Exercise plus Solution – Quick overview
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1.21.41.61.82.02.22.42.62.83.03.23.4 ppm
1
H
298.13
305.53
312.89
718.37
725.65
733.05
740.44
Hz
1
H NMR spectrum
measured at 250.13 MHz
C
3
H
5
ClO measured in CDCl
3
Deduce the structure!
1.21.41.61.82.02.22.42.62.83.03.23.4 ppm
1
H
298.13
305.53
312.89
718.37
725.65
733.05
740.44
Hz
First Considerations
The proton spectrum consists of two signal groups.
There are two chemically distinguishable protons
whose number is not yet known, i.e. n H
a
and m H
b
.
We need the values of n and m.
The molecular formula results in one double bond equivalent.
The structural fragments H-Cl, H-H and H
2
O are not
possible. In each case we would end in a mixture of at
least two components.
n H
a
m H
b
The structural fragment –OH is not excluded.
1.21.41.61.82.02.22.42.62.83.03.23.4 ppm
1
H
298.13
305.53
312.89
718.37
725.65
733.05
740.44
Hz
δ =
ν
Sample
− ν
Refere𝑛𝑐𝑒
ν
Refere𝑛𝑐𝑒
Chemical shifts
In order to estimate the chemical shifts more accurately than is
possible just by looking at the scale below the spectrum, we
need the peak labels given in Hz together with the well-known
formula
ν
Reference
is 250.13 MHz according to the problem
description, the peak labels do not represent absolute
frequencies, but rather the difference from the
reference signal (ν
Sample
− ν
Reference
).
1.21.41.61.82.02.22.42.62.83.03.23.4 ppm
1
H
298.13
305.53
312.89
718.37
725.65
733.05
740.44
Hz
δ
a
=
740.44 Hz + 718.37 Hz
2 ∗ 250.13 MHz
= 𝟐. 𝟗𝟐 𝐩𝐩𝐦
δ
b
=
312.89 Hz + 298.13 Hz
2 ∗ 250.13 MHz
= 𝟏. 𝟐𝟐 𝐩𝐩𝐦
Please remember: ppm is not a unit of
measurement, but the dimensionless number,
10
-6
, which is used like a unit of measurement.
2.92 ppm 1.22 ppm
Chemical shifts
1.21.41.61.82.02.22.42.62.83.03.23.4 ppm
1
H
298.13
305.53
312.89
718.37
725.65
733.05
740.44
Hz
2.92 ppm 1.22 ppm
Integration
The area of both signal groups contains an unknown coefficient of
proportionality, which is identical for each signal group.
61 pieces of gold 92 pieces of gold
And how do we extract the
coefficient of
proportionality?
A possible coefficient of proportionality for example might be
pieces of gold/proton
Let's measure our integrals using gold pieces.
1.21.41.61.82.02.22.42.62.83.03.23.4 ppm
1
H
298.13
305.53
312.89
718.37
725.65
733.05
740.44
Hz
2.92 ppm 1.22 ppm
Integration
61 pieces of gold 92 pieces of gold
And how do we extract the
coefficient of
proportionality?
The molecular formula (C
3
H
5
ClO) includes 5 protons.
The height of both integrals together is 153 pieces of gold.
153 gold pieces ≙ 5 protons
1 gold piece ≙ 0.033 protons
61 gold pieces ≙ 1.99 protons
92 gold pieces ≙ 3.01 protons
2H
3H
1.21.41.61.82.02.22.42.62.83.03.23.4 ppm
1
H
298.13
305.53
312.89
718.37
725.65
733.05
740.44
Hz
2.92 ppm 1.22 ppm
Struktural fragments
2H
3H
HH
C
H
Taking into account the available atoms, only carbon is able to
have thee hydrogens attached.
For two equivalent protons there are two choices:
-CH
2
- or H
2
O.
But we already excluded H
2
O by other reasons.
H H
C
A =CH
2
group is excluded because of the chemical
shift. Protons attached to sp
2
-hybridized carbon
atoms show a chemical shift of about 5 ...7 ppm.
1.21.41.61.82.02.22.42.62.83.03.23.4 ppm
1
H
298.13
305.53
312.89
718.37
725.65
733.05
740.44
Hz
2.92 ppm 1.22 ppm
2H
3H
HH
C
H
H H
C
A short inventory
molecular formula - C
3
H
5
ClO
already known fragments - C
2
H
5
missing atoms - C, Cl, O
what else is missing - 1 double
bond
equivalent
Cl
O
C
There is only one possibility for the missing
pieces.
