Introduction to Mass Spectrometry (MS)

A mass spectrometer produces a spectrum of masses based on the structure of a molecule.

The x-axis of a mass spectrum represents the masses of ions produced (m/z)

The y-axis represents the relative abundance of each ion produced

The pattern of ions obtained and their abundance is characteristic of the structure of a particular molecule

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Ionization (the formation of ions) A molecule is bombarded with a beam of high

energy electrons An electron is dislodged from the molecule by

the impact, leaving a positively charged ion with an unpaired electron (a radical cation)

This initial ion is called the molecular ion (M+.) because it has the same molecular weight as the analyte

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Fragmentation

Excess vibrational energy is imparted to the molecular ion by collision with the electron beam - this causes fragmentation

The fragmentation pattern is highly characteristic of the structure of the molecule

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Common Isotope Abundances

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3 Classes of Isotopes

• M - only a single isotope– EX: F, P, I

• M+1 - two isotopes with significant relative abundance differing by 1 mass unit– EX: H, C, N

• M+2 - two isotopes with significant relative abundance differing by 2 mass units– EX: Cl, S, O

Determination of Molecular Formulas & Weights The Molecular Ion and Isotopic Peaks

The presence of heavier isotopes one or two mass units above the common isotope yields small peaks at M+.+1 and M+.+2

Example: In the spectrum of methane one expects an M+.

+1 peak of 1.17% based on a 1.11% natural abundance of 13C and a 0.016% natural abundance of 2H

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Four Basic Rules If M+ is even, then the unknown contains an even

number of Nitrogen atoms (zero is an even number)

The abundance of M++1 indicates the number of Carbon atoms:

# of C = relative abundance/1.1

The abundance of the M++2 peak indicates the presence of S (4.4%), Cl (33%) or Br (98%)

The remaining unknown mass can be attributed to Hydrogen

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m/z intensity % abundance

78 (M+) 10.00 100%

79 .3 3%

80 3.3 33%

1. Is the molecular ion even?

Yes, there must be either an even number of N, or no Nitrogen atoms.

2. How many Carbon atoms are there?

# Carbons = 3 / 1.1 ≈ 3 carbon atoms

3. Is a S, Cl or Br present?

A M++2 peak of 33% indicates the presence of chlorine

4. How many Hydrogen atoms are there?

78 = (1 * 35) + (3 * 12) + (H * 1)

78 = 71 + H

# of Hydrogen atoms = 7 C3H7ClC – C – C - Cl

H HH

H H H

H

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m/z intensity % abundance

86 (M+) 10.00 100%

87 .56 5.6%

88 .04 .4%

1. Is the molecular ion even?Yes, there must be either 0, 2, 4 … Nitrogen atoms

2. How many Carbon atoms are there?# Carbons = 5.6 / 1.1 ≈ 5 carbon atomsif there are 2 N atoms then the FW would be (5*12) + (2*14) =

88Therefore, there are no nitrogen atoms

3. Is a S, Cl or Br present?A M++2 peak of .4% indicates no S, Cl or Br

4. How many Hydrogen atoms are there?86 = (5 * 12) + (H * 1)86 = 60 + H# of Hydrogen atoms = 26 C5H26 ~not possible~

What if there is Oxygen in the molecule?86 = (1 * 16) + (5 * 12) + (H * 1)86 = 76 + H# of Hydrogen atoms = 10

C5H10O8

C – C – C – C – C = O

H HH

H H H

H

HH

HH

86 (M+)

87 (M+1)

88 (M+2)

%Intensity

(M-1)

Determination ofMolecular Formula

distinguish between compounds of same MW

C5H10O4 or C10H14

Determination ofMolecular Formula

distinguish between compounds of same MW

C5H10O4

13C 5 * 1.11% = 5.55%2H 10 * 0.016% = 0.16%17O 4 * 0.04% = 0.16%

-------135peak/134peak 5.87%

Determination ofMolecular Formula

distinguish between compounds of same MW

C10H14

13C 10 * 1.11% = 11.1%2H 14 * 0.016% = 0.22%

-------135peak/134peak 11.32%

The Numbers Approach

• If compound with formula CwHxNyOz , relative intensities of M, M+1, and M+2 ions will be given by:

]%2.0200/[2

)%04.038.0016.011.1(1

)11.1(2

zM

M

zyxwM

M

w +=+

+++=+

High-Resolution Mass Spectrometry Low-resolution mass spectrometers

measure m/z values to the nearest whole number

High-resolution mass spectrometers measure m/z values to three or four decimal places

The high accuracy of the molecular weight calculation allows accurate determination of the molecular formula of a fragment

Example One can accurately pick the molecular formula

of a fragment with a nominal molecular weight of 32 using high-resolution MS

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The exact mass of certain nuclides is shown below

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http://www.cem.msu.edu/~reusch/VirtualText/Spectrpy/MassSpec/masspec1.htm

Fragmentation by Cleavage at a Single Bond

Cleavage of a radical cation gives a radical and a cation but only the cation is observable by MS

In general the fragmentation proceeds to give mainly the most stable carbocation

In the spectrum of propane the peak at 29 is the base peak (most abundant) 100% and the peak at 15 is 5.6%

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Fragmentation Equations The M+. Ion is formed by loss of one of its

most loosely held electrons If nonbonding electron pairs or pi electrons are

present, an electron from one of these locations is usually lost by electron impact to form M+.

In molecules with only C-C and C-H bonds, the location of the lone electron cannot be predicted and the formula is written to reflect this using brackets

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Example: The spectrum of hexane

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Example: spectrum of neopentane

Fragmentation of neopentane shows the propensity of cleavage to occur at a branch point leading to a relatively stable carbocationThe formation of the 3o carbocation is so favored that almost no molecular ion is detected

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Carbocations stabilized by resonance are also formed preferentially

Alkenes fragment to give resonance-stabilized allylic carbocations

Carbon-carbon bonds next to an atom with an unshared electron pair break readily to yield a resonance stabilized carbocation

Z=N, O, or S R may be H

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Carbon-carbon bonds next to carbonyl groups fragment readily to yield resonance stabilized acylium ions

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Alkyl substituted benzenes often lose a hydrogen or alkyl group to yield the relatively stable tropylium ion

Other substituted benzenes usually lose their substitutents to yield a phenyl cation

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Fragmentation by Cleavage of 2 Bonds

The products are a new radical cation and a neutral molecule

Alcohols usually show an M+.-18 peak from loss of water

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Cycloalkenes can undergo a retro-Diels Alder reaction (section 13.11) to yield an alkadienyl radical cation

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