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Mass spectral interpretation

Mass spectral interpretation is a science topic covered in the lgStudy science library. This page brings together a partial reference excerpt, illustrations, worked examples, real-world applications and a short study plan, so you can understand Mass spectral interpretation rather than just read about it. In short: Mass spectral interpretation is the method employed to identify the chemical formula, characteristic fragment patterns and possible fragment ions from the mass spectra. Mass spectra is a plot of relative abundance against mass-to-charge ratio.

Mass spectral interpretation — main illustration
Mass spectral interpretation — illustration

Key takeaways

  • Mass spectral interpretation belongs to science; place it in that map before memorising details.
  • Learn the definition first, then one example that makes the definition concrete.
  • Connect Mass spectral interpretation to a quantity you can measure, compute or draw — that is where exam questions come from.
  • Reproduce the core statement of Mass spectral interpretation from memory before moving on to harder problems.

Reference excerpt

Mass spectral interpretation is the method employed to identify the chemical formula, characteristic fragment patterns and possible fragment ions from the mass spectra. Mass spectra is a plot of relative abundance against mass-to-charge ratio. It is commonly used for the identification of organic compounds from electron ionization mass spectrometry. Organic chemists obtain mass spectra of chemical compounds as part of structure elucidation and the analysis is part of many organic chemistry curricula.

Mass spectra generation

Electron ionization (EI) is a type of mass spectrometer ion source in which a beam of electrons interacts with a gas phase molecule M to form an ion according to

M + e − → M + ∙ + 2 e − {\displaystyle M+e^{-}\to M^{+\bullet }+2e^{-}}

with a molecular ion M + ∙ {\displaystyle M^{+\bullet }} . The superscript "+" indicates the ion charge and the superscript "•" indicates an unpaired electron of the radical ion. The energy of the electron beam is typically 70 electronvolts and the ionization process typically produces extensive fragmentation of the chemical bonds of the molecule. Due to the high vacuum pressure in the ionization chamber, the mean free path of molecules are varying from 10 cm to 1 km and then the fragmentations are unimolecular processes. Once the fragmentation is initiated, the electron is first excited from the site with the lowest ionization energy. Since the order of the electron energy is non-bonding electrons > pi bond electrons > sigma bond electrons, the order of ionization preference is non-bonding electrons > pi bond electrons > sigma bond electrons.

The peak in the mass spectrum with the greatest intensity is called the base peak. The peak corresponding to the molecular ion is often, but not always, the base peak. Identification of the molecular ion can be difficult. Examining organic compounds, the relative intensity of the molecular ion peak diminishes with branching and with increasing mass in a homologous series. In the spectrum for toluene for example, the molecular ion peak is located at 92 m/z corresponding to its molecular mass. Molecular ion peaks are also often preceded by an M-1 or M-2 peak resulting from loss of a hydrogen radical or dihydrogen, respectively. Here, M refers to the molecular mass of the compound. In the spectrum for toluene, a hydrogen radical (proton-electron pair) is lost, forming the M-1 (91) peak. Peaks with mass less than the molecular ion are the result of fragmentation of the molecule. Many reaction pathways exist for fragmentation, but only newly formed cations will show up in the mass spectrum, not radical fragments or neutral fragments. Metastable peaks are broad peaks with low intensity at non-integer mass values. These peaks result from ions with lifetimes shorter than the time needed to traverse the distance between ionization chamber and the detector.

Molecular formula determination

Nitrogen rule The nitrogen rule states that organic molecules that contain hydrogen, carbon, nitrogen, oxygen, silicon, phosphorus, sulfur, or the halogens have an odd nominal mass if they have an odd number of nitrogen atoms or an even mass if they have an even number of nitrogen atoms are present. The nitrogen rule is true for structures in which all of the atoms in the molecule have a number of covalent bonds equal to their standard valency, counting each sigma bond and pi bond as a separate covalent bond.

Rings rule From degree of unsaturation principles, molecules containing only carbon, hydrogen, halogens, nitrogen, and oxygen follow the formula

Rings + π -Bonds = u = C − H 2 − X 2 + N 2 + 1 {\displaystyle {\text{Rings}}+\pi {\text{-Bonds}}=u=C-{\frac {H}{2}}-{\frac {X}{2}}+{\frac {N}{2}}+1\,}

where C is the number of carbons, H is the number of hydrogens, X is the number of halogens, and N is the number of nitrogen.

