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Analytical Chemistry 3
A4: Mass Spectroscopy
Text Reference: Higher Level Chemistry p. 444 - 447
1
IB Assessment Statements
A.4.1
Determine the molecular mass of a compound from
the molecular ion peak.
A.4.2
Analyse fragmentation patterns in a mass spectrum to find
structure of a compound.
Examples of fragments should include:
• (Mr − 15)+ loss of CH3
• (Mr − 17)+ loss of OH
• (Mr − 29)+ loss of C2H5 or CHO
• (Mr − 31)+ loss of CH3O
• (Mr − 45)+ loss of COOH.
2
Recall: Basic Operation of a Mass Spectrometer
high speed
electrons collide
with particles,
removing an
electron and
2. IONISATION
forming + ions
X(g) + e– → X+(g) + 2e–
3. ACCELERATION
charged particles moving
through an electric field speed
up to the same speed
1.VAPORISATION
heating of liquids/
solids to separate
atoms/molecules
vacuum to remove air
or other particles
4. DEFLECTION
charged particles are deflected by a
magnetic field - higher m/Z ratio
particles deflect less
5. DETECTION
detector measures mass and relative
amounts of ions present
http://www.chemguide.co.uk/analysis/masspec/howitworks.html
3
Recall: Determination of Relative Atomic Mass
Calculate the relative atomic mass of zirconium using this mass spectrum for
zirconium.
Most ions produced in the
mass spectrometer have a
charge of +1, so this m/z
ratio is usually also the mass
of each isotope.
five isotopes - assume all ions have a charge of +1 so mass = m/Z
90
Zr
51%
91
Zr
11%
92
Zr
18%
94 Zr
18%
96
Zr
4%
Ar = (90)(0.51) + (91)(0.11) + (92)(0.18) + (94)(0.18) + (96)(0.04)
= 93.23
4
Mass Spectroscopy and Molecular Formulas
Recall that problems on MOLECULAR FORMULAS usually included a “known”
molar mass for the compound under analysis.
This value is obtained from mass spectroscopy experiments.
Example 1: An unknown compound was analyzed and found to be 40.0% carbon, 6.7% hydrogen and
53.3% oxygen by mass. The largest mass on a mass spectrum for this compound corresponded to a
relative molecular mass of 60. Determine the molecular formula and structure for this compound.
1. Find the empirical formula.
Assume 100 g of compound.
1 mol
moles C = 40.0 g x
= 3.33 mol
12.01 g
1 mol
moles H = 6.7 g x
= 6.63 mol
1.01 g
1 mol
moles O = 53.3 g x
= 3.33 mol
16.00 g
∴ lowest mole ratio = 1:2:1
2. Find the molecular formula.
M (actual)
ratio of relative molecular masses =
M (EF)
60
=
(12 + 2 +16)
=2
∴ MF = (EF)2
= (CH2O)2
= C2H4O2
structure: 2 possibilities
∴ EF = CH2O
5
Mass Spectroscopy and Organic Compounds
Mass spectroscopy of organic compounds is based on the same principles.
In addition to providing information about the relative molecular mass of an
unknown, it can also reveal information about its structure.
During ionization, the compound loses an electron to produce a radical cation.
M + e– → •M+ + 2e–
This is called the molecular ion (M+).
The peak (line) with the highest
m/z value is most likely the
molecular ion M+.
100"
rela%ve'abundance'
90"
80"
70"
60"
50"
40"
30"
46
20"
46
10"
0"
0"
C2H5OH
10"
20"
30"
mass'/'charge''(or'm/Z)'
40"
Since the charge is 1+, the m/z
value for the molecular ion is
the relative molecular mass of
the compound.
50"
C2H5OH + e– → •C2H5OH+ + 2e–
6
Mass Spectroscopy and Organic Compounds
The molecular ion M+ will break down in the mass spectrometer
to produce smaller FRAGMENTS.
M+ → X+ + •Y
Charged “X+” fragments (with a + charge) will also be deflected and detected,
producing the other peaks on the mass spectrum.
Uncharged “•Y” fragments (free radicals) will not be detected and will be
removed by the vacuum conditions.
