1. Alkanes undergo extensive fragmentation
CH3—CH2—CH2—CH2—CH2—CH2—CH2—CH2—CH2—CH3
Relative intensity 43 57
100 80 60 40 20
Decane 71 85 99
142
m/z 13

2. More stable carbocations will be more abundant.

3. CH3 |
CH3 - CH2 -- C -- CH3 m/e = 71 +
+CH3 m/e = 15
CH3 |
+C - CH3 m/e = 57
CH3 - CH2 + m/e = 29
CH3 |
+C - H m/e = 43

4. MS of 2,2,4-trimethylpentane
Branched alkanes have small or absent M+

5. CH2—CH2CH3 Propylbenzene fragments mostlyat the benzylic position 100
Relative intensity
100 80 60 40 20 91
CH2—CH2CH3
120
m/z + 91 14

7. Cycloalkanes Loss of side chain Loss of ethylene fragments
Ionization followed by fragmentation
Ejection of electron
fragmentation
Radical/cation
Splitting out of ethylene.

8. MS of methylcyclopentane:

9. Alkene Fragmentation Fairly prominent M+
Fragment ions of CnH2n+ and CnH2n-1+
Terminal alkenes lose allyl cation if possible to form resonance-stabilized allylic cations

10. Cycloalkenes + Prominent molecular ion Retro Diels-Alder cleavage•
Observed!
Typically you see both.
More stable cation will predominate
Also works for hetero-substituted (e.g. make enol)
Both EI (shown) and in CI. (protonated molecular ion, cleave, then reprotonation)

11. Cyclohexenes give a 1,3-diene and an alkene, a process that is the reverse of a Diels-Alder reaction.

12. Mass spectrum of 4-terpineol as a good example for Retro Diels Alder fragmentation
EI Mass Spectrum + + +
4-terpineol(MW 154)
mz 86
mz 68

13. Alkyne Fragmentation Molecular ion readily visible
Terminal alkynes readily lose hydrogen atom
Terminal alkynes lose propargyl cation if possible
(resonance-stabilized propargyl cation or a substituted propargyl cation).

14. Aromatic Hydrocarbon Fragmentation
Molecular ion usually strong
Alkylbenzenes cleave at benzylic carbon tropylium ion formation

15. Some molecules undergo very little fragmentation
Benzene is an example.
The major peak corresponds to the molecular ion.
Relative intensity
100 80 60 40 20
m/z = 78
m/z 9

16. Benzene Compounds - - - - 7 memberd ring Benzonium ion O H C H + O H ·2 O H - CO - H +
m/z
107
m/z
108
7 memberd ring + H H H H - H 2 + H H
Benzonium ion H
Phenyl ion
m/z 79
m/z 77 [C H ] + 6 7 [C H ] + 6 5
CH3 -
CH3 + H + + H
CH3
CH3
Tropylium ion
m/z 91

17. Mass Spectrum of n-Octylbenzene
Tropylium ion

18. Mass Spectrum of benzyl bromide
Tropylium ion
Bromine pattern

19. Alcohol Fragmentation
One of the most common fragmentation patterns of alcohols is loss of H2O to give a peak which corresponds to M-18 and no M peak.
Molecular ion strength depends on substitution
primary alcohol weak M+
secondary alcohol VERY weak M+
tertiary alcohol M+ usually absent
Another common pattern is loss of an alkyl group from the carbon bearing the OH (-Cleavage) to give a resonance-stabilized oxonium ion and an alkyl radical (largest R group lost as radical).

20. MS of 1-butanol 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)

22. Mass Spectrum of 3-methyl-1-butanol

23. Mass Spectrum of 3-methyl-2-butanol

24. Carbonyl Compounds Fragmentation
Dominant fragmentation pathways:
a-cleavage
b-cleavage
McLafferty rearrangement

25. Characteristic fragmentation patterns are:
Aldehydes and Ketones
Characteristic fragmentation patterns are:
Cleavage of a bond to the carbonyl group (-cleavage).
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.

26. McLafferty rearrangement.
Splitting out an alkene (neutral molecule) and producing a new radical/cation.

27. Mass Spec of 2-octanone displays both a cleavage and McLafferty
CH3CH2CH2CH2CH2CH2CO+ resulting from a cleavage.
CH3CO+ resulting from a cleavage.
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.

