Chapter 12 (Mass Spectrometry (MS) and Infrared Spectroscopy (IR))

Introduction to Mass Spectrometry

Mass spectrometry is an analytical technique used in organic chemistry to find information about a compound’s mass, structure, and elements present. It is especially useful for identifying unknown compounds. Mass spectrometry is considered a destructive technique because the molecule breaks apart during the analysis. Unlike infrared spectroscopy or NMR spectroscopy, mass spectrometry does not use light.

Mass spectrometry separates particles based on their mass-to-charge ratio, which is written as m/z. For most organic molecules, the charge is +1, so the m/z value is approximately equal to the mass of the fragment.

How Mass Spectrometry Works

First, the sample is heated so that it turns into a gas. This gaseous sample is then passed into an ionization chamber. Inside the chamber, a high-energy electron strikes the molecule and removes one of its electrons. As a result, the molecule becomes positively charged.

This positively charged molecule is called the molecular ion, written as M⁺·. The molecular ion is a radical cation, meaning it has both a positive charge and an unpaired electron. Molecular ions are very unstable and contain a lot of energy, so they usually break apart into smaller pieces. This breaking apart is called fragmentation.

Only fragments that carry a positive charge are detected by the mass spectrometer. Neutral fragments are not detected.

The Mass Spectrum

A mass spectrum is a graph produced by the mass spectrometer. The x-axis shows the m/z value, which represents the mass-to-charge ratio. The y-axis shows the relative intensity, which tells us how abundant each fragment is compared to the most abundant one.

The tallest peak in the spectrum is always assigned a relative intensity of 100%. All other peaks are compared to this peak.

Important Peaks in a Mass Spectrum

The molecular ion peak (M⁺) corresponds to the mass of the entire molecule before it breaks apart. This peak usually appears at the highest m/z value in the spectrum. However, sometimes the molecular ion peak is weak or even missing if the molecule fragments very easily.

The base peak is the tallest peak in the spectrum. It represents the most stable and most abundant fragment formed during fragmentation. The base peak is not necessarily the molecular ion peak.

The M+1 peak appears one unit higher than the molecular ion peak. This peak is mainly caused by the presence of the carbon-13 isotope (¹³C). Carbon-13 occurs naturally at about 1.1% abundance.

Calculation: Estimating the M+1 Peak

The height of the M+1 peak can be estimated using the number of carbon atoms in the molecule. Each carbon atom contributes approximately 1.1% to the M+1 peak.

For example, if a molecule contains 6 carbon atoms, the expected M+1 peak height is approximately:

6 × 1.1% = 6.6% of the M⁺ peak

This type of calculation is sometimes tested on exams.

Low-Resolution vs High-Resolution Mass Spectrometry

Most mass spectra you will see in this course are low-resolution spectra. Low-resolution mass spectrometry gives the nominal mass, which means the mass is reported as a whole number or with very few decimal places.

High-resolution mass spectrometry gives the exact mass of a molecule. This allows chemists to determine the exact molecular formula, especially when multiple formulas could have the same nominal mass. High-resolution MS is more advanced and usually tested conceptually rather than in calculations.

The Nitrogen Rule

The nitrogen rule is a very important rule used in mass spectrometry. Nitrogen has an even atomic mass but forms an odd number of bonds. Because of this unusual behavior, nitrogen affects whether the molecular ion mass is odd or even.

If the molecular ion peak has an odd m/z value, the compound contains an odd number of nitrogen atoms (such as 1 or 3). If the molecular ion peak has an even m/z value, the compound contains zero or an even number of nitrogen atoms.

This rule only applies to compounds containing carbon, hydrogen, nitrogen, oxygen, and halogens.

Isotope Peaks and Halogens

Some elements have naturally occurring isotopes that appear clearly in mass spectra. These isotopes create extra peaks called isotope peaks.

Chlorine has two common isotopes, chlorine-35 and chlorine-37. Because chlorine-35 is much more abundant, a compound containing one chlorine atom shows two peaks separated by two mass units, with a 3:1 intensity ratio. This means the M peak is about three times taller than the M+2 peak.

