Transcript Organic Spectroscopy

Organic Spectroscopy
Mass Spectrometry
General
The mass spectrum is a plot of ion abundance versus m/e ratio (mass/charge ratio).
The most abundant ion formed in the ionization chamber gives to rise the tallest
peak in the mass spectrum, called the base peak.
By using one of the many ionization methods, the simple removal of an electron
from a molecule yields a positively charged radical cation, known as the
molecular ion and symbolized as [M]+.
After formation of molecular Ion in the ionization chamber, excess energy causes
further fragmentation of the molecular ion. Various positively charged masses
(and/or positively charged radical cations) show up in the spectrum.
The mass of the molecular ion can be determined to an accuracy of ± 0.0001 of a
mass unit, which yields a high resolution (hi-res) mass spectrum.
Ionization Methods
The most common method of ionization involves Electron impact (EI) in which
molecule is bombarded with high-energy electrons. This is the strongest ionization
method which usually causes further fragmentation. Since the molecular ion gets
fragmented, the molecular ion usually produces a small peak with this method.
Mass Spectrum of Isobutyrophenone
105
(base peak)
C6H5CO+
O
molecular
weight = 148
77
C6H5+
molecular ion, M
(148)
M+1
Facts Concerning the Molecular Ion Peak
1. The peak must correspond to the ion of highest mass excluding the
usually much smaller isotopic peaks that occur at M+1, M+2, etc.
2. To be a molecular ion, the ion must contain an odd number of electrons.
One electron is lost, forming a radical-cation.
To determine this, calculate the IHD. It must be a whole number.
Consider an ion at m/z = 112. A possible molecular formula is C6H8O2.
The IHD = 3 (a whole number), so this could be the molecular ion.
However, an ion at 105 could correspond to a molecular formula of
C7H5O. The IHD is 5.5 (not a whole number), so this can’t be the
molecular ion.
3. The ion must be capable of producing smaller, fragment ions by loss
of neutral fragments of predictable structure.
The Lifetimes of Various Molecular Ions
Aromatics > conjugated alkenes > alicyclic compounds> organic sulfides >
unbranched hydrocarbons > mercaptans > ketones > amines > esters >
ethers > carboxylic acids > branched hydrocarbons > alcohols
Therefore, alcohols produce small or non-existent molecular ions because
their lifetimes are too short. They fragment before they can be detected.
CH3CH2CH2CH2OH
MW = 74
No M
visible
The Nitrogen Rule
The nitrogen rule states that an odd number of nitrogen atoms will form
a molecular ion with an odd mass number. An even number of nitrogen
atoms (or none at all) will produce a molecular ion with an even mass
number. This occurs because nitrogen has an odd-numbered valence.
Examples:
C6H5CH2NH2 MW = 107
H2NCH2CH2NH2 MW = 60
Determining Possible Molecular Formulas
from the Molecular Ion: Rule of 13
• Rule of Thirteen: Based upon the assumption that CnHn
and its mass of 13 is present in most organic compounds.
• Divide the molecular ion by 13. This gives a value for n
and any remainder (R) = additional H’s.
• For a M+ = 106, n = 8(106/13) with a R of 2. A possible
molecular formula for this ion is C8H8+2 = C8H10
• For each CH there are heteroatom equivalents.
Rule of 13 Heteroatom Equivalents
Element
CH
Equivalent
Element
CH
Equivalent
1H
12
C
31P
C2H7
16O
CH4
32S
C2H8
14N
CH2
16O32S
C4
16O14N
C2H6
35Cl
C2H11
19F
CH7
79Br
C6H7
28Si
C2H4
127I
C10H7
Candidate molecular formulas for M+. = 108
108/13 = 8, R = 4  C8H12
O = CH4; therefore, C8H12 – CH4 + O = C7H8O
or C8H12 – 2(CH4) + 2O = C6H4O2
or C8H12 – C6H7 + Br = C2H5Br
Calculate three candidate molecular formulas for C10H18.
When an odd amu M+. is seen, suspect one nitrogen or an odd
multiple. Candidate molecular formulas for a M+. = 121 are:
121/13 = 9, R = 4  C9H13; IHD = 3.5 so it can’t be M+.
