Chem 3211

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Chem 3211
Mass Spec Lecture 1
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Mass Spectrometer
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Ions can be separated using a magnetic field.
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Ions can also be separated using a quadrapole.
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Ion traps are often used with gas chromatography
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Time-of-flight is also an option.
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Typical Mass Spectrum

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Molecular weight considerations M
Molecules containing only C, H, O have even
number molecular weights.
C5H10O MW 86
Nitrogen has an even number atomic weight and
forms an odd number of bonds so molecules
containing an odd number of N atoms will have an
odd numbered molecular weight.
C4H9N MW 71
C4H10N2 MW 86
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The Rule of 13
A CH contributes 13 units to the molecular mass
of a molecule. We can use this to identify
possible molecular formulas.
Suppose that we have an M at 110 amu.
110/13 8 with a remainder of 6. This might be
C8H14.
Consider the possibility of an oxygen. Then
(110 16)/13 7 with a remainder of 5. This
gives C7H12O.
What if we had two nitrogen atoms. Then
(110 28)/13 6 with a remainder of 4. This
gives C6H10N2 .
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It is important to understand that the masses
used to calculate molecular weights for mass
spectrometry are not the average atomic masses
found on the periodic table.
The mass spectrometer looks at individual
molecules (or ions) so the mass registered for a
particular molecule depends on the specific
isotopes present.
This means that a particular molecular species
will produce signals at different m/e values
depending on the particular isotopes it contains.
This is the origin of the M 1 and M 2 peaks.
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We can Use the Ratio of the M/M1 peaks to
distinguish between alternative structures with
the same molecular weight
Lets calculate the anticipated size of the M 1
peak for our three previous molecules compared to
an M peak of 100.
C8H14 C7H12O C6H10N2
8 x 1.08 8.64
7 x 1.08 7.56
(6 x 1.08) (2 x .38) 7.24
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Problem 4 An unknown hydrocarbon has a
molecular ion peak at m/e 84, with a relative
intensity of 31.3. The M 1 peak has a relative
intensity of 2.06, and the M 2 peak has a
relative intensity of 0.08. What is the
molecular formula of this compound?
2.08 x (100/31.3) 6.65 (M 1 relative to M
100)
0.08 x (100/31.3) 0.26 (M 2 relative to M
100)
6.65/1.08 6.16 (6 carbons)
82 (6 x 12) 10 (10 hydrogens)
Molecular formula is C6H10
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The M2 peak can be useful to determine the
molecular formula

• Ratio MM2 3510

• C3H7ClO

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What about Bromine?
This is the MS for CH3-O-CH2CH2Br
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See Page 403 for expected patterns when the
compound has multiple halogen atoms
M 2
M 4
M 2
M
M 2
relative intensity
relative intensity
relative intensity
M
M
M 6
m/e
m/e
m/e
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High Resolution Mass Spectrometry

• The molecular formula of a compound can often be
  determined using a High Resolution Mass Spectrum
  that gives the molecular weight of the molecular
  ion peak to a very high degree of precision.
• Use right hand column (precise nuclide mass!) for
  these calculations
• Try Problem 1 page 448.

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Problem 1 A low resolution mass spectrum of the
alkaloid, vobtusine, showed the molecular weight
to be 718. This molecular weight is correct for
the molecular formulas C43H50N4O6 and C42H46N4O7.
A high resolution mass spectrum provided a
molecular weight of 718.3743. Which of the
possible molecular formulas is the correct one
for vobtusine?
43(12.0000) 50(1.00783) 4(14.0031)
6(15.9949) 718.3733
42(12.0000) 46(1.00783) 4(14.0031)
7(15.9949) 718.3369
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Problem 1 A low resolution mass spectrum of the
alkaloid, vobtusine, showed the molecular weight
to be 718. This molecular weight is correct for
the molecular formulas C43H50N4O6 and C42H46N4O7.
A high resolution mass spectrum provided a
molecular weight of 718.3743. Which of the
possible molecular formulas is the correct one
for vobtusine?
43(12.0000) 50(1.00783) 4(14.0031)
6(15.9949) 718.3733
42(12.0000) 46(1.00783) 4(14.0031)
7(15.9949) 718.3369
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The mass spectrum of an unknown liquid shows a
molecular ion peak at m/e 78, with a relative
intensity of 23.6. The relative intensities of
the isotopic peaks are as follows.
m/e 79 (0.79) 80 (7.55) 81
(0.25)
3.35 32.0 1.06
These are the relative intensities based on the
M as 100.
The M 2 peak (80) tells us this contains Cl.
The M 1 peak (79) tells us this contains 3
carbons.
78 35 3(12) 7 There are 7 hydrogen
atoms
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Making use of the molecular formula Index of
hydrogen deficiency
Problem 6 (page 11) Determine the index of
hydrogen deficiency for each of the following.
A. C8H7NO
(2 x 8) 2 1 19 H if no double bonds or rings
19 7 12 Hs deficiency
12/2 6 rings and/or double bonds
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Index of hydrogen deficiency
C9H8O4
2 (9) 2 20
(20 8) / 2 6
(16 14) / 2 1
C7H11Cl3
2 (7) 2 16
2 (8) 2 1 19
(19 13) / 2 3
C8H13N
C10H12BrNO2
2 (10) 2 1 23
23 13 4