Chapter G, Problem G.9P

Interpretation Introduction

Interpretation:

The base peak in the mass spectrum of $\text{hexanamide}$ appears at m/z = $\text{59}$. The ion that corresponds to this mass peak is to be drawn, and how it is produced from the molecular ion is to be shown.

Concept introduction:

The mass of $\text{hexanamide}$ is $\text{115}$ u. The peak at m/z = $\text{59}$ is due to the loss of a $\text{but-1-ene}$, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH=CH}}_{\text{2}}$, and the formation of the enol radical cation ${\text{CH}}_{\text{2}}{\text{=C(}}^{\text{+}}{\text{OH)NH}}_{\text{2}}$. This is the McLafferty rearrangement of the amide’s molecular ion. A carbonyl-containing compound can undergo a McLafferty rearrangement if an alkyl group attached to the carbonyl carbon possesses a γ-carbon with at least one hydrogen atom. A $\text{H}$- atom from the $\text{γ}$-carbon shifts to the carbonyl $\text{O}$-atom, and the bond joining the $\text{α}$ and $\text{β}$ carbons is broken. The result is the ejection of an uncharged alkene molecule. The McLafferty rearrangement is not limited to just ketones, it is characteristic of several compound classes containing the carbonyl group, including aldehydes, carboxylic acids, esters, and amides.