CH 908: Mass Spectrometry
Lecture 4
Interpreting Electron Impact
Mass Spectra – Continued…
Recommended: Read chapters
8-9 of McLafferty
Prof. Peter B. O’Connor
outline
•
•
•
•
•
•
Ring fragmentations
Radical site migration
H-rearrangement – saturated
H-rearrangement – unsaturated
2H rearrangement
Displacement (similar to a long-range alpha
cleavage)
• Elimination
• PE diagrams
Cyclic species require ≥2 bond
cleavages to generate fragments.
Retro Diels-Alder
Retro Diels-Alder
Example: Predict the fragmentations
for this molecule: p-dioxane
p-dioxane
28
88
58
Fragmentation of Aromatic
Molecular Ions
Many simple aromatic molecular ions fragment
by elimination of a small, unsaturated molecule
by breaking the aromatic ring but giving a
further, stable cyclic ion as a product.
 Examples of small molecules lost include: 

Benzene, C2H2, Pyridine, HCN, Thiophene, HCS,
Furan, HCO, Phenols, CO, Anilines, HCN
M+.
[M-C2H2]+.
[M-HCN]+.
M+.
[M-HCO]+
M+.
[M-C2H2]+.
[M-C3H3]+
[M-CHS]+
M+.
H – rearrangements
H• migration
Types of Hydrogen Rearrangements
Double H• migration
“Mclafferty +1” rearrangement
Long-range rearrangements
Displacements
Eliminations
Even-Electron Ions, CI,
Even-Electron Ions - 1

Under EI conditions, M+. ions are formed and a major
fragmentation process is the loss of a radical, R., producing an
even-electron ion.

Once a radical has been lost, all subsequent fragmentations
involve the loss of a molecule to form further even-electron
ions.

Under CI conditions, an even-electron ion, such as MH+, is
formed; subsequent fragmentations involve the loss of a
molecule to form further even-electron ions.
Sites of Protonation

In order to rationalise the fragmentation of MH+ ions, one must
consider at which sites in the sample molecule the proton is
attached. The spectrum may then be rationalised in terms of
the fragmentation of the different types of MH+ ions.

In general, protonation occurs on heteroatoms having lone
pairs of electrons, such as O, N and Cl. This frequently
followed by charge-induced elimination of a molecule
containing the hetero-atom. Other possible protonation sites
are aromatic rings and regions of unsaturation.
Even Electron Ions


Ephedrine ionised by methane CI may protonate for
example on the O atom of the OH group:

Protonation on the N atom leads to the loss of CH3NH2 by
a similar mechanism, yielding an ion of m/z 135. Both m/z
148 and 135 are observed in the CI spectrum, indicating the
presence of OH and HNCH3 groups in the molecule.
Ephedrine EI and CI Spectra
•
Ephedrine, RMM 165,
gives an EI spectrum
dominated by the m/z 58
fragment ion and no M+.
ion giving an RMM.
Methane CI gives an MH+
ion at m/z 166 and
fragments at m/z 148, 135
and 58 due to protonation
on the OH and NHCH3
groups or on the aromatic
ring respectively
General Hints for Solving Spectra - 1

Aromatic or aliphatic? Provisionally identify M+.

Check that proposed neutral losses are sensible

Is N present? Is assignment of M+. incorrect?

Check for isotope peaks for Cl, Br, S, heavy metals

Use I([M+1]+)/I([M]+.) to estimate number of C atoms present

Postulate a molecular formula and estimate the double bond
equivalents
General Hints for Solving Spectra - 2

Inspect higher mass ions, possibly formed from M+. in one step,
e.g. even mass fragments formed in a rearrangment process

Look for characteristic neutral losses such as 16 Da, O from
ArNO2 or NH2 from an amide, 30 Da, CH2O from ArOCH3 and
characteristic ions, m/z 30, amines, m/z 74 methyl esters,
105/77/51, 91/65/39 for benzoyl and alkyl benzene compounds

Do not assume adjacent peaks are due to sequential losses of
neutrals; two or more charge sites lead to competing
fragmentation routes.
General Hints for Solving Spectra - 3

Do not try (initially) to interpret every small peak, especially
those at low m/z which result from sequential fragmentation

Never postulate the loss of a radical from an even electron ion
without very good reason

Use negative evidence as well as positive evidence: e.g. if
there is no peak at m/z 91, the sample is unlikely to be an alkyl
benzene.
Practical problems with “real world” spectra - 1

Beware of spurious peaks such as the following:

Background peaks from previous samples, pump oil or from an
air leak, e.g. m/z 40, 32, 28, 18 etc.

Peaks arising from incomplete removal of common solvents
such as m/z 83, 85, 87 from CHCl3, m/z 58, 43 from acetone

Peaks present due to incomplete reaction leaving traces of
starting materials in the sample

Peaks due to homologues, e.g. at 14 m/z units above or below
the true molecular ion peak
Practical problems with “real world” spectra - 1
• Compare your spectrum with that of an authentic sample
obtained by use of the same ionisation technique (probably
from a database) but remember that an exact match of relative
intensities is unlikely to be found because of varying mass
discrimination effects.
Example 1: Assign each peak
Example 2: Assign each peak
Example 3: Assign each peak
Self Assessment
• Explain how C60 can lose 24 Da. How can benzene lose
26 Da?
• Hydrogen atom rearrangments are usually promoted
by…
• Can multiple hydrogen atoms rearrange to generate the
observed fragments? How?
• Can you generate neutral radical losses from an even
electron ion? Why?
• Is loss of H2 common?
• What is a ‘distonic ion’?
• In MS of oligosaccharides, it’s common to lose several
residues due to an internal rearrangement. Why is this a
problem in interpreting the spectra?
Fini.
[-C2H5]+
Does this assignment
make sense?
-C2H6
M+.