[curiouscat]

Below is a typical explanation. The point that confuses me is that the ZPE must reduce in both the Reactant and Transition state vibrations, right? So how do we trivially claim the barrier always reduces? Won't this depend on which ZPE decreased by more?


" The mass of an atom affects the vibrational frequency. Heavier atoms will lead to lower vibration frequencies and will have lower zero-point energy (ZPE). With a lower zero-point energy, more energy must be supplied to break the bond, resulting in a higher activation energy for bond cleavage, which in turn lowers the measured rate"

[Babcock_Hall]

A good place to start might be Frank Westheimer's 1961 review.

The Magnitude of the Primary Kinetic Isotope Effect for Compounds of Hydrogen and Deuterium.
F. H. Westheimer
Chem. Rev., 1961, 61 (3), pp 265–273
Publication Date: June 1961 (Article)
DOI: 10.1021/cr60211a004

"But this [incorrect] argument, as stated, is in direct conflict with the theory of absolute rates (18). The derivation of Eyring’s equation requires that the vibration in question become translation, and no compromise with this point is possible which will preserve the essential outlines of the transition-state theory...The bond which is present in the transition state, and which contributes to its zero-point energy, is one which does not exist in either the reactants or the products; it is a vibration peculiar to the activated complex."  I am far enough outside my comfort zone to want to turn the question over to physical chemists at this point.

[curiouscat]

A good place to start might be Frank Westheimer's 1961 review.

The Magnitude of the Primary Kinetic Isotope Effect for Compounds of Hydrogen and Deuterium.
F. H. Westheimer
Chem. Rev., 1961, 61 (3), pp 265–273
Publication Date: June 1961 (Article)
DOI: 10.1021/cr60211a004

"But this [incorrect] argument, as stated, is in direct conflict with the theory of absolute rates (18). The derivation of Eyring’s equation requires that the vibration in question become translation, and no compromise with this point is possible which will preserve the essential outlines of the transition-state theory...The bond which is present in the transition state, and which contributes to its zero-point energy, is one which does not exist in either the reactants or the products; it is a vibration peculiar to the activated complex."  I am far enough outside my comfort zone to want to turn the question over to physical chemists at this point.

Thanks for the tip. I had read something like that (but not this piece) before but I never understood what they mean. 

[Babcock_Hall]

One simple generalization is that the heavier isotope enriches in the stiffer bond.  For secondary isotope effects, this sometimes produces inverse effects, where the heavier isotope leads to a faster reaction.  I am not sure whether this helps or not.

[curiouscat]

One simple generalization is that the heavier isotope enriches in the stiffer bond.

What do you mean by "enriches" in this context? More higher freq. modes? Yes, but in both the Initial, and Transition states? The barrier is only based on the difference which could either go up or down.

Maybe I misread your statement.

For secondary isotope effects, this sometimes produces inverse effects, where the heavier isotope leads to a faster reaction.  I am not sure whether this helps or not.

I can see how the effect happens, no doubt. But predicting whether it will be direct or inverse is where I'm stuck at. Yet, people seem to state in an obvious manner that r_H > r_D , at least for the simpler reactions. They must be right but I don't get it yet.

[Babcock_Hall]

I may have more time this evening to do this topic more justice.  By enrich, I meant preferentially populate the ground state versus the transition state.  If the heavier isotope enriches in the ground state, the isotope effect will be normal.  If the heavier isotope enriches in the transition state, the isotope effect will be inverse.

[Babcock_Hall]

In "Mechanism and Theory in Organic Chemistry" Lowry and Richardson obtain expressions for the zero point energies of the reactants and transition state on pp. 222-223.  "The transition state sum omits the reaction coordinate degree of freedom since it is not a bound vibration and does not contribute to the zero-point energy in the transition state."  It seems to me that one vibration is disappearing in the transition state, but the t.s. may have unrelated vibrations of its own.

On page 211, they discuss secondary isotope effects.  When the hybridization of a carbon atom changes from sp² to sp³, the hydrogen/deuterium isotope effect should be inverse.  I would also predict that when the ground state contains a carbon-oxygen single bond and the transition state has a carbon-oxygen double bond, that the C-13 and the O-18 isotope effects will be inverse.

[curiouscat]

Thanks! I'm still thinking it over.