Embed Size (px)
Citation preview
Interpretation of Mass Spectra Part 4
ObjectivesTo describe the main features of EI, CI, ESI spectra of organic compoundsTo indicate how to identify the M+., MH+ or [M+nH]n+ ion and to suggest a possible molecular formulaTo review the major types of fragmentation mechanisms and how to recognise them in the mass spectra of different classes of compoundsTo give some practice in the interpretation of mass spectra of simple unknown organic samples
•
Relative Molecular Mass and Isotopes
Relative Molecular MassMany non-radioactive elements exist in more than one isotopic form. The lightest isotopes of the common light elements are usually the most abundant e.g. 1H, 12C, 14N, 16O, 32S, 35Cl, 79Br The masses of these isotopes, rounded to the nearest whole number, are used to calculate the nominal RMM of a compound. The monoisotopic RMM of a compound is that calculated using the accurate atomic weights of the most abundant isotopes of the constituent elements.
•
Relative Molecular MassAs the number of H atoms increases, the difference between the nominal and the mono-isotopic RMM slowly rises and the nominal RMM is no longer useful, e.g. above about 600 Da.
For example:
Mass: C20H30 C40H60 C60H90
Nominal 270 540 810 Monoisotopic 270.235 540.470 810.705
Whole Number 270 540 811
Common isotopes of the lighter elements
Isotope Mass Relative
Abundance
Isotope Mass Relative
Abundance
1H 1.00782 100 2H 2.01410 0.016
12C 12.000 100 13C 13.0033 1.12
14N 14.0031 100 15N 15.0001 0.36
16O 15.9949 100 18O 17.9992 0.2
32S 31.9721 100 33S 32.9715 0.78
34S 33.9679 4.39
35Cl 34.9689 100 37Cl 36.9659 32.7
79Br 78.9183 100 81Br 80.9163 97.5
Relative Molecular MassesNaturally occurring compounds contain isotopes in their natural abundances. The inclusion of the less common isotopes leads to a higher average RMM.Hence we can define three different RMMs:
Nominal RMMMonoisotopic RMMAverage RMM
Which do we measure by mass spectrometry?
Measurement of RMMFor RMMs of < 500, normally the nominal RMM is obtained at low resolving power
The monoisotopic RMM can be measured if the resolving power is high enough to resolve isotope peaks and any other interfering peaks.
If the resolving power is insufficient to resolve isotope peaks, the average RMM is measured. If interfering peaks are not resolved, an accurate RMM cannot be measured.
Isotope Peaks - 1Ions containing elements that have more than one isotope exhibit isotope peaks on the high mass side of the peak due to the lightest ion and may indicate the elements present in an ion.
The probability that any one carbon atom is a 13C isotope is 1.1%
If there are n carbon atoms in an ion, the probability that at least one carbon atom is a 13C isotope is n x 1.1% so that I[(M+1)+]/I[(M+)] = (n x 1.1)/100
Isotope Peaks - 2For relatively low RMM samples, this ratio indicates the number of carbon atoms in the ion providing that there is no contribution from MH+ ions, e.g. for alcohols and amines at high sample pressures.For example, if M+. is at m/z 112, this could be due to C8H16
+., C7H12O+. or C6H8O2+. for which the ratio
I[(M+1)+]/I[(M+)] would ideally be 0.09, 0.08 and 0.07 respectively. This ratio exceeds unity once the number of C atoms in the molecule exceeds 85-90.
Molecular Ion Regions
C80H160 C80H160S3 C100H200
1121 1217 1402
Calculation of Isotope Patterns
Exemplified for Cl and Br atoms in C6H4OClBr
Write out isotopic abundances of Cl and Br isotopes (including zero abundances) across the top and down the side of a square.
Construct a matrix by multiplying each member of the row with each member of the column.
Sum the diagonals (top right to bottom left) to produce relative abundances.
Isotopic Peaks due to Cl + Br
3 0 1 35Cl, 36Cl, 37Cl abundances 1 | 3 0 1
0 | 0 0 0 79Br, 80Br, 81Br abundances 1 1 | 3 0 1 (vertical)
Relative Abundance. 3 : 0 : 4 : 0 : 1
RMM 114 116 118
Comp. 35Cl79Br 35Cl81Br 37Cl81Br 37Cl79Br
1-Bromo-4-Chlorobenzene Mass Spectrum
The 3:4:1 pattern in M+ region (with 13C peaks between). The fragment ion at m/z 111, 113 shows the 3:1 pattern indicating the presence of a Cl atom and that the Br atom has been lost.