M+ = C2H5OH+
100"
rela%ve'abundance'
90"
(Mr = 46)
fragments of M+
80"
70"
60"
C2H5OH+ → C2H5O+ + •H
50"
40"
30"
46
20"
10"
0"
0"
C2H5OH
10"
20"
30"
mass'/'charge''(or'm/Z)'
40"
50"
(Mr = 45)
C2H5OH+ → C2H5+ + •OH
(Mr = 29)
7
Mass Spectroscopy and Organic Compounds
C2H5OH fragments (major)
Label each fragment and identify its relative molecular mass (Mr)
C2H5OH+
CH3CH2+
(C2H5+)
Mr = 46
Mr = 29
C2H5O+ Mr = 45
CH3+ Mr = 15
CH2OH+ Mr = 31
8
Mass Spectroscopy and Organic Compounds
Fragmentations produce one cation and one uncharged particle.
The fragment that gives the most stable ion forms.
Only the charged species is detected.
Example:
2 possibilities for splitting the C-C bond in the ethanol molecular ion
C2H5OH+ → CH3+ + CH2OH
peak at 15
C2H5OH+ → CH3 + CH2OH+
peak at 31
Example:
Splitting the C-O bond in the ethanol molecular ion
C2H5OH+ → C2H5+ + OH
C2H5OH+ → C2H5 + OH+
peak at 29
unstable ion - does not form
so NO peak at 17
9
Analysing Mass Spectra
1. Find the molecular ion (M+) and find the molecular mass.
2. Try to find the molecular formula of the substance.
3. Look at the mass differences of the
fragments to determine what is lost
from the molecular ion M+.
fragment
loss of
(Mr - 15)+
CH3
(Mr - 17)+
OH
(Mr - 29)+
C2H5 or CHO
(Mr - 31)+
CH3O
(Mr - 45)+
COOH
4. Determine the structural formula.
10
Analysing Mass Spectra
1. molecular peak M+ @ 60 ∴ Mr = 60
120"
B
100"
rela%ve'abundance'
Example 1: A substance with the empirical
formula CH2O has this simplified mass
spectrum. Deduce the molecular formula
and the structure.
80"
A
C
60"
40"
20"
0"
2.
M (actual)
ratio of relative molecular masses =
M (EF)
60
=
(12 + 2 +16)
=2
0"
10"
20"
30"
40"
50"
60"
70"
m/Z'
∴ molecular formula = C2H4O2
3. fragment analysis
A = 45
mass diff = 15 ∴ CH3 lost
COOH+ present
B = 43
mass diff = 17 ∴ OH lost
C2H3O+ present
C = 15
mass diff = 45 ∴ COOH lost
CH3+ present
11
Analysing Mass Spectra
Example 2: These two mass spectra are for two isomers of C3H6O.
Deduce the structure of each one.
Spectrum A
Spectrum B
120"
120"
29
28
80"
58
60"
57
40"
80"
60"
40"
20"
20"
0"
0"
0"
10"
20"
30"
43
100"
rela%ve'abundance'
rela%ve'abundance'
100"
40"
50"
60"
70"
15
0"
10"
58
20"
m/Z'
30"
40"
50"
60"
70"
m/Z'
(analysis for B on next slide)
58
57
= M+ = C3H6O+
mass diff = 15 ∴ H lost
29
mass diff = 29 ∴ C2H5 or CHO lost ∴ fragment = CHO+ or C2H5+
28
mass diff = 1 (from peak 29)
∴ fragment = C3H5O+
∴ fragment = CO+
12
Analysing Mass Spectra
Example 2 continued
Spectrum A
Spectrum B
120"
120"
29
28
80"
43
100"
rela%ve'abundance'
rela%ve'abundance'
100"
58
60"
57
40"
20"
80"
60"
40"
15
20"
0"
58
0"
0"
10"
20"
30"
40"
50"
60"
70"
0"
10"
20"
m/Z'
30"
40"
50"
60"
70"
m/Z'
58
= M+ = C3H6O+
43
mass diff = 15 ∴ CH3 lost
∴ fragment = C2H3O+
15
mass diff = 43 ∴ C2H3O lost
∴ fragment = CH3+
13
Further Analysis of Mass Spectra - Carbocation Stability
Fragmentation can produces different types of “carbocations”
- primary, secondary and tertiary.
primary carbocations
secondary carbocations
tertiary carbocations
+ charge is on a
carbon atom attached
to only one other
carbon atom
+ charge is on a
carbon atom attached
to two other carbon
atoms
+ charge is on a
carbon atom attached
to three other carbon
atoms
C+H2CH3
CH3C+HCH3
CH3C+(CH3)CH3
Order of stability of carbocations:
primary < secondary < tertiary
More stable carbocations produce higher peaks on a mass spectrum.
This can help you distinguish between structural isomers.