O = CH4; N = CH2
or C9H13 – CH2 + N = C8H11N; IHD = 3, so it may be M+.
or C9H13 – 2(CH2) + 2N = C7H9N2; IHD = 4.5 so it can’t be M+.
or C9H13 – 3(CH2) + 3N = C6H7N3; IHD = 5, so it may be M+.
or C9H13 – (CH2) – (CH4) + N + O = C7H7NO IHD = 5,
so it may be M+.
Determining the Molecular Formula from the
Molecular Ion: Isotope Ratio Data
In this method, the relative intensities of the peaks due to the molecular ion
and related isotopic peaks are examined.
Advantage:
• Does not require an expensive high-res MS instrument.
Disadvantages:
• Isotopic peaks may be difficult to locate.
• Useless when the molecular ion peak is very weak or does not appear.
3-pentanone
Relative Abundances of Common Elements and Their isotopes
Isotope
Relative
Abundance
Isotope
Relative
Abundance
Carbon
12C
100
13C
1.08
Hydrogen
1H
100
2H
0.016
Nitrogen
14N
100
15N
0.38
Oxygen
16O
100
17O
Sulfur
32S
100
33S
Chlorine
35Cl
Bromine
79Br
Element
Isotope
Relative
Abundance
0.04
18O
0.20
0.78
34S
4.40
100
37Cl
32.5
100
81Br
98.0
Example: 3-pentanone, C5H10O
%(M + 1) = 100 (M + 1)/M = 1.08 x # C atoms + 0.016 x # H atoms + 0.04 x # O atoms
= 1.08 x 5 + 0.016 x 10 + 0.04 x 1= 5.60
Actual spectrum: [1% (M +1)/17.4% (M)] x 100 = 5.75
Relative Intensities of Isotope Peaks for Bromine and Chlorine
Halogen
Br
Br2
Br3
Cl
Cl2
Cl3
BrCl
Br2Cl
Cl2Br
M
100
100
100
100
100
100
100
100
100
M+2
97.7
195.0
293.0
32.6
65.3
97.8
130.0
228.0
163.0
M+4
M+6
95.4
286.0
93.4
10.6
31.9
31.9
159.0
74.4
3.47
31.2
10.4
Br
156
MW = 156
158
236
Br
Br
MW = 234
234
238
Cl
112
MW = 112
114
Cl
192
Br
MW = 190
190
194
Determining the Molecular Formula from
the Molecular Ion: High Resolution MS (HRMS)
Using low resolution (LR) MS, you could not distinguish between the following
molecular formulas, each of which has a mass of 60:
C3H8O = (3 x 12) + (8 x 1) + 16 = 60
C2H8N2 = (2 x 12) + (8 x 1) + (2 x 14) = 60
C2H4O2 = (2 x 12) + (4 x 1) + (2 x 16) = 60
CH4N2O = 12 + (4 x 1) + (2 x 14) + 16 = 60
However, they can be distinguished using HRMS.
Precise Masses of Some Common Elements
Element
Hydrogen
Carbon
Nitrogen
Oxygen
Atomic Weight
Isotope
1.0097
1H
1.00783
2H
2.01410
12C
12.0000
13C
13.00336
14N
14.0031
15N
15.0001
16O
15.9949
17O
16.9991
18O
17.9992
12.01115
14.0067
15.9994
Mass
Fluorine
18.9984
19F
18.9984
Silicon
28.086
28Si
27.9769
29Si
28.9765
30Si
29.9738
30.9738
Phosphorus
30.974
31P
Sulfur
32.064
32S
31.9721
33S
32.9715
34S
33.9679
35Cl
34.9689
37Cl
36.9659
79Br
78.9183
81Br
80.9163
127I
126.9045
Chlorine
Bromine
Iodine
35.453
79.909
126.904
Using precise masses:
C3H8O = 60.05754
C2H8N2 = 60.06884
C2H4O2 = 60.02112
CH4N2O = 60.03242
Fragmentation Patterns
Most common: one-bond cleavage to produce an odd-electron neutral fragment,
(radical, which is not detected) and an even-electron carbocation. Ease of fragmentation to form cations follows the scheme below:
CH3+ < RCH2+ < R2CH+ < R3C+ < CH2=CH-CH2+ < C6H5-CH2+
Difficult
Easy
Radical (not detected)
Fragmentation Patterns (cont.)
