Exercise plus Solution – Quick overview
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1
H NMR spectrum
recorded at 250.13 MHz
The Knoevenagel reaction between a malonic acid diester and an aldehyde yields a product
having the 250MHz
1
H NMR spectrum shown below. The products contains C, H, and O only.
Determine the structural formulae of the starting materials and the product. Assign as many
signals as possible. What is the solvent?
O
O
C
C
C
H
H
O
O
R
2
R
1
O C
R
3
H
O
O
C
C
C
O
O
R
2
R
1
C
H
R
3
Solution
+ H
2
O+
The question is about this chemical reaction.
O
O
C
C
C
H
H
O
O
R
2
R
1
O C
R
3
H
O
O
C
C
C
O
O
R
2
R
1
C
H
R
3
Solution
+ H
2
O+
We are looking for the reaction product.
The question is about this chemical reaction.
O
O
C
C
C
O
O
R
2
R
1
C
H
R
3
≈7 ppm
We are looking for the reaction product.
The question is about this chemical reaction.
The proton signal should appear at about 7 ppm, we don't know about the
R
1
, R
2
and R
3
residues at the moment.
100
1H
511
5H
418
4H
612
6H
Inte
-
gral
O
O
C
C
C
O
O
R
2
R
1
C
H
R
3
≈7 ppm
O
O
C
C
C
O
O
R
2
R
1
C
H
R
3
≈7 ppm
The proton signal should appear at about 7 ppm, we don't know about the
R
1
, R
2
and R
3
residues at the moment.
100
1H
511
5H
418
4H
612
6H
Inte
-
gral
O
O
C
C
C
O
O
R
2
R
1
C
H
R
3
≈7 ppm
Let's start with the easiest part: the solvent.
The residual proton signal (the non-deuterated portion) of chloroform
appears at about 7.25 ppm.
Here, this signal is hardly visible, but at low sample concentration
and/or poor quality of the solvent, the signal might dominate the
spectrum.
100
1H
511
5H
418
4H
612
6H
Inte
-
gral
O
O
C
C
C
O
O
R
2
R
1
C
H
R
3
≈7 ppm
Let's start with the easiest part: the solvent.
A second typical signal comes from traces of water in the solvent.
Depending on the origin of the water, it can be either H
2
O or HOD.
Spectroscopically, this is not distinguishable.
HOD/H
2
O
100
1H
511
5H
418
4H
612
6H
Inte
-
gral
O
O
C
C
C
O
O
R
2
R
1
C
H
R
3
≈7 ppm
Besides the solvent signals, the spectrum shows four signal groups. The
approximate chemical shifts are sufficient to answer our spectroscopic
question.
7.7 ppm
7.4 ppm
4.3 ppm
1.3 ppm
The total number of protons is unknown, but the smallest signal must
correspond to at least one proton.
100
1H
511
5H
418
4H
612
6H
Inte
-
gral
O
O
C
C
C
O
O
R
2
R
1
C
H
R
3
≈7 ppm
7.7 ppm
7.4 ppm
4.3 ppm
1.3 ppm
The total number of protons is unknown, but the smallest signal must
correspond to at least one proton.
The total number of protons is unknown, but the smallest signal must
correspond to at least one proton.
This would give us a proportionality factor of
100 a.u. (arbitrary unit) = 1H
and we can convert the integrals of the remaining three signal groups into
integer proton values with only minimal rounding.
Number of protons
O
O
C
C
C
O
O
R
2
R
1
C
H
R
3
≈7 ppm
Number of protons
7.4 ppm / 5H 4.3 ppm / 4H 1.3 ppm / 6H7.7 ppm / 1H (m) (m)
7.7 ppm
1H
5H
7.4 ppm
4.3 ppm
4H 6H
1.3 ppm
The essential information consists of four chemical shifts and the
corresponding number of protons.
Let's record these 8 pieces of information on four small sticky notes and
additionally note that the clean multiplets of the signal groups at 4.3 ppm
and 1.3 ppm probably deserve a closer look.
