Chapter 16, Problem 16.36P

Interpretation Introduction

Interpretation:

The value of $\text{m/z}$ expected of the ${\text{M}}^{+}$ peak of ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{SH}$ is to be determined. The expected value of $\text{m/z}$ of the $\text{M+2}$ peak is to be determined. If the ${\text{M}}^{+}$ peak is also the base peak, the expected intensity of the $\text{M+2}$ peak is to be determined.

Concept introduction:

In a mass spectrometer, the molecules of the compound are bombarded with high energy electrons. This generally results in one electron from the molecule being knocked out. The resulting ion is a radical ion, with an odd number of electrons, called the molecular ion. Since the mass of an electron is negligible, the mass of the molecule and the molecular ion are essentially the same.

The molecular ion is a high energy species and breaks into a number of fragments, one of which carries the charge. Another source of fragmentation is the bombarding electrons themselves. These electrons often have energies sufficient to not just knock out an electron but also enough to break bonds.

The only species detected in the mass spectrometer are the charged species, and their separation is based on the $\text{m/z}$ ratio, where m is the mass (amu) and z the charge.

The mass spectrum shows the distribution of the relative abundance of various ion fragments as a function of their $\text{m/z}$ ratio. In most cases, the ions generated have a +1 charge, which means the mass m of each ion fragment is the same as its $\text{m/z}$ value.

This means the $\text{m/z}$ value for the molecular ion (${\text{M}}^{+}$) peak corresponds to the molecular mass.

Sulfur has three isotopes of mass number 32, 33, and 34. The abundance of ${}^{33}\text{S}$ is very low, so only ${}^{32}\text{S}$ and ${}^{34}\text{S}$ are likely to show up in a mass spectrum. Their relative abundance ratio is 22.5:1.