Chapter 14
Mass Spectroscopy
Mass Spectrometer
Positive ions are detected. Neutral species
are undetected.
p.545
Parent peak due to
molecular radical
cation.
Figure 14.2, p.546
Detecting nitrogen, N
Consider some simple molecules and their nominal mass.
CH4
16
CH3NH2
29
CH3OH
32
CH3F
34
CH3Cl
50, 52
CH3SH
48
What is unusual about the N compound?
The parent peak should have an odd mass for an odd number of nitrogens.
p.546
One way to distinguish between molecules having the same about the same mass is to
measure their mass very precisely.
About the same mass for these species formed from the most common isotopes.
In these cases we can actually determine the molecular formula from high
resolution mass spectroscopy.
p.547
35.453 = (100 * 34.9689 + 31.98 * 36.9659) / 131.98
Recall that the atomic weight is the average mass for all isotopes found in nature.
Table 14.1, p.548
For example chlorine….
Low resolution mass spec does not involve itself with precise mass
measurements. Low resolution examines the various peaks produced.
First consider the parent radical cation: if an element has two naturally
occurring isotopes then two peaks will be produced.
Bromine has two isotopes 79Br and 81Br in about equal amounts.
Obtain two peaks at 122 and 124.
Figure 14.4, p.550
Further comments on presence of chlorine and bromine.
Both Cl and Br have two common isotopes separated by two mass units.
Given the natural abundances we may calculate the ratio of the M and M+2 peaks
for various combinations of Cl and Br being present.
The presence of peaks at X,
X+2… for the molecular ion or
Ratio offragment
peaks calculated
hopefully as
with close
to&the
expected
ratio37isCltaken
35Cl79Br
37Cl79Br
35Cl
81Br
81Br
as indication of Cl or Br.
1.00 *1.00 .324 *1.00+1.00 *.979
.324*.979
Ratio of 1.00
peaks calculated as .243
.767
35Cl
35Cl37Cl & 37Cl35Cl
37Cl
2
2
1.00*1.00
1.00*.324+.324*1.00
.324*.324
Molecular Peaks, M+1
Have seen that for Cl and Br, having two
common isotopes, two radical cation
peaks produced. What about other
elements having more than one isotope?
We know what the isotopes are and their
natural occurrence.
For the M+1 peak, one atom must be using
an isotope heavier by one.
Here is the data. We will use isotopic occurrence data for H, C, O for the M + 1 peak.
The M+2 peak
Recap: The M+1 peak has contributions from one atom being a heavier isotope
by 1.
The M+2 peak can have contributions from
•One atom being a heavier isotope by 2.
•Two atoms being heavier by 1 each.
M+2 peak, contributions from one atom and two atoms.
Recap:
The M+1 peak has contributions from one atom being a heavier isotope by 1.
(M+1)/M = ca. 1.1% * no. of C atoms + 0.36% * no. of N atoms
The M+2 peak can have contributions from two sources
•One atom being a heavier isotope by 2. Mainly O (excluding S, Cl and Br)
•Two atoms being heavier by 1 each. Mainly C atoms.
(M+2)/M = ca. (0.20% * no. of O atoms) + (1.1 * no. of C atoms)2/200%
Example 1: C5H5N
[(A + 1)+]/[A+] = 5 x 1.1% + 1 x 0.36% =
5.9%
[(A + 2)+]/[A+] = 5.52/200 % = 0.15%
Example 2: C7H5O
[(A + 1)+]/[A+] = 7 x 1.1% = 7.7%
[(A + 2)+]/[A+] = 7.72/200 % + 0.20% = 0.50%
Technique to obtain molecular formula using intensities of M, M+1, M+2 peaks.
Example. Given the data.
Peak
Intensity
150 (M)
100
151 (M+1)
10.2
152 (M+2) 0.88
Looking at M+2 there is no
Br, Cl or S. There could
be oxygen.
Even mass for M means
there could only be even
number of Nitrogen
Consider the M+1 peak, nominal mass + 1.
If we know the formula we should be able to calculate the relative intensity of that peak
due to the contributions from each of the atoms present. Here are the major
contributors to M+1.
Here are major contributors to M+2.
Technique to obtain molecular formula using intensities of M, M+1, M+2 peaks.
Example. Given the data.
Peak
Intensity
150 (M)
100
151 (M+1)
10.2
152 (M+2)
0.88
Equations
M+1: (1.11% x # of C) +
(0.38 x # of N+ small
contributions from O
M+2: (0.20 x # of O) +
(1.1 x # of C)2/200
Examine reasonable
formulae. Calculate M+1,
M+2 peaks M+1
M+2
We can have 0 or 2 nitrogens. Even number.
C7H10N4
9.25
0.38
We can have 0,1,2,3,4 oxygens. 0.88/0.2 < 5
Can have 0,1,2,3,4,5,6,7,8,9 carbons. 10.2/1.11 <10
C8H10N2O
9.61
0.61
C9H10O2
9.96
0.84
Find molecular formulas having reasonable M+1 peaks
C9H14N2
10.71
0.52
Return to Fragmentation of Molecular Radical Cation, M+
The highly energetic radical/cation can undergo fragmentation.
First distinguish between some species
•Radical/cation produced by ejection of electron from structure. It contains all the
atoms of the original molecule just minus one electron. Example C H +.
4
10
•Carbocation which is not a radical, is electron deficient and is a fragment of a
stable molecule. Example C4H9+
•Radical is not charged and is a fragment of a stable molecule. Example C4H9.
