1. Introduction to Mass Spectrometry (MS)
A mass spectrometer produces a spectrum of masses based on the structure of a molecule.
The x-axis of a mass spectrum represents the masses of ions produced (m/z)
The y-axis represents the relative abundance of each ion produced
The pattern of ions obtained and their abundance is characteristic of the structure of a particular molecule 1

3. Ionization (the formation of ions)
A molecule is bombarded with a beam of high energy electrons
An electron is dislodged from the molecule by the impact, leaving a positively charged ion with an unpaired electron (a radical cation)
This initial ion is called the molecular ion (M+.) because it has the same molecular weight as the analyte 2

4. Fragmentation
Excess vibrational energy is imparted to the molecular ion by collision with the electron beam - this causes fragmentation
The fragmentation pattern is highly characteristic of the structure of the molecule 3

5. Fragmentation by Cleavage at a Single Bond
Cleavage of a radical cation gives a radical and a cation but only the cation is observable by MS
In general the fragmentation proceeds to give mainly the most stable carbocation
In the spectrum of propane the peak at 29 is the base peak (most abundant) 100% and the peak at 15 is 5.6% 11

6. Fragmentation Equations
The M+. Ion is formed by loss of one of its most loosely held electrons
If nonbonding electron pairs or pi electrons are present, an electron from one of these locations is usually lost by electron impact to form M+.
In molecules with only C-C and C-H bonds, the location of the lone electron cannot be predicted and the formula is written to reflect this using brackets 12

7. Example: The spectrum of hexane13

8. Example: spectrum of neopentane
Fragmentation of neopentane shows the propensity of cleavage to
occur at a branch point leading to a relatively stable carbocation
The formation of the 3o carbocation is so favored that almost no
molecular ion is detected 14

9. Carbocations stabilized by resonance are also formed preferentially
Alkenes fragment to give resonance-stabilized allylic carbocations 15

10. Carbocations stabilized by resonance are also formed preferentially
Alkenes fragment to give resonance-stabilized allylic carbocations
Carbon-carbon bonds next to an atom with an unshared electron pair break readily to yield a resonance stabilized carbocation
Z=N, O, or S R may be H 15

11. Carbocations stabilized by resonance are also formed preferentially
Alkenes fragment to give resonance-stabilized allylic carbocations
Carbon-carbon bonds next to an atom with an unshared electron pair break readily to yield a resonance stabilized carbocation
Z=N, O, or S R may be H 15

12. Carbon-carbon bonds next to carbonyl groups fragment readily to yield resonance stabilized acylium ions 16

13. Carbon-carbon bonds next to carbonyl groups fragment readily to yield resonance stabilized acylium ions 16

14. Alkyl substituted benzenes often lose a hydrogen or alkyl group to yield the relatively stable tropylium ion
Other substituted benzenes usually lose their substitutents to yield a phenyl cation 17

15. Fragmentation by Cleavage of 2 Bonds
The products are a new radical cation and a neutral molecule
Alcohols usually show an M+.-18 peak from loss of water 18

16. Cycloalkenes can undergo a retro-Diels Alder reaction (section 13
Cycloalkenes can undergo a retro-Diels Alder reaction (section 13.11) to yield an alkadienyl radical cation 19

23. Ions with no nitrogen or an even # N atoms
The masses of molecular and fragment ions also reflect the electron count, depending on the number of nitrogen atoms in the species.
Ions with no nitrogen or an even # N atoms
Molecular Ion Fragment Ions
odd-electron ions even-electron ions
even-number mass odd-number mass
Ions having an odd # N atoms
odd-number mass even-number mass

24. Ions with no nitrogen or an even # N atoms
The masses of molecular and fragment ions also reflect the electron count, depending on the number of nitrogen atoms in the species.
Ions with no nitrogen or an even # N atoms
Molecular Ion Fragment Ions
odd-electron ions even-electron ions
even-number mass odd-number mass
Ions having an odd # N atoms
odd-number mass even-number mass

