Chemistry 118B Workshop Jim Hollister

Learning Skills Center

UC Davis

MASS SPECTROMETRY

I. MASS SPECTROMETRY (Radiation is not used, so strictly speaking, it is

not spectroscopy,)

A) Mass spectrometry data is used to determine the molecular mass of a sample (See
Section III. below) . Using the molecular mass and determining the number of
hydrogens from a 1H-NMR spectrum, as well as the number of carbons from a 13C-NMR
spectrum, the molecular formula of a compound can be determined.

B) Mass spectrometry data is also used to tell if the sample molecule has chlorine,
bromine, or an odd number of nitrogens in it. (See Example in IV. A. 2 i) below.)

C) Also, when there is a choice between two different molecular structures, for example,
when two different esters could have each produced a particular 1H-NMR spectrum, a
McLafferty Rearrangement fragment can help one decide on or confirm the proper ester
structure that would yield the 1H-NMR spectrum. (See Example in IV. G below.)

D) Examine Figure 11.18 in the fifth edition of Vollhardt and Schore to see how a mass
spectrometer generates radical cations of various sizes by knocking off an electron from a
neutral molecule. These radical cation fragments are attracted to a magnet and separated
based on the different masses of the fragments. The separation data is plotted by the
spectrometer. The mass to charge ratio, m/z (or m/e), is plotted on the x axis and the
height of the line for each peak, plotted on the y axis, represents the relative amount of
that particular ion versus other ions. Since the charge ,z, is usually +1, the m/z ratio is
equal to the mass of the fragment. . So, the heaviest peak is only missing one electron;
it is a radical cation. It usually breaks into a radical and a cation. Cations also
show on the spectrum.

II. Symbols used in mass spectrometry:

M = neutral molecule

M+. = molecular ion = radical cation = parent ion. The parent ion has an m/z ratio which
is the molecular mass of the molecule in question. The parent ion is usually the most
heavy ion*; it is not necessarily the highest line(peak). It is always a radical cation (but
other peaks may be just cations).

The base peak is the highest peak and is given a value of 100% (or 1.00) relative to the
height of the other peak lines.

* This is not counting the slightly heavier and usually much shorter line peak representing parent
fragments which contains the 13C isotope, as the [M + 1]+. described in your text. Ignore this peak and the
detail regarding the 13 C isotope in your text unless your professor says otherwise.
III. Mass spectrum of Methane, CH4.
Note: the molecular weight of methane must be16, because the parent ion's m/z ratio is 16.

16
100 M+.
Mass
Spectral The base peak is
80 Pattern for given a 100%
CH 4 value relative
to the height
Relative 60 of the other peaks.
Abundance In this case, the parent ion,
of Peaks in M+. , is also the base peak.
Per Cent 40
(M+1) +.
20 Ignore unless
your professor
discusses it.

5 m/z 20
The higher a peak in the mass fragmentation pattern, the more easily the fragment forms.

Tabulated
mass spectrum data
m/z @ relative height (%)
17 @ 1.1
16 @ 100.0 (base peak)
15 @ 85.0
14 @ 9.2
13 @ 3.9
12 @ 1.0
Often the only mass spectrum information given on a test problem is the tabulated data,
rather than the actual spectrum.

IV. Other Information from Mass Spectrometry

A. Usually in this course we will not be concerned with peaks from the isotopes 13C,
15N, or 2H. However, we can use information of the isotopes of chlorine and bromine to
determine when these elements are present.

1. When there is a mass difference of two between the last two major fragment
peaks of a sample, this may indicate that there is either a chlorine or bromine in the
sample.
2. Look at the base peak ratio of these last two peaks.

i) When there is a ratio of about 3:1, for example, 0.30 to 0.10 or 0.12 to 0.04,
it is an indication that the sample has chlorine in the molecular formula. The
reason is that the natural abundance, or percentage of the two chlorine isotopes in
nature, is about a 3:1 ratio of 35Cl and 37Cl, respectively. (There is a mass
difference of two between 37Cl and 35Cl, so one peak is for a parent ion that
contains 35Cl, and the other peak is for a parent ion that contains 37Cl.)
 Below, the two major peaks, 120 and 118, differ by two, and their height ratio,
0.12 to 0.04, is a 3:1 ratio, so the compound contains chlorine.
Tabulated
mass spectrum data
m/z @ relative height
120 @ 0.04
118 @ 0.12
105 @ 0.08
103 @ 0.25 ii) When there is a ratio of about 1:1,
83 @ 1.00 (base peak) for example, 0.15 to 0.15, it is an
indication that the sample has a
bromine in the molecular formula. The reason is that the natural abundance, or
percentage of the two bromine isotopes in nature, is about a 1:1 ratio of 79Br and
81Br, respectively. (There is a mass difference of two between 79Br and 81Br, so
one peak is for a parent ion that contains 79Br, and the other peak is for a parent
ion that contains 81Br.) See Figure 11.20 in text (Fifth Ed.).

