Determination of structure
Determination of structure (discovering the structure of a molecule) is a
very important part of organic chemistry. The problem is that the molecules
cannot be 'seen' as such and indirect evidence for the structure must be gleaned
by a variety of methods. The final 'proof' is often the synthesis of the molecule
from simpler units, producing a compound with identical characteristics to
the unknown substance.
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Mass spectrometry
The first stage to find out is the relative molecular mass of the substance
in question. Nowadays, this may be done accurately using a mass spectrometer.
In the past it was necessary to find the percentage composition of each component
element to find the empirical formula and then to find the molecular mass
by one of several physical techniques.
Once the relative molecular mass and chemical formula are known, structural
information may be obtained by both electronic and 'wet' chemical means.
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The molecular ion
The highest m/e
peak in the mass spectrum always consists of the molecular ion; i.e. the ion
that has been formed by dislodging one electron from the molecule under investigation.
Mass spectrometry m/e values can be read to many decimal places. As all atomic
isotopes have slightly different masses, this allows the molecular formula
to be obtained directly from the m/e value of the molecular ion.
Mass calculator from formula: http://www.bmrb.wisc.edu/metabolomics/mol_mass.php
Example: The two compounds ethenol and propane both have the same relative masses:
C
2H
3OH
- 2 x C = 24
- 1 x O = 16
- 4 x H = 4
Total relative mass = 44
C
3H
8
Total relative mass = 44
However, using accurate values for the relative masses of the atoms
we get:
C
2H
3OH
- 2 x C = 24.0000
- 1 x O = 1 x 15.994915
- 4 x H = 4 x 1.007825
Total = 44.026215
C
3H
8
- 3 x C = 3 x 12.0000
- 8 x H = 8 x 1.007825
Total = 44.062600
Clearly, these two values are not the same and can be differentiated
in the mass spectrometer.
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Fragmentation
Once the molecular ion has been formed in the high energy beam of electrons,
it is likely that this ion wlll break apart into fragments. Each fragment
is either an ion itself, or a neutral species. The ions formed by fragmentation
can also be detected in the mass spectrometer trace, but neutral fragments
are neither accelerated, deflected nor detected.
The likelihood of a specific ion forming depends on the bond energy of the
bond that must be broken and the stability of the fragment formed.
Typically, alkyl groups, acyl groups and allyl groups are most easily formed
and often appear in mass spectra.
Alkyl fragments
Acyl fragments
Allyl fragments
[CH3]+
[CH3CO]+
[CH2=CH2]+
[C2H5]+
[C2H5CO]+
[CH3CH=CH2]+
[C3H7]+
[C3H7CO]+
As each of these fragments has a specific m/e value, an experienced analyst
recognises the fragments as they are formed and uses this information to help
build a structure of the fragmenting molecule. It's like being given pieces
of a jigsaw puzzle and fitting them together to get the final picture.
The spectrum, showing several different peaks due to fragmentation, gives
rise to a fragmentation pattern, which is the suggested way that a specific
molecule has broken apart.
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Isotopes
The presence of isotopes in the sample may cause confusion as regards the
molecular ion. For example, a compound containing a chlorine atom will show
two peaks due to the presence of 35Cl and 37Cl.
The intensity (height) of the peaks containing isotopes is always in the
ratio of the natural abundance of the isotopes. In the example above, chlorine
occurs in nature in the ratio 75% of the 35-Cl isotope to 25% of the 37-Cl
isotope. This is a ratio of 3:1.
The heights of the peaks due to C3H735Cl
and C3H737Cl, i.e. the C3H735Cl
peak is three times higher than the C3H737Cl,
peak.
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Deducing the molecular formula
The highest m/e
peak in the mass spectrum always consists of the molecular ion; i.e. the ion
that has been formed by dislodging one electron from the molecule under investigation.
Mass spectrometry m/e values can be read to many decimal places. As all atomic
isotopes have slightly different masses, this allows the molecular formula
to be obtained directly from the m/e value of the molecular ion.
Mass calculator from formula: http://www.bmrb.wisc.edu/metabolomics/mol_mass.php
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Fragmentation patterns
There are two approaches to fragmentation.
Start from the molecular ion and calculate the mass of the particles that
have been subtracted from the m/e value of the molecular ion to give the peaks
seen.