Bromine also has two common isotopes, bromine-79 and bromine-81. These isotopes are present in almost equal amounts. As a result, a compound containing one bromine atom shows two peaks separated by two mass units with a 1:1 intensity ratio. The two peaks are almost the same height.

Fluorine and oxygen do not produce strong, recognizable M+2 patterns in mass spectrometry.

Fragmentation Patterns (Conceptual Understanding)

Fragmentation in mass spectrometry occurs through radical mechanisms, where electrons move one at a time. These mechanisms are often shown using fish-hook arrows.

Alcohols commonly fragment through two pathways. One pathway is alpha cleavage, where the bond next to the oxygen atom breaks, forming a resonance-stabilized ion. The second pathway is dehydration, where the molecule loses water and forms an alkene ion.

Amines also commonly fragment through alpha cleavage, similar to alcohols.

Carbonyl compounds frequently undergo alpha cleavage to form a resonance-stabilized acylium ion.

Some carbonyl compounds undergo a special fragmentation called the McLafferty rearrangement. This occurs when there is a hydrogen atom on the gamma carbon. During this rearrangement, a neutral alkene is formed along with a stabilized radical cation.

Introduction to Infrared (IR) Spectroscopy

Infrared spectroscopy is a technique that measures how molecules absorb infrared radiation. IR spectroscopy is used to identify functional groups present in a molecule. It does not give the complete structure of a molecule, but it provides very useful clues.

IR spectroscopy works because bonds within molecules are constantly vibrating. When infrared radiation has the same energy as a bond’s vibration, that radiation is absorbed.

Energy and Wavenumber

The energy of light is related to its frequency and wavelength. In IR spectroscopy, energy is usually expressed as wavenumber, with units of cm⁻¹. Higher wavenumbers correspond to higher energy.

The IR Spectrum

An IR spectrum is plotted with percent transmittance on the y-axis and wavenumber on the x-axis. A low percent transmittance means strong absorption.

Not every peak needs to be interpreted. Peaks below 1000 cm⁻¹ are part of the fingerprint region and are not tested in this course.

Important IR Absorptions

Alkane C–H (sp³) bonds absorb just below 3000 cm⁻¹. Alkene C–H (sp²) bonds absorb just above 3000 cm⁻¹. Alkyne C–H (sp) bonds absorb near 3300 cm⁻¹ and appear as sharp peaks.

Alcohol O–H bonds absorb near 3300 cm⁻¹ and produce broad, strong peaks. Carboxylic acid O–H bonds produce very broad and strong peaks that range from about 2500 to 3300 cm⁻¹.

Amine N–H bonds absorb near 3300 cm⁻¹ and are sharper and weaker than O–H peaks.

Carbonyl (C=O) bonds produce very strong peaks between 1650 and 1850 cm⁻¹. These peaks are extremely important because carbonyl groups are very common and easy to identify in IR spectra.

Aldehydes show two small C–H absorption peaks near 2800 cm⁻¹ and 2700 cm⁻¹, in addition to the carbonyl peak.

Calculation & Exam-Style Questions (With Explanations)

Question 1:
A compound has a molecular ion peak at m/z = 121. Does it contain nitrogen?

Answer: Yes. The molecular ion mass is odd, which means the compound contains an odd number of nitrogen atoms.

Question 2:
A molecule has 8 carbon atoms. Estimate the expected M+1 peak intensity.

Answer:
8 × 1.1% = 8.8%

Question 3:
A mass spectrum shows two peaks at m/z = 156 and 158 in a 3:1 ratio. What element is present?

Answer: Chlorine, because chlorine shows a 3:1 M : M+2 pattern.

Question 4:
A mass spectrum shows two peaks at m/z = 170 and 172 with equal height. What element is present?

Answer: Bromine, because bromine produces a 1:1 M : M+2 pattern.

Question 5:
An IR spectrum shows a very strong peak at 1715 cm⁻¹. What functional group is present?

Answer: A carbonyl group (C=O).