Two-bond cleavage: The odd-electron molecular ion produces an odd-electron
fragment ion and an even-electron neutral fragment (not detected).
Not detected
McLafferty Rearrangement
O
H
R
+
+
O H
R
+
H3C
O
H3C
O
71
CH3
Cleavage at branch points
H3C
C
CH2CH3
MW = 86
CH3
57
57
71
-cleavage to hetero atoms
87
H3C
43
O
CH3
MW = 102
CH3
CH3
cleavage  to hetero atoms
CH3-CH=OH+
43
87
91
-cleavage to aromatic ring
MW = 134
91
92 (from McLafferty
Rearrangement)
CH2
43
Cleavage  to carbonyl groups
O
H3C
43
MW = 86
CH3
60
O
McLafferty rearrangement
carboxylic acids
OH
60
OH
CH2
C OH
74
McLafferty rearrangement
esters
O
O
CH3
MW = 102
OH
CH2
C OCH3
74
Hexane
C6H14
MW = 86.18
Molecular ion peaks are present, possibly with low intensity.
The fragmentation pattern contains clusters of peaks 14 mass
units apart (which represent loss of (CH2)nCH3).
3-Pentanol
C5H12O
MW = 88.15
An alcohol's molecular ion is small or non-existent.
Cleavage of the C-C bond next to the oxygen usually
occurs. A loss of H2O may occur as in the spectrum below.
3-Phenyl-2-propenal
C9H8O
MW = 132.16
Cleavage of bonds next to the carboxyl group
results in the loss of hydrogen (molecular ion less 1)
or the loss of CHO (molecular ion less 29).
3-Methylbutyramide
C5H11NO
MW = 101.15
Primary amides show a base peak
due to the McLafferty rearrangement.
n-Butylamine
C4H11N
MW = 73.13
Molecular ion peak is an odd number.
Alpha-cleavage dominates aliphatic amines.
n-Methylbenzylamine
C8H11N
MW = 121.18
Another example is a secondary amine shown below.
Again, the molecular ion peak is an odd number.
The base peak is from the C-C cleavage adjacent to the C-N bond.
Naphthalene
C10H8
MW = 128.17
Molecular ion peaks are strong due to the stable structure.
2-Butenoic acid
C4H6O2
MW = 86.09
In short chain acids, peaks due to the loss of OH
(molecular ion less 17) and COOH (molecular ion less 45)
are prominent due to cleavage of bonds next to C=O.
Fragments appear due to bond cleavage next to C=O
(alkoxy group loss, -OR) and hydrogen rearrangements.
Ethyl acetate
C4H8O2
MW = 88.11
Ethyl methyl ether
C3H8O
MW = 60.10
Fragmentation tends to occur alpha to the
oxygen atom (C-C bond next to the oxygen).
1-Bromopropane
C3H7Br
MW = 123.00
The presence of chlorine or bromine atoms is
usually recognizable from isotopic peaks.
4-Heptanone
C7H14O
MW = 114.19
Major fragmentation peaks result from cleavage
of the C-C bonds adjacent to the carbonyl.
Which structure supports the following mass spectrum?
MW = 100
MW = 100
MW = 100
MW = 114
MW = 86
Which structure supports the following mass spectrum?
NH2
NH2
NH
MW = 101
MW = 87
MW = 87
NH2
NH2
MW = 87
MW = 73
Which structure supports the following mass spectrum?
O
O
O
O
O
MW = 116
O
MW = 130
O
MW = 116
O
O
MW = 116
O
MW = 102
An unknown compound has the mass spectrum shown below. The IR spectrum
shows peaks in the 3100-3030 and the 2979-2879 cm-1 ranges and a strong
absorption at 1688 cm-1. Suggest a structure consistent with this data.
An unknown compound has the mass spectrum shown below. The IR
spectrum shows peaks in the 2963-2861 cm-1 range and a strong
absorption at 1718 cm-1. Suggest a structure consistent with this data.