[Author © Dr Phil Brown PhD: Doc Brown's advanced level organic chemistry exam revision notes suitable for students of UK A level chemistry courses & US K12 grade 11, grade 12 and AP honors chemistry courses: Molecular spectroscopy analysis of 1-iodopropane (mass spectra) [spectra page updated Mar 30th 2026 *]

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Introductory note on the mass spectrum of 1-iodopropane

Students and teachers please note my explanation of the mass spectrum of 1-iodopropane is designed for advanced, but pre-university, chemistry courses.

If M represents the 1-iodopropane molecule, the initial ionisation to give the molecular ion is:

M_(g) + high KE e^-  ==> [M•]⁺_(g) + 2e^- and fragmentation equations assume [M]⁺ is the start of the processes and all species are in a gaseous state.

I've not usually shown an unpaired electron on e.g. an ion or a non-ionised alkyl radical R e.g.

[M•]⁺ ==> [X]⁺  +  R•, but you should be aware this is a more accurate depiction of some processes.

I've used simplified equations to show how some of the ions might be formed in the fragmentation pattern for the mass spectrum of 1-iodopropane.

I've included stick diagram and table of m/z ions for the mass spectrum of 1-iodopropane and conducting the mass spectrum analysis under standard conditions, a database can be built up based complex fingerprint patterns, often involving relative intensities of many fragment ions, that can be used to identify compounds including 1-iodopropane.

Interpreting the fragmentation pattern of the mass spectrum of 1-iodopropane

[M]⁺ is the parent molecular ion peak (M) with an m/z of 170 corresponding to [C₃H₇I]⁺, the original 1-iodopropane molecule minus an electron, [CH₃CH₂CH₂I]⁺

The small M+1 peak at m/z 171, corresponds to an ionised 1-iodopropane molecule with one ¹³C atom in it i.e. an ionised 1-iodopropane molecule of formula [¹³C¹²C₂H₇I]⁺

Carbon-13 only accounts for ~1% of all carbon atoms (¹²C ~99%), but the more carbon atoms in the molecule, the greater the probability of observing this ¹³C M+1 peak.

1-iodopropane has 3 carbon atoms, so on average, ~1 in 33 molecules will contain a ¹³C atom.

The most abundant ion of the molecule under mass spectrometry investigation (1-iodopropane) is usually given an arbitrary abundance value of 100, called the base ion peak, and all other abundances ('intensities') are measured against it.

The base ion peak for the mass spectrum of 1-iodopropane is the m/z 43 ion [C₃H₇]⁺

Identifying the species giving the most prominent peaks (apart from M) in the fragmentation pattern of 1-iodopropane.

Unless otherwise indicated, assume the carbon atoms in 1-iodopropane are the ¹²C isotope.

Some of the possible positive ions, [molecular fragment]⁺, formed in the mass spectrometry of 1-iodopropane.

The parent molecular ion of 1-iodopropane m/z 170: [CH₃CH₂CH₂I]⁺

Analysing and explaining the principal ions in the fragmentation pattern of the mass spectrum of 1-iodopropane

PLEASE NOTE I have found it difficult to find 'authentic' equations to explain mass spectra fragmentation patterns and it is complex chemistry! I've identified the formulae of the ionised fragments on the mass spectrum diagram, but the equations are from the internet or my conjecture as to how the ions might be formed - please take care in using the information, especially for assignments at university or pre-university level.

Atomic masses: H = 1;  C = 12 (~1% 13);  I = 127

Bond enthalpies = kJ/mol: C-C = 348;  C-H = 412;  C-I = 238

Possible equations to explain some of the most abundant ion peaks of 1-iodopropane (tabulated above)

Formation of m/z 128 ion:

[CH₃CH₂CH₂I]⁺  ===>  [HI]⁺  +  C₃H₆

Elimination of hydrogen iodide from the parent molecular ion, mass change 170 - 42 = 128, very low probability, as there is only a tiny peak for the m/z 42 [C₃H₆]⁺ ion.

Note that iodine has one stable isotope, so no double peak complications you get with organo-chlorine or organo-bromine compounds.

Formation of m/z 127 ion:

[CH₃CH₂CH₂I]⁺  ===>  [I]⁺  +  C₃H₇

C-I bond scission in the parent molecular ion, loss of the alkyl group, mass change 170 - 43 = 127, to give a positively ionised iodine atom.

Both the m/z 127 and 128 ions have a low probability of formation, evidenced by relatively small peaks of very low abundance.

Formation of m/z 43 ion:

[CH₃CH₂CH₂I]⁺  ===>  [C₃H₇]⁺  +  I

C-I bond scission in the parent molecular ion, loss of the iodine atom,

mass change 170 - 127 = 43 (M-127 ion peak), to give a positively ionised iodine atom.

The C-I bond is the weakest bond in the molecule, therefore the most easily broken and not surprisingly leads to the formation of the base ion peak of highest abundance in the mass spectrum of 1-iodopropane.

The m/z 43 ion is the base peak ion, the most abundant and 'stable' ion fragment.

The m/z 43 ion can lose protons to give m/z ions from 42 down to 39.

The m/z ion 44 is similar probably formed by the same process, but one of the carbon atoms has carbon-13 isotope in it i.e. [¹³C¹²C₂H₇]⁺

Formation of m/z 27 and 15 ion:

Two possibilities, and probably others too.

[C₃H₇]⁺  ===>  [C₂H₅]⁺  +  CH₂

or

[C₃H₇]⁺  ===>  [CH₃]⁺  +  C₂H₄

These could be formed by C-C bond scission of the m/z 43 ion, either fragment could be ionised, the m/z 27 ion seems to be the more likely formed judging from their relative abundances.

A comparative footnote on the mass spectra of organic iodine compounds like 1-iodopropane

Iodine consists of 100% of the isotope ¹²⁷I, and therefore the mass spectra of organic iodine compounds does not show the complexity of the mass spectra of organo bromine or organo chlorine compounds where you are dealing with twin molecular ion/fragment peaks from due to the two isotopes:

i.e. ³⁵Cl : ³⁷Cl (3:1) and ⁷⁹Br : ⁸¹Br (1:1).

The chemistry of HALOGENOALKANES (haloalkanes) revision notes INDEX

The infrared spectrum of 1-iodopropane (n-propyl iodide)

The H-1 NMR spectrum of 1-iodopropane (n-propyl iodide)

The C-13 NMR spectrum of 1-iodopropane (n-propyl iodide)

Mass spectrometry index

ALL SPECTROSCOPY INDEXES

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