28. Carboxylic Acids
Characteristic fragmentation patterns are:
-cleavage to give the ion [CO2H]+ of m/z 45.
McLafferty rearrangement if g H present.
Loss of water, especially in CI
Loss of 44 is the loss of CO2

29. MS of butanoic acid
Figure Mass spectrum of butanoic acid. Common fragmentation patterns of carboxylic acids are -cleavage to give the ion [COOH]_ of m/z 45 and McLafferty rearrangement.

30. Fatty Acids

31. Esters
-cleavage and McLafferty rearrangement:

32. MS of methyl butanoate
Figure Mass spectrum of methyl butanoate. Characteristic fragmentation patterns of esters are -cleavage and McLafferty rearrangement.

33. Ether Fragmentation a-cleavage
C-O cleavage (requires stable cation to lose)
Ion rearrangement
Ms of 2-Chloroethylphenyl ether 94
100 C l H
107 80 O 94 60
Intensity 40
156
107 20 20 40 60 80
100
120
140
160

34. Amines
The most characteristic fragmentation pattern of 1°, 2°, and 3° aliphatic amines is -cleavage

35. Cyclic amines will lose adjacent H•, form iminium ion
In CI, NH+ can eliminate adjacent alkene, reprotonate

36. Halide Fragmentation
Loss of halogen atom
Elimination of HX
a-cleavage

38. •+
Chlorobenzene •+ + 77
M+2 = 114
M+ = 112

40. Sulfur Compounds
Fortunately there is an [M+2]+ of 4% for the natural abundance of 34S. This is diagnostic for S vs 2x16O
Aliphatic thiols can split out H2S, [M-34]
a-cleavage at carbon bearing the sulfur in thiols, thioethers, similar to ethers, etc.

41. Nitroaromatics m/z = 93 Loss of •N=O Loss of CO Aromatic! m/z=65
(this can form from lots of different origins)
Loss of •N=O N + O• O
Loss of CO
Aromatic!
m/z=65
Good test for aryloxy

42. Clusters of Ions m/z Spaced by unit mass
Each peak is for the same molecular formula
Different peaks because there are some molecules with 13C, 2H etc.
Especially significant for Cl, Br
The Nominal mass is m/z of the lowest member of the cluster. This is the isotopomer that has all the C’s as 12C, all protons as 1H, all N’s as 14N, etc.
m/z

43. Analyzing Ion Clusters: a way to rule candidate structures
Mass spectrometry “sees” all the isotopomers as distinct ions
An ion with all 12C is one mass unit different from an ion with one 13C and the rest 12C
Since the isotope distribution in nature is known for all the elements (13C is 1.1%), the anticipated range and ratios of ions for a given formula can be predicted and calculated

44. H H H 79 79 78 93.4% 6.5% 0.1% all H are 1H and all C are 12C
Isotopic Clusters H H H 79 79 78
93.4%
6.5%
0.1%
all H are 1H and all C are 12C
one C is 13C
one H is 2H 10

45. visible in peaks for molecular ion
Isotopic Clusters in Chlorobenzene
35Cl
37Cl
visible in peaks for molecular ion
Relative intensity
100 80 60 40 20
112
114
m/z 11

46. + H Isotopic Clusters in Chlorobenzene
no m/z 77, 79 pair; therefore ion responsible for m/z 77 peak does not contain Cl
Relative intensity
100 80 60 40 20 77
m/z 11

47. Isotopic Abundance
81Br

48. Molecules with Heteroatoms
Isotopes: present in their usual abundance.
Hydrocarbons contain 1.1% C-13, so there will be a small M+1 peak.
If Br is present, M+2 is equal to M+.
If Cl is present, M+2 is one-third of M+.
If iodine is present, peak at 127, large gap.
If N is present, M+ will be an odd number.
If S is present, M+2 will be 4% of M+.