Example:
If the molecular ion appears at m/e = 58, and the next lowest peak
appears at m/e = 43, then a fragment has been lost that corresponds
to 58 - 43 = 15 mass units.
This corresponds to a methyl ion fragment,
[CH3]+.
The second approach involves looking at the fragments at the low m/e end
of the spectrum.
Example
If a fragment appears at m/e = 29, then this could be due to a ethyl
ion fragment, [C2H5]+.
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Rearrangement
One complication that arises during molecule fragmentation is that the bonds
don't simply break giving fragments, but they can also reform and fragments
themselves can rearrange to give more stable structures.
Full treatment of this is beyond the course, but it should be noted, as fragments
often appear that cannot be explained by simply breaking bonds.
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Worked examples
Q1133-01 Propanal, CH
3CH
2CHO,
(Mr=58) undergoes complete fragmentation is a mass spectrometer. What is the
m/z value of the molecular ion line in its mass spectrum?
- 15
- 29
- 43
- 58
Answer
|
The molecular ion has the same mass/chrge ratio as the relatgive
mass of the molecule before fragmentation, in this case 58.
It is an ion that has lost only one electron and consequently has
a positive charge.
|
Q1133-02 The mass spectrum
of CH
3COOC
2H
5 is not expected to show a major
ion peak at which m/e ratio?
- 88
- 32
- 29
- 15
Answer
|
All of the m/e ratios correspond to likely fragments from the molecule.
[CH3]+ corresponds to m/e=15, [C2H5]+
corresponds to m/e=29, [M]+ corresponds to the molecular ion. The
m/e=32 does not correspond to any
fragment.
|
Q1133-03 The mass spectrum
below corresponds to an organic compound. What is it's relative mass?
.gif)
Answer
|
The relative mass is found by looking at the m/e ratio of the peak
with the highest m/e value - in this case,
74
|
Q1133-04 The mass spectrum
in question 03 has a line at 73 and a line at 57. The relative mass of the molecule
is 74, suggest an identity for the species that has been lost leaving a fragment
of m/e=57.
Answer
|
The difference between the molecular ion and m/e 57 is 17 mass units.
This possibly corresponds to loss of one OH+
group.
|
Q1133-05 Using the spectrum
in question 03, the intense line that appears at m/e=45 suggests the loss of
which two possible organic groups?
Answer
|
The difference between the m/e of the molecular ion (74) and the
intense line at m/e=45 is (74-45)=29 mass units. A common organic
group with this mass is the C2H5+
(ethyl) group of atoms. Another possible answer is the CHO+
(aldehyde) group of atoms. Both of these could be lost as fragments
from the end of a molecule.
|
Q1133-06 Using the spectrum
shown below, suggest the fragment that has been lost from the molecular ion
to leave a particle at m/e 43.
.gif)
Answer
|
The difference between the m/e of the molecular ion (58) and the
intense line at m/e=43 is (58-43)=15 mass units, probably corresponding
to loss of a CH3+
(methyl) group.
|
Q1133-07 Which of the following
compounds will give a molecular ion with an odd number m/z ratio?
- CH3CH2OH
- CH3CH2CH2CH3
- CH3CO2H
- CH3CH2NH2
Answer
|
Summing all of the relative masses we find that CH3CH2NH2
adds up to an odd number (45). This is characteristic of nitrogen
containing compounds with an odd number of nitrogen atoms.
|
Q1133-08 A halogen containing
organic compound has a mass spectrum wth lines at m/e 156, 127 and 29 only.
Suggest an identity for the compound.
Answer
|
The halogen iodine has a relative mass of 127. Loss of an iodine
atom would leave a fragment with an m/e ratio of 156-127 = 29. This
fragment could be C2H5, making the compound
iodoethane, C2H5I.
|
Q1133-09 An unknown hydrocarbon
gave the mass spectrum shown below. Find the relative mass of the compound,
hence suggest a molecular formula and suggest a fragment that has broken off
leaving the line that appears at m/e 55.
.gif)
Answer
|
The molecular ion appears at m/e = 70,
hence this is the relative molecular mass. The molecule is a hydrocarbon
and therefore contains only carbon and hydrogen. The only possible
combination is C5H10.
The line at m/e = 55 corresponds to a loss of 70 - 55 = 15 mass units.