(m a in lib ) Be n ze n e , 1-b ro m o -4-c h lo ro -10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200
0
50
100
12 2438
44
50
55 61
75
85 96104
111
128 141 155
192
Br
C l
(m a in lib ) Be n ze n e , 1-b ro m o -4-c h lo ro -10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200
0
50
100
12 2438
44
50
55 61
75
85 96 104
111
128 141 155
192
Br
C l
(m a in lib ) Be n ze n e , 1-b ro m o -2-c h lo ro -20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200
0
50
100
2537
50
55 61
75
8591
96104
111
128 140 155
192
C l
Br
The Molecular Ion RegionFor compounds containing only C, H, O and N atoms, 13C isotopes control the isotope pattern. In large molecules, such as proteins, few ions contain less than 2 or 3 heavy isotopes; the peak due to ions containing only light isotopes is of very low intensity.
The presence of S, Cl and Br atoms leads to characteristic peaks at M+2, M+4, etc.
Electron Ionization Spectra
Production of EI Spectra70 eV electrons bombard the sample vapour and deposit typically 0 - 20 eV in a molecule.Typically 8 - 10 eV is needed to produce M+. in its ground state so M+. ions are produced with internal energies of approximately 0 - 10 eVThese undergo one or more fragmentations in the ion source and products ejected after about 1 ms are collected to produce the mass spectrum. Discrimination effects differ between instruments so that the spectrum of a given sample varies somewhat from instrument to instrument.
Initial Inspection of the Spectrum - 1
Look at the overall appearance of the spectrum: try to identify the molecular ion, M+. and obtain information from any isotope peaks present.
If the major peaks are at low m/z and M+. is under 20% of the most intense peaks, the sample is probably aliphatic.
The more intense M+. is, the greater the degree of unsaturation is present (alkene, carbonyl compound).
Hexane
(m a in lib ) He xa n e0 10 20 30 40 50 60 70 80 90 100
0
50
100
2 1215
27
28
29
30
39
41
42
43
55
57
58 7184
86
M+.
[M-C2H5]+
2-Hexene
(m a in lib ) 2-He xe n e10 20 30 40 50 60 70 80 90 100
0
50
100
15 26
27
28
29
30 38
39
40
41
42
43
44 5153
55
56
57 61 63 65 67
69
70 74 77 79 81 83
84
85
M+.
[M-C2H4]+.
[M-CH3]+
Cyclohexane
(m a in lib ) C yc lo h e xa n e10 20 30 40 50 60 70 80 90 100
0
50
100
1315 26
27
28
30 38
39
40
41
42
43
44 5154
55
56
5761 63 65 67
69
70 74 77 79 8183
84
85
M+.
[M-CH3]+
[M-C2H4]+.
Cyclohexene
(m a in lib ) C yc lo h e xe n e10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90
0
50
100
14 1626
27
29 3238
39
40
41
42 49
51 53
54
55 6365
67
6874
77 7980
81
82
83
M+.
[M-CH3]+
[M-C2H4]+.
1,3-Cyclohexadiene
(m a in lib ) 1,3-C yc lo h e xa d ie n e10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90
0
50
100
15
27
2838
39
40 49
5152
54 63 65 74 76
77
78
79
80
81
M+.
Benzene
(m a in lib ) Be n ze n e10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90
0
50
100
15 26 28
39
4049
51
53 6163 74 76
77
78
79
M+.
1,5-hexadien-3-yne
(m a in lib ) 1,5-He xa d ie n -3-yn e10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90
0
50
100
12 14 2427
29 31 3438
39
40 42 44 4749
50
51
52
53 55 6163
64 67 6973
7476
77
78
79
M+.
[M-C2H2]+.
Initial Inspection of the Spectrum - 2
If peaks due to M+. and other high mass ions dominate the spectrum, the sample is probably aromatic.
A large number of peaks often indicates a large number of H atoms are present.
The lack of any dominant peaks suggests the absence of a hetero-atom.
The simpler the spectrum, the more symmetry is likely to be present in the sample molecule.
1-Napthalenol
(m a in lib ) 1-Na p h th a le n o l10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160
0
50
100
18 26 39 43 51 58 63 72 8993 97 101
115
125
144
O H
M+.
[M-CO]+.
