Chapter 15Mass Spectrometry

Mass Spectrometry

A beam of electrons causes molecules to ionize and fragment.

C C C C + e

C C C C

ionization

fragmentation

radical cation

radical carbocation

Diagram of a Mass Spectrometer

A beam of electrons causes molecules to ionize and fragment. The mixture of ions is accelerated and passes through a magnetic field, where the paths of lighter ions are bent more than those of heavier atoms. By varying the magnetic field, the spectrometer plots the abundance of ions of each mass.

Diagram of a Mass Spectrometer

The exact radius of curvature of an ion's path depends on its mass-to-charge-ration, symbolized by m/z. In this expression, m is the mass of the ion (in amu) and z is its charge. The vast majority of ions have a +1 charge, so we consider their path to be curved by an amount that depends only on their mass.

Diagram of a Mass Spectrometer

r = mvzB

A Typical Mass Spectrum

Peaks of interest arise from cations with +1 charge.

The heaviest peak with a reasonable abundance is usually the molecular ion.

The m/z value of the molecular ion is the same as the molar mass of the sample compound.

The base peak is the peak caused by the most stable or kinetically accessible cation in the ion stream.

It typically is significantly more abundant than the molecular ion.

Mass SpectrometryAnalyzing the (M) Peak+

Mass Spectrometry-Analyzing the (M) PeakThe Nitrogen Rule

+

An odd molecular weight indicates an odd number of nitrogens in a compound.

CH4 NH3 H2O CH3OH 16 15 18 32

NH2NH2 NH2C(O)NH2

32 60

Mass SpectrometryAnalyzing the (M) Peak

Determining a Reasonable Formula+

A molecule containing only C,H, and O shows a molecular ion at 86 amu. Propose a reasonable molecular formula for the compound.

86 amu -----> 7(12) + 2(1) -------> C7H2

C=12 H=1 O=16

86 amu -----> 6(12) + 14(1) -------> C6H14

86 amu -----> 5(12) + 10(1) +1(16) -------> C5H10O86 amu -----> 4(12) + 6(1) +2(16) -------> C4H6O2

Low-abundance peaks heavier than the molecular ion arise due to the presence of isotopes of various atoms in many of the molecules being analyzed.

A Typical Mass Spectrum

Mass SpectrometryAnalyzing the (M+x) Peaks+

Low-abundance peaks heavier than the molecular ion arise due to the presence of isotopes.

Most heavier elements do not consist of a single isotope but contain heavier isotopes in varying amounts.

The heavier isotopes give rise to small peaks at higher mass numbers than the M+ molecular ion peak.

The height of the M+, M+1, and M+2 peaks will depend on the isotopic composition of the element.

Mass SpectrometryAnalyzing the (M+1) Peak+

Isotope Relative mass (in amu)

Abundance in nature (%)

Isotope Relative mass (in amu)

Abundance in nature (%)

1H 1.0078 99.99 16O 15.9949 99.76

2H 2.0141 0.01 17O 16.9991 0.04

3H 3.0161 <0.01 18O 17.992 0.20

12C 12.0000 98.93 35Cl 34.9689 75.78

13C 13.0034 1.07 37Cl 36.9659 24.22

14C 14.0032 <0.01 79Br 78.9183 50.69

14N 14.0031 99.63 81Br 80.9163 49.31

15N 15.0001 0.37

Relative Atomic Mass and Abundance of Several Elements

1.1% of all carbons are C-13.For a molecule containing 10 carbon atoms there is a

11% chance of a C-13 atombeing present.

The relative percentage of theM+1 peak compared to the

M peak can be used to calculate the # of carbonatoms in the molecule.

Relative Abundance of M+1 Peak for Various # of Carbons

100carbons

50carbons

40carbons

20carbons

10carbons

5carbons

A molecule containing only C,H, and O shows a molecular ion at 86 amu with a relative abundance of 20.1% and a (M+1) peak with a relative abundance of 1.1%. Propose a reasonable molecular formula for the compound.

