Understanding Isotopic Enrichment
Introduction
Advances in mass spectrometry have led to increased interest in materials labeled with stable isotopes. More researchers have become interested in the relative abundances of molecules enriched in stable isotopes. Questions arise about nomenclature and conventions used in stable isotope chemistry. Here we will describe our specifications for enrichment and how they are related to calculations of theoretical distributions for isotopologuesa.
Isotopic Distributions
Let’s say, we want to know how many molecules will have a certain number of labeled atoms. When we compare the relative amounts of molecules with different numbers of labeled atoms we have the distribution of isotopologues. We need to know the maximum number of atoms that may be labeled. This is the number of labeled atom “sites” in the molecule.
Enrichment
We need to know the probability of finding a labeled atom at any single site. This is the “enrichment”. Enrichment is specified as a percentage. The enrichment is not the same as probability that a molecule will be fully labeled.
Examples
Glycerol (D5, 99%) has five labeled sites. In this case, the subscript 5 stands for the number of sites. Up to five of the hydrogen atoms in glycerol may be replaced by deuterium atoms. The chance of finding a deuterium atom at any of these sites is 99%.
Benzene (13C6, 99 %) has 6 labeled sites. The probability that any carbon atom will be 13C is 99%.
L-tryptophan (15N2, 98 %) has 2 labeled sites. L-tryptophan has 2 nitrogen atoms. There is a 98% chance that either of these N atoms will be 15N.
L-Tryptophan (15N2, 98%)
Atomic Combinations
We may also want to know how many L-tryptophan molecules have 2 labels (“N2”), how many have 1 label (“N1”), and how many have no labels (“N0”). Ignoring carbon, hydrogen, and oxygen we see the combinations of nitrogen isotopes that may occupy the 2 sites in tryptophan, 15N2, 15N14N, and 14N2, respectively. In this case, the subscripts within the quotation marks stand for the number of labeled atoms. This can be confusing. We need to know the context to tell if the subscripts refer to the number of sites or atoms occupying the sites.
We can determine a relative distribution where some percentage will be N2, some will be N1, and some will be N0. There are 3 possible combinations for 2 labelsb. The percentage with both N atoms labeled is not the same as the enrichment of L-tryptophan-D2, although common notation may suggest this. In most cases, the enrichment per atom site is greater than the percentage of fully labeled molecules.
In terms of probability, two options are possible for each stable isotope we have discussed (e.g., 13C or 12C). For example, two “outcomes” (labels) are possible at each site. This leads us to apply the binomial theorem to the problem of calculating the probability that a particular combination of labeled and unlabeled atoms will occur. Labeling a site can be considered a “trial” in statistical terms.
The binomial coefficient
The binomial coefficient calculates the number of ways to pick k unordered outcomes, from n possibilities. This is sometimes referred to as “n choose k”.
For positive integers, n, the binomial theorem gives equation 2. The standard notation has been defined in terms of the problem of isotopic distributions.
x is the enrichment as a decimal number (1 = 100 %)
a is the probability of having an unlabeled atom at a given site (1 – x)
n is the number of labeled sites
k is the number of labeled atoms
The binomial expansions below describe cases with (n = ) 2, 3, and 6 labeled atom sites, respectively.
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Each term in the summations above corresponds to the probability of a different combination of atoms. Applying equation 2, we use the enrichment per atom site as x. The variable a depends upon x. If x is known, a is known.
Applying the above equation we determine the following distribution pattern. The complete table has been abbreviated to only show up to a maximum of 6 labeled sites. Note this is the distribution where x, the enrichment per atom site, is 99%
When we apply these calculations in terms of the 13C6-benzene with 99% isotopic enrichment, we find the following actual enrichment for the 7 isotopologues:
Note that the abundance becomes extremely low when more than 2 or 3 labeled atoms are “missing” from their respective sites. The bar graph below is similar to a mass spectrum. The ion intensities would correspond to the height of the bars here.
By comparison, when isotopic enrichment is decreased by just 1%, for example 98% instead of 99%, a significant change occurs in the relative abundance for each isotopologue.
Therefore, in general terms it is desirable to have the highest isotopic enrichment possible in order to obtain the greatest relative abundance of the fully labeled isotopologue.
Conclusions
These isotopologue distribution tables show a number of points made earlier. The “width” of the distribution depends upon the number of labeled sites in the compound. The percentage of fully labeled molecules does not equal the enrichment. The percentage of labeled atoms per atom site is greater than the percentage of fully labeled molecules.
aIsotopologues are molecular species that differ only in isotopic substitution (e.g., H2O and D2O. Isotopologues may have different numbers of substituted atoms (D for H, 13C for 12C, etc.). They can easily be confused with isotopomers, which have the same number of isotopic atoms, but differ in their placement of isotopic atoms.
bFrom a chemical viewpoint, the two nitrogen atoms are different. Exchanging the 15N in the amine group with the 14N of the ring would result in isotopomers that could be distinguished, one from another. In that sense, there are four possible combinations. For our purposes here we consider 15N14N and 14N15N to be indistinguishable.