(m a in lib ) 2-Na p h th a le n o l10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160
0
50
100
15 2739
4351 57
6374 79 87
89
92 97 102
115
126
144
HO
(m a in lib ) 1-Na p h th a le n o l10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160
0
50
100
18 2639
43 5058 63 72
7789
97 102
115
125
144
O H
Nonanal
(m a in lib ) No n a n a l10 20 30 40 50 60 70 80 90 100 110 120 130 140 150
0
50
100
15 19
27
29
31
41
57
60
70
79
82
8591
95
98
114124
141
O
M+.
[M-C2H5]+
C6H13+
[M-C5H10]+
[M-C6H13]+
C3H7+
C2H5+
M+.[M-C3H6]+.
[C4H9CO]+
[M-C3H6-C3H6]+.
M+.
[M-C2H5O]+
C5H7+
CH2NH2+
Base peak of primary amines
Found in all amine spectra and in spectra of amides
M+.
Identifying the Molecular Ion - 1
The common isotopes of elements C, O, S have even relative atomic masses and even valencies whereas the common isotopes of H, F, Cl, Br, P and Na have odd relative atomic masses and odd valencies. Hence organic compounds that contain only these elements (i.e. no nitrogen atoms) have an even relative molecular mass (RMM) so that M+. occurs at an even value of m/z or the MH+ ion appears at an odd value of m/z .
Identifying the Molecular Ion - 2
The one common element that is an exception to this rule is nitrogen, the most common isotope of which has a relative atomic mass of 14 and a valency of 3.Hence an odd relative molecular mass results when the molecule contains an odd number of N atoms. Thus if M+. has an odd m/z, it suggests a possible amide, amine, nitrile or N-heterocyclic compound.
Uses of Isotope PeaksCommon elements that give M+2 isotope peaks:
35Cl:37Cl rel. ab. ~ 3 : 1 79Br:81Br rel. ab. ~ 1 : 1 32S:34S rel. ab. ~ 100 : 4 28Si:30Si rel. ab. ~ 100 : 3.4
Hence peaks at M+2, M+4, etc. indicate the presence of Cl, Br, S, Si; the absence of these peaks indicates the absence of these elements.Common elements that give rise to M+1 isotope peaks are C and N but only C isotope peaks need be considered:
12C:13C rel. ab. ~ 100 : 1.1
So that I([M+1]+)/I([M+]) = n x 1.1/100 for an ion containing n C atoms.
Approximate ratio
27:27:9:1
Approximate ratio
9:6:1
Approximate
ratio 3:1
81:108:54:12:1
27:27:9:1
9:6:1
1:4:6:4:1
1:3:3:1
C12H22S2+.
m/z 230 231 232 Intensities 100 13 9
[M-(C6H10)]+
m/z 148 149 150 Intensities 100 7 9
Is it Really the Molecular Ion?
Check that the supposed M+. loses neutral species that are sensible, e.g. radicals such as alkyl radicals or OH. or molecules such as alkenes, CO, HCl, H2O, etc.
If there are losses that cannot be explained, e.g. 3 - 13, 21 - 25 Da, the assignment should be re-examined.
If it appears that M+. loses 3 Da, this could arise from losses of CH3 and H2O giving peaks due to the ions [M-15]+ and [M-18]+..
Is it Really the Molecular Ion?
Try to identify the main species lost by M+.. These often indicate the type of compound to which the sample belongs.
Rearrangement ions formed by loss of a molecule are often particularly informative. If no nitrogen is present, these appear at an even value of m/z.
Identify ions characteristic of a compound type: m/z 105, 77, 51 for benzoyl compounds, m/z 91, 65, 39 for alkylbenzenes, m/z 30 for amines, etc.
M+.
absent
[M-CH3]+
[M-H2O]+.
[M-C3H7]+
[M-C4H9]+
M+.
C3H7CO+
C2H5CO+
[M-CH3]+ M+.
[M-C3H6]+.
[CH3CO]+
M+.
C7H7+
C5H5+
C3H3+
How to Work Out the Molecular Formula - 1
Start with the RMM and the value of I([M+1])/I([M]) which gives an indication of the number of C atoms present.
Suppose the RMM is 136. The maximum possible number of C atoms is found by dividing this by 12. This gives 11 but C11H4 is very unlikely to be correct. Try 10 C atoms, converting the other 12 Da to H atoms, giving C10H16 (e.g. pinene or limonene, etc).
