
2
H
3
O
2
with 1 DBE
We no longer need the information-poor spectra.
C
O
Let's write down the half-molecule fragment we are looking for
and the little known information on sticky notes.
This is, of course, a carbonyl group.
A small interim calculation:
- wanted: -C
2
H
3
O
2
+ 1 DBE
- found: -CO- + 1 DBE
- missing: CH
3
O without DBE
At first glance, this looks like either a methoxy group or a single
oxygen atom plus a methyl group. Let's try the second version. A look
at the overview tables of the chemical shifts let us expect the
measured values in the carbon as well as in the proton range.
O
CH
3
But …
C : 36.92 ppm C : 185.20 ppmH : 2.39 ppmwanted: -C
2
H
3
O
2
with 1 DBE
O O
CH
3
C
C CH
3
O
O
... that was only half the molecule and after adding the mirror-
imaged second half of the molecule, you get a richly explosive-
looking compound..
Of course, "explosive" is not a spectroscopically relevant
argument. Nevertheless, maybe we should check out one obvious
and two less obvious permutation possibilities first.
The easiest way seems to be the exchange of – O – und – CO –.
Let's try this out.
C
O
O
CH
3
C : 36.92 ppm C : 185.20 ppmH : 2.39 ppm
H
3
C
CH
3
Doing this intra-molecular exchange, we get oxalic acid dimethyl
ester, a very commonly used compound.
C
O
C
O
OO
C C
O
O
OH
3
C
O CH
3
Unfortunately, this compound does not match the measured
chemical shifts. A quick check using an overview table of chemical
shifts, found in many books on NMR spectroscopy, suggests a
chemical shift of 3.5 ... 4.0 ppm for the protons of the methoxy
group..
That’s a significant difference to the measured value of 2.39 ppm.
But we can still exchange some more atoms, even if
this is not obvious at first sight. First of all, let’s rewrite
the methyl groups a little bit. For sure, normally you
wouldn’t write a methyl group this way.
expected:
3.5 … 4.0 ppm
measured:
2.39 ppm
C : 36.92 ppm C : 185.20 ppmH : 2.39 ppm
Using this somewhat strange notation, the fragments – CH
2
– and
– O – now offer themselves for mutual exchange.
C C
O
O
H
H
C
H
H
C
H
H
O
O
O
OO C
H
H
C C
O
H
C
H
H
H
But ... now there are two different proton groups in the integral
ratio 2 : 1?
At this point we should revisit all available pieces of information.
There is a small detail, nearly always without of interest.
Solvent: D2O!
What does this mean?
The concentration of the sample might be 10 mMol, so that the
concentration of OH groups is 20 mMol. The concentration of OD
groups in D
2
O is 110 moles (not “milli”).
Because of the large excess of deuterium, the chemically
exchangeable OH protons are almost completely converted into
OD groups, which show no signal in the proton spectrum.
Translation in progress
Last processing status: 7
th
of September 2022
C : 36.92 ppm C : 185.20 ppmH : 2.39 ppm
Of course, the OH protons have not disappeared. They are now
part of the HOD signal at about 4.7 ppm and thus not in the range
shown here.
O
OO C
H
H
C C
O
H
C
D
H
D
The Schoolery rule can be used to estimate the chemical shift of
the methylene protons. The calculation gives us
δ(CH
2
) = 1.70 ppm + 2.56 ppm = 4.26 ppm
and thus deviates almost 2 ppm from the measured value (2.39
ppm, see sticky note).
But there is a last permutation possibility.
C : 36.92 ppm C : 185.20 ppmH : 2.39 ppm
O
O
D
D
C
H
H
H
C
H
O
C
O
C
A new estimation of the chemical shift of the methylene protons
using the Schoolery rule now gives
δ(CH
2
) = 0.47 ppm + 1.55 ppm = 2.12 ppm
which is very close to the measured value (2.39 ppm, see sticky
note).
2.39
C : 36.92 ppm C : 185.20 ppm
It should be easy to assign the carbon signals.
C
C C
H
H
H
H
C
O
O
D
O
O
D
2.39
36.92
185.20
Because of the chemical exchange between succinic acid and the
solvent there is no assignments possible for the OH/OD groups.
