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Introductory note on the mass spectrum of chlorobenzene

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

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

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

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 chlorobenzene, but the mass spectrometer software does!

chlorobenzene, C₆H₅Cl,     monosubstituted benzene compound

The molecular structure and naming of aromatic compounds

Interpreting the fragmentation pattern of the mass spectrum of chlorobenzene

[M]⁺ are the molecular ion peaks with an m/z of 112 and 114 corresponding to

the M ion [C₆H₅³⁵Cl]⁺ and the M+2 ion [C₆H₅³⁷Cl]⁺

i.e. the original chlorobenzene molecule minus an electron, [C₆H₅Cl]⁺.

There are two molecular ion peaks because of the two most abundant isotopes of chlorine in the chlorobenzene molecule.

Chlorine is composed of ~75% chlorine-35, ³⁵Cl AND ~25% of chlorine 37, ³⁷Cl (~ratio 3:1).

Therefore you expect a ratio of ~3:1 in the twin peaks for any molecular ion or fragment ion that contains chlorine, and you can see this in the mass spectrum diagram of chlorobenzene above.

The small M+1 and M+3 peaks at m/z 113 and 115, corresponds to an ionised chlorobenzene molecule with one ¹³C atom in it i.e. an ionised chlorobenzene molecule of formula [¹³C¹²C₅H₅Cl]⁺

You can also see they are also in a ratio of 3:1.

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.

Chlorobenzene has 6 carbon atoms, so on average, ~1 in 17 parent molecule or fragment ions will contain a ¹³C atom.

The most abundant ion of the molecule under mass spectrometry investigation (chloroethane) 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 ion peak and molecular ion peak are the same m/z 112 ion [C₆H₅³⁵Cl]⁺

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

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

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 (~1% 13);  Cl = 35 or 37 (~ratio 3:1)

Bond enthalpies kJ/mol:   = 518;  C-H = 412; C-Cl = 338

Suggested equations to explain the most abundant ion peaks of chlorobenzene

Unusually, the m/z 112 ion, is the molecular ion [C₆H₅³⁵Cl]⁺, and base peak ion, the most abundant and 'stable' ion fragment.

Formation of m/z 77 ion:

[C₆H₅Cl]⁺  ===>  [C₆H₅]⁺  +  Cl

C-Cl bond scission in the parent molecular ion.

mass changes: 112 - 35 = 77  or  114 - 37 = 77

The phenyl cation is a characteristic fragment in the mass spectra of benzene compounds due to the stability of the benzene ring.

The m/z 77 ion can lose hydrogen atoms to give ions from m/z 76 down to m/z 73 [C₆H₁]⁺

The m/z 78 ion is probably formed in the same way, but contains a 13C atom i.e. its formula is [¹³C¹²C₅H₅]⁺ rather than the highly unlikely formation of [C₆H₆]⁺

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:

For m/z 78: [¹³C¹²C₅H₅]⁺ = 78.0424  and  [C₆H₆]⁺ = 78.0468, a difference of 0.0044 in relative ion mass.

Formation of lower value m/z ions:

These would be formed by loss of H atoms, H₂ molecules or C_xH_y fragments, initially from the m/z 77 ion e.g.

[C₆H₅]⁺  ===>  [C₄H₃]⁺  +  C₂H₂

_{mass loss 77 - 26 = m/z 51 ion}

[C₆H₅]⁺  ===>  [C₄H₂]⁺  +  C₂H₃

_{mass loss 77 - 27 = m/z 50 ion}

[C₆H₅]⁺  ===>  [C₃H₂]⁺  +  C₃H₃

_{mass loss 77 - 39 = m/z 38 ion}

_{but there are lots of possibilities for sequences by losing C and in particular H atoms e.g.}

[C₆H₅]⁺  _==>  [C₄H₃]⁺  _==>  [C₄H₂]⁺  _==>  [C₃H₂]⁺

The infrared spectrum of chlorobenzene

The H-1 NMR spectrum of chlorobenzene

The C-13 NMR spectrum of chlorobenzene

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