Homolytic and heterolytic acleavage

100-

chooh

El mass spectrum of ethanol.

Under EI conditions the analyte develops a positive charge through the loss of one electron. If there is an electronegative atom in the structure of the molecule such as nitrogen or oxygen, this positive charge will be on the electronegative atom(s). If an electronegative atom is absent the charge is more difficult to locate with certainty. Figure 9.4 shows the EI spectrum of ethanol which provides an example of two types of fragmentation. The process is as follows:

(i) Homolytic a-cleavage (Fig. 9.5) is promoted by the presence of a hetero atom such as oxygen, nitrogen or sulphur and in molecules containing a hetero atom it often gives rise to the most abundant ion in the mass spectrum (the base peak).

CH3-ch-oh -

Homolytic a-cleavage of ethanol.

m/z 46

m/z 45

-». ch2=oh + ch3-

m/z 46

m/z 31

(ii) One electron in the bond broken goes to the radical and the other combines with the unpaired electron on the hetero atom to produce a double bond; the hetero atom becomes positively charged.

(iii) Loss of the largest possible radical is most favoured. In the case of ethanol, loss of CH3- gives rise to the base peak in the mass spectrum at m/z 31.

For many drug molecules this type of fragmentation dominates their mass spectra.

A minor ion in the spectrum of ethanol results from heterolytic a-cleavage (Fig. 9.6).

As is illustrated in Figure 9.7, homolytic a-cleavage is the major fragmentation mechanism for chains containing hetero atoms.

CH3-CH2-OH

Heterolytic a-cleavage of ethanol.

m/z 46

m/z 29

CH3-CH2-CH-CH3 -CH3-CH2 ■ + CH-CH3 of homolytic a-cleavage.

Self-test 9.1

Indicate the type(s) of cleavage and the fragments lost which gives rise to the major ions in the spectrum of butan-2-ol shown in Figure 9.8.

sso| m|M aBeAeap-n 3HA|OLUOH

(!!!) ',eHD sso| Lj}|M a6eAe3p-n 3nA|OUJOH (!!) i-H sso| qj|M 36eAeap-x> dijA|Ouioh (!) \sj3msuv

Figure 9.9 shows the spectrum of n-butanol. In this case homolytic a-cleavage, which gives rise to the ion at m/z 31, does not completely dominate the spectrum and the spectrum produced is more complex as a result. Loss of the neutral fragment H20 occurs via a 1,4 elimination (Fig. 9.10); this produces an ion at m/z 56 which, since it is still a radical cation, gives rise to the fragment at m/z 41 via loss of CH3 and followed by loss of H2 as a neutral fragment to give m/z 39.

100-

CH3CHOH

El mass spectrum of butan-2-ol.

a^jJ

20 40 60 80

40 60

El mass spectrum of n-butanol.

40 60

C ^ r

Fig. 9.10

Loss of water via 1,4

elimination.

,L 1 * ¿H2 ch,

ch ch2

ch2

m/z 56

1,4 -elimination

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