Structure Determination (Mass Spectrometry)
0 Pages | Leaving School | 20/11/2024

Mass SpectrometryMass SpectrometryMass Spectrometry

Mass Spectrometry


Introduction

For AS level you would have used mass spectrometry to figure out the relative abundances of different isotopes within a sample of an element, to determine relative atomic masses, and to deduce molecular formulae.

However, mass spectrometry can also be used to work out the structure of organic molecules. In certain ways molecules and atoms behave in similar ways in a mass spectrometer. But, there are a number of important differences.

An electron is removed when a molecule is ionised. Molecules tend to contain only paired electrons which means that this removal results in a species which contains an unpaired electron or a free radical. For example, take ethane:

C2H6 ? [C2H6]+ + e

The species which results is a positively charged ion or free radical. In this case it is called the molecular or parent ion.

Once this molecular ion has been formed there are two possible routes it can take:

  • It can pass through the mass spectrometer intact and then onto the detector. Here it is detected as possessing an m/z ratio (where z = 1 and m = relative mass of molecular ion).
  • Fragmentation can occur in which the molecular ion can breaks up and becomes two smaller and more stable species. One species is a free radical and the other is a positively charged ion. The detected species will have m/z ratios lower than that of the molecular ion.

This means that for an organic molecule you will observe different peaks in the mass spectrum:

  • The molecular ion corresponds to the peak which has the highest m/z ratio.
  • Fragmentation of the molecular ion correspond to the peaks which have the lower m/z ratios. As different molecules fragment in different ways and others are more stable, more information about the structure of the molecules can be taken from these peaks.

Fragmentation

In fragmentation free radical molecular ions are broken up. Due to the fact that the molecular ion is composed of an odd number of electrons, one of the fragmented species will be a positive ion (which contains an even number of electrons) and the other a free radical (which contains an odd number of electrons). For example, take ethane:

[C2H6]+ ? [CH3] + [CH3]+

Only [CH3]+ (the positive electron) will be detected because it is charged. The free radical, which is neutral, will be neither deflected nor accelerated and so will not be detected.

Therefore, fragmentation can be represented by the following general equation:

R+? X+ + Y

In which:

  • R+ = the molecular ion (which is detected as long a not all of them fragment)
  • X+ = the fragment ion (which is detected)
  • Y = the fragment radical (which is not detected)

Stable ions

However, not all fragments are detected. This is because some form themselves into stable ions.

The majority of stable cations are carbocations, like CH3+ or CH3CH2+, and acylium ions, like HCO+ and CH3CO+. As forming a stable ion is the higher chance scenario, these fragments are more likely to be detected. In a mass spectrum they produce the most intense peaks.

Therefore, in the mass spectrum of an organic compound the main peaks you will find are:

  • molecular ion peak
  • peaks for the most stable ions which can form due to fragmentation of the molecular ion

Determining structure using mass spectra

Using the molecular ion peak it is possible to deduce the relative molecular mass of a compound.

The other peaks provide the masses of the fragments and, therefore, possible identities. This is important for determining the structure of the unfragmented molecule. The presence or absence of fragments provides important clues to which features are present and which are not present in the molecule. For example:

  • a peak at 29 suggests CH3CH2+ is present
  • a peak at 43 suggests either CH3CH2CH2+, CH3CHCH3+ or CH3CO+ is present
  • a peak at 57 suggests CH3CH2CO+ or CH3COCH2+ is present

This can be useful for distinguishing between different isomers. For example:

  • C4H10:

? butane gives peaks at 15, 29, 43 and 58

? methylpropane gives peaks at 15, 43 and 58 but not at 29

  • C3H6O:

? propanal gives peaks at 15, 29, 43 and 58

? propanone gives peaks at 15, 43 and 58 but not at 29

ADVERTISEMENTS

ADVERTISEMENTS