Question

(a) Show how fragmentation occurs to give the base peak at m/z 58 in the mass spectrum of ethyl propyl amine (N-ethylpropan-1-amine), shown below.

(b) Show how a similar cleavage in the ethyl group gives an ion of m/z72.

(c) Explain why the peak at m/z 72 is much weaker than the one at m/z 58.

Short answer

(a)

(b)

(c) The fragmentation in part (a) occurs more often than the one in part (b) because of the stability of radicals produced along with the iminium ion. Ethyl radical is much more stable than methyl radical so pathway (a) is preferred.

Step-by-step solution

Explanation of part (a):

In mass spectrum, the vertical axis denotes the relative abundance of ions. The most intensive peak in a spectrum is called as base peak. In the mass spectrum, the relative intensity of each ion is normally found using the peak with the highest intensity as standard or base peak. The heaviest ion or the one with greatest m/z value is the molecular ion.

N-ethylpropan-1-amine of m/z value 87, undergoes fragmentation in the form of alpha-cleavage to produce two fragments which have m/z values 58 and 29. The fragment with m/z value 58 is formed due to stabilisation of carbocation by +M-effect of nitrogen atom. Fragment with m/z 29 is the ethyl radical.

Explanation of part (b):

In mass spectrum, the vertical axis denotes the relative abundance of ions. The most intensive peak in a spectrum is called as base peak. In the mass spectrum, the relative intensity of each ion is normally found using the peak with the highest intensity as standard or base peak. The heaviest ion or the one with greatest m/z value is the molecular ion.

N-ethylpropan-1-amine undergoes alpha cleavage to produce the fragment which has m/z value of 72. This fragment is stable as nitrogen though having positive charge is stable as it has complete octet. Methyl radical is also produced as a fragment which has m/z value of 15.

Explanation of part (c):

The fragmentation in part (a) occurs more often than the one in part (b) because of the stability of radicals produced along with the iminium ion. Ethyl radical is much more stable than methyl radical as radicals are electron deficient species, thus, CH₃ in CH₃CH₂ stabilises the radical by inductive effect, so pathway (a) is preferred.