Struktural fragments
1.21.41.61.82.02.22.42.62.83.03.23.4 ppm
1
H
298.13
305.53
312.89
718.37
725.65
733.05
740.44
Hz
2.92 ppm 1.22 ppm
2H
3H
HH
C
H
H H
C
Cl
O
C
O
Cl
C
H
H H
H
C
C
H
HH
C
H
C
C
H
H
O
Cl
We have two possibilities to
put the four fragments
together.
Which one is correct?
Struktural fragments
1.21.41.61.82.02.22.42.62.83.03.23.4 ppm
1
H
298.13
305.53
312.89
718.37
725.65
733.05
740.44
Hz
2.92 ppm 1.22 ppm
2H
3H
HH
C
H
H H
C
Cl
O
C
HH
C
H
C
C
H
H
O
Cl
HH
C
H
C
C
H
H
O
Cl
O
Cl
C
H
H H
H
C
C
H
O
Cl
C
H
H H
H
C
C
H
Struktural fragments
1.21.41.61.82.02.22.42.62.83.03.23.4 ppm
1
H
298.13
305.53
312.89
718.37
725.65
733.05
740.44
Hz
2.92 ppm 1.22 ppm
Final structures
2H
3H
O
Cl
C
H
H H
H
C
C
H
HH
C
H
C
C
H
H
O
Cl
First, let us try to apply
Shoolery‘s rule for the
methylene protons.
base value 0.23
-CH
3
0.47
-COR 1.70
sum 2.40
base value 0.23
-Cl 2.53
-COR 1.70
sum 4.43
1.21.41.61.82.02.22.42.62.83.03.23.4 ppm
1
H
298.13
305.53
312.89
718.37
725.65
733.05
740.44
Hz
2.92 ppm 1.22 ppm
2H
3H
O
Cl
C
H
H H
H
C
C
H
HH
C
H
C
C
H
H
O
Cl
base value 0.23
-Cl 2.53
-COR 1.70
sum 4.43
Furthermore, in this
compound the protons of the
CH
2
group are 4 bonds away
from the protons of the CH
3
group. Instead of multiplets,
we would see two singlets.
1.22 ppm
2.92 ppm
Final structures
1.21.41.61.82.02.22.42.62.83.03.23.4 ppm
1
H
298.13
305.53
312.89
718.37
725.65
733.05
740.44
Hz
2.92 ppm 1.22 ppm
Finale structure
2H
3H
O
Cl
C
H
H H
H
C
C
H
1.22 ppm
2.92 ppm
O
Cl
C
H
H H
H
C
C
H
1.22 ppm
2.92 ppm
In this structure the protons of the methyl
group at 1.22 ppm are three bonds away
from the two equivalent protons of the
methylene group at 2.92 ppm and appear
according to the n+1 rule as a triplet in the
intensity ratio 1 : 2 : 1.
1.21.41.61.82.02.22.42.62.83.03.23.4 ppm
1
H
298.13
305.53
312.89
718.37
725.65
733.05
740.44
Hz
2.92 ppm 1.22 ppm
2H
3H
O
Cl
C
H
H H
H
C
C
H
1.22 ppm
2.92 ppm
On the other hand, the protons of the methylene group
at 2.92 ppm are also three bonds away from the three
equivalent protons of the methyl group at 1.22 ppm.
According to the n+1 rule, this results in a quartet for
the protons of the methylene group with the intensity
ratio 1 : 3 : 3 : 1.
Finale structure
1.21.41.61.82.02.22.42.62.83.03.23.4 ppm
1
H
298.13
305.53
312.89
718.37
725.65
733.05
740.44
Hz
2.92 ppm 1.22 ppm
2H
3H
O
Cl
C
H
H H
H
C
C
H
1.22 ppm
2.92 ppm
The coupling constant might either be read directly
from two neighbouring lines or averaged over a
whole multiplet in the interest of better accuracy.
𝐽 =
740.44 Hz − 718.37 Hz
3
= 𝟕. 𝟑𝟔 𝐇𝐳
𝐽 = 305.53 Hz − 298.13 Hz = 𝟕. 𝟒𝟎 𝐇𝐳
7.37 Hz
(Mittelung über beide Multipletts)
Finale structure
