On this page, you will learn how fragmentation patterns are formed when organic molecules are put into a mass spectrometer, and how you can extract information from the mass spectrum.
The origin of fragmentation patterns
The formation of molecular ions.
When the vaporized organic sample enters the ionization chamber of a mass spectrometer, it is bombarded by a stream of electrons. These electrons have a high enough energy to remove an electron from an organic molecule and form a positive ion. This ion is called the molecular ion or sometimes the parent ion.
Note: If you're not sure how to create a mass spectrum, it might be worth taking a quick look at the description page.how does a mass spectrometer work.
The molecular ion is usually given the symbol M+o M⋅ - the dot in this second version represents the fact that there is a single unpaired electron somewhere in the ion. That's half of what was originally a pair of electrons; the other half is the electron that was removed by ionization.
fragmentation
Molecular ions are energetically unstable and some of them break down into smaller fragments. The simplest case is when a molecular ion splits into two parts, one of which is another positive ion and the other an uncharged free radical.
{\text{M}}{\bullet}^+ \longrightarrow \text{X}^+ + {\text{Y}}{\bullet}
Note: A free radical is an atom or group of atoms that contains a single unpaired electron. The most complicated separations are beyond the scope of A Level programs.
The released free radical does not produce a line in the mass spectrum. The mass spectrometer only accelerates, deflects, and detects charged particles. These discharged particles are simply lost in the machine; finally, the vacuum pump removes them.
The ion, X+, it passes through the mass spectrometer like any other positive ion and creates a line on the bar graph.
All kinds of fragmentations of the parent molecular ion are possible, and that means you get a bunch of lines in the mass spectrum. For example, the mass spectrum for pentane looks like this:
Note: All mass spectra on this page were obtained using data from the Spectral Database System for Organic Compounds (SDBS) at the National Institute of Materials and Chemical Research in Japan. They have been simplified by omitting all secondary lines with maximum heights of 2% or less of the base peak (the highest peak).
It is important to realize that the pattern of lines in the mass spectrum of an organic compound says something quite different than the pattern of lines in the mass spectrum of an element. For an element, each line represents a different isotope of that element. In a compound, each line represents a different fragment that forms when the molecular ion breaks.
Note: If you are interested in themass spectrum of the elements, you can follow this link.
The molecular ion peak and the base peak.
In the bar graph showing the mass spectrum of pentane, the line produced by the heaviest ion passing through the machine (at m/z = 72) is due to the molecular ion.
Note: Care must be taken here, because in some cases the molecular ion is so unstable that each one splits and none goes through the machine to register in the mass spectrum. It is highly unlikely that you will find this case at level A.
The highest line on the bar graph (at m/z = 43 in this case) is called the base peak. Usually it is given an arbitrary height of 100 and the height of everything else is measured against it. The base peak is the highest peak because it represents the most frequently formed fragment ion, either because it can form in different ways when the original ion fragments, or because it is a particularly stable ion.
Using fragmentation patterns
This section ignores the information you can get from the molecular ion (or molecular ions). This is covered in three other pages accessed from the mass spectrometry menu. There is a link at the bottom of the page.
Find out which ion produces which line
This is usually the simplest thing you can be asked to do.
The mass spectrum of pentane.
Let's look again at the mass spectrum of pentane:
What causes the line at m/z = 57?
How many carbon atoms does this ion contain? There can't be 5 because 5 × 12 = 60. What about 4? 4 × 12 = 48. That leaves 9 for a total of 57. How about C?4H9+So?
C4H9+would be [CH3CH2CH2CH2]+, and this would be produced by the following fragmentation:
[\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3]{\bullet}^+ \longrightarrow [\text{CH}_3\text{ CH}_2\text{CH}_2\text{CH}_2]^+ + {\bullet}\text{CH}_3
The methyl radical produced is simply lost in the machine.
The line at m/z = 43 can be calculated in a similar way. If you play with the numbers, you'll find that this corresponds to a fission that creates a 3-carbon ion:
[\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3]{\bullet}^+ \longrightarrow [\text{CH}_3\text{ CH}_2\text{CH}_2]^+ + {\bullet}\text{CH}_2\text{CH}_3
The line at m/z = 29 is typical for an ethyl ion, [CH3CH2]+:
[\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3]{\bullet}^+ \longrightarrow [\text{CH}_3\text{ CH}_2]^+ + {\bullet}\text{CH}_2\text{CH}_2\text{CH}_3
The other lines in the mass spectrum are more difficult to explain. For example, lines with m/z values of 1 or 2 less than one of the single lines are mainly due to the loss of one or more hydrogen atoms during the fragmentation process. It is highly unlikely that you will only have to explain the most obvious cases on a high school exam.
