Interpreting the mass spectrum of an organic molecule

Mass spectrometry is a really powerful tool for helping us to piece together the structure of an organic molecule. Ionisation of molecules causes them to fragment and these ionised fragments are recorded as peaks on the mass spectrum. If we know the molecular formula and the nature of the functional groups in the molecule (from infra-red spectroscopy), then a little detective work is all that is needed.

The first step is to ionise the sample:

X_(g) + e^– ⇾ X⁺_(g) + 2e^–

If our sample is comprised of molecules, ionisation causes many of the molecules in the sample to fragment. Each fragment that is a positively charged ion will produce a peak ion the mass spectrum.

If methyl benzoate is ionised and an electron is knocked out of the molecule we would expect to see a peak at m/z 136. This is known as the M⁺ or molecular ion peak (it will be the peak with the highest m/z ratio, bar minuscule peaks that are the result of there being ¹³C isotopes present in the molecule, which I’ll talk about later).

C₆H₅COOCH₃ + e^– ⇾ C₆H₅COOCH₃⁺ + 2e^–

The base peak is due to the fragment ion which is most commonly formed during ionisation and has a relative intensity of 100. Molecules fragment at obvious places when we look at the structure and this is where we can use a little detective work to understand the spectrum, assigning peaks to fragment ions as well as determining the groups of atoms lost when molecules fragment.

If the fragment is responsible for a peak on the spectrum it must be positively charged. The atoms lost are radicals (•). However, we need to appreciate that millions of molecules are present in the sample that generated a mass spectrum, so if one molecule fragments to form the [C₆H₅]⁺ ion and the COOCH₃^• group is lost, another may form the [COOCH₃]⁺ ion and C₆H₅^• is lost. If we look at the spectrum above, there is no visible peak at m/z 59 for the [COOCH₃]⁺ ion, so this is clearly not a stable outcome of fragmentation in this molecule.

Why are there peaks for fragments with a higher m/z than the molecular ion?

1. If you notice there is a peak at m/z 137 (and m/z 138) on the mass spectrum for methyl benzoate. This is the M⁺¹ peak and is due to the presence of a ¹³C isotope in the molecule.

Approximately 1.1% of all carbon atoms are ¹³C and we can use the ratio of the height of the M⁺¹ : M⁺ peaks, simply measured in mm, to tell us how many carbon atoms there are in the molecule, which is handy.

2. Molecules with chlorine and bromine will have M⁺² peaks that correspond to the presence of either ³⁵Cl / ³⁷Cl or ⁷⁹Br / ⁸¹Br isotopes. The relative height of the M⁺ and M⁺² peaks reflects the natural abundance of each isotope (75%:25% for chlorine and 50%:50% for bromine).

High resolution mass spectrometry

High resolution MS has the ability to measure masses to 4 decimal places and so we can use this technique to identify the molecular formula for a parent ion when there are multiple possibilities.

E.g. C₁₀H₁₆O₄, C₁₁H₄O₄ and C₁₁H₂₀O₃ all have a M_r of 200, but high resolution MS gives a M+ peak with a m/z ratio of 200.0110. Given that the accurate atomic mass of carbon is 12.0000, oxygen is 15.9949 and hydrogen is 1.0078, determine which formula is correct.

Answer: C₁₁H₄O₄

Practice questions

1. Identify the molecular ion peak and the base peak for butanone, CH₃COCH₂CH₃. Suggest a possible fragment ion for the base peak.

2. (a)  Identify the molecular ion peak for benzaldehyde, C₆H₅CHO.

      (b)  Suggest possible fragment ions for the peaks at m/z ratio 105 and 77.

3. The mass spectrum of a hydrocarbon with the molecular formula C₄H₁₀ is shown below.

(a) Draw two possible structures for C₄H₁₀

(b) Which structure has given rise to the spectrum? Explain your reasoning with reference to the fragment ions that have a m/z ratio of 58, 43, 29 and 15.

4. Compounds C and D are isomers with the molecular formula, C₃H₆O, and both contain a carbonyl group. 

(a) Draw out two possible structures for C₃H₆O

(b) Identify the group of atoms lost when the fragment ion with a m/z ratio of 43 forms in spectrum C

(c) Identify the group of atoms lost when the fragment ions with m/z ratios of 29 and 57 form in spectrum D

(d) Suggest a formula for the ion of mass 43 in spectrum C 

(e) Suggest formulae for the ions of mass 28, 29 and 57 in spectrum D

(f) Name the molecules responsible for each spectrum 

 

---

Answers

1. Molecular ion peak is at 72, base peak is at 43.  Fragment ion responsible for the base peak is CH₃CO⁺

2. (a) 106

       (b) 105 is C₆H₅CH⁺ and 77 is C₆H₅⁺

3.   (a)   CH₃-CH₂-CH₂-CH₃       and  CH₃-CH(CH₃)-CH₃

      (b) 58 [C₄H₁₀]⁺    43 [C₃H₇]⁺    29 [C₂H₅]⁺    15 [CH₃]⁺ 

The structural formula of the hydrocarbon must be CH₃–CH₂–CH₂–CH₃. Fragmentation of butane by breaking C–C bonds leads to the four ions above. The branched isomer, methylpropane, would not produce the C₂H₅+ ion by breaking C–C bonds but the other three ions will also appear in the spectrum of this isomer. 

4.   (a)  CH₃-CH₂CH-O and CH₃-(C=O)-CH₃
       (b)  58-43 = 15 (CH₃)  –  note no positive charge, this is not the fragment ion but the atoms / group lost

     (c)  58-57 = 1 (H);  57-29 = 28 (CO or C₂H₄) –  note no positive charge, these are not the fragment ions but the atoms / groups lost

      (d) CH₃CO⁺

      (e) 28 is CO⁺ or C₂H₄⁺;  29 is CH₃CH₂⁺; 57 is CH₃CH₂CO⁺

      (f) C is propanone and D is propanal

Ready to take it to the next level? Download a booklet of exam style questions with perfectly structured, fully explained answers and exam tips. Preparation is the key to success!