B. When the molecular weight of a compound has an odd numbered value, it

contains an odd number of nitrogen atoms. If the parent ion value is odd, instructors
almost always have only one nitrogen in the unknown molecule, not another odd
number of nitrogens, as three or five. An even number of nitrogens is also not a problem
usually seen in this course. Do Exercise 11.13 (Fifth Ed., Vollhardt & Schore).

C. The higher a peak in the mass spectral fragmentation pattern, the more easily the
fragment forms. Fragmentation is more likely at highly substituted centers. Intense
fragment peaks usually result from either the loss of relatively stable neutral species or
are due to the formation of relatively stable cations. The neutral fragment does not
appear in the spectrum, but its existence can be determined by noticing the difference in
the mass of a fragment ion subtracted from the mass of the original ion.** In the
fragmentation below, 72 - 43 = 29, so the neutral fragment must have been the neutral
radical, .CH2CH3.
CH3 CH3
CH CH
H3 C C CH2 CH3 minus 2 3 H3 C C
H H
m/z = 72 minus 29 m/z = 43
The above could be abbreviated as M+. minus .CH2CH3 = (M - CH2CH3)+ = 43: The
fragment cation that causes the peak at 43 is represented as (M - CH2CH3)+. The neutral
radical .CH2CH3 seen above does not cause a peak at 29. (The peak seen at 29 on the
mass spectrum [see figure 11.22 in Vollhardt and Schore] is caused by the cation,
(CH2CH3)+, which is made from other complex reorganizations that we need not worry
about here.) See problem 11.40 (Fifth Ed.) of your solution manual for more examples.

**From Organic Chemistry, Vollhardt and Schore, 1999, and Introduction to

Spectroscopy, Pavia, Lampman, Kriz. 1979.
 1. A useful approach to remember the mass of some alkyl fragments:

FRAGMENT A MULTIPLE of 15 MINUS A SMALL

INTEGER = ( MASS IN BOLD TYPE)
CH3- 15 - 0 = 15
C2H5- (same as, CH3CH2-) 30 - 1 = 29
C3H7- as CH3CH2CH2- or branched 45 - 2 = 43
C4H9- as CH3CH2CH2CH2- or branched 60 - 3 = 57
C5H11- as CH3CH2CH2CH2CH2- or branched 75 - 4 = 71

D. Alcohol Fragmentation: Alpha -Cleavage [(α) -Cleavage] and Dehydration: Did

your professor lecture on these? It is covered in your text.

E. Amine Alpha -Cleavage: Did your professor lecture on this? It is covered in your
text.

F. Alpha -Cleavage [(α) -Cleavage] of Carbonyl Groups: Did your professor lecture on
this? Here is an example:
O
!-cleavage
R C O on right of
R C R' !-cleavage
on left of O C R'
carbonyl; carbonyl;
- R' alkyl - R alkyl
radical radical O C R'
RC O
acylium ion acylium ion
yields a mass yields a mass
spec. peak spec. peak
G. McLafferty Rearrangement: can occur in ketones, aldehydes, carboxylic acids and
esters. (Also amides, nitriles, and benzene rings with propyl or larger groups, but it is
very unlikely you will see these in rearrangements in this course.)

Carbonyl compounds that have a hydrogen on the atom that is three atoms away (the
gamma, γ, carbon) from the carbonyl group can undergo a β-cleavage called a
McLafferty Rearrangement.

Below, both esters have the same molecular mass and very similar 1H-NMR spectra, but
the McLafferty Rearrangement fragments from each differ.

# H HO
H2 C O HCH
+ C
CH C CH2 H2 C O CH2 CH CH2
" 2 O CH2 CH CH2
CH2 28
! (does not
m/z = 128 show in mass m/z = 100
ester spectrum) McLafferty Rearrangement fragment
CH3
H CH3 HO
HC O CH
#
+ C
"CH2O C
CH2 CH CH2
CH2 O CH2 CH CH2
! 42
( does not m/z = 86
m/z = 128 show in mass
ester McLafferty Rearrangement fragment
spectrum)

V. Skip High Resolution Mass Spectrometry unless your professor discussed or

assigned problems on it.