Chapter 16, Problem 16.34P

Interpretation Introduction

Interpretation:

The number of carbon atoms present in the compound that will produce the mass spectrum shown 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 up 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. Since carbon (as well as other elements) naturally occurs as a mixture of isotopes, the molecular ion often gives rise to two peaks. Naturally occurring carbon consists of two isotopes, ${}^{\text{12}}{\text{C and }}^{\text{13}}\text{C}$. The molecular ion peak, therefore, will show up as a pair of peaks at ${\text{M}}^{+}$ and $\text{M+1}$.

The number of carbon atoms in a molecule can be estimated on the basis of the relative abundances of the ${\text{M}}^{+}$ and $\text{M+1}$ peaks as

$\text{Number of C atoms = }\frac{\text{intensity of M+1 peak}}{{\text{intensity of M}}^{\text{+}}\text{ peak}}\text{×}\frac{\text{100\%}}{\text{1}\text{.1\%}}$