Exercise plus Solution – Quick overview
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1
H NMR spectrum
recorded at 250.13 MHz
C
6
H
12
O
2
measured in CDCl
3
Get the constitution!
(In the case of an ambiguous solution,
please refer to the Schoolery rules)
C
C
H
CH
3
CH
3
O O
CH
2
H
3
C
Solution at a glance
There is no step by step solution available so far.
Solution at a glance
There is no step by step solution available so far.
Isobutyric acid ethyl ester
Solution
C
6
H
12
O
2
Basics
Double bond equivalents,
number of signal groups,
integration
𝒏
𝐃𝐁𝐄
=
𝟐𝒏
𝐂
− 𝒏
𝐇
+ 𝟐
𝟐
From 12 protons, 6 carbon atoms and two oxygen atoms we
determine according to the formula
One double equivalent (DBE).
1 DBÄ
We need the value later and keep it on a small sticky note.
C
6
H
12
O
2
Two signal groups are easy to recognise.
1 DBÄ
With the colour marking made here, the two high-field signal groups
are also easy to recognise.
Not having this aid, it sometimes helps to partially hide spectral
ranges that are close together.
A doublet is now clearly visible.
Basics
Double bond equivalents,
number of signal groups,
integration
Solution
Solution
C
6
H
12
O
2
When the grey rectangle is removed, a triplet remains.
1 DBÄ
The separation of the two closely neighbouring multipletts should
also be marked in the integral line for the next step.
Basics
Double bond equivalents,
number of signal groups,
integration
C
6
H
12
O
2
Of course, one could evaluate integrals absolutely.
1 DBÄ
However, we are missing a whole range of factors for this, for
example
C
6
H
12
O
2
• the sample concentration,
• the diameter of the sample tube,
• various properties of the device electronics and
• the geometry of the display medium.
Only the quotient of integrals can be used meaningfully. We measure
the integrals themselves in arbitrary units (a.u.), for example the
filling height of glass tubes.
Basics
Double bond equivalents,
number of signal groups,
integration
30%
15%
45%
89%
1 DBÄ
C
6
H
12
O
2
Using the simple rule of three,
one can now distribute the 12
protons proportionally according
to the four filling levels.
2H
1H
3H
6H
Basics
Double bond equivalents,
number of signal groups,
integration
NMR parameters
Chemical shifts, multiplicity,
coupling constants
1 DBÄ
C
6
H
12
O
2
The multiplets look pretty well. They seem to be „pure“
multiplets. Perhaps this cursory observation is wrong,
but let's start with the simplest possibility. If the
assumption is wrong, we have to revisit everything later.
2H
1H
3H
6H
1 DBÄ
C
6
H
12
O
2
Let's start with the quartet at deepest field.
2H
1H
3H
6H
1011.96
1019.0
8
1026.2
1
1033.35
Hz
4.09 ppm
quartet
J = 7.13 Hz
δ =
1033.35 Hz + 1011.96 Hz
2 ∗ 250.13 MHz
= 𝟒. 𝟎𝟗 𝐩𝐩𝐦
𝐽 =
1033.35 Hz − 1011.96 Hz
3
= 𝟕. 𝟏𝟑 𝐇𝐳
NMR parameters
Chemical shifts, multiplicity,
coupling constants
1 DBÄ
C
6
H
12
O
2
Because of the very weak intensity of the two
outermost lines, the multiplet at approx. 2.5 ppm can
only be recognised as a septet when magnified.
Note: In routine measurement mode, these two lines
might disappear in the noise.
2H
1H
3H
6H
δ =
644.77 Hz + 602.81 Hz
2 ∗ 250.13 MHz
= 𝟐. 𝟒𝟗 𝐩𝐩𝐦
𝐽 =
644.77 Hz − 602.81 Hz
6
= 𝟕. 𝟏𝟑 𝐇𝐳
4.09 ppm
quartet
J = 7.13 Hz
602.81
609.75
616.72
623.71
630.71
637.72
644.77
Hz
2.49 ppm
septet
J = 6.99 Hz
NMR parameters
Chemical shifts, multiplicity,
coupling constants
1 DBÄ
C
6
H
12
O
2
The calculations for the remaining triplet and doublet at
high field (low values of the chemical shifts) are done
analogously.
2H
1H
3H
6H
4.09 ppm
quartet
J = 7.13 Hz
2.49 ppm
septet
J = 6.99 Hz
278.92
285.91
297.88
305.01
312.14
Hz
1.13 ppm
quartet
J = 6.99 Hz
1.22 ppm
triplet
J = 7.13 Hz
NMR parameters
Chemical shifts, multiplicity,
coupling constants
Building blocks
1 DBÄ
C
6
H
12
O
2
2H
1H
3H
6H
4.09 ppm
quartet
J = 7.13 Hz
2.49 ppm
septet
J = 6.99 Hz
1.13 ppm
quartet
J = 6.99 Hz
1.22 ppm
triplet
J = 7.13 Hz
C
C
H
H
In each of the four multipletts, the coupling constant –
independent of the multiplicity – is about 7 Hz.
