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Introductory note on the mass spectrum of propylbenzene

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

If M represents the propylbenzene molecule, the initial ionisation to give the molecular ion is:

M_(g) + high KE e^-  ==> [M•]⁺_(g) + 2e^- and for 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 that might be formed in the fragmentation pattern for the mass spectrum of propylbenzene and only the formation of singly charged positive are considered for the mass spectrum of propylbenzene.

I've included a stick diagram and table of m/z ions for the mass spectrum of propylbenzene and doing the mass spectrum analysis under standard conditions, databases can be compiled based on complex fingerprint patterns, often involving the relative intensities of many fragment ions, and used to identify compounds including propylbenzene.

In selected cases, where two different fragment ions have the same integer m/z value, I've pointed out that modern mass spectrometers can measure relative ion mass to four decimal places. So, using accurate isotopic masses, I've calculated the accurate ion masses, BUT strictly speaking, 0.0005 should be deducted for singly charged ions to account for the loss of the electron in their formation. I have NOT done this for propylbenzene, but the mass spectrometer software does!

propylbenzene, 1-phenylpropane, C₉H₁₂ , C₆H₅CH₂CH₂CH₃ ,   ,  

Interpreting the fragmentation pattern of the mass spectrum of propylbenzene

[M]⁺ is the molecular ion peak (M) with an m/z of 120 corresponding to [C₉H₁₂]⁺, the original propylbenzene molecule minus an electron, [C₆H₅CH₂CH₂CH₃]⁺

The small M+1 peak at m/z 121, corresponds to an ionised propylbenzene molecule with one ¹³C atom in it i.e. an ionised propylbenzene molecule of formula [¹³C¹²C₈H₁₂] (see also m/z 92 ion)

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.

Propylbenzene has 9 carbon atoms, so on average, ~1 in 11 molecules will contain a ¹³C atom.

The most abundant ion of the molecule under mass spectrometry investigation (propylbenzene) 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 propanal is the m/z 91 ion [C₇H₇]⁺

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

The parent molecular ion for propanal is the m/z 120 ion [C₆H₅CH₂CH₂CH₃]⁺  or   [C₉H₁₂]⁺

Analysing and explaining the principal ions in the fragmentation pattern of the mass spectrum of propylbenzene

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 (13 for ~1 in 100)

Bond enthalpies = kJ/mol: = 518 (benzene ring); C-C = 348;  C-H = 412

Possible suggested equations to explain the most abundant ion peaks of the propylbenzene mass spectrum

There are many fragments and lots of possible fragmentation process, some are more favoured than others, but this becomes university level analysis!

Formation of m/z 105 ion

[C₆H₅CH₂CH₂CH₃]⁺  ===>  [C₈H₉]⁺  +  CH₃

C-C bond fission in the parent molecular ion and loss of methyl group from the end of the alkyl side-chain.

The m/z 105 ion is the base peak ion, the most abundant and 'stable' ion fragment, formed by the loss of a methyl group from the parent molecular ion.

Mass change 120 - 15 = 105 (M=15 ion peak)

Formation of the m/z 91 ion

[C₆H₅CH₂CH₂CH₃]⁺  ===>  [C₆H₅CH₂]⁺  +  CH₂CH₃

or more simply (since structures might not be quite the same as in the original molecule)

[C₉H₁₂]⁺  ===>  [C₇H₇]⁺  +  C₂H₅

C-C bond fission in the parent molecular ion and loss of ethyl group from the end of the alkyl side-chain.

mass change 120 - 29 = 91 (M-29 ion peak)

The m/z 92 ion theoretically is the [C₇H₈]⁺ ion, but I think it is more likely to be the [¹³C¹²C₆H₇]⁺ ion, formed by the process described above since around 1 in 11 molecules or fragment ions has a crabon-13 atom in it.

Note that an accurate mass spectrometer can sort out ions with the same integer m/z value because they can measure relative fragment ion masses to four decimal places e.g. using accurate relative isotopic masses:

¹H = 1.0078  ¹²C = 12.0000  ¹³C = 13.0034, from which you can calculate (predict) that the accurate relative ion masses are:

For m/z 92: [C₇H₈]⁺ = 92.0264 or  [¹³C¹²C₆H₇]⁺ = 92.0580, a relative ion mass difference of 0.0316.

Formation of the m/z 77 ion

[C₆H₅CH₂CH₂CH₃]⁺  ===>  [C₆H₅]⁺  +  CH₂CH₂CH₃

C-C bond fission with loss of the alkyl side chain.

Mass change 120 - 43 = 77 (M-43 ion peak)

This is a phenyl cation and very characteristic in the mass spectrum of aromatic benzene compounds like propylbenzene.

It could also be formed from the m/z 91 ion by C-C bond scission and loss of CH₂

[C₇H₇]⁺  ===>  [C₆H₅]⁺  +  CH₂

mass change 91 - 14 = 77

The m/z 78 ion can be formed involving proton rearrangement in larger fragment - apparently a characteristic ion in the mass spectra of aromatic benzene compounds like propylbenzene.

Formation of m/z ion 65  ?

[C₇H₇]⁺  ===>  [C₅H₅]⁺  +  C₂H₂

Elimination of ethyne from the m/z 91 ion

mass change 91 - 26 = 65

Formation of m/z 51 ion

Loss of CH₂ from the m/z 65 ion?

[C₅H₅]⁺  ===>  [C₄H₃]⁺  +  CH₂

mass change 65 - 14 = 51

Formation of m/z 39 ion

[C₅H₅]⁺  ===>  [C₃H₃]⁺  +  C₂H₂

Elimination of ethyne from the m/z 65 ion

mass change 65 - 26 = 39

but other possibilities too.

The infrared spectrum of propylbenzene

The H-1 NMR spectrum of Propylbenzene

TheC-13 NMR spectrum of Propylbenzene

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Website content © Dr Phil Brown 2000+. All copyrights reserved on revision notes, images, quizzes, worksheets etc. Copying of Doc Brown's pre-university advanced level chemistry website material is NOT permitted. Exam revision summaries & references to science course specifications are unofficial. These organic chemistry revision notes on spectroscopy (on the infrared spectrum of propylbenzene, 1-phenylpropane) are suitable for use of pre-university students studying AQA advanced level chemistry, Edexcel advanced level chemistry, OCR advanced level chemistry, IB advanced level chemistry, WJEC (Eduqas) advanced level chemistry, CIE advanced level chemistry, CCEA advanced level chemistry, US grade 11-12 AP honors chemistry courses and they will also prove useful to 1st year undergraduate students of chemistry.