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Interpreting the fragmentation pattern of the mass spectrum of bromomethane

[M]⁺ is the molecular ion peak (M) with an m/z of 94 and 96 corresponding to [CH₃Br]⁺, the original bromomethane molecule minus an electron.

There are two molecular ion peaks because bromine as two isotopes, 50.5% ⁷⁹Br and 50.5% ⁸¹Br.

Their average relative mass is ~80, so the relative molecular mass for bromomethane is ~95.

However, this means any fragment carrying a bromine atom should show up as twin peaks, two mass units apart and approximately of equal height (intensities).

e.g. the twin peaks of the molecular ion correspond to m/z 94 [CH₃⁷⁹Br]⁺ and m/z 96 [CH₃⁸¹Br]⁺.

The tiny M+1 peak at m/z 97, corresponds to an ionised bromomethane molecule with a ¹³C atom in it i.e. an ionised bromomethane molecule of formula [¹³CH₃⁸¹Br]⁺

There is also a nearly equal probability of m/z 95 ion [¹³CH₃⁷⁹Br]⁺ being formed, but this m/z values applies to another fragment ion too (see table below), which is why the m/z 95 peak is higher than the m/z 97 peak.

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.

Bromomethane has1 carbon atom, so on average, ~1 in 100 molecules will contain a ¹³C atom.

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

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

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

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

Atomic masses: C = 12;  H = 1; O = 16;  Cl = 35/37;  Br = 79/81; N = 14

Bond enthalpies kJ/mol:  C-H = 412; C-Br 276

Equations to explain the most abundant ion peaks of bromomethane

Formation of the base ion:

The m/z 94 ion [CH₃⁷⁹Br]⁺ is the base peak ion, the most abundant and 'stable' ion fragment.

It also happens to be one of the parent molecular ion peaks formed in the initial ionisation, the other molecular ion being m/z ion 96 [CH₃⁸¹Br]⁺.

Formation of m/z 95 and 93 ions:

[CH₃Br]⁺  ===>  [CH₂⁷⁹Br]⁺  or  [CH₂⁸¹Br]  +  CH₂

Proton loss from the parent molecular ion.

Further proton losses yield ions with m/z values of  94 to 91 - you will get peak overlap with different fragments with the same m/z value.

Formation of m/z 80 and 82 ions:

[CH₃Br]⁺  ===>  [H⁷⁹Br]⁺  or  [H⁸¹Br]  +  CH₂

Elimination of hydrogen bromide from the parent molecular ion.

Formation of m/z 79 and 81 ions:

[CH₃Br]⁺  ===>  [⁷⁹Br]⁺  or  [⁸¹Br]  +  CH₃

C-Br bond scission, where the bromine atom carries the positive charge.

Formation of m/z 15 ion:

[CH₃Br]⁺  ===>  [CH₃]⁺  +  Br

C-Br bond scission, where the methyl group carries the positive charge.

The chemistry of HALOGENOALKANES (haloalkanes) revision notes INDEX

Mass spectroscopy index

ALL SPECTROSCOPY INDEXES

All Advanced Organic Chemistry Notes

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