Quite a basic QM quesiton:
I have an isolated system (in vacuum). The nitrogen in the schiff base of my reactant will become protonated from a neighbouring water molecule. My product is thus a R-C=NH+-R with a neighbouring OH-
The charge and spin for my reactants are 0 and 1, respectively. The charge and spin for my product is 0 and 3. (odd electron on the OH- and from the transfered proton to the nitrogen)
My question: should the spin multiplicity be identical between the reactant and product? I'm now just wondering how do I define the spin for my Transition State.
Which of the product states (single or triplet) has lower free energy? If in fact the multiplicity changes between the ground state of the reactant and that of the products, you may have a spin non-adiabatic reaction on your hands, and you have to calculate the probability of the spin flipping at the crossing point of the two surfaces.
Dear Anthony,
The the spin multiplicity in reactants, intermediates, transition sate and product should be the same, 1.
The charge is a different story, for the protonated entity, in your case the imminium ion the charge should be 1. In the reactant, imine, and product the charge should be zero.
Hoping this will be helpful,
Rafik
Dear Anthony,
The the spin multiplicity in reactants, intermediates, transition sate and product should be the same, 1.
The charge is a different story, for the protonated entity, in your case the imminium ion the charge should be 1. In the reactant, imine, and product the charge should be zero.
Hoping this will be helpful,
Rafik
Multiplicity 3 means, that you are breaking some bond via radical path. Though If you are transfering hydrogen atom (proton+electron), this would be still singlet, but multiconfigurational singlet (say hi to casscf). Transfer of proton wouldn't change anything, it would be still singlet
Dear Rafik,
Many thanks for your help. I understand that both states (reactant and product) have a zero charge i.e., reactants: neutral water + neutral schiff base, products: OH- and protonated schiff base (-N+(H)=C).
I also understand that in the above case the spin for the reactant will be a singlet. However, I don't understand how the spin in the case of the product could be a singlet. The charge of the product system is 0 (+1 on the imini and -1 on the hydroxide ion, give a net 0 charge), which comes to a spin triplet (one unpaired electron on the oxygen and one unpaired on the proton bonding with the nitrogen - giving a triplet).
I've tried running on a geometry optimisation on the product as a singlet and as a triplet spin. The triplet spin produces a OH & -N(H)=C structure. But the singlet, the product reproduces the reactants i.e., HOH & -N=C.
I appreciate your thoughts.
Many thanks
Anthony
Anthony,
(OH)- have charge -1 and no unpaired electrons, so does protonated product - charge +1 and no unpaired electrons. No unpaired electrons - singlet.
Try optimization with PCM, it will stabilize charged particles a bit.
Dear Anthony,
Anions and cations all are in a singlet state. In the case of OH- the two spins are in opposite direction. On the other hand, carbenes which have two electrons can exist in a singlet or triplet state; in the singlet sate the direction of two spins is one opposite to the other; in the case of triplet the two spins have the same direction.
If needed I can send you some publications on the singlet and triplet states.
Hoping this will be helpful,
Rafik
Dear Rafik,
Those publications on singlet and triplet state would be extremely helpful.
Many thanks for your help.
Mr. Nash,
Please pay attention to the attachment showing a fast preliminary computations (not the whole set), using R = CH3, which however repeat your data for GP. There is an indication for a H-transfer HOH...to...N (optimization step 4), but your picture becomes TS (attachment also). The data in solution (opt=gdiis, scrf=(pcm,solvent=water); attachment) however have shown H-transfer HOH...to...N and HOH...to...OH(...H) (second figure):
------------- VARIATIONAL PCM RESULTS -------------
(Polar.Solute)-Solvent (kcal/mol) = -25.49
------------------------------------------------------
Cavitation energy (kcal/mol) = 21.55
Dispersion energy (kcal/mol) = -19.90
Repulsion energy (kcal/mol) = 4.07
Total non electr. (kcal/mol) = 5.72.
So, may be the nature of R-should be varied within a large scale of possible substituents with different M+ and I- properties, including conjugation, in order to obtain a realistic information about the mechanism of this reaction by quantum chemistry, along with GP and solution computations (a comparative analysis).
Which of the product states (single or triplet) has lower free energy? If in fact the multiplicity changes between the ground state of the reactant and that of the products, you may have a spin non-adiabatic reaction on your hands, and you have to calculate the probability of the spin flipping at the crossing point of the two surfaces.
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