The McLafferty rearrangement (Figure 19), is a classic example of a rearrangement reaction. This rearrangement results in formation of an intact neutral molecule and a radical ion with an even mass to charge ratio. This reaction is significantly different from the cleavage reactions discussed previously. The McLafferty rearrangement is often observed for carbonyl compounds that contain a linear alkyl chain. If this alkyl chain is long enough, a six-membered ring forms from the carbonyl oxygen to the hydrogen on the fourth carbon. This spacing allows the hydrogen to transfer to the carbonyl oxygen via a six membered ring. The McLafferty rearrangement is energetically favorable because it results in loss of a neutral alkene and formation of a resonance stabilized radical cation.

The products from the McLafferty rearrangement are observed in the mass spectra of C₆H₁₂O isomers (Figure 20). The mass spectrum of n-hexanal contains two even mass ions. C₂H₄O⁺ (m/z 44) is produced by the McLafferty rearrangement and C₄H₈⁺ (m/z 56) is the McLafferty compliment. The McLafferty compliment is produced when the charge is transferred to the alkene fragment during the rearrangement. The mass spectrum of 2-hexanone is easily distinguished from n-hexanal because the McLafferty rearrangement breaks a different C-C bond. This results in loss of C₃H₆ and produces C₃H₆O⁺ (m/z 58). The mass spectrum of 3-hexanone does not have any major even mass fragment ions so apparently the McLafferty rearrangement is not favorable. If the McLafferty rearrangement did occur, where would the ion be observed in this mass spectrum? What fragmentation would produce the major ions observed in the mass spectrum of 3-hexanone (m/z 29, 43, 57, and 71)? Based upon the discussion so far you should be able to identify many of the other fragments in these three mass spectra. Spend some time with a piece of scratch paper and see what you come up with.