# How to calculate m/e value for the base peak in mass spectroscopy?

My working out:

Propan-1-ol : CH3CH2CH2OH

This can be broken up into:

• CH3 + CH2CH2OH
• CH3CH2 + CH2OH
• CH3CH2CH2 + OH

Base peak = most abundant fragment formed (highest peak)

How do i know which fragment is the "most abundant?"

After I know the fragment structure, i can work out the molar mass, and then deduce the m/e value on the assumption of a (+1 charge).

Here is the exam paper solution:

• The answer is right in front of you. What is the total mass of propan-1-ol? What is the mass of (propan-1-ol -18)?
– LDC3
Apr 12, 2014 at 2:56
• molar mass of propan-1-ol = 60.10 ?? Apr 12, 2014 at 3:05
• Now subtract 18.
– LDC3
Apr 12, 2014 at 3:19
• 42 ? why am i subtracting 18 ?? Apr 12, 2014 at 3:22
• You are subtracting the mass of water from the alcohol. Water is likely to be lost when the molecule is fragmented.
– LDC3
Apr 12, 2014 at 3:33

I would say that the question is not quite well formulated and that the answer that you quote is even further misleading in giving the proper answer.

To start, let's consider fragmentation of primary alcohols, which is the question at hands here. First of all, can we rule out any of these answers?

Propan-1-ol undergoes complete fragmentation in a mass spectrometer.

This sentence means that we can rule out the molecular ion $M^{\bullet +}$ as a base peak in the spectrum. Propan-1-ol has a monoisotopic peak at $m/z$ 60 so it rules out answer D.

Now we apply the standard fragmentation rules for alcohols: consider ionization on the oxygen lone pair, we have two possible $\alpha$-cleavage leading to $m/z$ 31 and $m/z$ 59. An additionnal side reaction could also be hydrogen transfer to the oxygen atom, followed by water loss. This would lead to an ion at $m/z$ 42.

In the options above, should we thus completely rule out $m/z$ 29? I think that in the spirit of those who asked the exam question, the answer is yes, and we should rule out answer A. But see below...

We are now left with two options which can be explained by the fragmentation chemistry. Which should we choose? Technically, the problem here is that now we rely on abundances and not on the simple occurence of these fragments anymore. This is much more tricky, as there are no absolute rules. If we have a look at good textbooks, we find that loss of the largest radical is favored. Therefore, we have here to compare loss of $H^{\bullet}$ vs loss of $CH_{3}CH_{2}^{\bullet}$. Thus the answer is B.

# Is the above answer complete?

The problem in such a question (and in the answer that is given below) is that it assumes that in mass spectrometry one could correctly predict fragment ion abundances based on very simple rules. And, as I tell my students, currently, even if you were able to map out the complete potential energy surface of the fragmentation pathway and do some RRKM calculations on the transition states, you would probably be off by a factor of 2 from the experimental results in terms of abundances. So asking for predicting the base peak (i.e. the highest abundance peak in a spectrum) can be quite wrong when two fragmentation channels are quite close to each other.

Let's have a look at the experimental electron ionization spectrum. Well OK, the base peak is indeed $m/z$ 31, so we are not too bad.

But let's have a deeper look. Remember we crossed-out $m/z$ 29 as a possible base peak, and it happens to be the second most abundant peak in the spectrum, ahead of $m/z$ 42 and $m/z$ 59 which followed the basic rules!

$m/z$ 29 could be either $HCO^{+}$ or $C_{2}H_{5}^{+}$. The former could be a secondary fragmentation of $m/z$ 31 after $H_{2}$ loss, whereas the latter could be produced as a heterolytic bond cleavage of the C-C bond. (In fact it is the same bond elongation leading to $m/z$ 31, except that they electron remains on the opposite side.) Energetically, these two channels differ by the difference in ionization energy between the $C_{2}H_{5}^{\bullet}$ and $^{\bullet}CH_{2}OH$ radicals, i.e. 8.11 eV vs 7.56 eV. Although it amounts to almost 100 kJ/mol, with the fairly large energy distribution in electron ionization, this is still possible.

And for the second rule that we applied, is it always true that the largest radical if formed? This empirical rule works out quite well, even more so when the comparison is between a hydrogen radical (which is quite difficult to produce) and a carbon atom centered radical.

# And how about the exam paper solution

Here I am very highly surprised by what they seem to affirm. OK as we just saw above, $m/z$ 31 is a quite characteristic peak of primary alcohols. And loss of water is also characteristic of alcohols having side chains longer than 3 carbons. But does it mean that these should be the base peaks?

Let's have a look at the series of small primary alcohols:

Alcohol              m/z 31        Water loss           Other base peak
Ethanol              100            10
Propan-1-ol          100            15
Butan-1-ol            95           100
2-methylpropan-1-ol   38            10                  m/z 43
Pentan-1-ol           38            50                  m/z 42


So I do not see the rational behind this answer. It starts failing as soon as the small alcohol sizes reaches four carbon atoms. Of course there is a rational behind each of these base peak (for instance, in the case of 2-methylpropan-1-ol, the ionization energy of the isopropyl radical is of 7.37 eV, i.e. lower than that of $^{\bullet}CH_{2}OH$ and thus it becomes the most abundant ion, with exactly the same bond cleaved), but you have to go through the mechanisms, the thermochemistry and sometimes even the transition states to get to the global picture.