How to Work Out the Molecular Formula - 2
Repeating the process for 9 C atoms gives C9H28 which is unacceptable. Convert 16 H atoms to one O atom giving C9H12O as a possibility, e.g. benzyl ethyl ether. Repeating this for 8 C atoms gives C8H8O2 which could be an aromatic acid or ester.Knowing the number of C atoms present, one can suggest a molecular formula. In this case, suppose C8H8O2 is suggested.
Rings + Double Bonds: 1For a molecule of molecular formula CxHyNzOa the number of double bond equivalents (DBEs) is given by the expression
DBEs = x - (y/2) + (z/2) + 1
A DBE is a unit of unsaturation, e.g. an alkene or carbonyl group is 1 DBE, a saturated ring is 1 DBE, an aromatic ring is 4 DBEs, a triple bond is 2 DBEs. The number of DBEs is independent of the number of O atoms present. Halogen atoms are counted as H atoms, S and P atoms are counted as O and N atoms respectively.
Rings + Double Bonds: 2If the formula C8H8O2 is suggested, the number of RDBs is
8 - (8/2) + (0/2) + 1 = 5.
This immediately suggests the presence of an aromatic ring (4 RDBs) and in view of the presence of two O atoms, the other is almost certainly a carbonyl group.
A formula such as C9H12O would have
9 - (12/2) + (0/2) + 1 = 4
This suggests, for example, an aromatic ring (4 RDBs) and a saturated substituent.
Suggesting Possible Structures
C6H5COOCH3 A benzoyl compound so m/z 105, 77 and 51 should be prominent features.
C6H5CH2COOH A benzyl compound so m/z 91, 65 and 39 should be prominent features.
CH3C6H4COOH The o-, m- and p-isomers all give a prominent peak at m/z 119 (loss of OH), least intense for the m-isomer. The o-isomer gives an m/z 118 ion (loss of H2O formed by the OH of the -COOH group and H from the methyl group. Comparison of the spectra with those of authentic samples would confirm the identification.
M+.
[M-OH]+
[M-COOH]+
M+.
[M-H2O]+.
C7H7+
[M-H2O-CO]+.
[M-OH]+
Fragmentation of M+.
Ions
Unknown mass Spectrum
Odd m/z suggests a N atom present
m/z 105, 77, 51 suggests benzoyl
Since N accounts for 14 of the other 16 Da, 2 H atoms are present
Hence the sample is benzamide, C6H5CONH2.
Common Neutral Losses of Diagnostic Value - 1
15 CH3 Alkyl branching if intense peak, otherwise neglect
16 O Nitroaromatic, oxime, sulfoxide
16 NH2 RCONH2
18 H2O Alcohol, (ketone, aldehyde, less common)
20 HF Alkyl fluoride
26 C2H2 Aromatic hydrocarbon
27 HCN ArCN, N-heterocylic compounds, ArNH2 rarely
27 C2H3 Ethyl ester (low abundance)
28 CO Quinones, some phenols
28 C2H4 n-Propyl ketones, ethyl esters, ArOC2H5
29 C2H5 Ethyl ketones, Ar - n-C3H7 compounds
30 CH2O Aromatic methyl esters
31,32 CH3O,CH3OH Methyl esters of carboxylic acids
33,34 SH, H2S RSH
Common Neutral Losses of Diagnostic Value - 2
41 C3H5 Propyl ester
42 C3H6 n-butyl ketone
CH2CO RCOCH3, ArOCOCH3, ArNHCOCH3
43 C3H7 RCOC3H7, Ar-n-C4H9 compounds
44 CO2 Anhydrides, esters
45 COOH RCOOH
OC2H5 Ethyl esters of carboxylic acids
46 NO2 Aromatic nitrocompounds
48 SO Aromatic sulfoxide
55 C4H7 Butyl ester of carboxylic acid
56 C4H8 RCOC5H11, ArOC4H9, Ar-C5H11 (n- or i-)
57 C4H9 RCOC4H9
C2H5CO RCOC2H5
60 CH3COOH Acetate
M+.
[M-O]+.
[M-NO]+
[M-NO2]+
“McLafferty+1” peak given by ethyl esters
[M-C2H3]+
[M-C6H12]+. [CH2C(OH)OC2H5]+.
Characteristic of ethyl ester of long-chain carboxylic acid
M+.
