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Advanced Organic Chemistry: Mass spectrum of 3-ethylpentane

The mass spectrum of 3-ethylpentane

Doc Brown's Chemistry Advanced Level Pre-University Chemistry Revision Study Notes for UK IB KS5 A/AS GCE advanced A level organic chemistry students US K12 grade 11 grade 12 organic chemistry courses involving molecular spectroscopy analysing mass spectra of 3-ethylpentane

C7H16 mass spectrum of 3-ethylpentane fragmentation pattern of m/z m/e ions for analysis and identification of 3-ethylpentane image diagram doc brown's advanced organic chemistry revision notes 

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Interpreting the fragmentation pattern of the mass spectrum of 3-ethylpentane

[M]+ is the molecular ion peak (M) with an m/z of 100 corresponding to [C7H16]+, the original 3-ethylpentane molecule minus an electron, [(CH3CH2)3CH]+.

The very tiny M+1 peak at m/z 101, corresponds to an ionised 3-ethylpentane molecule with one 13C atom in it i.e. an ionised 3-ethylpentane molecule of formula [13C12C6H16]+

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.

3-ethylpentane has 7 carbon atoms, so on average, ~1 in 14 molecules will contain a 13C atom.

Identifying the species giving the most prominent peaks (apart from M) in the fragmentation pattern of 3-ethylpentane.

m/z value of [fragment]+ 85 71 70 57 56 55
[molecular fragment]+ [C6H13]+ [C5H11]+ [C5H10]+ [C4H9]+ [C4H8]+ [C4H7]+
m/z value of [fragment]+ 43 42 41 39 29 27
[molecular fragment]+ [C3H7]+ [C3H6]+ [C3H5]+ [C3H3]+ [C2H5]+ [C2H3]+

Analysing and explaining the principal ions in the fragmentation pattern of the mass spectrum of 3-ethylpentane

Atomic masses: C = 12  H = 1

Equations to explain the most abundant ion peaks of 3-ethylpentane

Formation of m/z 85 ion:

[(CH3CH2)3CH]+  ===>  [C6H13]+  +  CH3

Most fragmentation in alkanes arises from C-C bond scission (C-C bond weaker than C-H).

In this case and end methyl group is broken off, mass loss = 100 - 15 = 85.

However, this peak is small, and the m/z ion probably readily loses another methyl group to give the m/z 70 ion (see below).

Note in the equations below that both fragments are capable of being ionised, but only one at a time.

Many other fragments are formed by proton loss so you get sequences like 71 => 70, 57 => 56 => 55, 43 => 42 => 41 => 40 => 39 and 29 => 28 => 27 etc.

Ethene is often eliminated to give a smaller fragment e.g. the m/z ion 85 gives the m/z ion 57

[C6H13]+  ===>  [C4H9]+  +  C2H4

Mass change 85 - 28 = 57.

Formation of m/z 70 ion:

[C6H13]+  ===>  [C5H12]+  +  CH3

Mass change 85 - 15 = 70

or proton loss from m/z 71 ion too?

[C5H11]+  ===>  [C5H10]+  +  H

Formation of m/z 71 ion:

[(CH3CH2)3CH]+  ===>  [C5H11]+  +  CH2CH3

Here an end ethyl group is broken off, mass loss = 100 - 29 = 71.

Formation of m/z 57 ion: (see also above)

[(CH3CH2)3CH]+  ===>  [C4H9]+  +  C3H7

Loss of C3H7 group, mass loss = 100 - 43 = 57

Formation of m/z 43 ion:

[(CH3CH2)3CH]+  ===>  [C3H7]+  +  C4H9

Loss of a C4H9 group, mass loss = 100 - 57 = 43

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

Loss of a C4H9 group, mass loss = 100 - 57 = 43

Will also be formed by

Formation of m/z 29 ion:

[(CH3CH2)3CH]+  ===>  [CH2CH3]+  +  C5H11

Here an end ethyl group is broken off, mass loss = 100 - 71 = 29.


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