Mass Spectrometers and Molecular Mass and Structure

Mass spectrometry is an analytical method in which a molecule is ionized and the resulting fragments and ions are analyzed. Mass spectrometry looks at the ions that a molecule produces in its gas phase. Mass spectrometry is commonly conducted by bombarding a molecule with high-energy electrons. Electron impact is the method for producing positive ions in mass spectrometry, resulting from the bombardment of a molecule with high-energy electrons. When a high-energy electron hits a molecule, one of the molecule's electrons is dislodged, creating a cation radical, which is a positively charged molecular ion containing an odd number of electrons. A radical species is also produced, but it is not detected by the mass spectrometer. A molecular ion, sometimes called a parent ion, is an ion formed by removal or addition of an electron without compromising the molecular structure. The molecular ion has the same mass as the original molecule, having only lost a single electron, which is of negligible mass.

The M⁺ peak is the peak on a mass spectrum indicating the molecular ion, sometimes called the parent ion. Peaks in the mass spectrum, including the molecular ion ${\rm{M^+}}$ peak and the ${\rm {M}^+{+}\;1}$ and ${\rm {M}^+{+}\;2}$ peaks, give clues as to which atoms are present in the molecule and in what quantity. The ${\rm {M}^+{+}\;1}$ peak is used to determine the number of carbon atoms present in a molecule. If the ${\rm {M}^+}$ is an odd number, nitrogen is present in the molecule. An ${\rm {M}^+{+}\;2}$ peak may also be generated and is used to determine whether Cl, S, or Br is present in the molecule.

Fragmentation patterns are analyzed to understand the structure and composition of the molecule. Alkanes submitted to mass spectrometry produce numerous fragments because their ${\rm {C{-}C}}$ bonds are susceptible to cleavage. The mass spectra of alcohols often lack an ${\rm{M^+}}$ peak because of their easy fragmentation at the ${\rm {C{-}C}}$ bond next to the O. Amines show alpha cleavage on the mass spectrum, and ketones and aldehydes undergo a McLafferty rearrangement, in which the alpha carbon and carbonyl are eliminated. Mass spectrometry uses electrons with enough energy (about 70 eV) to ionize the molecule and break its chemical bonds, causing the molecule to dissociate into fragments. When a cation radical is dissociated, a neutral fragment and a positively charged fragment result. In a mass spectrometer, the positively charged ions that dissociate from the sample are sent down an analyzer tube that is surrounded by a magnet. The magnet forces the ions to follow a circular path that adopts a characteristic radius, depending on their m/z ratio. The m/z ratio is the ratio of an ion's mass (m) to its charge (z). Ions with a small m/z ratio follow a more divergent path than those with a larger m/z. Ions are separated by their specific m/z ratio by adjusting the strength of the magnetic field or the ion acceleration through the analyzer. The individual ions pass through a slit in the machine and into a detector, where they are counted. Scanning all of the ions gives a distribution of m/z values, called a mass spectrum, which is distinct for particular compounds. Because radicals are neutral, the magnet does not move them down the pathway, and they are not included. The data collected from a mass spectrometer are outputted in different ways, but are most commonly analyzed as a bar graph in which relative intensity is plotted on the $y$-axis, and the m/z is plotted on the $x$-axis. For example, in benzene, C₆H₆, the most intense peak, or the base peak, is the molecular ion ${\rm {M}^+}$ peak; it is assigned a relative intensity of 100, with $m/z=78$. Benzene does not extensively fragment, and the other fragment ions are much less intense than the molecular ion. The smaller peaks on a mass spectrum show the possible ways a molecule can fragment—each fragment type has a characteristic peak relating to its m/z. More stable fragments are more abundant and result in a larger bar, or peak, in the mass spectrum.

An M⁺ + 1 peak is a peak on a mass spectrum that is used to determine the number of carbon atoms in a molecule. ${\rm {M}^+{+}\;1}$ is the next peak after ${\rm {M}^+}$, and it corresponds to the molecular ion that has a mass one unit higher than the molecular ion ${\rm {M}^+}$ peak because it contains a heavier isotope (1 mass unit heavier). For example, benzene contains isotopes in which one ¹²C is replaced with a ¹³C, bringing the m/z up to 79. This will appear as a small peak on the mass spectrum. Similarly, some of the benzene will contain ²H in place of a ¹H, also with an m/z ratio of 79. In benzene, the molecular ion in which all carbon atoms are ¹²C will comprise 93.4%, molecular ions with one ¹³C will comprise 6.5%, and molecular ions with a ²H will amount to 0.1%.

The ${\rm {M}^+{+}\;1}$ is used to determine the number of carbon atoms in a molecule. If the ${\rm {M}^+}$ peak is scaled to 100%, $({\rm{M}^+{+}\;1} \text{ intensity})/1.1\%$ gives the number of carbon atoms in the compound. In benzene, the ${\rm {M}+1}$ intensity is 6.5%, and $6.5/1.1=5.9$, or 6 carbon atoms. The natural abundance of carbon-13 is about 1.1%, meaning that in a sample of carbon, about 1.1% of that sample is carbon-13.

Furthermore, the m/z value indicates how many nitrogen atoms are present in the molecule. If the m/z for ${\rm {M}^+}$ is even, there are an even number of nitrogen atoms, which includes zero nitrogen atoms. If the ${\rm {M}^+}$ peak is odd, the molecule has an odd number of nitrogen atoms (and there must be at least one). Ethylamine (CH₃CH₂NH₂) has an m/z of 45. Since the m/z is odd, there must be at least 1 nitrogen, but there could be 3, 5, 7, 9, and so on nitrogen atoms.

Another peak that may appear on the mass spectrum, the ${\rm {M}^+{+}\;2}$, shows isotopes with a mass that is two units higher than ${\rm {M}^+}$. Three elements contribute to the ${\rm {M}^+{+}\;2}$ peak: ³⁴S, ³⁷Cl, and ⁸¹Br. The presence of S, Cl, and Br is determined by looking at the $( {\rm {M}^++}\;2)/{\rm {M}}^+$ ratio. If the ratio is 100:4.4, there is one sulfur (S) in the compound. If the ratio is 100:31.9 (roughly one-third the height of ${\rm {M}^+}$), there is one chlorine (Cl) in the compound. If the ratio is 100:97.2, roughly equal heights, there is one bromine (Br) in the compound.

Ethyl chloride (CH₃CH₂Cl) has an ${\rm {M}^+}$of 64 with a relative intensity of 50% and an ${\rm {M}^+{+}\;2}$ with a relative intensity of 1.6%. The ratio of ${\rm {M}^+{+}\;2}/{\rm {M}^+}$ is $50/16=31.25$, so the compound must contain a chlorine atom.