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Interpreting the fragmentation pattern of the mass spectrum of 1-bromopropane

[M]⁺ is the molecular ion peak (M) with an m/z of 122 and 124 corresponding to [C₃H₇Br]⁺, the original 1-bromopropane molecule minus an electron, [CH₃CH₂CH₂Br]⁺.

There are two 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-bromopropane is ~123.

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).

The very tiny M+1 peaks at m/z 123 and 124, corresponds to an ionised 1-bromopropane molecule with one ¹³C atom in it i.e. an ionised 2-bromopropane molecule of formula [¹³C¹²C₂H₇Br]⁺

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 these ¹³C M+1 peaks.

1-bromopropane has 3 carbon atoms, so on average, ~1 in 33 molecules will contain a ¹³C atom.

The most abundant ion of the molecule under mass spectrometry investigation (1-bromopropane) 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 1-bromopropane.

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

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

Atomic masses: C = 12;  H = 1; O = 16;  Br = 79 or 81

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

Equations to explain the most abundant ion peaks of 1-bromopropane

Formation of m/z 107 and 109 ions:

[CH₃CH₂CH₂Br]⁺  ===>  [C₂H₄Br]⁺  +  CH₃

C-C bond scission to lose the end methyl group.

Mass loss = 124 - 15 = 109  and  122 - 15 = 107

Very small peaks since C-C bond scission is much less likely than C-Br scission - the latter having a significantly lower bond enthalpy.

Formation of m/z 93 and 95 ions:

[CH₃CH₂CH₂Br]⁺  ===>  [CH₂⁷⁹Br]⁺  or  [CH₂⁸¹Br]⁺  +  C₂H₅

C-C bond scission.

Mass loss = 124 - 29 = 95  and  122 - 29 = 93

Again, very small peaks since C-C bond scission is much less likely than C-Br scission - the latter having a significantly lower bond enthalpy.

Formation of m/z 80 and 82 ions:

[CH₃CH₂CH₂Br]⁺  ===>  [H⁷⁹Br]⁺  or  [H⁸¹Br]⁺  +  C₃H₆

Elimination of hydrogen bromide from the parent molecular ion.

Formation of m/z 79 and 81 ions:

[CH₃CH₂CH₂Br]⁺  ===>  [⁷⁹Br]⁺  or  [⁸¹Br]⁺  +  C₃H₇

C-Br bond scission, but alkyl group is more likely to carry the positive charge (see below).

Mass loss = 124 - 43 = 81  and  122 - 43 = 79

Formation of m/z 43 ion:

[CH₃CH₂CH₂Br]⁺  ===>  [C₃H₇]⁺  +  Br

C-Br bond fission to give the ionised alkyl ion (bond weaker than C-C).

Mass loss 124 - 81 = 43  and  122 - 79 = 43

The m/z 43 ion is the base peak ion, the most abundant and 'stable' ion fragment.

It undergoes success proton loss e.g. 43 => 42 => 41 => 40 => 39

Formation of m/z 29 ion:

[CH₃CH₂CH₂Br]⁺  ===>  [C₂H₅]⁺  +  CH₂Br

C-C bond scission.

Mass loss 124 - 95 = 29  and  122 - 93 = 29

Formation of m/z 27 ion:

[CH₃CH₂CH₂Br]⁺  ===>  [C₂H₃]⁺  +  CH₂Br

Formation of m/z 15 ion:

[CH₃CH₂CH₂Br]⁺  ===>  [CH₃]⁺  +  C₂H₄Br

C-C bond scission.

Mass loss 124 - 109 = 15  and  122 - 107 = 15

The chemistry of HALOGENOALKANES (haloalkanes) revision notes INDEX

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