14
Further Analysis of Mass Spectra - Carbocation Stability
Example 3: Structural Isomers of C5H12
simplified mass spectrum of pentane
first isomer: pentane
72
= M+ = C5H12+
57
mass diff = 15 ∴ CH3 lost
∴ fragment = C4H9+ = [CH3CH2CH2C+H2]
(less stable primary cation ∴ low peak)
43
mass diff = 29 ∴ C2H5 lost
∴ fragment = C3H7+ = [CH3CH2CH2]+
29
mass diff = 43 ∴ C3H7 lost
∴ fragment = C2H5+ = [CH3CH2]+
15
Further Analysis of Mass Spectra - Carbocation Stability
Example 3: Structural Isomers of C5H12 (continued)
Second Isomer: 2-methylbutane
72
simplified mass spectrum
= M+ = C5H12+
57
mass diff = 15 ∴ CH3 lost
43
mass diff = 29 ∴ C2H5 lost
∴ fragment = C4H9+ = [+CH(CH3)CH2CH3]
(secondary carbocation ∴ more stable and higher
peak than in pentane)
∴ fragment = C3H7+ = [CH3C+H(CH3)]
(secondary carbocation ∴ stable )
29
mass diff = 43 ∴ C3H7 lost
∴ fragment = C2H5+ = [CH3CH2]+
(primary carbocation)
16
Further Analysis of Mass Spectra - Carbocation Stability
1. molecular peak M+ @ 58 ∴ Mr = 58
2.
M (actual)
ratio of relative molecular masses =
M (EF)
58
=
24 + 5
=2
120"
43
100"
rela%ve'abundance'
Example 4: This mass spectrum is for a
compound with an empirical formula of
C2H5. Identify the fragments formed and
deduce the structure.
80"
29
60"
40"
15
58
20"
0"
0"
10"
20"
30"
40"
50"
60"
70"
m/Z'
∴ molecular formula = C4H10
3. fragment analysis
58
43
29
= M+ = C4H10+
mass diff = 15 ∴ CH3 lost
C3H7+ present
mass diff = 29 ∴ C2H5 lost
C2H5+ present
15
mass diff = 43 ∴ C3H7 lost
CH3+ present
But there are two isomers for C4H10. Which one would give this MS?
17
Further Analysis of Mass Spectra - Carbocation Stability
Example 4: continued
120"
butane
rela%ve'abundance'
There are 2 possible isomers for C4H10
43
100"
80"
29
60"
40"
15
58
20"
0"
0"
10"
20"
30"
40"
50"
60"
70"
m/Z'
2-methylpropane
3. fragment analysis
43
29
15
mass diff = 15 ∴ CH3 lost
mass diff = 29 ∴ C2H5 lost
mass diff = 43 ∴ C3H7+ lost
C3H7+ present = [+CH(CH3)CH3]
high peak suggests it is a secondary carbocation
(more stable AND produced more frequently 3 ways to produce it)
C2H5+ present
So the structure is
CH3+ present
18
A Few Fine Details about Mass Spectra
Most mass spectra examples given in this presentation are “simplified”
to show the MAIN peaks for the molecular ion and the fragments.
Actual mass spectra have many more lines with lower or higher peaks.
Consider these two mass spectra for ETHANOL:
ethanol
simplified mass spectrum
100"
31 =
CH2OH+
rela%ve'abundance'
90"
80"
70"
15
26, 27,
28, 29
60"
50"
15 =
CH3+
40"
30"
20"
29 =
C2H5+
45 =
C2H5O+
46 = M+
10"
47
0"
0"
10"
20"
30"
40"
50"
mass'/'charge''(or'm/Z)'
47 = M+ + 1
peaks with a m/Z value ONE higher than an expected fragment contain
13C atoms instead of 12C
26, 27, 28, 29
peaks with a m/Z value lower than an expected fragment may have H
atoms removed
19
A Few Fine Details about Mass Spectra
Compounds containing CHLORINE will also have extra peaks at higher m/z ratios.
- two common isotopes of chlorine - 35Cl and 37Cl
- present in the ratio 3:1 so many molecules contain the heavier 37Cl atoms
- peaks at m/Z ratios TWO greater than the expected values (in a 3:1 ratio)
Example: chloroethane (Mr = 63.45)
Compounds containing CHLORINE will also have extra peaks at higher m/z ratios.
64 = M+ = CH3CH235Cl+
66 = CH3CH237Cl+
49 = (M+- 15) = CH235Cl+
51 = CH237Cl+
29 = (M+- 35) = C2H5+
20