Even electron rule The even electron rule states that ions with an even number of electrons (cations but not radical ions) tend to form even-electron fragment ions and odd-electron ions (radical ions) form odd-electron ions or even-electron ions. Even-electron species tend to fragment to another even-electron cation and a neutral molecule rather than two odd-electron species.

Stevenson's rules The more stable the product cation, the more abundant the corresponding decomposition process. Several theories can be utilized to predict the fragmentation process, such as the electron octet rule, the resonance stabilization and hyperconjugation and so on.

Rule of 13 The Rule of 13 is a simple procedure for tabulating possible chemical formula for a given molecular mass. The first step in applying the rule is to assume that only carbon and hydrogen are present in the molecule and that the molecule comprises some number of CH "units" each of which has a nominal mass of 13. If the molecular weight of the molecule in question is M, the number of possible CH units is n and

M 13 = n + r 13 {\displaystyle {\frac {M}{13}}=n+{\frac {r}{13}}\,}

where r is the remainder. The base formula for the molecule is

C n H n + r {\displaystyle C_{n}H_{n+r}\,}

and the degree of unsaturation is

… excerpt ends here. Continue reading the full article.

Illustrations

Mass spectral interpretation: Electron ionization mass spectrum of the steroid alcohol brassicasterol
Electron ionization mass spectrum of the steroid alcohol brassicasterol
Mass spectral interpretation: The mass spectrum for toluene has around 30 peaks.
The mass spectrum for toluene has around 30 peaks.
Mass spectral interpretation: Several toluene fragmentation peaks can be rationalized in this fragmentation pattern
Several toluene fragmentation peaks can be rationalized in this fragmentation pattern
Mass spectral interpretation: fragmentation at heteroatom
fragmentation at heteroatom
Mass spectral interpretation: An example of the McLafferty rearrangement
An example of the McLafferty rearrangement

Worked examples

Example 1 — a first encounter with Mass spectral interpretation

Start with the simplest possible case. Write down what Mass spectral interpretation claims or describes in one sentence, then invent the smallest concrete situation in which that sentence is true. In science, the smallest case is usually a single object, a single equation or a single measurement. Check that every symbol or term in your sentence has a meaning in that case.

Example 2 — changing one variable

Take the situation from Example 1 and change exactly one quantity: double it, halve it, or set it to zero. Predict what should happen to Mass spectral interpretation before you calculate. Comparing your prediction with the result is the fastest way to find out whether you understand the idea or only the words.

Example 3 — an exam-style question

Typical questions about Mass spectral interpretation ask you to (a) state it precisely, (b) apply it to given data, and (c) explain a limitation. Practise writing all three answers in under five minutes; the third part is what separates a full-mark answer from an average one.

Applications of Mass spectral interpretation

In research
Mass spectral interpretation appears in science research whenever the underlying quantities have to be modelled precisely. Papers usually cite it as a starting assumption and then explore where it breaks down.
In technology and industry
Engineering practice reuses Mass spectral interpretation in design rules, simulations and safety margins. Knowing the idea lets you read a specification sheet and understand why the numbers look the way they do.
In the classroom
Mass spectral interpretation is common in secondary-school and first-year university syllabi. It links to neighbouring topics Mass spectrometry, so understanding it makes those chapters shorter.
In everyday life
Look for Mass spectral interpretation outside the textbook — in sport, cooking, traffic, electronics or the sky above you. An example you found yourself is remembered far longer than one you were given.
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How to study Mass spectral interpretation in 20 minutes

  1. Read the reference excerpt below once, without taking notes.
  2. Close the page and write down what Mass spectral interpretation means in your own words.
  3. Compare your version with the excerpt and mark what you missed.
  4. Work through the three examples above with pen and paper.
  5. Explain Mass spectral interpretation out loud to somebody else — or to Teacher Smith in the lgStudy chat.

Frequently asked questions

What is Mass spectral interpretation in simple terms?

Mass spectral interpretation is the method employed to identify the chemical formula, characteristic fragment patterns and possible fragment ions from the mass spectra. Mass spectra is a plot of relative abundance against mass-to-charge ratio.

Why does Mass spectral interpretation matter?

Because it connects several science ideas at once: it gives you a definition you can apply, a quantity you can calculate, and a way to check whether a result is plausible.

How should I study Mass spectral interpretation?

Read the excerpt, restate it from memory, then work through the examples and applications listed on this page. The five-step study plan above takes about twenty minutes.

What does this page cover?

It gives you a compact reference excerpt plus original lgStudy explanations, examples, applications and study material on Mass spectral interpretation.

Tags

  • Mass spectrometry

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