The mass spectrum of Pentan-3-one
This time the base peak (the highest peak and therefore the most common fragment ion) is at m/z = 57. However, this is not produced by the same ion as the same m/value peak. z in pentane.
If you recall, the m/z = 57 peak in pentane was obtained from [CH3CH2CH2CH2]+. If you look at the structure of the pentan-3-one, it's impossible to get that particular fragment out.
Work your way through the molecule, mentally cutting pieces until you reach something that adds up to 57. With a little patience you will eventually find [CH3CH2CO]+– as a result of this fragmentation:
[\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_2\text{CH}_3]{\bullet}^+ \longrightarrow [\text{CH}_3\text{ CH}_2\text{CO}]^+ + {\bullet}\text{CH}_2\text{CH}_3
You would get exactly the same products no matter which side of the CO group the molecular ion cleaves on.
The m/z = 29 peak is generated by the ethyl ion, which in turn can be formed by molecular ion cleavage on either side of the CO group.
[\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_2\text{CH}_3]{\bullet}^+ \longrightarrow [\text{CH}_3\text{ CH}_2]^+ + {\bullet}\text{COCH}_2\text{CH}_3
Peak heights and ion stability
The more stable an ion is, the more likely it is to form. The more of a given type of ion is formed, the greater its peak height. We will look at two common examples of this.
Examples with carbocations (carbon ions)
Important! If you don't know what aCarbokation (oder Carboniumion)or because different types have different levels of stability, be sure to follow this link before proceeding.
Order of carbocation stability primary < secondary < tertiary
Note: The symbol "<" means "is less than". So this means that primary ions are less stable than secondary ions, which in turn are less stable than tertiary ions.
Applying this logic to fragmentation patterns means that a division that produces a secondary carbocation will be more successful than one that produces a primary. A split that creates a tertiary carbocation will be even more successful.
Let's look at the mass spectrum of 2-methylbutane. 2-Methylbutane is an isomer of pentane: isomers are molecules with the same molecular formula but a different spatial arrangement of atoms.
First look at the very strong peak at m/z = 43. This is caused by a different ion than the corresponding peak in the pentane mass spectrum. This increase in 2-methylbutane is caused by:
[\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_3]{\bullet}^+ \longrightarrow \text{CH}_3{\stackrel {+}\text{C}}\text{HCH}_3 + {\bullet}\text{CH}_2\text{CH}_3
The ion formed is a secondary carbocation: it has two alkyl groups attached to the positively charged carbon. As such, it is relatively stable.
The peak at m/z = 57 is much higher than the corresponding line in pentane. Again a secondary carbocation is formed, this time by:
[\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_3]{\bullet}^+ \longrightarrow \text{CH}_3{\stackrel {+}\text{C}}\text{HCH}_2\text{CH}_3 + {\bullet}\text{CH}_3
Of course you would get the same ion if the CH is on the left.3The group split up instead of the bottom one when we drew them.
In these two spectra, this is probably the most dramatic example of the added stability of a secondary carbocation.
Examples with acylium ions, [RCO]+
Positively charged ions on the carbon of a carbonyl group, C=O, are also relatively stable. This can be clearly seen in the mass spectra of ketones such as pentan-3-one.
The base peak at m/z = 57 is due to [CH3CH2CO]+Ion. We have already discussed the fragmentation this creates.
Note: There are many other examples of positive ions that have additional stability and as a result are produced in large numbers in a mass spectrometer. Without making this article even longer than it already is, it is impossible to cover all possible cases. Check past reviews to see if you are likely to need to know about other opportunities. If you do not have previous documents, follow the linkstudy programspage to find out how to get them.
Using Mass Spectra to Distinguish Compounds
Suppose you need to find a way to distinguish between pentan-2-one and pentan-3-one based on their mass spectra.
Each of these is likely to split to produce ions with a positive charge in the CO group.
In the case of Pentan-2-one, there are two different ions like this:
- [CH3CO]+
- [PUTREFACTION2CH2CH3]+
This would give you strong lines at m/z=43 and 71.
With pentan-3-one you would get such an ion:
- [CH3CH2CO]+
In this case, you would get a strong line at 57.
You don't have to worry about the other lines on the spectrum - lines 43, 57 and 71 give you a big difference between the two. Lines 43 and 71 are absent from the spectrum for pentan-3-one and line 57 is absent from pentan-2-um.
Note: Do not confuse the line at m/z = 58 in the pentane-2 µm spectrum. This is due to a complicated reorganization that could not have been foreseen in Tier A.
The two spectra look like this:
Computer comparison of mass spectra
As you have seen, even the mass spectra of very similar organic compounds are quite different due to the different fragmentation that can occur. As long as you have a computer database of mass spectra, any unknown spectrum can be analyzed by computer and easily compared to the database.
Questions to test your understanding.
Questions about fragmentation patternsAnswer
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