ESPECTROMETRIA DE MASAS
TEMA 4
4.1. Introduccíon
La espectrometría de masas (EM) es una técnica de análisis basada sobre la separación de acuerdo a sus razones masa/carga de las especies cargadas formadas a partir de la ionización de una muestra.
Se trata de una técnica extremadamente sensible, de gran versatilidad y cuyos campos de aplicación experimentan un crecimiento vertiginoso en nuestros días.
La EM suministra información muy valiosa sobre los compuestos químicos: la masa molecular, la fórmula global y, a partir del patrón de fragmentaciones, la estructura molecular
Los fundamentos de la espectrometría de masas pueden ejemplifican mediante la técnica de ionización electrónica.
Electrones acelerados a través de un campo eléctrico adquieren energía cinética considerable y pueden interactuar con moléculas (M) para originar especies cargadas:
Principios fisicoquimicos
M + e- →M.+ + 2e-
En este caso M.+ es un catión-radical molecular. Normalmente M .+ se forma en un estado excitado y puede sufrir fragmentación, que puede ser de diferentes formas:
Masa molecular en baja y alta resolutuion
Se pueden distinguir unidades de masa Ejemplo 28 D. de CO2, N2, CH2N, C2H6
Indice de insaturación
CxHyNzOn Index = x – ½ y + ½ z +1
4.2. Instrumentación
4.3. Espectro de masas
Ion Molecular
Molecular Formula Determination
The molecular ion gives the nominal molecular weight of the substance. To determine the chemical formula (or a series of possible chemical formulas), one can use the rule of 13.The Rule of Thirteen is so named because, to generate a base formula (containing only carbon and hydrogen); the molar mass of the substance is divided by 13. The numerator from division gives the number of carbon atoms in the base formula. The remainder, when added to the numerator, gives the number of hydrogens in the base formula
Molecular formula: RULE of 13Consider CH unit 13 amu
If we divide the mass by 13 we can establish easily possible formula:
Example: Molecular ion => 152
152 / 13 = # carbons = 11
Mass = 12 * 11 = 132 therefore H = 152 – 132 = 20
Basic formula C11 H20
If Oxygen is present : mass = 16 remove CH4
If Nitrogen is present : mass = 14 remove CH2
If one oxygen : C11 H20 – CH4 + O = C10 H16 O
If second oxygen : C10 H16 O – CH4 + O = C9 H12 O2
If third oxygen : C9 H12 O2 – CH4 + O = C8 H8 O3
Molecular formula: RULE of 13
After establishing the basic formula with only Carbon/hydrogen,Other element can be introduced by substracting the proper hydrocarbonvalue
16O => CH4
14N => CH2
19F => CH71H12 => C
28Si => C2H4
31P => C2H7
32S => C2H8
35Cl => C2H11
79Br => C6H7
127I => C10H7
Table 1 shows the carbon/hydrogen equivalents for some of the more common elements.
Accurate Mass
Modern high-resolution mass spectrometers are capable of determining very precise and accurate molecular weights for substances. Whereas typical chemical methods (vapor density, cryoscopy, vapor pressure osmometry, and neutralization equivalent) are accurate to within 0.1 to 1% (two or three significant figures), certain mass spectrometers have an accuracy of 0.0005% or better.
High resolution:
With sufficient accuracy, unique molecular formula can be determine
e.g. Distinguish CO, N2, CH2N and C2H4 (all having m/z 28)
CO 12C 12.000016O 15.9949
27.9949
N214N 14.003114N 14.0031
28.0062
CH2N12C 12.00001H x2 2.0156 14N 14.0031
28.0187
C2H412C x2 24.00001H x4 4.0312 28.0312
Exact Mass can provide Molecular Formula
Exact Mass can provide Molecular FormulaConsider the following 2 formulas which have m/z 287:
C15H10NO3Cl C14H8N2O3Cl
Mass Defect of C15H10NO3Cl:(10 x .0078) + (.0031) + (3 x -.0051 ) + (-.0311 ) = .0347
Mass Defect of C14H8N2O3Cl:(8 x .0078) + (2 x .0031) + (3 x -.0051 ) + (-.0311 ) = .0202
With sufficiently high resolution MS, it is possible to propose a unique empirical formula for an ion.
Returning to the example of benzamide, the accurate masses for the three most likely formulas ascertained from the molecular ion are: 121.08923 (C8H11N), 121.06411 (C6H7N3), and 121.05281 (C7H7NO). Selection of the correct formula is straightforward if the accurate mass is determined by high resolution mass spectrometry.