Common Characteristic Ions
m/z 105 + 77 + 51 Benzoyl compounds
m/z 91 + 65 + 39 Alkyl benzenes, benzyl compounds
m/z 30 Base peak RNH2 otherwise other amines
m/z 44, 58, 72, . . . Amines, amides
m/z 31 Primary alcohol; low intensity, other alcohols, ethers
m/z 31, 45, 59, . . . Ethers
m/z 74 Methyl esters of carboxylic acids
m/z 60 Straight chain carboxylic acids
m/z 77 or 76 Mono- or di-substituted benzene (low intensity)
[CH2NH2]+
Suggests amine or amide
[CH3CO]+
[C2H4NH2]+ [M-C3H6]+.
[M-CH3]+ M+.
[M-C8H16]+. [CH2C(OH)OCH3]+.
Characteristic of methyl ester of a long-chain carboxylic acid
M+.
CH3(CH2)8CO.OCH3
[M-OCH3]+
Characteristic H-rearrangement ions of straight chain methyl esters
143
The Odd-Even Electron Rule - 1
For molecules that do not contain an odd number of N atoms, M+. is an even mass, odd electron ion. If it loses a molecule in a rearrangment process, the resulting fragment ion is again an even mass, odd electron ion. If it loses a radical, which is an odd mass, odd electron species, this produces an odd mass, even electron fragment ion.
The Odd-Even Electron Rule - 2
Once a radical has been lost to produce an even electron, closed shell ion, further fragmentations can occur only by the loss of molecules to produce further odd mass, even electron ions.Successive loss of two radicals NEVER occurs.Do not assume that an ion is always formed from the next highest mass fragment ion. Ions may fragment by several routes so that adjacent peaks may not belong to ions of the same fragmentation sequence.
(CH3)2CH-C6H4COOH
All fragment ions are odd mass, even electron ions M+.
Charge Localisation - 1Although the charge on a molecular ion may be delocalised, it is useful to consider it formally as localised. Where on the molecular ion is the charge located?Which is the easiest (lowest energy) electron to remove?These are usually
(a) lone pair electrons on heteroatoms
(b) -electrons in unsaturated systems
Charge Localisation - 2If there is a choice of electrons that could be removed, the formal charge may be placed on one of several atoms.
Hence, formally, one can think of M+. ions as consisting of a mixture of ions with the formal charge being on one of several possible sites.
Each type of molecular ion can give rise to a different type of fragmentation and the spectrum observed will be the weighted sum of the products of these.
Examples of Charge Localisation
Carbonyl compounds are assumed to lose a lone pair electron from the carbonyl oxygen
Ionized toluene is assumed to have lost a ring
-electron
Charge Localisation at Several Sites
NH2+.
O
R
NH2
O+.
R
NH2
O
R
+.
Here there are three possible sites for charge localisation. Fragmentation may be rationalised in terms of the decomposition of three different types of molecular ion
[M-CH3]+ -cleavage
[M-CH3CO]+
inductive cleavage M+.
Factors Influencing Ion Abundance - 1
Eint required for decomposition: in general, low energy processes will predominate but different ionization methods yield different internal energy distributions and hence different mass spectra from a particular sample.
Stability of the product ion
Factors Influencing Ion Abundance - 2
Stability of the neutral productDelocalization of electron e.g. in allyl radical
Placing of electron on electronegative atom e.g. .OH
Loss of small stable molecule containing multiple bonds, e.g. CO, C2H2, HCN
Stevenson’s Rule
AB+. A+ + B. or A. + B+
Preference for formation of ion from fragment having lower IE (except largest R. is lost preferentially)
Radical Site Initiation -Cleavage) - 1
This is particularly important for ions that contain N or O atoms. Electron pairing occurs by the transfer of an odd electron from the bond alpha to the atom carrying the charge and the transfer of the odd lone pair electron to form a new bond. The remaining electron from the alpha bond is lost on the radical that is eliminated as a result of the bond breaking.
Radical Site Initiation -Cleavage) - 2
In this type of fragmentation, the charge site does not move but the radical site moves as a result of the alpha bond breaking. In the example below, C - X is the alpha bond broken. An example is the fragmentation of carbonyl compounds:
R - C - X RCO+ + X.
|| O+.
(X = H, R, OH, OR, Cl, NH2) aldehydes, ketones, acids, esters, acid halides, amides
Mechanisms of Alpha Cleavage
X = H, R, OH, OR, Cl, NH2 for aldehydes, ketones, acids, esters, acid halides and amides.
m/z 30 is the base peak in primary amine spectra
Primary alcohols give the m/z 31, ethers often give m/z 45, 59 . . by this reaction
The ease of loss of alkyl groups is R1>R2>R3 where this is also the order of decreasing size.
M+.
