I've synthesised a protic and deuterated version of a specific molecule where there are two deuterium atoms in the latter. I then go about and measure the kinetic isotope effect (KIE) for a reaction involving these two molecules. However, in many papers where there are multiple deuteriums per molecule it is common to quote a KIE per deuterium in addition to the observed KIE.

This is usually achieved by taking the nth root of the observed KIE where n= no. of deuteriums (see attachment for example: 10.1039/ b405075d).

My questions are:

  1. Why the nth root? What is the basis for this? No paper explains why they chose the nth root.
  2. Should one use the observed KIE or the KIE per deuterium when discussing a mechanistic proposal?

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  • 2
    $\begingroup$ It seems intuitive that if one deuterium changes the rate constant by $k_\ce{D}/k_\ce{H}$, $n$ deuteriums will change the rate constant by $(k_\ce{D}/k_\ce{H})^n$; but that seems way too simple to be true (it feels like there must be some more subtle considerations), and I'm interested to see an answer to this too. $\endgroup$ Mar 6, 2021 at 14:36
  • $\begingroup$ @orthocresol I'm not sure it's intuitive (to me) as surely the location of said deuteriums will change their relevant contributions to the rate? Consider, the rate of Z-E thermal isomerisation of d2-stilbene: in one case you have both alkene C-Hs substit. for Ds but in another case one D is at the alkene position, another on the ring. Would it still be make sense to take the sqr. root and report the KIE as per D? I'm not sure... $\endgroup$
    – Hazinga
    Mar 6, 2021 at 14:46
  • 1
    $\begingroup$ Hmm, well, I was thinking about $\ce{CH3} \to \ce{CD3}$ substitutions such as those in the table you showed. I think in the situation you raised the $k_\ce{D}/k_\ce{H}$ for each position will in general be different, as you said. $\endgroup$ Mar 6, 2021 at 14:57
  • $\begingroup$ @orthocresol As a separate question, for such an asymmetric case, are you aware of a way to partition out the contributions on a theoretical basis? $\endgroup$
    – Hazinga
    Mar 6, 2021 at 16:42
  • $\begingroup$ Nope, not at all. A speculative suggestion would be to deuterate one position at a time. You shouldn't take me too seriously, though: I'm not a phys org chemist in any way. $\endgroup$ Mar 6, 2021 at 16:44


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