Bromine and Chlorine
Calculating M+1 and M+2
Isotope peaksUsually [M+2] peak is very small
Except for:
Sulfur : 32S : 100 34S : 4.4
Chlorine : 35Cl : 100 37Cl : 32.5
Bromine : 79Br : 100 81Br : 98
Silicon : 28Si: 100 29Si: 5.2 30Si : 3.35
M M+1 M+2
Isotope peaks : CH3Br
[12CH379Br]+
[12CH381Br]+
[12CH279Br]+
[12CH281Br]+
[13CH379Br]+
M - H
[13CH381Br]+
Bromine
2 intense peaks for the molecular ion, spaced by 2 daltons.79Br and 81Br
79Br
79Br 81Br
81Br
Chlorine
35Cl (P= .75)37Cl (P= .25) Ratio 3:1
Probability : [M+] / [M + 2] = 0.75 / 0.25 = 3 / 1
100 : 33
Calculation of isotope pattern: 2 Cl
1 Cl
2 Cl
C2H235Cl2
C2H235Cl 37Cl
C2H2 37Cl2
P (2 35Cl) = (0.75 )2 = 0.563
P (35Cl 37Cl) + P (37Cl 35Cl ) = (0.75 ) (0.25 ) + (0.25 ) (0.75 ) = 0.375
P (2 37Cl) = (0.25 )2 = 0.063
[M+] / [M + 2] / [M + 4] = 100 / 66 / 11
C2H237Cl 35Cl
Isotopic abundances for Carbon containing compounds
M+ => p(M) = p(n 12C) = (0.989)n
[M+1]+ => p(M+1) = p(n -1) 12C + 1 13C = n p[n-1] 12C x p(13C)
The probability of finding 1 13C among n carbons is:
= n(0.989)n-1 (0.011)
Relative ratio: [M+1]+ / [M+ ] = n(0.989)n-1 (0.011) / (0.989)n
= n (0.011) / (0.989)
= n (0.0111)
In percentage: n x 1.1 %
Calculating Peak Intensities from Isotopic abundances
Metastable ions
Some ions have so short lifetimes that they dissociate while moving through the spectrometer.
•An ion of mass m1 is accelerated after initial ionization •But a different ion m2 (daughter ion) passes the magnetic analyzer•The resulting peak is neither m1 or m2 but appear at m* (metastable)
m* = (m2)2/m1
These metastable ions are formed during 10-5 s (time spent between electrostatic and magnetic analyzer) are quite broad but provide direct information about ion reactions
Nitrogen rule
Molecular ion => 100
100 / 13 = # carbons = 7
Mass = 12 * 7 = 84 therefore H = 16Basic formula C7 H16
1 Oxygen: C7 H16 - CH4 + O
C6 H12 O
Molecular ion => 99 odd!
Basic formula C7 H16
1 Nitrogen: C7 H16 – CH2 + N
C6 H14 N
OH NH2
Nitrogen rule
Molecular ion => 26
26 / 13 = # carbons = 2
Mass = 12 * 2 = 24 therefore H = 2Basic formula C2 H2
C C HH
Molecular ion => 27 odd!
27 / 13 = # carbons = 2
Basic formula C2 H3
One Nitrogen C2 H3 - CH2 + N
C NH
4.3. Fragmentación
C
O
NH2
105
77 44
- NH2C6H5C=O +
m/z 105
- COC6H5
+
Benzamide: EIB+
M+
m/z 77
3-methyl-6-i-PropylCyclohex-2-ene-1-one
MW = 152
Isotopic cluster M, M+1, M+2
Molecular formula M+ (molecular ion) m/z
M mass of the most abundant isotope
e.g C7H7NO : 7 x 12C = 847 x 1H = 71 x 14N = 141 x 16O = 16
m/z 121
Isotope peaks: (% respect to M)
M+1 due to (13C) or 15N (17O) or 2H are negligeable)
%(M+1) = (1.1 * #C) + (.38 * #N) = 8.08 %
%(M+2) = (1.1 * #C)2/200 + (.2 * #O) = 0.5%
For C, H, N,O, F, P composition
Neutral losses and Ion series
M-1 H
M-15 CH3
M-16 O (rare) , NH2
M-17 OH , NH3 (rare)
M-18 H2O
M-19 F
M-20 HF (very rare)
M-26 HCCH , CN
M-27 HCN
M-28 H2C=CH2 , CO
M-29 CH3CH2 , HCO
M-30 NO (Nitro compounds), H2 CO (anisoles)
M-31 CH3O
M-32 CH3OH
M-35 Cl
M-36 HCl
M-42 CH2=C=O, CH2=CH-CH3
M-43 CH3CO , C3H7
M-44 CO2
M-45 CH3CH2O , CO2H