Mass Spectrometry: Isotope Effects
- Page ID
- 330
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The example above is simple, but the same methods can be applied to determine isotope peaks in more complicated molecules as well. The molecule C4Br1O2H5 has several isotope effects: 13C, 2H, 81Br, 17O, and 18O all must be taken into account. First we will look at the (M+1)+ peak in comparison with the M+ peak. Only isotopes that will increase the value of M by 1 must be taken into consideration here – since 81Br and 18O would both increase M by 2, they can be ignored (the most abundant isotopes for Br and O are 79Br and 16O). Like the previous example, there are 1.08 13C atoms for every 100 12C atoms. However, there are 4 carbon atoms in our molecule, and any one of them being a 13C atom would result in a molecule with mass (M+1). So it is necessary to multiply the probability of an atom being a 13C atom by the number of C atoms in the molecule. Therefore, we have:
4C * 1.08 = 4.32 = molecules with a 13C atom per 100 molecules
We can repeat this analysis for 2H and 17O:
5H * 0.015 = 0.075 = molecules with a 2H atom per 100 molecules
2O * 0.04 = 0.08 = molecules with a 17O atom per 100 molecules
Any of the three isotopes, 13C, 2H, or 17O occurring in our molecule would result in an (M+1)+ peak. To get the ratio of (M+1)+/M+, we need to add all three probabilities:
4.32 + 0.075 + 0.08 = 4.475 = (M+ 1)+ molecules per 100 M+ molecules
We can say then that the (M+1)+ peak is 4.475% as high as the M+ peak.
A similar analysis can be easily repeated for (M+2)+:
1Br * 98 = 98 = molecules with an 81Br molecule per 100 molecules
2O * 0.2 = 0.4 = molecules with an 18O molecule per 100 molecules
98 + 0.4 = 98.4 = (M+2)+ molecules per 100 M+ molecules
The (M + 2)+ peak is therefore 98.4% as tall as the M+ peak.
This method is useful because using isotopic differences, it is possible to differentiate two molecules of identical mass numbers.
Contributors and Attributions
- Morgan Kelley (UCD)