Example. Identify this molecule
m/e
Abundance
1
<0.1
16
1.0
17
21
18
100
19
0.15
20
0.22
H2O
Ejection of an H
Molecular radical ion
Due to heavier isotopes
Example 2
m/e
Abundance
12
3.3
13
4.3
14
4.4
15
0.07
16
1.7
Oxygen
28
31
H ejection
29
30
31
32
100
89
1.3
0.21
CH2O
carbon
parent
Heavier isotopes
Fragmentation of the radical/cation can lead to
1. A radical and carbocation as separate species. Usually a bond is
split. Choice of bond to split is frequently controlled by the
carbocation stability.
Carbocation,
CH3CH2+
Radical/
cation,
.
C4H10+.
Radical,
CH3CH2.
Remember only the positive species are detected. Neutral species are invisible.
2. Rearrangements can occur including elimination of neutral
molecules to produce a different radical/cation.
Radical/
cation,
H2O +
C5H12O+.
OH
Radical/
cation,
C5H10+.
H
p.550
How to think about it…
For this
.
fragmentation
Think of it this way…
One electron in this
bond. When it splits we
get a carbocation
(observable in MS) and a
radical (not observable in
MS).
This C becomes +
The
fragmentation
is written this
way.
This C bears the
.
In fragmentation to produce a carbocation stability of the carbocation is an
important factor in determining where the fragmention will occur.
p.551
For simple linear alkanes fragmentation will occur towards the middle of the
chain.
Figure 14.5, p.552
Caution. Sometimes a peak will occur which cannot be explained
such as the ethyl peak below.
Figure 14.6, p.552
Technique: recall that the neutral molecules split out do not produce a peak.
The mass of the neutral particle (invisible) may sometimes be obtained by
subtracting the mass of the newly formed positive ion (detected) from the mass
of the original radical carbocation.
In this example the parent molecule has mass of 84. The mass of a
positive fragment is 56.
The peak at 56 is subtracted from the mass of the original ion, 84, yielding
28, the mass of ethylene which is taken as the invisible neutral molecule.
Figure 14.7, p.553
Ionization followed by fragmentation
.
Ejection of
electron
Splitting out of ethylene.
fragmentation
Radical/cation
p.553
Alkenes can yield allylic stabilized carbocations by fragmentation,
splitting out a radical.
.+ CH
3
Difference
is 15, the
methyl
radical
Alcohols have several characteristic fragmentation patterns.
1. An alcohol radical/cation can undergo a fragmentation to produce a
radical and a resonance stabilized carbocation.
.
+
R
R
R'
C
O
H
R'
R"
.
C
R"
O
H
R'
C
O
H
R"
The one “electron bond”
p.555
2. Alcohol radical/cations can split out water to produce a new alkene
radical/cation which may be detected.
Here is an example demonstrating both processes.
Elimination of water.
Elimination of propyl radical.
74 – 56 = 18 (water).
74 – 31 = 43 (C3H7)
Figure 14.10, p.555
Aldehydes and Ketones
Several characteristics reactions the radical/cations
1. a cleavage: break bond to carbonyl group
Note that an a cleavage of an aldehyde could produce a peak at M – 1 by
eliminating H atom.
This is useful in distinguishing between aldehydes and ketones.
2. McLafferty Rearrangement: splitting out an alkene (neutral molecule) and
producing a new radical/cation.
Note that the process involves a six membered ring for a transition state.
p.557
Mass Spec of 2-octanone displays both a cleavage and McLafferty
The “invisible” radical C6H13
“Invisible” pent-1-ene
CO+
CH3
resulting from
a cleavage at
this bond.
CH3 radical
CH3CH2CH2CH2CH2CH2CO+
resulting from a cleavage here.
Carboxylic Acids can also undergo a cleavage and McLafferty rearrangement.
The peak at 60 is usually prominent
for a carboxylic acid.
Likewise for esters, 2 a cleavages (around C=O) and McLafferty
C
An unexpected observation.
p.559
Figure 14.14, p.559
Some general principles
The relative height of the parent radical cation peak is greatest for straight chain
compounds and decreases for branched structures. Easier cleavage.
Cleavage is favored at branched carbons due to increased stability of
substituted carbocations.
Double bonds or aromatic rings stabilize the parent radical cation increasing the
size of its peak.
Double bonds (aromatic rings) favor cleavage yielding allylic (benzylic or
tropylium) carbocations.
R
CH2
H
R
H
Bonds beta to a hetero atom having lone pairs are frequently cleaved.
Elimination of small, stable molecules (water, alkenes, etc.)
can occur to yield a new radical cation.
N
H
H
H
Example
Given spectra for an unknown compound: IR, MS, and NMR. Identify the compound.
First, let’s do the MS
Mass Spectrum
Now the
IR….
Base peak, the largest, others are relative to
it.
Want M+1 and
M+2 relative to
M not base
peak.
Observations:
1. Parent mass is even.
Even number of
nitrogens: 0, 2, 4
2. No Br or Cl since M+2
is much too small.
3. At most about 9 carbons
since 9.9/1.1 =9.
What molecular formulas could
we have with good values for
M+1 between 9 and 11?
Molecular peak, parent peak.
Best fit for
both M+1
and M+2.
Maybe
aromatic!
Infra Red. Assumed formula C9H10O2
Carbonyl, no
OH
C-O stretch
Looks as if it may be an ester with an aromatic group.
Next the NMR….
NMR . Assumed formula C9H10O2
First the hydrogen
counts.
Chemical Arithmetic. Assumed formula C9H10O2
Consistent with the lack of
splitting could be either.
C9H10O2
O
O
- C6H5
- CH2
- CH3
Tentatively
identified
parts,
Subtract to get CO2
O
O