4.2.4(f)-(g) - Organic mass spectrometry

Reading the spectrum

A mass spectrum is a graph. The horizontal axis is $m/z$, which means mass-to-charge ratio. The vertical axis is relative intensity, so the tallest peak is set to 100 and the other peaks are compared with it.

OCR limits this work to ions with single charges. That is a helpful boundary: if the charge is $+1$, the $m/z$ value is the relative mass of the ion.

The tallest peak is called the base peak. It is the most intense peak, but it is not automatically the molecular ion peak. In organic spectra, the base peak is often a stable fragment ion.

[DIAGRAM: asset_name: Lesson 4.2.4f-g: Organic mass spectrometry - diagram 01; asset_slug: 04_02_04c_organic_mass_spectrometry__diagram_01; recommended_method: drawn_chem; description: Labelled stick mass spectrum showing fragment peaks at lower m/z, the base peak, the molecular ion peak M+ at m/z 58, and a small M+1 peak from carbon-13.]

Worked example: a spectrum has a molecular ion peak at $m/z = 72$. The ion is singly charged, so the molecular ion has relative mass 72. The compound therefore has $M_r = 72$.

Molecular ion and m+1

To find the molecular ion peak, look for the highest significant $m/z$ peak for the main molecule. There is one important caution: an organic spectrum may contain a very small peak one unit higher than the molecular ion peak. This is the $M+1$ peak.

The $M+1$ peak is caused by the small proportion of carbon atoms that are carbon-13 instead of carbon-12. Carbon-13 is one mass unit heavier than carbon-12, so a molecule containing one carbon-13 atom gives a molecular ion one unit heavier than the usual molecular ion.

This lesson only needs carbon-13 as an awareness point. Do not treat $M+1$ as the molecular ion peak, and do not use halogen isotope patterns here: OCR does not expect you to interpret mass spectra of organic halogen compounds at this stage.

Worked example: a spectrum has peaks at $m/z = 58$ and a much smaller peak at $m/z = 59$. The peak at 59 is labelled as a small $M+1$ peak. The molecular ion peak is therefore at 58, so $M_r = 58$.

Fragmentation peaks

After the molecular ion forms, some molecular ions break into smaller pieces. If a piece is positively charged, it can be detected and gives a peak. If a piece is neutral, it is not detected by the mass spectrometer.

Fragment analysis is partly arithmetic and partly structure. For example:

• $\ce{CH3+}$ has mass $12 + 3 = 15$, so it may give a peak at $m/z = 15$.
• $\ce{C2H5+}$ has mass $(2 \times 12) + 5 = 29$, so it may give a peak at $m/z = 29$.
• $\ce{CHO+}$ also has mass $12 + 1 + 16 = 29$, so the same $m/z$ value can sometimes fit more than one fragment.
• $\ce{C3H7+}$ has mass $(3 \times 12) + 7 = 43$, so it may give a peak at $m/z = 43$.

An oxygen-containing fragment can have the same nominal mass as a hydrocarbon fragment. For example, $\ce{CH3CO+}$ has mass $2(12)+3+16=43$, just like $\ce{C3H7+}$. Write connectivity as well as atom counts when asked for a structure. The ethanoyl fragment can be represented as $\mathrm{CH_3{-}C^{+}{=}O}$, and the hydrocarbon ion may be $\mathrm{CH_3CH_2CH_2^{+}}$ or $\mathrm{(CH_3)_2CH^{+}}$, depending on the parent.