A coupling constant of 7 Hz is very common. This is a coupling
via three bonds (technical term: vicinal coupling) along the
chain H – C – C – H under the condition of a free rotation
around the C – C single bond.
1 DBÄ
C
6
H
12
O
2
2H
1H
3H
6H
4.09 ppm
quartet
J = 7.13 Hz
2.49 ppm
septet
J = 6.99 Hz
1.13 ppm
quartet
J = 6.99 Hz
1.22 ppm
triplet
J = 7.13 Hz
With this assumption, one immediately obtains three
structural fragments.
C
H
CH
3
CH
2
A CH
6
group for the doublet at 1.13 ppm is impossible, of course.
This have to be either two chemically equivalent methyl or three
chemically equivalent methylene groups.
Let's try 3 methylene groups first.
Building blocks
1 DBÄ
C
6
H
12
O
2
2H
1H
3H
6H
4.09 ppm
quartet
J = 7.13 Hz
2.49 ppm
septet
J = 6.99 Hz
1.13 ppm
quartet
J = 6.99 Hz
1.22 ppm
triplet
J = 7.13 Hz
All 6 CH
n
fragments would add up to C
6
H
12
.
C
H
CH
3
CH
2
3 * >CH
2
Comparing with the molecular formula and the small sticky
note, only two oxygen atoms and, above all, one double bond
equivalent would now have to be assigned. This is impossible.
Thus, the 6 protons of the doublet at 1.13 pm can come only
from two equivalent methyl groups.
Building blocks
1 DBÄ
C
6
H
12
O
2
2H
1H
3H
6H
4.09 ppm
quartet
J = 7.13 Hz
2.49 ppm
septet
J = 6.99 Hz
1.13 ppm
quartet
J = 6.99 Hz
1.22 ppm
triplet
J = 7.13 Hz
The fragments now add up to C
5
H
12
. One carbon, two oxygen
atoms and a double bond fragment would still have to be
assigned. This is possible, for example, as a carboxyl group.
C
H
CH
3
CH
2
H
3
C
H
3
C
Building blocks
Building blocks
Linking
1 DBÄ
C
6
H
12
O
2
2H
1H
3H
6H
4.09 ppm
quartet
J = 7.13 Hz
2.49 ppm
septet
J = 6.99 Hz
1.13 ppm
quartet
J = 6.99 Hz
1.22 ppm
triplet
J = 7.13 Hz
If the assumption with the "pure" multiplets is correct, there
cannot be a chain of the type – CH
x
– CH
y
– CH
z
–.
In such a chain, a multiplet of the type doublet of triplets or
similar would be present in the middle at CH
y
.
C
H
CH
3
CH
2
H
3
C
H
3
C
The fragments can only be linked in pairs and the pairs
separated have tzo be separated from each other. Oxygen
might separate the pairs.
1 DBÄ
C
6
H
12
O
2
2H
1H
3H
6H
4.09 ppm
quartet
J = 7.13 Hz
2.49 ppm
septet
J = 6.99 Hz
1.13 ppm
quartet
J = 6.99 Hz
1.22 ppm
triplet
J = 7.13 Hz
In a fragment of type – CH
x
– CH
y
– the coupling constant
observed in the multiplet of CH
x
must be identical to that of
multiplet CH
y
.
C
H
CH
3
CH
2
H
3
C
H
3
C
At both 4.09 and 1.22 ppm we observe the same coupling
constant of 7.13 Hz.
Building blocks
Linking
1 DBÄ
C
6
H
12
O
2
2H
1H
3H
6H
4.09 ppm
quartet
J = 7.13 Hz
2.49 ppm
septet
J = 6.99 Hz
1.13 ppm
quartet
J = 6.99 Hz
1.22 ppm
triplet
J = 7.13 Hz
Identical coupling constants of 6.99 Hz each exist in the
multiplet at 2.49 ppm and 1.13 ppm. The doublet at 1.13 ppm
belongs to two equivalent methyl group.
C
H
CH
3
CH
2
H
3
C
H
3
C
CH
2
CH
3
4.09
ppm
1.22
ppm
7.13 Hz
Building blocks
Linking
1 DBÄ
C
6
H
12
O
2
2H
1H
3H
6H
4.09 ppm
quartet
J = 7.13 Hz
2.49 ppm
septet
J = 6.99 Hz
1.13 ppm
quartet
J = 6.99 Hz
1.22 ppm
triplet
J = 7.13 Hz
Identical coupling constants of 6.99 Hz each exist in the
multiplet at 2.49 ppm and 1.13 ppm. The doublet at 1.13 ppm
belongs to two equivalent methyl group.
