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Chem 3211
Mass Spec Lecture 1
Mass Spectrometer
M + e-M + 2 e-
Ions can be separated using a magnetic field. Ions can also be separated using a quadrapole.
Ion traps are often used with gas chromatography Time-of-flight is also an option.
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Mass Spectrometry: Understanding Ions, Molecular Weights, and Formula Identification - Pro, Study notes of Organic Chemistry
An introduction to mass spectrometry, including the role of mass spectrometers, typical mass spectra, molecular weight considerations, and the use of the rule of 13 for identifying molecular formulas. It also covers the significance of m+1 and m+2 peaks and the use of relative intensities for structure distinction.
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Chem 3211
Mass Spec Lecture 1
Mass Spectrometer
M + e-^ M + 2 e-
Ions can be separated using a magnetic field. Ions can also be separated using a quadrapole.
Ion traps are often used with gas chromatography Time-of-flight is also an option.
Typical Mass Spectrum
M+1+
Molecular Ion Peak, M+
Base Peak
Molecular weight considerations: M+
Molecules containing only C, H, & O have even number molecular weights.
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.
C 5 H 10 O: MW = 86
C 4 H 9 N: MW = 71 C 4 H 10 N 2 : MW = 86
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 C 8 H14. Consider the possibility of an oxygen. Then
(110 – 16)/13 = 7 with a remainder of 5. This gives C 7 H 12 O.
What if we had two nitrogen atoms. Then
(110 – 28)/13 = 6 with a remainder of 4. This gives C 6 H 10 N 2.
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.
Iodine^127 I 100
Bromine^79 Br 100 81 Br 98.
Chlorine^35 Cl 100 37 Cl 32.
Sulfur^32 S 100 33 S 0.78^34 S 4.
Phosphorus^31 P 100
Silicon^28 Si 100 29 Si 5.10^30 Si 3.
Fluorine^19 F 100
Oxygen^16 O 100 (^17) O 0.04^18 O 0.
Nitrogen^14 N 100 (^15) N 0.
Carbon^12 C 100 (^13) C 1.
Hydrogen^1 H 100 (^2) H 0.
ELEMENT RELATIVE ABUNDANCE
We can Use the Ratio of the M/M+1 peaks
to distinguish between alternative structures
with the “same” molecular weight
Let’s calculate the anticipated size of the M + 1 peak for our three previous molecules compared to an M+^ peak of 100.
C 8 H 14
C 7 H 12 O
C 6 H 10 N 2
8 x 1.08 = 8.
7 x 1.08 = 7. (6 x 1.08) + (2 x .38) = 7.
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 C 43 H 50 N 4 O 6 and C 42 H 46 N 4 O 7. 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.
42(12.0000) + 46(1.00783) + 4(14.0031) + 7(15.9949) = = 718.
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 C 43 H 50 N 4 O 6 and C 42 H 46 N 4 O 7. 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.
42(12.0000) + 46(1.00783) + 4(14.0031) + 7(15.9949) = = 718.
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)
multiplying each of these by (^) gives
These are the relative intensities based on the M+^ as 100.
The M + 1 peak (79) tells us this contains 3 carbons.
The M + 2 peak (80) tells us this contains Cl.
78 – 35 – 3(12) = 7 There are 7 hydrogen atoms
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. C 8 H 7 NO (2 x 8) + 2 + 1 = 19 H if no double bonds or rings
19 – 7 = 12 H’s deficiency
12/2 = 6 rings and/or double bonds
Index of hydrogen deficiency:
C 9 H 8 O 4 2 (9) + 2 = 20 (20 – 8) / 2 = 6
C 7 H 11 Cl 3 2 (7) + 2 = 16 (16 – 14) / 2 = 1
C 8 H 13 N 2 (8) + 2 + 1 = 19^ (19 – 13) / 2 = 3
C 10 H 12 BrNO 2 2 (10) + 2 + 1 = 23 23 – 13 = 4