O
O
C
C
C
O
O
R
2
R
1
C
H
R
3
≈7 ppm
7.4 ppm / 5H 4.3 ppm / 4H 1.3 ppm / 6H7.7 ppm / 1H (m) (m)
7.7 ppm
1H
7.4 ppm 5H 4.3 ppm 4H
1.3 ppm 6H
O
O
C
C
C
O
O
R
2
R
1
C
H
R
3
≈7 ppm
In the centre of the spectrum (4.3 ppm) a
clean quintet is eye catching. It seems easiest
to start the further evaluation at this point.
The full spectrum is not necessary, let's focus
on that multiplet.
1060.02
1067.15
1074.75
1081.97
1089.17
Hz
4.3 ppm
7.4 ppm / 5H 4.3 ppm / 4H 1.3 ppm / 6H7.7 ppm / 1H (m) (m)
7.7 ppm
1H
7.4 ppm 5H 4.3 ppm 4H
1.3 ppm 6H
O
O
C
C
C
O
O
R
2
R
1
C
H
R
3
≈7 ppm
The full spectrum is not necessary, let's focus
on that multiplet.
1060.02
1067.15
1074.75
1081.97
1089.17
Hz
The full spectrum is not necessary, let's focus
on that multiplet.
A coupling constant of 7.28 Hz ( 1089.17 Hz –
1060.02 Hz) / 4 ) occurs very frequently between
neighbouring protons bound to sp
3
-hybridised
carbon atoms.
According to the n+1 rule, a quintet is formed by
four equivalent neighbouring protons.
On our sticky notes, we see an integral of
4. But … these are the protons of the
quintet itself, not neighbouring protons.
? ??
4.3 ppm
Something is not right. Let's put this
quintet on hold for now and try an
analysis of the multiplet at 1.3 ppm.
7.4 ppm / 5H 4.3 ppm / 4H 1.3 ppm / 6H7.7 ppm / 1H (m) (m)
7.7 ppm
1H
7.4 ppm 5H 4.3 ppm 4H
1.3 ppm 6H
O
O
C
C
C
O
O
R
2
R
1
C
H
R
3
≈7 ppm
1060.02
1067.15
1074.75
1081.97
1089.17
Hz
4.3 ppm
Something is not right. Let's put this
quintet on hold for now and try an
analysis of the multiplet at 1.3 ppm.
Something is not right. Let's put this
quintet on hold for now and try an
analysis of the multiplet at 1.3 ppm.
307.51
314.64
319.53
321.78
326.65
333.79
Hz
1.3 ppm
The multiplet might look a little bit confusing
at a first glance. Hopefully this changes if we
colour three of the six lines differently.
7.4 ppm / 5H 4.3 ppm / 4H 1.3 ppm / 6H7.7 ppm / 1H (m) (m)
7.7 ppm
1H
7.4 ppm 5H 4.3 ppm 4H
1.3 ppm 6H
O
O
C
C
C
O
O
R
2
R
1
C
H
R
3
≈7 ppm
307.51
314.64
319.53
321.78
326.65
333.79
Hz
1.3 ppm
The multiplet might look a little bit confusing
at a first glance. Hopefully this changes if we
colour three of the six lines differently.
307.51
314.64
319.53
321.78
326.65
333.79
Hz
Now a "red" and a "green" triplet are clearly
visible.
The multiplet might look a little bit confusing
at a first glance. Hopefully this changes if we
colour three of the six lines differently.
The coupling constant in both triplets is
7.13 Hz. For the vicinal coupling constant
in the structural fragment – CH
2
– CH
2
–
this is a perfect textbook value.
But are these now two independent
triplets or a doublet of triplets?
J = 7.13 Hz J = 7.13 Hz
7.4 ppm / 5H 4.3 ppm / 4H 1.3 ppm / 6H7.7 ppm / 1H (m) (m)
7.7 ppm
1H
7.4 ppm 5H 4.3 ppm 4H
1.3 ppm 6H
O
O
C
C
C
O
O
R
2
R
1
C
H
R
3
≈7 ppm
1.3 ppm
307.51
314.64
319.53
321.78
326.65
333.79
Hz
But are these now two independent
triplets or a doublet of triplets?