[M-H]+ C6H5CO+
C6H5+
C4H3+
[M-R.]+ -cleavage, 57+, 71+
[R]+ inductive cleavage 29+, 43+
M+.
[M-16]+.
characteristic of amides
M+.
M+. absent
[M-H2O]+.
[M-H2O-C3H6]+.[CH2OH]+
M+.
[C4H9OCH2]+
3 examples of [CH2OR]+ ions at m/z 31, 59 and 87
[C2H5OCH2]+
[CH2OH]+
[CH2NH2]+ base peak of primary amines
M+.
Loss of the Largest Alkyl Radical -1
Where there is a choice of alkyl radicals that can be lost in an alpha-cleavage, the largest radical is lost preferentially followed by the next largest, etc.
Loss of the Largest Alkyl Radical - 2
The loss of H or an alkyl radical gives rise to the following series of ions for different classes of compoundsAldehydes and ketones m/z 29, 43, 57, 71 . . . Aldehydes usually give a fairly prominent m/z 29 peak (CHO+) and a weaker [M - 1]+ peak.
Methyl ketones often give m/z 43 (CH3CO+) as the base peak.
Loss of the Largest Alkyl Radical - 3Alcohols and ethers m/z 31, 45, 59, 73 . . . Primary alcohols typically give a m/z 31 peak (CH2OH+), often of fairly low intensity.Methyl ethers give m/z 45 peaks, again often of low intensity (charge induced fragmentation usually predominates).Amines m/z 30, 44, 58, 72 . . . n-alkyl amines often give m/z 30 as the base peak whereas secondary and tertiary amines often give m/z 44 and 58 as base peaks together with m/z 30
Loss of the Largest Alkyl Radical – 4
(amines)
[A] C3H7 loss predominates to give an intense m/z 30 peak; little H loss evident, m/z 72 negligible; M+. m/z 73 (10%)
[B] CH3 loss is preferred to H loss giving m/z 58 as the base peak; some H loss giving m/z 72 (20%); M+. m/z 73 (30%)
[C] CH3 loss is only possible alpha-cleavage so that m/z 58 is again the base peak but there is no peak at either m/z 72 (H loss) or 73, M+.
C3H7HC
H
NH2 H3CHC
H
HN
HC CH3
H
H3C C
CH3
CH3
NH2
A B C
[CH2NHC2H5]+
M+.
-C2H4
M+.
[M-CH3]+
[M-C3H7]+
[M-C5H11]+
M+. absent
CH3 loss
C2H5 loss
C3H7 loss
Allylic Cleavage
This is a major type of fragmentation for alkenes leading to alkyl radical loss. Unfortunately, migration of the double bond occurs before fragmentation so that the observation of ions of this type is of little structural value. The mass spectra of many alkenes, especially polyenes, tend to be independent of the position of the double bond so that isomers cannot be distinguished.The m/z 41 ion is the most common ion observed in the mass spectra of aliphatic compounds, together with homologues of m/z 55, 69, 83 . . .
Allylic ions at m/z 41, 55, 69, 83
M+.
Charge Site Initiation (Inductive Cleavage)
[C2H5OCH2]+
M+.
[HOCH2]+
C3H7+
C2H5+
M+.
[CH2SC3H7]+
[CH2SC2H5]+
[C3H7]+
[C2H3]+
Rearrangement Reactions
McLafferty Rearrangement - 1
McLafferty Rearrangement - 2
Deuterium labelling studies show that a -H atom is transferred quite specifically through a six-membered transition state. If there are no -H atoms, the rearrangement does not occur.
Note that substituents on the -carbon atom are retained in the ion but substituents on the - and -carbon atoms are lost as part of the alkene - useful in locating site of branching in an alkyl chain.
No alkyl chain 3 or more C atoms long so no McLafferty Rearrangement leading to alkene loss
M+.
[C2H5CO]+
[C2H5]+
C2H4 loss from C3H7 alkyl chain
CH3 loss
M+.
[C3H7]+, [CH3CO]+
Loss of C4H8 from C5H11 alkyl chain (McLafferty Rearrangement)
[M-OCH3]+
M+.
[M-C5H11]+
Loss of C4H8 from
C5H11 alkyl chain
[M-NH2]+
M+.
McLafferty Rearrangement - 3
If an even mass fragment ion is found which could be formed from the molecular ion by the loss of 28, 42, 56, 70, . . . Da., always suspect that it results from a McLafferty rearrangement or of a related process.The driving force for the rearrangement is the formation of a strong bond between the H atom and the unsaturated heteroatom carrying the charge. Similar reactions occur with H-transfer to other heteroatoms.
