1. Chapter 14
Mass Spectroscopy

2. Positive ions are detected. Neutral species are undetected.
Mass Spectrometer
Figure 14.1 Schematic diagram of an electron ionization mass spectrometer (EI MS).
Positive ions are detected. Neutral species are undetected.

3. p.545

4. Parent peak due to molecular radical cation.
Figure 14.2 A partial mass spectrum of dopamine showing all peaks with intensity equal to or greater than 0.5% of the base peak.
Parent peak due to molecular radical cation.
Figure 14.2, p.546

5. 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.

6. p.546

7. 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

8. = (100 * * ) /
Recall that the atomic weight is the average mass for all isotopes found in nature.
For example chlorine….
Table 14.1, p.548

9. Bromine has two isotopes 79Br and 81Br in about equal amounts.
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.
Figure 14.4 Mass spectrum of 1-bromopropane.
Bromine has two isotopes 79Br and 81Br in about equal amounts.
Obtain two peaks at 122 and 124.
Figure 14.4, p.550

10. 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 fragment hopefully with close to the expected ratio is taken as indication of Cl or Br.
Ratio of peaks calculated as
35Cl79Br
37Cl79Br & 35Cl81Br
37Cl81Br
1.00 *1.00
.324 * *.979
.324*.979
.767
1.00
.243
Ratio of peaks calculated as
35Cl2
35Cl37Cl & 37Cl35Cl
37Cl2
1.00*1.00
1.00* *1.00
.324*.324

11. 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.

12. Here is the data. We will use isotopic occurrence data for H, C, O for the M + 1 peak.

13. One approach to obtain the molecular formula needing very high resolution of mass.
Suppose you have a mass spec with a Molecular Ion peak at 28 and a much smaller one at 29.
Conclude approximate molecular weight is 28 and peak at 29 is due to heavier isotopes.
The compound could be CO, C2H4, N2, etc. Limit our consideration to CO and C2H4.
Using the natural occurrence data we can calculate the expected intensity ratio of the 28 and 29 peaks for both CO and C2H4.
For CO, the 29 peak could be produced by either the C or the O being the heavier isotope. For C2H4, either C or H. CO
I29/I28 = 1.11/ /100.0 =
C2H4
I29/I28 = 2 * 1.11/ * 0.016/100=0.0229
13C16O
12C17O
13C12C1H4
12C2H1H3
Can tell which we have but we do not proceed further with this technique.

14. 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.

15. 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 % * 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 % % = 0.50%

16. 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
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.

17. 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
C7H10N4
9.25
0.38
C8H10N2O
9.61
0.61
C9H10O2
9.96
0.84
C9H14N2
10.71
0.52
We can have 0 or 2 nitrogens. Even number.
We can have 0,1,2,3,4 oxygens /0.2 < 5
Can have 0,1,2,3,4,5,6,7,8,9 carbons /1.11 <10
Find molecular formulas having reasonable M+1 peaks

18. 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 C4H10+.
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.

19. Example. Identify this molecule
m/e
Abundance 1
<0.1 16
1.0 17 21 18
100 19
0.15 20
0.22
Ejection of an H
Molecular radical ion
Due to heavier isotopes
H2O

20. Example 2
m/e
Abundance 12
3.3 13
4.3 14
4.4 15
0.07 16
1.7 28 31 29 30 32
100 89
1.3
0.21
carbon
Oxygen
H ejection
parent
Heavier isotopes
CH2O

21. 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, C5H12O+.
Radical/ cation, C5H10+.
p.550

22. How to think about it… For this fragmentation Think of it this way…
This C becomes +
One electron in this bond. When it splits we get a carbocation (observable in MS) and a radical (not observable in MS).
The fragmentation is written this way.
This C bears the .

23. In fragmentation to produce a carbocation stability of the carbocation is an important factor in determining where the fragmention will occur.
p.551

24. For simple linear alkanes fragmentation will occur towards the middle of the chain.
Figure 14.5 Mass spectrum of octane.
Figure 14.5, p.552

25. Caution. Sometimes a peak will occur which cannot be explained such as the ethyl peak below.
Figure 14.6 Mass spectrum of 2,2,4- trimethylpentane. The peak for the molecular ion is of such low intensity that it does not appear in this spectrum.
Figure 14.6, p.552

26. 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.
Figure 14.7 Mass spectrum of methylcyclopentane.
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

27. Ionization followed by fragmentation
Ejection of electron
fragmentation
Radical/cation
Splitting out of ethylene.
p.553

28. Difference is 15, the methyl radical
Alkenes can yield allylic stabilized carbocations by fragmentation, splitting out a radical.
Figure 14.8 Mass spectrum of 1-butene.
Difference is 15, the methyl radical 28

29. p.554 29

30. Alcohols have several characteristic fragmentation patterns.
1. An alcohol radical/cation can undergo a fragmentation to produce a radical and a resonance stabilized carbocation.
The one “electron bond”
p.555 30

31. Here is an example demonstrating both processes.
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.
Figure Mass spectrum of 1-butanol.
Elimination of water.
74 – 56 = 18 (water).
Elimination of propyl radical.
74 – 31 = 43 (C3H7)
Figure 14.10, p.555 31

32. 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. 32

33. 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 33

34. Mass Spec of 2-octanone displays both a cleavage and McLafferty
The “invisible” radical C6H13
CH3 radical
“Invisible” pent-1-ene
CH3CH2CH2CH2CH2CH2CO+ resulting from a cleavage here.
CH3CO+ resulting from a cleavage at this bond.
Figure Mass spectrum of 2-octanone. Ions of m/z 43 and 113 result from -cleavage. The ion at m/z 58 results from McLafferty rearrangement. 34

35. The peak at 60 is usually prominent for a carboxylic acid.
Carboxylic Acids can also undergo a cleavage and McLafferty rearrangement.
The peak at 60 is usually prominent for a carboxylic acid. 35

36. Likewise for esters, 2 a cleavages (around C=O) and McLafferty36

37. An unexpected observation.37

38. Figure 14. 14 Mass spectrum of toluene
Figure Mass spectrum of toluene. Prominent are the intense molecular ion peak at m/z 92 and the tropylium cation at m/z 91.
Figure 14.14, p.559 38

39. 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.
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.

40. Example
Given spectra for an unknown compound: IR, MS, and NMR. Identify the compound.
First, let’s do the MS

41. 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:
Parent mass is even. Even number of nitrogens: 0, 2, 4
No Br or Cl since M+2 is much too small.
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!

42. 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….

43. NMR . Assumed formula C9H10O2
First the hydrogen counts.

44. Chemical Arithmetic. Assumed formula C9H10O2
Consistent with the lack of splitting could be either.
C9H10O2
- C6H5
CH2
CH3
Tentatively identified parts,
Subtract to get CO2