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Mass spectrometry - spectra index  *  [privacy policy, cookies and disclaimer]

Introductory note on the mass spectrum of ethylbenzene

Students and teachers please note my explanation of the mass spectrum of ethylbenzene is designed for advanced, but pre-university, chemistry courses.

If M represents the ethylbenzene molecule, the initial ionisation to give the molecular ion is:

M_(g) + high KE e^-  ==> [M•]⁺_(g) + 2e^- and for fragmentation equations assume [M]⁺ is the start of the processes and all species are in a gaseous state.

I've not usually shown an unpaired electron on e.g. an ion or a non-ionised alkyl radical R e.g.

[M•]⁺ ==> [X]⁺  +  R•, but you should be aware this is a more accurate depiction of some processes.

I've used simplified equations to show how some of the ions that might be formed in the fragmentation pattern for the mass spectrum of ethylbenzene and only the formation of singly charged positive are considered for the mass spectrum of ethylbenzene.

I've included a stick diagram and table of m/z ions for the mass spectrum of ethylbenzene and doing the mass spectrum analysis under standard conditions, databases can be compiled based on complex fingerprint patterns, often involving the relative intensities of many fragment ions, and used to identify compounds including ethylbenzene.

In selected cases, where two different fragment ions have the same integer m/z value, I've pointed out that modern mass spectrometers can measure relative ion mass to four decimal places. So, using accurate isotopic masses, I've calculated the accurate ion masses, BUT strictly speaking, 0.0005 should be deducted for singly charged ions to account for the loss of the electron in their formation. I have NOT done this for ethylbenzene, but the mass spectrometer software does!

The molecular structure and naming of aromatic compounds

Interpreting the fragmentation pattern of the mass spectrum of ethylbenzene

[M]⁺ is the molecular ion peak (M) with an m/z of 106 corresponding to [C₈H₁₀]⁺, the original ethylbenzene molecule minus an electron, [C₆H₅CH₂CH₃]⁺

The small M+1 peak at m/z 107, corresponds to an ionised ethylbenzene molecule with one ¹³C atom in it i.e. an ionised ethylbenzene molecule of formula [¹³C¹²C₇H₁₀O]⁺

Carbon-13 only accounts for ~1% of all carbon atoms (¹²C ~99%), but the more carbon atoms in the molecule, the greater the probability of observing this ¹³C M+1 peak.

Ethylbenzene has 8 carbon atoms, so on average, ~1 in 13 molecules will contain a ¹³C atom.

The same argument applies to ionised fragments for the original ionised molecule of ethylbenzene.

The most abundant ion of the molecule under mass spectrometry investigation (ethylbenzene) is usually given an arbitrary abundance value of 100, called the base ion peak, and all other abundances ('intensities') are measured against it.

The base peak ion for ethylbenzene is the m/z 91 ion [C₇H₇]⁺

Some of the possible positive ions, [molecular fragment]⁺, formed in the mass spectrometry of ethylbenzene - identifying the species giving the most prominent peaks (apart from M) in the fragmentation pattern of the ethylbenzene mass spectrum.

Unless otherwise indicated, assume the carbon atoms in the ethylbenzene molecular ion and fragment ions are the ¹²C isotope.

m/z ion 92, [C₇H₈]⁺, could also be [¹³CC₆H₇]⁺ (see also m/z 107 ion).

Analysing and explaining the principal ions in the fragmentation pattern of the mass spectrum of ethylbenzene

PLEASE NOTE I have found it difficult to find 'authentic' equations to explain mass spectra fragmentation patterns and it is complex chemistry! I've identified the formulae of the ionised fragments on the mass spectrum diagram, but the equations are from the internet or my conjecture as to how the ions might be formed - please take care in using the information, especially for assignments at university or pre-university level.

Suggested equations to explain some of the most abundant ion mass spectrum peaks of ethylbenzene i.e. explaining the principal fragments of the mass spectrum of ethylbenzene.

Formation of m/z 105 ion:

[C₆H₅CH₂CH₃]⁺  ===>  [C₆H₅CH₂CH₂]⁺  +  H

C-H bond scission in the ethyl alkyl group.

mass change 106 - 1 = 105 (M-1 ion peak),

proton loss from the parent molecular ion of ethylbenzene.

Formation of m/z 91 ion:

[C₆H₅CH₂CH₃]⁺  ===>  [C₆H₅CH₂CH₂]⁺  +  CH₃

The m/z 91 ion is the base peak ion, the most abundant and 'stable' ion fragment.

Mass change 106 - 15 = 91 (M-1 ion peak)

loss of CH₃ from the parent molecular ion

or less likely, loss of CH₂ from the m/z 105 ion.

[C₆H₅CH₂CH₂]⁺  ===>  [C₆H₅CH₂CH₂]⁺  +  CH₂

mass change 105 - 14 = 91

Both involve C-C bond scission in the alkyl side-chain of ethylbenzene.

The m/z 92 ion is probably formed in the same way but contains a 13C atom i.e. it has the formula [¹³C¹²C₆H₇]⁺ rather than [C₇H₈]⁺

Note that an accurate mass spectrometer can sort out (resolve) pairs of ions with the same integer m/z value because they can measure relative fragment ion masses to four decimal places,

e.g. using accurate relative isotopic masses:

¹H = 1.0078  ¹²C = 12.0000   ¹³C = 13.0034: you can then calculate (predict) that the accurate relative ion masses are:

m/z 92: [C₇H₈]⁺ = 92.0264 or  [¹³C¹²C₆H₇]⁺ = 92.0580, a relative ion mass difference of 0.0316.

Formation of m/z 78 and m/z 79 ions:

(i) [C₆H₅CH₂CH₃]⁺  ===>  [C₆H₇]⁺  +  C₂H₃

mass change 106 - 27 = 79 (M-27 ion peak)

(ii) [C₆H₅CH₂CH₃]⁺  ===>  [C₆H₆]⁺  +  CH₂CH₂

Elimination of ethene from the parent molecular ion.

mass change 106 - 28 = 78 (M-28 ion peak)

Formation of m/z 77 ion:

[C₆H₅CH₂CH₃]⁺  ===>  [C₆H₅]⁺  +  CH₂CH₃

mass change 106 - 29 = 77 (M-29 ion peak)

ethyl group broken off,

or elimination of ethene from the m/z 105 ion

[C₆H₅CH₂CH₂]⁺  ===>  [C₆H₅]⁺  +  CH₂=CH₂

again, both involve C-C bond scission of the benzene ring from the alkyl side-chain of ethylbenzene.

The observation of an m/z ion of mass 77 is very characteristic of (aromatic) benzene compounds, though the peak can be much more prominent in other aromatic compound mass spectra.

It is actually a phenyl radical carrying a positive charge (the free radical cation, [C₆H₅•]⁺.

Formation of m/z 29 ions:

(ii) [C₆H₅CH₂CH₃]⁺  ===>  [CH₂CH₂C₆H₅]⁺  +  C₆H₅

mass change 106 - 77 = 29

Formation of other m/z <77 ions:

The m/z ion 77 can break down by loss of H and C atoms to give e.g. the m/z ion series of 67 to 64,  52 to 50, 39 and 27.

There are lots of possible equations.

The infrared spectrum of ethylbenzene

The H-1 NMR spectrum of ethylbenzene

The C-13 NMR spectrum of ethylbenzene

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