McLafferty Rearrangement - 4
Peaks due to alkene losses at m/z 112, 98, 84, 70, 56, 42
M+.
M+.
[M-C3H6]+.
M+.
[M-C2H4]+.
H-Atom Rearrangement to a Saturated Heteroatom -
1
M+.
[M-HCl]+
H-Atom Rearrangement to a Saturated Heteroatom - 2In these reactions, an unpaired electron on the heteroatom is donated to form a new bond with a H atom with cleavage of the original bond to that H atom.
A second radical site reaction leading to H2O+. formation is not favoured since this is an energetic process that does not lead to particularly stable products.
H-Atom Rearrangement to a Saturated Heteroatom - 3
Instead, a charge site reaction occurs leading to the loss of H2O (very common for primary alcohols, much less so for secondary and tertiary alcohols because of α-cleavage competition) followed by a further charge site reaction in which C2H4 is lost.
In general, the loss of a small neutral species is more energetically favorable than formation of a small ionic species.
Fragmentation of Aromatic Molecular Ions
Many simple aromatic molecular ions fragment by elimination of a small, unsaturated molecule by breaking the aromatic ring but giving a further, stable cyclic ion as a product.
Examples of small molecules lost include: -Benzene, C2H2, Pyridine, HCN, Thiophene, HCS, Furan, HCO, Phenols, CO, Anilines, HCN
M+.
[M-C2H2]+.
M+.
[M-HCN]+.
M+.
[M-HCO]+
M+.[M-C2H2]+.
[M-C3H3]+
[M-CHS]+
H-Atom Rearrangement in Phenols and Anilines
Phenols expel CO to give an [M-28]+. peak often accompanied by an [M-29]+ peak due to the loss of a H atom.Replacing the O atom by NH to give aniline, the loss of HCN to give an [M-27]+. ion can be similarly rationalized.Deuterium labeling shows considerable H/D scrambling indicates that the mechanisms are more complicated than indicated above.
•
Phenol
Aniline
M+.
M+.
[M-CO]+.
[M-HCN]+.
H-Atom Rearrangement to a Saturated Heteroatom - 4
THE ORTHO EFFECT requires a labile H atom (OH, NH2, CHO) and formation of a stable ionic product by charge migration, eliminating a small saturated molecule e.g. H2O, CH3OH, HCl. These products are more energetically favourable than those given by charge retention reactions.
2-Hydroxybenzaldehyde
(m a in lib ) Be n za ld e h yd e , 2-h yd ro xy-10 20 30 40 50 60 70 80 90 100 110 120 130
0
50
100
13 1729
39
42 4750 53
63
65
71 74
76
87
93
104
122
O H
O
M+.[M-H]+
[M-H2O]+.
[M-CO]+.
[M-CO-H2O]+.
2-Hydroxybenzoic acid ethyl ester
(m a in lib ) Be n zo ic a c id , 2-h yd ro xy-, e th yl e ste r20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180
0
50
100
27
39
4353
65
76 81
92
104 109
120
138
166
O H
O
O
Loss of C2H5OHLoss of CO from
m/z 120
M+.
Even-Electron Ions, CI,
Even-Electron Ions - 1Under EI conditions, M+. ions are formed and a major fragmentation process is the loss of a radical, R., producing an even-electron ion.
Once a radical has been lost, all subsequent fragmentations involve the loss of a molecule to form further even-electron ions.
Under CI conditions, an even-electron ion, such as MH+, is formed; subsequent fragmentations involve the loss of a molecule to form further even-electron ions.
Sites of ProtonationIn order to rationalize the fragmentation of MH+ ions, one must consider at which sites in the sample molecule the proton is attached. The spectrum may then be rationalized in terms of the fragmentation of the different types of MH+ ions.In general, protonation occurs on heteroatoms having lone pairs of electrons, such as O, N and Cl. This frequently followed by charge-induced elimination of a molecule containing the hetero-atom. Other possible protonation sites are aromatic rings and regions of unsaturation.
Even Electron Ions
Ephedrine ionized by methane CI may protonate on the O atom of the OH group:
Protonation on the N atom leads to the loss of CH3NH2 by a similar mechanism, yielding an ion of m/z 135. Both m/z 148 and 135 are observed in the CI spectrum, indicating the presence of OH and HNCH3 groups in the molecule.
