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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 relatively simple known 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 the weakest carbon-carbon bonds are expected to be the most likely to break. You might want to consult the tablebond dissociation energiestrying out problems related to the interpretation of mass spectra.
This page explains how fragmentation patterns are formed when organic molecules are introduced into a mass spectrometer and how mass spectral information can be obtained.
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 ionand is often given the M symbol+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 an electron pair originally was; the other half is the electron removed during ionization.
Molecular ions are energetically unstable and some of them break down into smaller pieces. The simplest case is that a molecular ion is broken into two parts, one of which is 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 uncharged 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'll have a lot of lines in the mass spectrum. For example, the mass spectrum of pentane looks like this:
Use
The line pattern in the mass spectrum of aorganic compostsays 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 of pentane, the line produced by the heaviest ion passing through the machine (at m/z = 72) is due to themolecularly pure. The highest line on the bar graph (in this case at m/z = 43) is calledPico-Basis. Usually this is given an arbitrary height of 100 and the height of everything else is measured against that. 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+Then?
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 solved in a similar way. If you play around with the numbers you'll find that this corresponds to a fission creating 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 of 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 in pentane.
If you recall, the m/z = 57 peak in pentane was obtained from [CH3CH2CH2CH2]+. If you look at the structure of pentan-3-one, it's impossible to get this particular fragment.
Go through the molecule, mentally cutting the pieces until you find something that adds up to 57. With a little patience you will eventually find [CH3CH2CO]+- resulting from this fragmentation:
You get exactly the same products no matter which side of the CO group you split 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 ions of a given species that form, the greater their peak height. We'll look at two common examples of this.
carbocations (carbon ions)
To summarize the main conclusion of the page on carbocations:
Order of stability of carbocations
primary < secondary < tertiary
Applying this logic to fragmentation patterns means that a cleavage that produces a secondary carbocation will be more successful than one that produces a primary. Even more successful is a cleavage that generates a tertiary carbocation. 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 with 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 positively charged alkyl groups attached to the 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 detached 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.
acyl ions, [RCO]+
The 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.
Use
The more stable an ion is, the more likely it is to form. The more a given ion is formed, the greater its maximum height.
Using mass spectra to distinguish compounds
Suppose you needed 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 will likely split to produce ions with a positive charge on the CO group. In the case of Pentan-2-one, 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 ion:
- [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 are absent from the pentane-3 µm spectrum and line 57 is absent from the pentane-2 µm spectrum.
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
Seek
After completing this section, you will be able to predict the expected fragmentation for common functional groups such as alcohols, amines, and carbonyl compounds.
key terms
Make sure you can define and use the following key terms in context.
- Alfa (A) Decollete
- McLafferty rearrangement
Much of the utility of MS electron ionization stems from the fact that the radical cations generated in the electron bombardment process tend to be degraded in a predictable manner. A detailed analysis of the typical fragmentation patterns of different functional groups is beyond the scope of this text, but some representative examples are worth exploring even if we don't try to understand the exact process by which the fragmentation occurs. For example, we have seen that the base peak in the mass spectrum of acetone is m/z=43. This is the result of cleavage at the "alpha" position; in other words, on the carbon-carbon bond adjacent to the carbonyl. . Alpha cleavage leads to the formation of an acylium ion (representing the base peak at m/z = 43) and a methyl radical, which is neutral and therefore not detected.
After the main peak and the base peak, the next largest peak is at m/z=15 with a relative abundance of 23%. This is expected to be the result of the formation of a methyl cation in addition to an acyl moiety (which is neutral and not detected).
A common fragmentation pattern for larger carbonyl compounds is reportedMcLafferty rearrangement:
The 2-hexanone mass spectrum shows a "McLafferty fragment" at m/z = 58, while the propene fragment cannot be seen since it is a neutral species (remember that MS only detects cationic fragments see are). The base peak in this spectrum is again an acylium ion.
When alcohols are subjected to MS electron ionization, the molecular ion is very unstable and therefore no main peak is usually detected. Often the base peak is from an "oxonium" ion.
Other functional groups also have predictable fragmentation patterns. By carefully analyzing the fragmentation information of a mass spectrum, an expert spectrometer can often “put the puzzle together” and make some very reliable predictions about the structure of the original sample.
Click hereExamples of compounds listed by functional group showing patterns seen in mass spectra of compounds ionized by electron impact ionization.
Example 12.3.1
The mass spectrum of an aldehyde shows clear peaks atm/z= 59 (12%, highest value ofm/zin spectrum), 58 (85%) and 29 (100%), among others. Propose a structure and identify the three types whosem/zValues have been listed.
Solution