Colourful Solutions > Data Processing > Mass spectrometry of organic compounds

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Mass spectrometry is a very useful tool in the identification of organic compounds.

Syllabus ref: S3.2.8

Structure 3.2.8 - Mass spectrometry (MS) of organic compounds can cause fragmentation of molecules. (HL)

  • Deduce information about the structural features of a compound from specific MS fragmentation patterns.

Guidance

  • Include reference to the molecular ion.
  • Data on specific MS fragments are provided in the data booklet

Tools and links

 

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:

C2H3OH
  • 2 x C = 24
  • 1 x O = 16
  • 4 x H = 4
Total relative mass = 44
C3H8
  • 3 x C = 36
  • 8 x H = 8

Total relative mass = 44

However, using accurate values for the relative masses of the atoms we get:

C2H3OH
  • 2 x C = 24.0000
  • 1 x O = 1 x 15.994915
  • 4 x H = 4 x 1.007825

Total = 44.026215

C3H8
  • 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, CH3CH2CHO, (Mr=58) undergoes complete fragmentation is a mass spectrometer. What is the m/z value of the molecular ion line in its mass spectrum?
  1. 15
  2. 29
  3. 43
  4. 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 CH3COOC2H5 is not expected to show a major ion peak at which m/e ratio?
  1. 88
  2. 32
  3. 29
  4. 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?

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.

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?
  1. CH3CH2OH
  2. CH3CH2CH2CH3
  3. CH3CO2H
  4. 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.

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.

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, CH3CH2CHO, (Mr=58) undergoes complete fragmentation is a mass spectrometer. What is the m/z value of the molecular ion line in its mass spectrum?
  1. 15
  2. 29
  3. 43
  4. 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 CH3COOC2H5 is not expected to show a major ion peak at which m/e ratio?
  1. 88
  2. 32
  3. 29
  4. 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?
  1. CH3CH2OH
  2. CH3CH2CH2CH3
  3. CH3CO2H
  4. 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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