86 amu -----> 7(12) + 2(1) -------> C7H2

C=12 H=1 O=16

86 amu -----> 6(12) + 14(1) -------> C6H14

86 amu -----> 5(12) + 10(1) +1(16) -------> C5H10O86 amu -----> 4(12) + 6(1) +2(16) -------> C4H6O2

The M+1 peak represents 1.1/20.1= 5.5% of the M peak.5.5%/1.1% = maximum of 5 carbons atoms

Mass SpectrometryAnalyzing the (M+2) Peak+

Isotope Relative mass (in amu)

Abundance in nature (%)

Isotope Relative mass (in amu)

Abundance in nature (%)

1H 1.0078 99.99 16O 15.9949 99.76

2H 2.0141 0.01 17O 16.9991 0.04

3H 3.0161 <0.01 18O 17.992 0.20

12C 12.0000 98.93 35Cl 34.9689 75.78

13C 13.0034 1.07 37Cl 36.9659 24.22

14C 14.0032 <0.01 79Br 78.9183 50.69

14N 14.0031 99.63 81Br 80.9163 49.31

15N 15.0001 0.37

Relative Atomic Mass and Abundance of Several Elements

Chlorine is a mixture of 75.5% 35Cl and 24.5% 37Cl.

The molecular ion peak M+ that contains 35Cl is 3 times more abundant than the M+2 peak that contains 37Cl.

Isotopic Effect of ChlorineMass Spectrum of 2-Chloropropane

Isotopic Effect of BromineMass Spectrum of 1-Bromopropane

Notice that the M+ and the M+2 peaks are about the same height.

Bromine is a mixture of 50.5% 79Br and 49.5% 81Br.

The molecular ion peak M+ that contains 79Br is about the same size as the M+2 peak that contains 81Br.

Isotopic Effect of BromineMass Spectrum of 1-Bromopropane

Relative intensities of ions expected for various combinations of Bromine and Chlorine.

Br Br2 Br3 Cl Cl2 Cl3 BrCl Br2Cl BrCl2

M+ 100 51 34 100 100 100 78 45 62

M + 2 94 100 100 31 65 95 100 100 100

M + 4 47 97 10 31 24 74 45

M + 6 31 3 14 6

Mass SpectrometryAnalyzing the Fragments

Common Fragmentation Patterns

C C C C

C C YY

CR C YRY

C H

C YC

C

H

Y

Carbon-Carbon Cleavage

Carbon-Hetero Cleavage

Alpha Bond Cleavage

Elimination

Mass Spectrum of Decane

What do 14 amu correspond to?What do 29 amu correspond to?What do 43 amu correspond to?

Mass Spectrum of 2-Methylpentane

Mass Spectrum of 2-chloropropane

Mass Spectrum of 1-bromopropane

Mass Spectrum of 2-pentanone

Mass Spectrum of 3-pentanone

86

57

29

The peak at 102 is M. The peak at m/z = 45 is due to loss of n-butyl, The peak at 84 is due to loss of water.The peak at 87 is due to loss of methyl.

Mass Spectrum of 2-hexanol

Chapter 15High Resolution Mass Spectrometry

A molecule shows a molecular ion at 60 amu. Propose a reasonable molecular formula for the compound.

A high-resolution mass spectrum of the molecule gives anexact mass of 60.0578. Which is the correct formula ?

60.0575

60.0211

60.0688

60.0323

60 amu -----> 4(12) + 12(1) -------> C4H12

C=12 H=1 O=16 N=14

60 amu -----> 3(12) + 8(1) + 1(16) -------> C3H8O

60 amu -----> 2(12) + 4(1) + 2(16) -------> C2H4O2

60 amu -----> 2(12) + 8(1) + 2(14) -------> C2H8N2

60 amu -----> 1(12) + 4(1) +1(16) + 2(14) -------> CH4N2O