The Molecular Ion Peak

Introduction

The single most valuable number a mass spectrum can give is the relative molecular mass of the compound. It comes from the molecular ion peak , the signal produced by molecules that have lost one electron but have not fallen apart. Knowing Mr instantly narrows the list of possible structures: a compound with Mr = 58 might be propanone or butane, but it cannot be ethanol or benzene. This page explains how to find the molecular ion, how to use it, and when to be cautious.

Core explanation

Forming the molecular ion. In electron impact ionisation, a fast electron removes one electron from a molecule:

M(g) + e⁻ → M⁺•(g) + 2e⁻

The product has the same mass as the molecule (the lost electron has negligible mass, about 1/1836 of a proton) and a single positive charge. Its m/z value therefore equals the relative molecular mass. The dot shows it has an odd number of electrons; many exam boards simply write M⁺.

Locating the peak. Fragments are always lighter than the ion they came from, so the molecular ion is normally the peak at the highest m/z value . There is one important exception: small peaks one or two units higher, labelled M+1 and M+2, arise from molecules containing heavier isotopes such as carbon-13, chlorine-37 or bromine-81. When reading a spectrum, look for the last substantial peak and check whether tiny peaks just beyond it are isotope peaks.

Using Mr. Once Mr is known, candidate molecular formulae can be tested. For Mr = 58, possibilities include C₄H₁₀ (58), C₃H₆O (58) and C₂H₂O₂ (58). Other evidence — an IR carbonyl band, for example — selects between them.

When the molecular ion is weak or missing. The molecular ion carries excess energy from ionisation. If it has an easy way to break apart, few survive long enough (about 10 microseconds) to reach the detector.

- Branched alkanes fragment readily at branch points, so M⁺ is small; in 2,2-dimethylpropane it is almost absent. - Alcohols lose water or cleave next to the C–O bond easily, so M⁺ is often weak. - Aromatic compounds have delocalised electrons that spread the charge and stabilise the ion, so M⁺ is strong; for benzene it is the tallest peak.

If the molecular ion cannot be seen, chemists switch to a soft technique such as electrospray, where the dominant ion is [M+H]⁺ at Mr + 1.

The nitrogen rule. Carbon and oxygen have even masses and even valencies; hydrogen and the halogens have odd masses but odd valencies, so they always pair up to give an even total. Nitrogen has an even mass (14) but an odd valency (3). The result: a compound containing only C, H, O, S and halogens always has an even Mr, and an odd Mr signals an odd number of nitrogen atoms. Methylamine, CH₃NH₂, has Mr = 31.

Step-by-step reasoning

To find Mr and a formula from an EI spectrum:

1. Scan to the right-hand end of the spectrum. 2. Identify the last significant peak, ignoring small M+1 and M+2 companions. 3. Read its m/z value; this is Mr. 4. Apply the nitrogen rule: odd Mr means an odd number of N atoms. 5. List formulae that fit Mr and use other data to choose between them.

Visual explanation

Imagine a spectrum of propanone. A tall line stands at m/z 43, a small one at m/z 15, and at the far right a moderate line at m/z 58 followed by a tiny line at m/z 59. The line at 58 is M⁺; the one at 59 is M+1 from carbon-13, not a heavier molecule.

Real-world analogy

Weighing a sealed parcel tells you its total mass before anyone unpacks it. The molecular ion is the unopened parcel; the fragment ions are the items that fall out when the parcel is dropped. If the parcel is flimsy, it may burst before reaching the scales, just as fragile molecular ions break before reaching the detector.

Real-world example

Pharmaceutical quality-control laboratories confirm that a newly synthesised batch contains the intended drug by checking its molecular ion. Paracetamol, C₈H₉NO₂, should show [M+H]⁺ at m/z 152 in electrospray mode; its odd Mr of 151 is consistent with its single nitrogen atom.

Why?

Why does the molecular ion appear at the highest m/z? Fragmentation can only remove atoms, never add them, because the ions travel through a vacuum with nothing to react with. Every fragment is therefore lighter than the ion it came from, leaving the intact ion as the heaviest species, apart from its own isotopic variants.

Common misconception

"If no peak appears at the expected Mr, the compound must be something else." The molecular ion may simply be too unstable to survive. Absence of M⁺ is common for alcohols and highly branched alkanes and does not by itself rule a structure out.

Worked example

Question: A neutral compound shows its molecular ion at m/z 72, with a tiny peak at m/z 73. Its IR spectrum has a strong absorption at 1715 cm⁻¹. Suggest a molecular formula.

Reasoning: Mr = 72 (m/z 73 is M+1). Even Mr, so zero or two N atoms. The IR band shows a C=O group. C₄H₈O = 48 + 8 + 16 = 72 fits.

Answer: C₄H₈O, for example butanone.

Quick check

1. A compound gives its molecular ion at m/z 59. What does this tell you about the nitrogen content of the molecule? Answer: Mr is odd, so the molecule must contain an odd number of nitrogen atoms, such as one.

Exam focus

State that M⁺ is at the highest m/z and gives Mr. Write the ionisation equation to show the molecular ion. When asked to "deduce Mr", mention that you ignored the small M+1 peak. Watch for halogen compounds, where the M and M+2 peaks are both significant.

Advanced insight

Low-resolution mass spectra report nominal masses based on the most abundant isotopes, which is not the same as the average Mr found in data tables. Chloromethane has a tabulated Mr of 50.5, yet no ion appears at m/z 50.5: instead there are peaks at 50 (with ³⁵Cl) and 52 (with ³⁷Cl). Always interpret m/z values in terms of individual isotopes.

Summary

The molecular ion M⁺• forms when a molecule loses one electron and has an m/z equal to Mr. It is normally the highest-m/z peak apart from small isotope peaks. It can be weak or absent for fragile molecules such as alcohols and branched alkanes, and strong for aromatics. The nitrogen rule links an odd Mr to an odd number of nitrogen atoms.

Practice questions

1. Write an equation for the formation of the molecular ion of ethanol. Answer: C₂H₅OH(g) + e⁻ → C₂H₅OH⁺•(g) + 2e⁻ 2. A hydrocarbon has its molecular ion at m/z 86. Suggest its molecular formula. Answer: C₆H₁₄, since 6 × 12 + 14 × 1 = 86. 3. Explain why the molecular ion peak of benzene is much larger than that of hexan-1-ol. Answer: The delocalised π system of benzene stabilises the ion so it rarely fragments, whereas the alcohol ion readily loses water or breaks next to the C–O bond. 4. Which of these formulae could give a molecular ion at m/z 45: C₂H₇N, C₂H₆O, CH₃NO? Answer: C₂H₇N (24 + 7 + 14 = 45) and CH₃NO (12 + 3 + 14 + 16 = 45) both fit, each with one nitrogen as the nitrogen rule predicts; C₂H₆O has Mr 46.