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Goals
After completing this section, you should be able to do so
- suggest possible molecular formulas for a compound, given them/zValue for the molecular ion or a mass spectrum from which this value can be obtained.
- predict the relative heights of the peaks M+·, (M + 1)+·, etc. in the mass spectrum of a compound, taking into account the natural abundance of isotopes of carbon and other elements present in the compound.
- interpret the fragmentation pattern of the mass spectrum of a known relatively simple compound (e.g. hexane).
- Use the fragmentation pattern in a given mass spectrum to identify a relatively simple unknown compound (e.g. an unknown alkane).
study notes
When interpreting fragmentation patterns, it can be helpful to know that, unsurprisingly, the weakest carbon-carbon bonds are the most likely to break. You can refer to the table of bond dissociation energies when trying to solve problems related to mass spectral interpretation.
This page explains how fragmentation patterns are formed when organic molecules are placed in a mass spectrometer and how you can obtain information from the mass spectrum.
The Origin of Fragmentation Patterns
When the vaporized organic sample enters the ionization chamber of a mass spectrometer, it is bombarded by a stream of electrons. These electrons have high enough energy to remove an electron from an organic molecule and form a positive ion. This ion is calledMolecular ion - or sometimes the parent ionforksoften marked with the symbol M+Ö
. The point 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 an electron pair; the other half is the electron removed during ionization.
Molecular ions are energetically unstable and some of them break into smaller pieces. The simplest case is when a molecular ion splits into two parts, one being another positive ion and the other an uncharged free radical.
The free radical is dischargedNOproduce a line in the mass spectrum. The mass spectrometer accelerates, directs and detects only charged particles. These downloaded 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 sorts 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:
monitoring
The line pattern in the mass spectrum of aBio-Abosays something completely different than the line pattern in the mass spectrum of aElement. For an element, each line represents a different isotope of that element. In a compound, each line represents a different fragment formed when the molecular ion is cleaved.
In the bar graph showing the mass spectrum for pentane, the line produced by the heavier ion passing through the machine (at m/z = 72) is due to the molecular ion. The highest line on the bar graph (in this case at m/z = 43) is calledBasis-Pico. This is usually 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 common fragment ion to form, either because there are multiple ways it can be generated during fragmentation of the original ion, 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 on three other pages accessed from the mass spectrometry menu. There is a link at the bottom of the page.
Example 12.2.1: 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 x 12 = 60. What about 4? 4 x 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:
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 split producing a 3-carbon ion:
The line at m/z = 29 is typical for an ethyl ion, [CH3CH2]+:
The other lines in the mass spectrum are more difficult to explain. For example, lines with m/z values 1 or 2 smaller than one of the simple lines are often due to the loss of one or more hydrogen atoms during the fragmentation process.
Example 12.2.2: 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/z value peak for 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 pick out that particular fragment.
Work your way through the molecule, mentally cutting pieces until you get to something that adds up to 57. With a little patience you will eventually find [CH3CH2CO]+- resulting from this fragmentation:
You would get exactly the same products no matter which side of the CO group you cleave the molecular ion on. The m/z = 29 peak is generated by the ethyl ion, which in turn can be formed by cleavage of the molecular ion on either side of the CO group.
Maximum heights and stability
The more stable an ion is, the more likely it is to form. The more of a given ion species is formed, the greater its peak height. We'll look at two common examples of this.
carbocations (carbon ions)
To recap the key points from the Carbs page:
Order of stability of carbocations
primary < secondary < tertiary
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 the atoms.
First note 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:
The ion formed is a secondary carbocation: it has two alkyl groups attached to the positively charged carbon. As such, it's 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:
You would of course get the same ion if the CH is on the left3The group split up instead of the ones below when we drew them. In these two spectra, this is probably the most dramatic example of the added stability of a secondary carbocation.
acyl 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 from [CH3CH2CO]+Ion. We've already discussed the fragmentation this creates.
monitoring
The more stable an ion is, the more likely it is to form. The more of a given ion that is formed, the greater the height of its peak.
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.
| Pentan-2-one | | CH3ROJO2CH2CH3 |
| Pentan-3-one | | CH3CH2ROJO2CH3 |
Each of these is likely to split to produce ions with a positive charge in the CO group. In the case of pentane-2-a, there are two different ions like this:
- [CH3CO]+
- [ROJO2CH2CH3]+
This would give you strong lines at m/z=43 and 71. With Pentan-3-one you would only get one of these ions:
- [CH3CH2CO]+
In this case you would get a strong line at 57. You don't have to worry about the other lines in the spectrum: lines 43, 57, and 71 offer a big difference between the two. Lines 43 and 71 of the spectrum for pentan-3-one are absent and line 57 of pentan-2-um is absent.
The two mass spectra look like this:
As you have seen, even the mass spectra of very similar organic compounds are quite different due to the different fragmentation patterns that can occur. As long as you have a computer database of mass spectra, any unknown spectrum can be computer analyzed and easily compared to the database.
exercises
employees and tasks
dr Dietmar KennepohlFCIC (Professor of Chemistry,athabasca university)
Professor Steven Farmer (Sonoma State University)
(Video) Mass spectrometry | Atomic structure and properties | AP Chemistry | Khan AcademyOrganic chemistry with a focus on biologyvonTim Soderberg(University of Minnesota, Morris)
Jim Clark (Chemguide.es)