3
$\begingroup$

When computational studies are performed on transition state geometries, the "thermodynamic" (that is, with no kinetic considerations) energy of the analyzed state is obtained.

I was wondering: if a (multistep) reaction begins in a biradical state, the kinetics of the interconversion from triplet to singlet is so slow that the reaction is likely to proceed in a radical fashion, or if a lower transition state exists for the ionic (singlet) counterpart, then it is likely to proceed in a ionic way, because the rate of interconversion is extremely fast?

I am sorry if the question is dumb, but all I could find was about transition metals, and I don't know if the same applies.

$\endgroup$

2 Answers 2

3
$\begingroup$

It depends on the organic molecule.

When spin-orbit coupling facilitates intersystem crossing, it can be fast: 100 femtoseconds for nitronapthalene (viz. Chem. Eur. J. 2018, 24, 5379 – 5387).

The Wikipedia article on intersystem crossing cites a timescale for radiative decay (phosphorescence) of triplet to singlet of 10−8 to 10−3 s.

$\endgroup$
2
$\begingroup$

Spin-orbit coupling will cause the spin flip needed. This is always present but becomes larger the heavier the atom are, iodine, xenon for example present either in solvent (such as iodobenzene) or as part of the molecule. Paramagnetic species also work in the same way, e.g. O2, NO as does an external magnetic field because, as the molecule tumbles, the apparent changing magnetic field the molecule experiences causes T-S crossing.

How fast intersystem crossing is will depend largely on the energy gap between singlet and triplet; the larger the gap the slower the process, and as the strength of spin-orbit coupling increases so does the rate constant. This 'energy gap law' has been confirmed by many experiments on excited states and of aromatic molecules and metal complexes and has the form $k=Ae^{-\beta\Delta E}$ where $A$ depends on the strength of the interaction and $\beta$ on the properties of the molecule, this varies as $\ln(\Delta E)$ so is almost constant as $\Delta E$ changes, typical values are 3 to 5.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.