J = 7.13 Hz J = 7.13 Hz
But are these now two independent
triplets or a doublet of triplets?
Let's just complete this multiplet to the doublet
of triplets on a trial basis.
J = 12.01 Hz
The 6 protons at 1.3 ppm are clearly bound
to an sp
3
-hybridised C atom. Vicinal
coupling constants of 12 Hz involving such
protons are only possible with fixed
geometry, for example in ring systems.
A ring system could not be ruled out with
certainty. But perhaps there is another
argument against a doublet of triplets.
7.4 ppm / 5H 4.3 ppm / 4H 1.3 ppm / 6H7.7 ppm / 1H (m) (m)
7.7 ppm
1H
7.4 ppm 5H 4.3 ppm 4H
1.3 ppm 6H
O
O
C
C
C
O
O
R
2
R
1
C
H
R
3
≈7 ppm
1.3 ppm
307.51
314.64
319.53
321.78
326.65
333.79
Hz
A ring system could not be ruled out with
certainty. But perhaps there is another
argument against a doublet of triplets.
A ring system could not be ruled out with
certainty. But perhaps there is another
argument against a doublet of triplets.
H
2
C
C
H
2
X
H
A doublet of triplets would be explained with this
kind of a structural fragment. X can be O or C. We
ignore the somewhat strange coupling constant of
12.01 Hz.
7.13 Hz
12.01 Hz
1.3 ppm /
6H (!)
But we need not 2, but 6 protons. For
that, this fragment would have to be
present three times.
7.4 ppm / 5H 4.3 ppm / 4H 1.3 ppm / 6H7.7 ppm / 1H (m) (m)
7.7 ppm
1H
7.4 ppm 5H 4.3 ppm 4H
1.3 ppm 6H
O
O
C
C
C
O
O
R
2
R
1
C
H
R
3
≈7 ppm
1.3 ppm
307.51
314.64
319.53
321.78
326.65
333.79
Hz
H
2
C
C
H
2
X
H
7.13 Hz
12.01 Hz
1.3 ppm /
6H (!)
H
2
C
C
H
2
X
H
7.13 Hz
12.01 Hz
1.3 ppm /
6H (!)
3x
But we need not 2, but 6 protons. For
that, this fragment would have to be
present three times.
7.4 ppm / 5H 4.3 ppm / 4H 1.3 ppm / 6H7.7 ppm / 1H (m) (m)
7.7 ppm
1H
7.4 ppm 5H 4.3 ppm 4H
1.3 ppm 6H
O
O
C
C
C
O
O
R
2
R
1
C
H
R
3
≈7 ppm
1.3 ppm
307.51
314.64
319.53
321.78
326.65
333.79
Hz
H
2
C
C
H
2
X
H
7.13 Hz
12.01 Hz
1.3 ppm /
6H (!)
3x
For this fragment, another 3 • 2 protons
for the neighbouring CH
2
group would
have to appear somewhere in the
spectrum. But another multiplet with the
integral 6 does not exist.
6H
7.4 ppm / 5H 4.3 ppm / 4H 1.3 ppm / 6H7.7 ppm / 1H (m) (m)
7.7 ppm
1H
7.4 ppm 5H 4.3 ppm 4H
1.3 ppm 6H
O
O
C
C
C
O
O
R
2
R
1
C
H
R
3
≈7 ppm
1.3 ppm
307.51
314.64
319.53
321.78
326.65
333.79
Hz
CH
2
H
3
C
H
3
C CH
2
7.13 Hz
7.13 Hz
Completely without a strange coupling
constant of 12.01 Hz, two independent
triplets with integral 6 can be explained as
two chemically distinguishable methyl
groups, which are part of ethyl fragments.
A coupling constant of 7.13 Hz is
absolutely typical for this fragment, and
the triplet structure results from the two
equivalent protons of the adjacent
methylene groups.