49. Molecular Formula as a Clue to Structure1

50. Molecular Weights
One of the first pieces of information we try to obtain when determining a molecular structure is the molecular formula.
We can gain some information about molecular formula from the molecular weight.
Mass spectrometry makes it relatively easy to determine molecular weights. 6

51. Mass spectrometry can measure exact masses.
Exact Molecular Weights
CH3CO O
CH3(CH2)5CH3
Heptane
Cyclopropyl acetate
Molecular formula
C7H16
C5H8O2
Molecular weight
100
100
Exact mass
Mass spectrometry can measure exact masses.
Therefore, mass spectrometry can be used to distinguish between molecular formulas. 6

52. The “Nitrogen Rule”
Molecules containing atoms limited to C,H,O,N,S,X,P of even-numbered molecular weight contain either NO nitrogen or an even number of N.
This is true as well for radicals as well.
Not true for pre-charged, e.g. quats, (rule inverts) or radical cations.
In the case of Chemical Ionization, where [M+H]+ is observed, need to subtract 1, then apply nitrogen rule.
Example, if we know a compound is free of nitrogen and gives an ion at m/z=201, then that peak cannot be the molecular ion.

53. The Nitrogen Rule NH2 93 138 NH2 O2N 183 NH2 O2N NO2
A molecule with an odd number of nitrogens has an odd molecular weight.
A molecule that contains only C, H, and O or which has an even number of nitrogens has an even molecular weight.
NH2 93
138
NH2
O2N
183
NH2
O2N
NO2 6

54. Index of Hydrogen Deficiency Degree of Unsaturation
Relates molecular formulas to multiple bonds and rings
For a molecular formula, CcHhNnOoXx, the degree of unsaturation can be calculated by:
index of hydrogen deficiency = ½ (2c h - x + n) 6

55. index of hydrogen deficiency =
Molecular Formulas
Knowing that the molecular formula of a substance is C7H16 tells us immediately that is an alkane because it corresponds to CnH2n+2
C7H14 lacks two hydrogens of an alkane, therefore contains either a ring or a double bond
Index of Hydrogen Deficiency
index of hydrogen deficiency = 1
(molecular formula of alkane – molecular formula of compound) 2 6

56. Index of hydrogen deficiency
Example 1
C7H14
Index of hydrogen deficiency 1 2 =
(molecular formula of alkane – molecular formula of compound) 1 2 =
(C7H16 – C7H14) 1 2 =
(2) = 1
Therefore, one ring or one double bond. 6

57. Example 2
C7H12 1 2 =
(C7H16 – C7H12) 1 2 =
(4) = 2
Therefore, two rings, one triple bond, two double bonds, or one double bond + one ring. 6

58. Index of hydrogen deficiency =
Oxygen has no effect
CH3(CH2)5CH2OH (1-heptanol, C7H16O) has same number of H atoms as heptane
Index of hydrogen deficiency = 1
(C7H16 – C7H16O)
= 0 2
No rings or double bonds 6

59. Index of hydrogen deficiency =
CH3CO O
Cyclopropyl acetate
Index of hydrogen deficiency = 1
(C5H12 – C5H8O2)
= 2 2
One ring plus one double bond 6

60. Treat a halogen as if it were hydrogen.
If halogen is present
Treat a halogen as if it were hydrogen. H Cl
C3H5Cl C
same index of hydrogen deficiency as for C3H6 H
CH3 6

61. Rings versus Multiple Bonds
Index of hydrogen deficiency tells us the sum of rings plus multiple bonds.
Using catalytic hydrogenation, the number of multiple bonds can be determined. 6

62. Interpretation of Mass Spectra
Select a candidate peak for the molecular ion (M+)
Examine spectrum for peak clusters of characteristic isotopic patterns
Test (M+) peak candidate by searching for other peaks correspond to reasonable losses
Look for characteristic low-mass fragment ions
Compare spectrum to reference spectra

64. Isotopic Abundances

65. Formula Matching Basics
Atomic weights are not integers (except 12C)
14N = amu; 11B = amu; 1H = amu
16O = amu; 19F = amu; 56Fe = amu
Sum of the mass defects depends on composition
Hydrogen increases mass defect, oxygen decreases it
Accurate mass measurements narrow down the possible formulae for a particular molecular weight
301 entries (150 formulae) in NIST’02 with nominal MW = 321
4 compounds within Da (5 ppm) of
Mass spectrum and user info complete the picture
Isotope distributions indicate/eliminate elements (e.g. Cl, Br, Cu)
User-supplied info eliminates others (e.g. no F, Co, I in reaction)
Isomers are not distinguished in this analysis