[Author ©  Dr Phil Brown GRIC, PhD: Doc Brown's advanced level organic chemistry exam revision notes suitable for students of UK A level chemistry courses & US K12 grade 11, grade 12 & AP honors chemistry courses: Molecular spectroscopy of benzene [spectrum page updated RE-EDIT]

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Introductory note on the mass spectrum of benzene

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

If M represents the benzene 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 benzene and only the formation of singly charged positive are considered for the mass spectrum of benzene.

I've included a stick diagram and table of m/z ions for the mass spectrum of benzene 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 benzene.

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 benzene, but the mass spectrometer software does!

The molecular structure and naming of aromatic compounds

Interpreting the mass spectrum of benzene C₆H₆

[M]⁺ is the molecular ion peak (M) with an m/z of 78 corresponding to [C₆H₆]⁺, the original molecule minus an electron.

Unless otherwise stated, C means a ¹²C atom, if not, the isotopic carbon atom ¹³C will be indicated.

The small M+1 peak at m/z 79, corresponds to an ionised benzene molecule with one ¹³C atom in it i.e. an ionised molecule of formula [¹³C¹²C₅H₆]⁺

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.

Benzene has 6 carbon atoms, so on average, ~1 in 17 molecules will contain a ¹³C atom.

You can see this in the intensity ratio of the m/z 78 : 79 peaks of ~17 : 1 for the two possible parent molecular ions [C₆H₆]⁺  and  [¹³C¹²C₅H₆]⁺, the latter containing one ¹³C atom.

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

In this case the base peak ion is also the molecular ion of benzene itself [C₆H₆]⁺

Identifying the species giving the most prominent peaks (apart from M) in the fragmentation pattern of benzene.

Unless otherwise indicated, assume the carbon atoms in benzene are the ¹²C isotope.

Some of the possible positive ions, [molecular fragment]⁺, formed in the mass spectrometry of benzene.

The parent molecular ion of benzene m/z 78: [C₆H₁₀]⁺ which, unusually, is also the base ion peak.

Data table of some of the ions formed in the fragmentation pattern of the mass spectrum of benzene

Explaining the principal fragments

Possible equations to explain some of the most abundant ion peaks in the mass spectrum of benzene (tabulated above)

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.

Atomic masses: H = 1;  C = 12 (13 for ~1 in 100)

Bond enthalpies = kJ/mol: = 518 (benzene ring);  C-H = 412

A details list of m/z values and possible molecular ion identities

The 'phenyl' fragment m/z 77 is formed by the loss a proton from the molecular ion.

[C₆H₆]⁺  ===>  [C₆H₅]⁺  +  H

Mass change 78 - 1 = 77 (M-1 ion peak)

This 'phenyl' fragment is very indicative of the presence of a benzene ring in the molecule and common in aromatic mass spectra of derivatives of benzene.

The m/z 77 ion can further lose a hydrogen atom/molecule to give m/z ions of 76, 75, 74 and 73.

The m/z ions 50 to 52

These involve the loss of an unionised two carbon fragment to give a C₄ fragment ion e.g. 78 - 26 = 52  and  78 - 28 = 50,

but proton loss from m/z 50 and 51 may happen too.

[C₆H₆]⁺  ===>  [C₄H₄]⁺  +  C₂H₂  (M-26 ion peak)

[C₆H₆]⁺  ===>  [C₄H₃]⁺  +  C₂H₃  (M-26 ion peak)

[C₆H₆]⁺  ===>  [C₄H₂]⁺  +  C₂H₄  (M-28 ion peak)

[C₄H₄]⁺  ===>  [C₄H₃]⁺  +  H  (52 - 1  = 51)

[C₄H₄]⁺  ===>  [C₄H₂]⁺  +  H₂  (52 - 2 = 50)

[C₄H₃]⁺  ===>  [C₄H₂]⁺  +  H₂  (51 - 1 = 50)

The m/z 39 ion

This is a common fragment, [C₃H₃]⁺ found in many organic mass spectra (I think its a triangle with a delocalised system stabilising the positive charge).

[C₆H₆]⁺  ===>  [C₃H₃]⁺  +  C₃H₃ (M-39 ion peak) ???

[?]⁺  ===>  [C₃H₃]⁺  +  ?

From any fragment ion from [C₆H₅]⁺ C-C bond breaking?

The H-1 NMR spectrum of Benzene