7.4 ppm / 5H 4.3 ppm / 4H 1.3 ppm / 6H7.7 ppm / 1H (m) (m)
7.7 ppm
1H
7.4 ppm 5H 4.3 ppm 4H
1.3 ppm 6H
R
2
R
1
O
O
C
C
C
O
O
C
H
R
3
≈7 ppm
1.3 ppm
307.51
314.64
319.53
321.78
326.65
333.79
Hz
H
3
C CH
2
7.13 Hz
CH
2
H
3
C
7.13 Hz
Could the two ethyl groups be the
fragments R
1
und R
2
?
The molecule is asymmetric, but the
chemical environment of R
1
and R
2
is
nevertheless very similar. This would
explain the small differences in the
chemical shifts of the two methyl groups.
Let's just try it.
7.4 ppm / 5H 4.3 ppm / 4H 1.3 ppm / 6H7.7 ppm / 1H (m) (m)
7.7 ppm
1H
7.4 ppm 5H 4.3 ppm 4H
1.3 ppm 6H
R
2
R
1
O
O
C
C
C
O
O
C
H
R
3
≈7 ppm
1.3 ppm
307.51
314.64
319.53
321.78
326.65
333.79
Hz
O
O
C
C
C
O
O
C
H
R
3
≈7 ppm
H
3
C CH
2
7.13 Hz
CH
2
H
3
C
7.13 Hz
H
3
C CH
2
7.13 Hz
CH
2
H
3
C
7.13 Hz
Let's just try it.
The Schoolery rules give an approximate value of 3.6 ppm for the
methylene protons. The agreement with the measured 4.3 ppm could
be slightly better, but the simple Schoolery rules does not take into
account the conjugated double bonds in this molecule.
The singlet at 7.7 ppm should be assigned to the proton
bound to the same carbon as R
3
. But what might R
3
be?
Let's just try it.
O
O
C
C
C
O
O
C
H
R
3
7.7 ppm
H
3
C CH
2
7.13 Hz
CH
2
H
3
C
7.13 Hz
7.4 ppm / 5H7.4 ppm 5H
1.3 ppm
4.3 ppm
1.3 ppm
4.3 ppm
The two ethyl groups are chemically distinguishable, but the exact
assignment of the slightly different chemical shifts is impossible with
the available data.
5 protons with a chemical shift of about 7 ppm? Since we have no
carbon spectrum, some speculations would be possible. From the
proton spectrum alone, only one very probable fragment can be
deduced: a phenyl group.
The common abbreviation Ø for the phenyl group is used here.
There is one last question.
O
O
C
C
C
O
O
C
H
Ø
7.7 ppm
H
3
C CH
2
7.13 Hz
CH
2
H
3
C
7.13 Hz
4.3 ppm
1.3 ppm
4.3 ppm
But what might be the last question?
4.3 ppm
1.3 ppm
At 4.3 ppm we expect the signals of the methylene protons. Because of
the neighbouring methyl group, a quartet and not a quintet should
appear there?
Because of the asymmetry of the molecule, the two methylene groups,
as well as the two methyl groups, are chemically distinguishable. As a
result, we obtain the expected quartet for each of the two methylene
groups. Coincidentally, in this compound, the chemical shifts of the two
quartets differ almost exactly by the vicinal coupling constant.
Let's change the colours a little for better illustration.
O
O
C
C
C
O
O
C
H
Ø
7.7 ppm
H
3
C CH
2
7.13 Hz
CH
2
H
3
C
7.13 Hz
4.3 ppm
1.3 ppm
4.3 ppm
4.3 ppm
1.3 ppm
3
1 1
3
3
1 1
3
1 4 6 4
1
Let's start with the quartet of protons marked in purple.
And now the methylene protons marked in green ...
… and add both quartets.
C
C
H
C
C
O O
O CH
2
CH
3
O
H
2
C
CH
3
might be reversed
might be reversed
protons of the
phenyl ring
Summary
