Isotopes in Mass Spectrometry

• Molecular ions of pentane and 2-methylbutane both have m/z values of

72, each spectrum shows a very small peak at m/z = 73 (Figures 2 and

3).

• This peak is called an M + 1 peak because the ion responsible for it is

one unit heavier than the molecular ion.

• The M + 1 peak owes its presence to the fact that there are two naturally

occurring isotopes of carbon: 98.89% of natural carbon is 12C and

1.11% is 13C.

• So 1.11% of the molecular ions contain a instead of a 12C and thus

appear at M + 1 1

• Compound responsible for a mass spectrum can be identified by

the help of peaks attributable to isotopes.

• E. g., if a compound contains five carbon atoms, the relative

intensity of the M + 1 ion should be 5(1.1%) = 5(.011), multiplied

by the relative intensity of the molecular ion.

• Meaning that the number of carbon atoms in a compound can be

calculated if the relative intensities of both the M and M + 1 peaks are known.

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• Isotopic distributions of several elements commonly found in organic

compounds are displayed in Table 1.

• From the isotopic distributions, it is seen why peak can be used to

determine the number of carbon atoms in a compound:

• It is because the contributions to the peak by isotopes of H, O, and

the halogens are very small or nonexistent.

• This formula does not work as well in predicting the number of

carbon atoms in a nitrogen-containing compound because the

natural abundance of 15N is relatively high. 3

Table 1. The Natural Abundance of Isotopes Commonly Found in Organic Compounds

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• Mass spectra can show M + 2 peaks as a result of a contribution from 18O or

from having two heavy isotopes in the same molecule (13C and/or 2H, two 13C’s).

• Often, the M + 2 peak is very small.

• Presence of a large M + 2 peak is evidence of a compound containing either

Cl or Br.

• Based on the natural abundance of the isotopes Cl and Br in Table 1, it can

be concluded that if the M + 2 peak is ⅓ the height of the molecular ion

peak, then the compound contains 1 Cl atom because the natural abundance

of 37Cl is ⅓ that of 35Cl

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• If the M and M + 2 peaks are about the same height, then the compound

contains one bromine atom because the natural abundances of 79Br and 81Br

are about the same.

• In calculating the molecular masses of molecular ions and fragments, the

atomic mass of a single isotope of the atom must be used (Cl = 35 or 37,

etc.) as mass spectrometry only measures the m/z value of an individual

fragment.

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Determination of Molecular Formulas

• The previous mass spectra shown were obtained with a low-resolution

mass spectrometer. Such spectrometers give the nominal molecular

mass of a fragment.

• High-resolution mass spectrometers can determine the exact molecular

mass of a fragment to an accuracy of 0.0001 amu.

• If the exact molecular mass of the molecular ion is known, the

compound’s molecular formula can be determined.

• E. g., the under listed compounds show a nominal molecular mass of

122 amu, but each of them has a different exact molecular mass.7

Some Compounds with a Nominal Molecular Mass of 122 amu and their Exact

Molecular Masses

Table 2. The Exact Masses of Some Common Isotopes

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Fragmentation at Functional Groups

Each functional group has a specific way of fragmentation pattern.

Fragmentation patterns of alkyl halides, ethers, alcohols, and ketones are

considered as examples.

Alkyl Halides

• Mass spectrum of 1-bromopropane, shown below has the relative heights of

the M and M + 1 peaks nearly equal, thus compound contains a bromine

atom.

• Bombardment with electrons frees a lone-pair of electron if the molecule has

any: holding on of electrons by molecule is tighter for lone-pair of electrons

than bonding electrons.

• Thus, electron bombardment frees one of bromine’s lone-pair of electrons.9

• Bond breaks heterolytically C-Br, both electrons move to more

electronegative atoms, forming a propyl cation and a bromine atom.

• Hence, the base peak in the mass spectrum of 1-bromopropane is at

m/z = 43 [M-79 or (M + 2)].

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Figure 1. The mass spectrum of 1-bromopropane.

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• Mass spectrum of 2-chloropropane is shown in Figure 2.

• The compound contains a chlorine atom, because the M + 2 peak is one-third

the height of the molecular ion peak.

• The base peak at m/z =43 results from heterolytic cleavage of the C-Cl bond.

• The peaks at m/z = 63 and m/z = 65 have a ratio of 3:1, indicates that

fragments contain a chlorine atom.

• This results from homolytic cleavage of a C-C bond at the carbon.

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Figure 2. The mass spectrum of 2-chloropropane.

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Ethers

Fragmentation pattern of an ether is similar to that of an alkyl halide.

• Electron bombardment dislodges one of the lone-pair electrons from oxygen.

• Fragmentation of the resulting molecular ion occurs in two principal ways:

a. C-O bond is cleaved heterolytically, with the electrons going to the more electronegative

oxygen atom.

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Figure 3. The mass spectrum of sec-butylisopropyl ether

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b. C-C bond is cleaved homolytically at the position which leads to

relatively stable cation in which the positive charge is shared by two atoms.

The most stable radical easily cleaved from alkyl group.

Thus, the peak at m/z = 87 is more abundant than the one at m/z = 101 though

the compound has 3 methyl groups bonded to carbons that can be cleaved to

produce a peak at m/z = 101, primary radical is more stable than a methyl

radical.

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Alcohols

• Molecular ions obtained from alcohols fragment is such that few of them

make it to the collector.

• As a result, the mass spectra of alcohols show small molecular ion peaks.

• Note the small molecular ion peak at m/z = 102 in the mass spectrum of 2-

hexanol (Figure 4).

• Like alkyl halides and ethers, alcohols undergo cleavage.

• Consequently, the mass spectrum of 2-hexanol shows a base peak at m/z = 45

(leads to more stable butyl radical) and a smaller peak at m/z = 87 (leads to a

less stable methyl radical). 19

Figure 4.The

mass spectrum

of 2-hexanol.

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• In the previous fragmentations examples, only one bond is broken.

• An important fragmentation occurs in alcohols, which involve breaking

of two bonds.

• Two bonds break because the fragmentation forms a stable water

molecule.

• The water that is eliminated comes from the OH group of the alcohol

and a γ hydrogen.

• Thus, alcohols show a fragmentation peak at m/z = M-18 because of

loss of water21

Note that alkyl halides, ethers, and alcohols have the following fragmentation

pattern in common:

1. A bond between carbon and a more electron egativeatom (a halogen or an

oxygen) breaks heterolytically.

2. A bond between carbon and an atom of similar electronegativity

(a carbon or a hydrogen) breaks homolytically.

3. The bonds most likely to break are the weakest bonds and those that lead

to formation of the most stable cation. 22

Ketones

• Mass spectrum of a ketone generally has an intense molecular ion peak.

• Ketones fragment homolytically at the C-C bond adjacent to the C = O

bond, which resulting in the formation of a cation with a positive charge

shared by two atoms.

• The alkyl group leading to the more stable radical is the one that is more

easily cleaved.

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• If one of the alkyl groups attached to the carbonyl carbon has a hydrogen,

a cleavage known as a McLafferty rearrangement may occur.

• This rearrangement affords homolytically cleavage between the carbon and the β carbon and a hydrogen atom from the carbon migrates to the oxygen atom.

• Fragmentation occurred again in a way that produces a cation with a positive charge shared by two atoms

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Figure 5. The mass spectrum of 4-methyl-2-pentanone

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