Interpreting Mass Spectra

Introduction

A mass spectrum can look like a forest of lines, but only a handful of them carry most of the information. An experienced analyst reads it in a fixed order: first the molecular ion, then the isotope peaks, then the biggest fragments and the gaps between them. This page brings together everything from the earlier mass spectrometry pages into one practical method for turning a spectrum into a proposed structure.

Core explanation

Step one: find the molecular ion. The molecular ion is the peak at the highest m/z, apart from its small isotope partners at M+1 and M+2. Its m/z equals the relative molecular mass, Mᵣ. Remember that it may be weak — alcohols and branched alkanes often give small molecular ions — so do not simply choose the tallest peak.

Step two: read the isotope peaks. The M+1 peak comes mainly from ¹³C, which makes up about 1.1% of natural carbon. The ratio of heights gives an estimate of the number of carbon atoms:

number of carbon atoms ≈ (height of M+1 ÷ height of M) × 100 ÷ 1.1

An M+2 peak one-third the height of M shows one chlorine atom; an M+2 peak equal to M shows one bromine atom. An odd molecular ion indicates an odd number of nitrogen atoms.

Step three: identify the base peak and other major fragments. Common diagnostic ions include:

m/z Likely ion --- --- 15 CH₃⁺ 29 C₂H₅⁺ or CHO⁺ 31 CH₂=OH⁺ 43 C₃H₇⁺ or CH₃CO⁺ 57 C₄H₉⁺ or C₂H₅CO⁺ 77 C₆H₅⁺ (phenyl) 91 C₇H₇⁺ (benzyl/tropylium)

Step four: calculate neutral losses. Subtract each important fragment from M. Losses of 15 (CH₃), 17 (OH), 18 (H₂O), 29 (C₂H₅ or CHO), 31 (OCH₃) and 45 (COOH) are common and help to identify groups at the edge of the molecule.

Step five: propose and test a structure. Suggest a structure with the right Mᵣ, then check that each major fragment can be formed by a sensible cleavage — usually one that gives a stabilised cation such as an acylium, oxonium or tertiary carbocation. Finally, ask whether any other isomer would give the same spectrum. If it would, you need further evidence, such as an IR or NMR spectrum.

Step-by-step reasoning

Summarised as a checklist:

1. Molecular ion: Mᵣ, and whether it is odd or even. 2. M+1 and M+2: approximate carbon count, presence of Cl or Br. 3. Base peak and large fragments: likely ions. 4. Neutral losses from M. 5. Candidate structures, tested against every major peak.

Visual explanation

Imagine the spectrum annotated with arrows: a long arrow from the molecular ion down to the base peak labelled with the neutral loss, and shorter arrows between fragments. The pattern of labelled gaps is like a family tree showing how each piece came from the parent ion.

Real-world analogy

Interpreting a mass spectrum is like reconstructing a broken vase from its fragments. The total weight tells you how big the vase was, the shapes of the large pieces show where it tends to crack, and missing pieces are identified by comparing what remains with the original total.

Real-world example

Environmental chemists screen river water with gas chromatography–mass spectrometry. Each separated compound gives a spectrum that is compared automatically with libraries of hundreds of thousands of reference spectra, but a trained analyst still checks the molecular ion, isotope pattern and key fragments before a pollutant is reported.

Why?

Why test every major peak against a proposed structure? A single fragment can often be explained by several structures, but a correct structure must explain all the large peaks together. Checking each one removes candidates that fit some of the evidence but not the rest.

Common misconception

"If two compounds have the same Mᵣ, a mass spectrum cannot tell them apart." Isomers usually fragment differently. Pentan-2-one gives a base peak at m/z 43 (CH₃CO⁺), while pentan-3-one gives its base peak at m/z 57 (C₂H₅CO⁺), even though both have M = 86.

Worked example

Question: A compound of C, H and O shows M at m/z 58 with an M+1 peak 3.3% of the height of M. The base peak is at m/z 43 and there is a peak at 15. Identify it.

Reasoning: Carbon atoms ≈ 3.3 ÷ 1.1 = 3. With Mᵣ 58 and three carbons, C₃H₆O fits (36 + 6 + 16 = 58). Butane, C₄H₁₀, also has Mᵣ 58 but would give an M+1 of about 4.4%. The loss 58 − 43 = 15 means CH₃ was lost, leaving CH₃CO⁺ at 43; CH₃⁺ appears at 15.

Answer: Propanone, CH₃COCH₃.

Quick check

1. The M+1 peak of a hydrocarbon is 6.6% of the height of M. How many carbon atoms does it contain? Answer: About 6.6 ÷ 1.1 = 6 carbon atoms.

Exam focus

Show your reasoning: state Mᵣ, give the formula of each fragment ion with a positive charge , and state the neutral species lost. Examiners often penalise fragment formulae written without a charge or written as radicals.

Advanced insight

The M+1 method assumes the ¹³C contribution dominates. For compounds containing many nitrogen atoms (¹⁵N, 0.37%) or silicon or sulfur, other isotopes contribute to M+1 and M+2, so a carbon count from M+1 becomes less reliable. High-resolution mass spectrometry avoids the problem by giving an exact mass that fixes the molecular formula directly.

Summary

Read a mass spectrum in order: molecular ion, isotope peaks, base peak and major fragments, then neutral losses. The M+1 ratio estimates the carbon count, M+2 reveals chlorine or bromine, and an odd M suggests nitrogen. Propose a structure and check that it explains every major peak; if isomers remain possible, use additional techniques.

Practice questions

1. Why is the tallest peak not always the molecular ion? Answer: The tallest peak is the base peak, the most abundant ion; the molecular ion may fragment readily and give only a small peak at the highest m/z. 2. A spectrum shows M at 86 and a base peak at 57. Calculate the neutral loss and suggest what it is. Answer: 86 − 57 = 29, a loss of C₂H₅· (or CHO·); for a ketone this fits pentan-3-one losing an ethyl radical to give C₂H₅CO⁺. 3. Molecular ion peaks appear at m/z 78 and 80 in a 3:1 ratio. What does this show? Answer: The compound contains one chlorine atom, with the ³⁵Cl and ³⁷Cl isotopes giving the 3:1 pattern. 4. Explain how the M+1 peak distinguishes propanone from butane. Answer: Propanone has three carbon atoms (M+1 about 3.3% of M) while butane has four (about 4.4%), so the relative size of the M+1 peak differs.