•
Ephedrine EI and CI Spectra
Ephedrine, RMM 165, gives an EI spectrum dominated by the m/z 58 fragment ion and no M+. ion. Methane CI gives an MH+ ion at m/z 166 and fragments at m/z 148, 135 and 58 due to protonation on the OH and NHCH3 groups or on the aromatic ring respectively
Field’s Rule - 1In the decomposition of even electron ions , if more than one neutral can be eliminated, the lower the proton affinity of the neutral, the greater the tendency to leave.Hence under CI conditions, a protonated chloroalkane has a high tendency to lose HCl (PA 5.8 eV), a protonated alcohol has a moderate tendency to lose H2O (PA 7.2 eV) and a protonated amine has a low tendency to lose NH3 (PA 9.0 eV), consistent with the [M-18]+ and [M-31]+ relative abundances in the CI mass spectrum of ephedrine.
Field’s Rule - 2In the case of unsaturated ions which could lose
unsaturated neutrals, the higher proton affinities of these species often result in other more complex H-rearrangement modes of fragmentation being more favourable. For example,
C2H5O+=CH2 CH2O + C2H5+ (9% CID) (inductive cleavage)
C2H5O+=CH2 C2H4 + HO+=CH2 (58% CID)
(H rearrangement) predominates since the PAs of C2H4 and CH2O are respectively 7.0 and 7.4 eV.
Charge Site Rearrangements - 1
Even-electron ions undergo H-transfer to the charge site; all electrons remain paired (low energy process), the charge site does not move and an unsaturated or cyclic neutral species is lost
Charge Site Rearrangements - 2
Very important in the fragmentation of amines. After an initial α-cleavage, this fragmentation requires the presence of at least one ethyl group or larger so that an alkene may be lost.
[M-CH3]+ -cleavage
M+.
-C2H4
[CH2NH2]+
Violations of Even Electron Rule - 1
Usually R - Y+ undergoes heterolytic cleavage to give R+ + Y, all species being even-electron species.
This is a much lower energy process than homolytic cleavage with charge retention giving R. + Y+. so producing two odd-electron species.
Violations of Even Electron Rule - 2
The most common exceptions are in the spectra of aromatic molecules containing electronegative substituents, Cl, CN, NO2, etc.
In the spectra of 1,3- and 1,4-dinitrobenzenes, the ions C6H4
+. and C6H3+ are of almost equal
abundance (losses of 2 NO2 and NO2 + HNO2 respectively) not only because IE(benzyne) < IE(NO2) but also because other reactions giving even-electron products are not energetically more competitive. Other examples of this behaviour are common.
M+.
[M-NO2]+
-NO
-O
-C2H2
NO+
Practical Problems - 1
Beware of spurious peaks such as the following:Background peaks from previous samples, pump oil or from an air leak, e.g. m/z 40, 32, 28, 18 etc.Peaks arising from incomplete removal of common solvents such as m/z 83, 85, 87 from CHCl3, m/z 58, 43 from acetonePeaks present due to incomplete reaction leaving traces of starting materials in the samplePeaks due to homologues, e.g. at 14 m/z units above or below the true molecular ion peak
Practical Problems - 2
• Compare your spectrum with that of an authentic sample obtained by use of the same ionisation technique but remember that an exact match of relative intensities is unlikely to be found because of varying mass discrimination effects.
General Hints - 1Aromatic or aliphatic? Provisionally identify M+.
Check that proposed neutral losses are sensible
Is N present? Is assignment of M+. incorrect?
Check for isotope peaks for Cl, Br, S, heavy metals
Use I([M+1]+)/I([M]+.) to estimate number of C atoms present
Postulate a molecular formula and estimate the double bond equivalents
General Hints - 2Inspect higher mass ions, possibly formed from M+. in one step, e.g. even mass fragments formed in a rearrangment processLook for characteristic neutral losses such as 16 Da, O from ArNO2 or NH2 from an amide, 30 Da, CH2O from ArOCH3 and characteristic ions, m/z 30, amines, m/z 74 methyl esters, 105/77/51, 91/65/39 for benzoyl and alkyl benzene compounds Do not assume adjacent peaks are due to sequential losses of neutrals; two or more charge sites lead to competing fragmentation routes.
General Hints - 3Do not try to interpret every small peak, especially those at low m/z which result from sequential fragmentation
Never postulate the loss of a radical from an even electron ion without very good reason
Use negative evidence as well as positive evidence: e.g. if there is no peak at m/z 91, the sample is unlikely to be an alkyl benzene.
