SN2 Mechanism Mode Specifity
Hey there,
So I was reading this new article on new simulations for different SN2-mechanisms in the gas phase (so simple model). One of the key facts they wanted
to find out, as far as I understood, was if the reaction uses specific vibrational modes.
In order to form a Pre-complex, where the Nucleophile has some sort of interaction (Ion-Dipole, H-bonds,...) translational energy of the Nuc has to be
transformed into the other free degrees of the molecule. Also if this reaction should happen this is required. If in the gas phase my Nuc enters with
a lot of kinetic energy I need to transfer the translational as in Nuc-CH3 the Nuc can't have a different translation than the CH3 or they will
dissociate.
The question was if these energy transfers into the vibrational modes of the molecule are specific. And the calculations proofed that if the energy is
transfered into the "umbrella"-deformation mode the reaction is favored. I think this is due to the similarity to the Walden-Inversion which happens
in the SN2 as well. So if we already put energy into this motion we favor the reaction. Especially as, if you look at the more complex and detailed
reaction coordinates for SN2 mechanisms, even the steps where the Hydrogen in CH3 need to bend from the tetrahedral angle to the trigonal plane
coordination in the transition state. Although the energy is quite small it exists and if I can already put energy into the system it might help to
add it to this motion.
The other, even better mode to excite would be the C-L (L=Leaving group) vibration. Somehow it seems logic to put more energy into this vibration so
it might break but can I really reduce it to such an easy thought?
I remember from Spectroscopy and non-harmonic oscillators that you cannot stretch a bond forever and putting energy into the system is just stretching
it even more. At some point it will face dissociation. So is this the reason why it is better to transfer the translational energy into the C-Y
vibrations ?
Thanks in advance for all answers,
Greetings,
fluorescence
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