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Goals
After completing this section, you should be able to.
- suggest possible molecular formulas for a compound, given them/zValue for the molecular ion or a mass spectrum from which this value can be derived.
- predict the relative heights of the peaks M+ , (M + 1)+ , etc. in the mass spectrum of a compound, given 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 (eg, hexane).
- Use the fragmentation pattern in a given mass spectrum to help identify a relatively simple and unknown compound (for example, an unknown alkane).
study notes
When interpreting fragmentation patterns, it can be helpful to know that the weakest carbon-carbon bonds are the most likely to break. You may want to refer to the table ofbond dissociation energiesfor problems with the interpretation of mass spectra.
This page is about the formation of fragmentation patterns when organic molecules are put into a mass spectrometer and how information can be extracted 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 a 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+o. 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 during ionization.
Molecular ions are energetically unstable and some of them break down into smaller fragments. The simplest case is that a molecular ion splits into two parts: one is another positive ion and the other is an uncharged free radical.
The discharged free radical is converted toNoproduce a line in the mass spectrum. The mass spectrometer only accelerates, deflects, and detects charged particles. These uncharged particles are simply lost in the machine; finally, the vacuum pump removes them.
The ion, X+, travels through the mass spectrometer like any other positive ion, creating a line on the bar graph. All kinds of fragmentations of the parent molecular ion are possible, and that means you get a few lines in the mass spectrum. For example, the mass spectrum for pentane looks like this:
observation
The pattern of lines in the mass spectrum of aorganic connectionsays something very different from the line pattern in the mass spectrum ofElement. 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 decays.
In the bar graph showing the mass spectrum of pentane, the line produced by the heavier ion passing through the machine (at m/z = 72) is due to thepure molecular. The highest line on the bar graph (in this case at m/z = 43) is denoted as thebase tip. Usually this 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 common fragment ion to form, either because there are several ways it can form during fragmentation of the original ion, or because it is a particularly stable ion.
Using fragmentation patterns
This section ignores the information you might get from the molecular ion (or molecular ions). This is covered in three other pages accessed from the Mass Spec menu. There is a link at the bottom of the page.
Example: 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 is equal to a fraction that creates 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 less than one of the single lines are often due to the loss of one or more hydrogen atoms during the fragmentation process.
Example: 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.
As you recall, the m/z = 57 peak in pentane was obtained from [CH3CH2CH2CH2]+. If you look at the structure of Pentan-3-um, it is impossible to get this particular fragment.
Work your way through the molecule, mentally cutting pieces, until you find something that does 57. With a little patience, you'll eventually find [CH3CH2CO]+- as a result of this fragmentation:
You will get exactly the same products no matter which side of the CO group the molecular ion cleaved off. 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.
maximum heights and stability
The more stable an ion is, the more likely it is to form. The more ions of a given type are formed, the greater the height of their peak. We will look at two common examples of this.
The carbocation (carbonium ion)
In short, the main takeaway from 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 carbocation. A cleavage that produces 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 different spatial arrangements of atoms.
First, consider 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 is relatively stable. The peak at m/z = 57 is much larger than the corresponding line in pentane. Again a secondary carbocation is formed, this time by:
Of course, you would get the same ion if the left CH3the group was divided instead of the lower one as we drew it. In these two spectra, this is probably the most dramatic example of the added stability of a secondary carbocation.
Aciliumion, [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.
observation
The more stable an ion is, the more likely it is to form. The more of a given ion is formed, the greater the height of its peak.
Using Mass Spectra to Distinguish Compounds
Suppose you have to find a way to distinguish between pentan-2-one and pentan-3-one based on their mass spectra.
pentan-2-one | CH3PUTREFACTION2CH2CH3 | |
pentan-3-one | CH3CH2PUTREFACTION2CH3 |
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 only get one of those ions:
- [CH3CH2CO]+
In that 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 show a big difference between the two. Lines 43 and 71 of the pentane-3-one spectrum are missing, and line 57 of the pentane-2-one spectrum is missing.
The two mass spectra look like this:
As you have seen, even the mass spectrum of very similar organic compounds is 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 analyzed by computer and easily compared to the database.
a practice
5.Caffeine has a mass spectrometry determination of 194.19 amu and contains C, N, H, O. What is the molecular formula of this molecule?
6.Below are the spectra of 2-methyl-2-hexene and 2-heptene whose spectra belong to the correct molecule. Explain.
A:
B:
Quelle: SDBSWeb:http://sdbs.db.aist.go.jp(National Institute of Advanced Industrial Science and Technology, December 2, 2016)
- Respondedor
-
5.C8H10norte4o2
C = 12 × 8 = 96
norte = 14 × 4 = 56
alto = 1 × 10 = 10
O = 2 × 16 = 32
96+56+10+32 = 194 g/mol
6.Spectrum (A) is 2-methyl-2-hexene and spectrum (B) is 2-heptene. Considering (A) the peak at 68m/zis the fractional molecule in which only the trisubstituted alkene is present. While (B) has a strong peak around 56m/z, which in this case is the remaining disubstituted alkene of the linear heptene.
Credits and Attributions
dr Dietmar KennepohlFCIC (Professor of Chemistry,athabasca university)
Profesor Steven Farmer (sonoma state university)
(Video) Mass Spectrometry - Interpretation Made Easy!Organic chemistry with a biological approachVontim soderberg(University of Minnesota, Morris)
jim clark (Chemguide.es)