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Interpreting the fragmentation pattern of the mass spectrum of 1,1-dibromoethane

[M]⁺ are the molecular ion peaks (M) with m/z values of 186, 188 and 190 corresponding to [C₂H₄Br₂]⁺, the original 1,1-dibromoethane molecule minus an electron.

The three possibilities (and in a 1:2:1 ratio, see below) are:

m/z 186 [CH₃CH⁷⁹Br⁷⁹Br]⁺, m/z 188 [CH₃CH⁷⁹Br⁸¹Br]⁺, and m/z 190 [CH₃CH⁸¹Br⁸¹Br]⁺

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

Their average relative isotopic mass is ~80, so the relative molecular mass for 1,1-dibromoethane is ~188.

There are four possible bromine permutations in the 1,1-dibromoethane molecule:

⁷⁹Br⁷⁹Br, ⁷⁹Br⁸¹Br,  ⁸¹Br⁷⁹Br and ⁸¹Br⁸¹Br (think of a 2 x 2 random probability square).

However, this means any fragment carrying a bromine atom should show up as three peaks, two mass units apart and approximately in a ratio of 1:2:1 in heights (intensities).

The very tiny M+1 peaks at m/z 187 and 189, corresponds to an ionised 1,1-dibromoethane molecule with one ¹³C atom in it i.e. an ionised 1,1-dibromoethane molecule of formula [¹³C¹²CH₄Br₂]⁺

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 peaks.

1,1-dibromoethane has 2 carbon atoms, so on average, ~1 in 50 molecules will contain a ¹³C atom.

The most abundant ion of the molecule under mass spectrometry investigation (1,1-dibromoethane) is usually given an arbitrary abundance 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 1,1-dibromoethane.

Unless otherwise indicated, assume the carbon atoms in 1,1-dibromoethane are the ¹²C isotope.

Analysing and explaining the principal ions in the fragmentation pattern of the mass spectrum of 1,1-dibromoethane

Atomic masses: H = 1; C = 12; Br = 79 or 81 (1:1)

Bond enthalpies kJ/mol: C-C = 348;  C-H = 412; C-Br 276

Equations to explain the most abundant ion peaks of 1,1-dibromoethane

The three molecular ions exist due to the permutations of the two bromine isotopes.

Their occurrence and ~ 1:2:1 ratio have already been described above.

Formation of m/z 171, 173 and 175 ions:

[CH₃CHBr₂]⁺  ===>  [CHBr₂]⁺  +  CH₃

C-C bond scission to release a methyl group.

Low probability due to the strong C-C bond, but note the approximate 1:2:1 ratio which fits in with ratio expected due to the two bromine isotopes.

[CH⁷⁹Br⁷⁹Br]⁺, [CH⁷⁹Br⁸¹Br]⁺, [CH⁸¹Br⁷⁹Br]⁺, and [CH⁸¹Br⁸¹Br]⁺

Formation of m/z 107 and 109 ions:

[CH₃CHBr₂]⁺  ===>  [C₂H₄⁷⁹Br]⁺  or  [C₂H₄⁸¹Br]⁺  +  Br

Scission of the C-Br bond, the weakest bond in the molecule.

The weaker C-Br bond is more likely to be broken than the stronger C-C or C-H bonds.

In fact, and as a consequence, apart from the m/z 27 ion peak, all other ion peaks, formation described below, are of a much lower intensity, but I offer some suggestions of how they may be formed.

The m/z 107 ion is the base peak ion, the most abundant and 'stable' ion fragment.

Formation of m/z 79 and 81 ions:

[CH₃CHBr₂]⁺  ===>  [⁷⁹Br]  or  [⁸¹Br]⁺  +  BrCH₂CH₂

C-Br bond scission of parent molecular ion (or other bromine containing fragment).

Formation of m/z 80 and 82 ions:

[CH₃CHBr₂]⁺  ===>  [H⁷⁹Br]⁺  or   [H⁸¹Br]⁺  +  C₂H₃Br

Elimination of hydrogen bromide from parent molecular ion.

Formation of m/z 28 ion:

[C₂H₄Br]⁺  ===>  [C₂H₄]⁺  +  Br

C-Br bond scission of the m/z 107 or 109 ions to give an ionised ethene molecule.

Formation of m/z 27 ion:

[C₂H₄Br]⁺  ===>  [C₂H₃]⁺  +  HBr

Elimination of hydrogen bromide from the m/z 107 and 109 ions.

The chemistry of HALOGENOALKANES (haloalkanes) revision notes INDEX

Mass spectroscopy index

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