28. Common Isotope Abundances5

29. 3 Classes of Isotopes M - only a single isotope
EX: F, P, I
M+1 - two isotopes with significant relative abundance differing by 1 mass unit
EX: H, C, N
M+2 - two isotopes with significant relative abundance differing by 2 mass units
EX: O, S, Cl, Br

30. Determination of Molecular Formulas & Weights
The Molecular Ion and Isotopic Peaks
The presence of heavier isotopes one or two mass units above the common isotope yields small peaks at M+.+1 and M+.+2
Example: In the spectrum of methane one expects an M+.+1 peak of 1.17% based on a 1.11% natural abundance of 13C and a 0.016% natural abundance of 2H 6

31. # of C = relative abundance/1.1
Four Basic Rules
If M+ is even, then the unknown contains an even number of Nitrogen atoms (zero is an even number)
The abundance of M++1 indicates the number of Carbon atoms:
# of C = relative abundance/1.1
The abundance of the M++2 peak indicates the presence of O (0.2%), S (4.4%), Cl (33%) or Br (98%)
The remaining unknown mass can be attributed to Hydrogen 7

32. C3H7Cl C – C – C - Cl Is the molecular ion even?
m/z
intensity
% abundance
78 (M+)
10.00
100% 79 .3 3% 80
3.3
33%
Is the molecular ion even?
Yes, there must be either an even number of N, or no Nitrogen atoms.
How many Carbon atoms are there?
# Carbons = 3 / 1.1 ≈ 3 carbon atoms
Is a O, S, Cl or Br present?
A M++2 peak of 33% indicates the presence of chlorine
How many Hydrogen atoms are there?
78 = (1 * 35) + (3 * 12) + (H * 1)
78 = 71 + H
# of Hydrogen atoms = 7
C – C – C - Cl H
C3H7Cl 8

33. C5H10O C – C – C – C – C = O H Is the molecular ion even?%
Intensity
(M-1)
m/z
intensity
% abundance
86 (M+)
10.00
100% 87
.56
5.6% 88
.02
.2%
Is the molecular ion even?
Yes, there must be either 0, 2, 4 … Nitrogen atoms
How many Carbon atoms are there?
# Carbons = 5.6 / 1.1 ≈ 5 carbon atoms
if there are 2 N atoms then the FW would be (5*12) + (2*14) = 88
Therefore, there are no nitrogen atoms
Is an O, S, Cl or Br present?
A M++2 peak of .2% indicates O
How many Hydrogen atoms are there?
86 = (5 * 12) + (O*1)+ (H * 1)
86 = 76 + H
# of Hydrogen atoms = 10
C – C – C – C – C = O H
C5H10O 8

34. Determination of Molecular Formula
distinguish between compounds of same MW
C5H10O4 or C10H14

35. Determination of Molecular Formula
distinguish between compounds of same MW
C5H10O4
13C 5 * 1.11% = 5.55%
2H 10 * 0.016% = 0.16%
17O 4 * 0.02% = 0.08%
135peak/134peak %

36. Determination of Molecular Formula
distinguish between compounds of same MW
C10H14
13C 10 * 1.11% = 11.1%
2H 14 * 0.016% = 0.22%
135peak/134peak %

37. The Numbers Approach
If compound with formula CwHxNyOz , relative intensities of M, M+1, and M+2 ions will be given by:

38. High-Resolution Mass Spectrometry
Low-resolution mass spectrometers measure m/z values to the nearest whole number
High-resolution mass spectrometers measure m/z values to three or four decimal places
The high accuracy of the molecular weight calculation allows accurate determination of the molecular formula of a fragment
Example
One can accurately pick the molecular formula of a fragment with a nominal molecular weight of 32 using high-resolution MS 9

39. The exact mass of certain nuclides is shown below10

40. Elemental Composition
Search Report:
Target Mass:
Target m/z = ± .003
Charge = +2
Possible Elements:
Element: Exact Mass: Min: Max: C H Cl N
Additional Search Restrictions:
None
Search Results:
Number of Hits = 1
m/z Delta m/z DBE Formula
C28H20Cl2N2+2