C
H
H
3
C
H
3
C
CH
2
CH
3
4.09
ppm
1.22
ppm
7.13 Hz
C
H
H
3
C
H
3
C
1.13
ppm
2.49
ppm
6.99 Hz
Building blocks
Linking
Building blocks
A little check
1 DBÄ
C
6
H
12
O
2
2H
1H
3H
6H
4.09 ppm
quartet
J = 7.13 Hz
2.49 ppm
septet
J = 6.99 Hz
1.13 ppm
quartet
J = 6.99 Hz
1.22 ppm
triplet
J = 7.13 Hz
Are the multiplet multiplicities correct?
CH
2
CH
3
4.09
ppm
1.22
ppm
7.13 Hz
C
H
H
3
C
H
3
C
1.13
ppm
2.49
ppm
6.99 Hz
Let us take the septet at 2.49 ppm as an example.
Along an H – C – C – H – chain, 6 equivalent protons of two
equivalent methyl groups are adjacent to the corresponding
poton via three bonds. According to the n+1 rule, the
observed septet results.
Finalisation
Two possibilities
1 DBÄ
C
6
H
12
O
2
The isopropyl and ethyl groups add up to C
5
H
12
.
CH
2
CH
3
4.09
ppm
1.22
ppm
7.13 Hz
C
H
H
3
C
H
3
C
1.13
ppm
2.49
ppm
6.99 Hz
Missing:
- one carbon atom,
- two oxygen atoms, and
- one double bond equivalent.
This is a carboxylic group that has to be placed between the two
fragments. Let's rearrange these two fragments a little bit.
1 DBÄ
C
6
H
12
O
2
CH
2
CH
3
4.09
ppm
1.22
ppm
7.13 Hz
C
H
H
3
C
H
3
C
1.13
ppm
2.49
ppm
6.99 Hz
CH
2
CH
3
4.09
ppm
1.22
ppm
7.13 Hz
C
H
H
3
C
H
3
C
1.13
ppm
2.49
ppm
6.99 Hz
And now we can tentatively insert the carboxylic group.
Finalisation
Two possibilities
CH
2
CH
3
4.09
ppm
1.22
ppm
7.13 Hz
C
H
H
3
C
H
3
C
1.13
ppm
2.49
ppm
6.99 Hz
O
C
O
Is this structure correct? There are no measurement available
that could directly confirm the chosen connectivity.
The expected chemical shift of methylene protons can be
estimated quite well with the help of a simple calculation. A
search for Schoolery's rule should quickly show the simple
calculation method.
For the methylene protons in propionic acid methyl ester shown
here, the estimation results in
δ = (0.23 + 0.47 + 1.55) ppm = 2.25 ppm
The predicted value deviates very strongly from the measured value.
Finalisation
Two possibilities
CH
2
CH
3
4.09
ppm
1.22
ppm
7.13 Hz
C
H
H
3
C
H
3
C
1.13
ppm
2.49
ppm
6.99 Hz
O
C
O
Let us duplicate our structure …
CH
2
CH
3
4.09
ppm
1.22
ppm
7.13 Hz
C
H
H
3
C
H
3
C
1.13
ppm
2.49
ppm
6.99 Hz
O
C
O
Finalisation
Two possibilities
CH
2
CH
3
4.09
ppm
1.22
ppm
7.13 Hz
C
H
H
3
C
H
3
C
1.13
ppm
2.49
ppm
6.99 Hz
O
C
O
CH
2
CH
3
4.09
ppm
1.22
ppm
7.13 Hz
C
H
H
3
C
H
3
C
1.13
ppm
2.49
ppm
6.99 Hz
O
C
O
… and invert the carboxylic group.
Using some minor cosmetics …
Finalisation
Two possibilities
CH
2
CH
3
4.09
ppm
1.22
ppm
7.13 Hz
C
H
H
3
C
H
3
C
1.13
ppm
2.49
ppm
6.99 Hz
O
C
O
CH
2
CH
3
4.09
ppm
1.22
ppm
7.13 Hz
C
H
H
3
C
H
3
C
1.13
ppm
2.49
ppm
6.99 Hz
O
C
O
C
O
O CH
2
CH
3
C
H
H
3
C
H
3
C
4.09
ppm
1.22
ppm
1.13
ppm
2.49
ppm
7.13 Hz
6.99 Hz
… all bonds are in place and ...
… the carbon atom symbol is the right way up again.
Finalisation
Two possibilities
CH
2
CH
3
4.09
ppm
1.22
ppm
7.13 Hz
C
H
H
3
C
H
3
C
1.13
ppm
2.49
ppm
6.99 Hz
O
C
O
C
O
O CH
2
CH
3
C
H
H
3
C
H
3
C
4.09
ppm
1.22
ppm
1.13
ppm
2.49
ppm
7.13 Hz
6.99 Hz
This time the estimation gives
δ = (0.23 + 0.47 + 3.13) ppm = 3.83 ppm
Not absolutely perfect, but much better than before for the left
structure.
And what about the prediction of the chemical shifts for the
protons of the methylene group now?
Finalisation
Two possibilities
C
C
H
CH
3
CH
3
O O
CH
2
H
3
C
Summary
Isobutyric acid ethyl ester
Summary
(Feel free to turn the 3D structure)