This is typical of loss of a CH3+
group.
|
Q1133-10 The mass spectrum
shown below is an isomer of the hydrocarbon in question 09. Suggest an identity
for the line that appears at m/e = 42.
.gif)
Answer
|
The line at m/e 42 is due to loss of a fragment with a mass of 70
- 42 = 28 mass units. This could correspond to a loss of C2H4
leaving a fragment [C3H6]+.
|
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Q1133-11 Propanal, CH
3CH
2CHO,
(Mr=58) undergoes complete fragmentation is a mass spectrometer. What is the
m/z value of the molecular ion line in its mass spectrum?
- 15
- 29
- 43
- 58
Answer
|
The molecular ion has the same mass/charge ratio as the relative
mass of the molecule before fragmentation, in this case 58.
It is an ion that has lost only one electron and consequently has
a positive charge.
|
Q1133-12 The mass spectrum
of CH
3COOC
2H
5 is not expected to show a major
ion peak at which m/e ratio?
- 88
- 32
- 29
- 15
Answer
|
All of the m/e ratios correspond to likely fragments from the molecule.
[CH3]+ corresponds to m/e=15, [C2H5]+
corresponds to m/e=29, [M]+ corresponds to the molecular ion. The
m/e=32 does not correspond to any
fragment.
|
Q1133-13 The mass spectrum
below corresponds to an organic compound. What is it's relative mass?
Answer
|
The relative mass is found by looking at the m/e ratio of the peak
with the highest m/e value - in this case,
74
|
Q1133-14 The mass spectrum
in question 03 has a line at 73 and a line at 57. The relative mass of the molecule
is 74, suggest an identity for the species that has been lost leaving a fragment
of m/e=57.
Answer
|
The difference between the molecular ion and m/e 57 is 17 mass units.
This possibly corresponds to loss of one OH
group.
|
Q1133-15 Using the spectrum
in question 03, the intense line that appears at m/e=45 suggests the loss of
which two possible organic groups?
Answer
|
The difference between the m/e of the molecular ion (74) and the
intense line at m/e=45 is (74-45)=29 mass units. A common organic
group with this mass is the C2H5
(ethyl) group of atoms. Another possible answer is the CHO
(aldehyde) group of atoms. Both of these could be lost as fragments
from the end of a molecule.
|
Q1133-16 Using the spectrum
shown below, suggest the fragment that has been lost from the molecular ion
to leave a particle at m/e 43.
Answer
|
The difference between the m/e of the molecular ion (58) and the
intense line at m/e=43 is (58-43)=15 mass units, probably corresponding
to loss of a CH3 (methyl) group.
|
Q1133-17 Which of the following
compounds will give a molecular ion with an odd number m/z ratio?
- CH3CH2OH
- CH3CH2CH2CH3
- CH3CO2H
- CH3CH2NH2
Answer
|
Summing all of the relative masses we find that CH3CH2NH2
adds up to an odd number (45). This is characteristic of nitrogen
containing compounds with an odd number of nitrogen atoms.
|
Q1133-18 A halogen containing
organic compound has a mass spectrum wth lines at m/e 156, 127 and 29 only.
Suggest an identity for the compound.
Answer
|
The halogen iodine has a relative mass of 127. Loss of an iodine
atom would leave a fragment with an m/e ratio of 156-127 = 29. This
fragment could be C2H5, making the compound
iodoethane, C2H5I.
|
Q1133-19 An unknown hydrocarbon
gave the mass spectrum shown below. Find the relative mass of the compound,
hence suggest a molecular formula and suggest a fragment that has broken off
leaving the line that appears at m/e 55.
Answer
|
The molecular ion appears at m/e = 70,
hence this is the relative molecular mass. The molecule is a hydrocarbon
and therefore contains only carbon and hydrogen. The only possible
combination is C5H10.
The line at m/e = 55 corresponds to a loss of 70 - 55 = 15 mass units.
This is typical of loss of a CH3
group.
|
Q1133-20 The mass spectrum
shown below is an isomer of the hydrocarbon in question 09. Suggest an identity
for the line that appears at m/e = 42.
Answer
|
The line at m/e 42 is due to loss of a fragment with a mass of 70
- 42 = 28 mass units. This could correspond to a loss of C2H4
leaving a fragment [C3H6]+.
|
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