Skip to content

MS Fragmentation and Molecular Formula

Fragmentation

Is primarily based on the strength of the associated fragments and their bonding energy

A stability/fragmentation hierarchy would look something like this (from most to least stable):

  • Carbonyls
  • Aromatics
  • Less saturated hydrocarbons
  • Unbranched hydrocarbons
  • Ketones
  • Amines
  • Esters
  • Esters
  • Carboxylic acids
  • Branched hydrocarbons
  • Alcohols

The more loosely bound the electrons, the less likely they will to be removed, e.g. LPEs will be removed before \(\delta\) and \(\pi\) electrons, before \(\sigma\) electrons.

When a cluster of peaks occur in a single grouping, it likely represents the same amount of carbons, with varying amounts of hydrogens attached. it stands to reason that some of these fragments will be more stable than others, resulting in the abundance ratios that we see.

Remember that the abundance of the fragments is directly proportional to their stability.

Alkyl Fragmenting

When an alkyl chain fragments, it fragments in a particular pattern with specific formulae:

\(\ce{C1}\) \(\ce{C2}\) \(\ce{C3}\) \(\ce{C4}\) \(\ce{C5}\) \(\ce{C6}\) …
\(\ce{C_{n}H_{2n{+1}}+}\) 15 29 43 57 71 85 …
\(\ce{C_{n}H_{2n{-1}}+}\) 13 27 41 55 69 83 …
\(\ce{C_{n}H_{{2n}}+}\) 14 28 42 56 70 84

The \(\ce{C_nH_{2n+1}+}\) fragments will be the most abundant, as they are the most stable, having a higher degree of saturation.

Fragmentation Formulae

As a general rule of thumb:

  • A fragment with an even m/z, that formed from an even mass parent will form from a rearrangement, or two bonds breaking
  • A fragment with an odd m/z, that formed from an even mass parent will form from a single bond breaking
  • A fragment with an even m/z that formed from an odd mass parent will form from a single bond breaking

Molecular Formula

While it’s no longer really necessary thanks to computational tools, there is a process to convert the molecular ion mass to a molecular formula:

  1. Rule of 13 - consider a \(\ce{CH}\) unit as 13 dalton

    • If we divide the mass by 13 we can get a starting point for the formua
    • e.g. \(152\div13=11\) carbons
      • \(11\times12\) dalton \(=132\) dalton
      • \(152-132=20\) hydrogens

  2. If another element is present, remove the amount of carbon and hydrogen atoms required to account for that mass

    • e.g. if there’s one oxygen, we remove a \(\ce{CH4}\)

      • \(\ce{C11H20 - CH4 + O} = \ce{C10H16O}\)
      Atom type \(\ce{CH}\) equivalents
      \(\ce{^14N}\) \(\ce{CH2}\)
      \(\ce{^16O}\) \(\ce{CH4}\)
      \(\ce{^19F}\) \(\ce{CH7}\)
      \(\ce{^28Si}\) \(\ce{C2H4}\)
      \(\ce{^31P}\) \(\ce{C2H7}\)
      \(\ce{^32S}\) \(\ce{C2H8}\)
      \(\ce{^35Cl}\) \(\ce{C2H11}\)
      \(\ce{^79Br}\) \(\ce{C6H7}\)
      \(\ce{^127I}\) \(\ce{C10H7}\)

Nitrogen Rule

Based on a quirk of the maths we can identify the presence of nitrogen.

A molecule with an even number of nitrogens must also have an even number of hydrogens and must have an even molecular weight. By the same logic, a molecule with an odd number of nitrogens must also have an odd number of hydrogens and thus must have an odd molecular weight.

This only works for \(\ce{^14N}\), as it’s the common isotope in which the mass is even, while its valence is odd. It will also work when the molecule contains oxygen, phosphorus, sulphur and halogens.