Mass Spectrometry

This lesson covers: 

1. How mass spectrometry works to produce mass spectra
2. Using molecular ion and fragmentation peaks to determine molecular mass and structure
3. Using M+1 and M+2 peaks
4. Differentiating compounds using fragmentation patterns

Mass spectrometers produce mass spectra

Mass spectrometry involves bombarding vaporised sample molecules with electrons, resulting in the production of charged fragments. These fragments are then separated based on their mass-to-charge ratio (m/z) and detected to create a mass spectrum.

Here are the key steps:

1. The sample is vaporised, turning it into a gas.
2. An electron beam bombards the gaseous molecules, knocking out an electron to form molecular ions (M⁺).
3. Some of these molecular ions break apart into smaller fragment ions.
4. The ions are then deflected through a magnetic field, which separates them according to their mass/charge ratio (m/z).
5. Finally, the separated ions are detected and their data is displayed as a mass spectrum.

A mass spectrum represents the abundance of each fragment against its m/z value, providing critical information about the sample's composition.

Molecular ion (M) peak indicates molecular mass

The molecular ion peak, or M peak, in a mass spectrum indicates the relative abundance of the intact M⁺ ions. Assuming the molecular ion carries a +1 charge, the m/z value of this peak directly corresponds to the compound's molecular mass. Thus, the m/z value of the M peak reveals the compound's relative molecular mass (M_r).

For instance, in the mass spectrum below, pentane (C₅H₁₂) has an M peak at m/z 72. This shows that pentane's relative molecular mass is 72 g mol^-1.

Using M+1 peak to calculate number of carbons (n)

The relative size of the [M+1] peak can be used to determine the number of carbon atoms (n) in an unknown organic compound.

The [M+1] peak is caused by molecules containing a ¹³C isotope atom instead of the more common ¹²C isotope. As carbon has a 1.1% natural abundance of ¹³C, the percentage [M+1] peak abundance relates to the number of carbon atoms.

The equation is:

n =1.1100​×abundance of M+ ionabundance of [M+1]+ ion​

Worked example 1 - Calculating the number of carbons using the M+1 peak

An unknown compound has an M peak with a relative abundance of 97.6% and an [M+1] peak with an abundance of 2.4%.

Calculate the number of carbon atoms in the molecule.

Step 1: Equation

n =1.1100​×abundance of M+ ionabundance of [M+1]+ ion​

Step 2: Substitution and correct evaluation

n =1.1100​×97.62.4​=2.24

As the number of carbon atoms must be an integer, the compound likely contains 2 carbon atoms.

The M+2 peak indicates the presence of chlorine or bromine

The presence and identity of halogen atoms in an organic compound can be determined by examining the [M+2] peak.

An [M+2] peak is caused by molecules containing a ³⁷Cl or ⁸¹Br isotope atom instead of the more common ³⁵Cl or ⁷⁹Br isotopes.

• If the [M+2] peak height equals the M peak height, there is one bromine atom per molecule.
• If the [M+2] peak is one-third of the height of the M peak, there is one chlorine atom per molecule.

This occurs because:

• Bromine has an equal abundance of ⁷⁹Br and ⁸¹Br isotopes.
• Chlorine has isotopes of ³⁵Cl (75%) and ³⁷Cl (25%).

So the relative [M+2] peak height reveals whether Br or Cl is present.

For example:

• Bromomethane (CH₃Br) will have a molecular ion peak at m/z = 94 corresponding to [CH₃⁷⁹Br]⁺ and an M+2 peak of equal intensity at m/z = 96 corresponding to [CH₃⁸¹Br]⁺.
• Chloromethane (CH₃Cl) will have a molecular ion peak at m/z = 50 corresponding to [CH₃³⁵Cl]⁺ and an M+2 peak at one third of the intensity at m/z = 52 corresponding to [CH₃³⁷Cl]⁺.

Fragmentation produces pattern related to structure

When molecular ions break apart, they form fragment ions that appear as distinct peaks on the mass spectrum.

These fragments can include:

• CH₃⁺ (m/z = 15)
• C₂H₅⁺ (m/z = 29)
• C₃H₇⁺ (m/z = 43)
• OH⁺ (m/z = 17)

The pattern of fragmentation offers clues about the molecule's structure, as the m/z values of these peaks can be matched to known fragments to deduce the structural formula.

For example, the mass spectrum of pentane shows:

• A peak at m/z 29 suggests the presence of a CH₃CH₂⁺ fragment, indicating an ethyl group within the molecule.
• A peak at m/z 43 suggests a CH₃(CH₂)₂⁺ fragment, indicating a propyl group.
• A peak at m/z 57 suggests a CH₃(CH₂)₃⁺ fragment, indicating a butyl group.

Mass spectra differentiate between compounds

Compounds with identical molecular formulas can have unique fragmentation patterns in their mass spectra. This uniqueness enables mass spectrometry to distinguish between closely related molecules, such as propanal and propane, despite them having the same formula, C₃H₈O. 

Propanal, for instance, can fragment to produce a C₂H₅⁺ ion, a fragmentation that is not possible for propane.

Computer databases can rapidly match an unknown compound's distinctive spectrum to identify its structure.