Abstract
AbstractThe fragmentation of n‐hexane, n‐nonane and n‐tetradecane under electron impact has been investigated, using 13C labelled compounds. The mechanism of the formation of the alkyl radical ions is quantitatively explained by using a method of calculation developed in an earlier publication for n‐heptane. It is assumed that these ions are formed either by a direct C‐C bond cleaveage or by a secondary olefin loss from an alkyl radical ion. In the latter case the probability for a particular carbon to be lost in the neutral fragment is assumed to be random. The probability for a direct cleavage to an alkyl ion is about 80% for an ion containing at least half of the number of carbon atoms of the molecular ion and 15% for the smaller ions. The [MH]+ ion seems to be a special case not yet clearly understood. Former results about the loss of methyl from the molecular ion are confirmed.
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