Chapter G, Problem G.5P

Interpretation Introduction

Interpretation:

Mass spectra were acquired for $\text{1, 4-diethylbenzene}$ and $\text{1-methyl-4-propylbenzene}$. The base peak of spectrum A is at m/z = 105, and that of spectrum B is at m/z = $\text{119}$. Each spectrum is to be matched to its compound.

Concept introduction:

The ionization process causes the loss of a single electron from a closed-shell, uncharged molecule, so the molecular ion must have an unpaired electron, making it a free radical. More specifically, it also carries a positive charge, so the molecular ion is a radical cation, and ${\text{M}}^{\text{+}}$ is more accurately depicted as ${\text{M}}^{\text{+ •}}$. Both $\text{1, 4-diethylbenzene}$ and $\text{1-methyl-4-propylbenzene}$ have a mass of $\text{134}$ u. The base peak in Spectrum A at m/z = $\text{1}0\text{5}$ is due to the loss of an ethyl fragment ($\text{29}$ u), and the base peak in Spectrum B at m/z = $\text{119}$ is due to the loss of a methyl fragment ($\text{15}$ u). The ${\text{M}}^{+}$ of A can lose a ${\text{CH}}_{\text{3}}$ radical via the cleavage of an original benzylic $\text{C-C}$ bond, and the ${\text{M}}^{+}$ of B can lose an ethyl radical via the cleavage of an original benzylic $\text{C-C}$ bond. Therefore, $\text{1, 4-diethylbenzene}$ produces Spectrum B, and $\text{1-methyl-4-propylbenzene}$ produces Spectrum A.