Question about the viability of a chemical reaction

[sgstudent]

For organic chemistry we typically look at the stability of the end product to determine if the reaction will proceed. For instance for an Sn2 reaction we would check if the product anion is more stable or less stable than the nucleophile attacking - if the product anion is more stable than the nucleophile then the reaction would proceed e.g. R₃COH+F^- --> R₃CF+OH^- is viable as OH- is more stable than F- so the reaction would occur.

1) however what I don't understand is why we don't compare the stability of the neutral species; in the example they were R3CF and R3COH.

2) this matters because in the case of polar protic solvents F- becomes more stable than OH-. So in this chemical reaction: R₃COH+F^- --> R₃F+OH^- in a polar protic solvent, shouldn't this reaction be considered viable since F- is more stable than OH-? However I thought this reaction is still not viable despite the increased stability of F-?

 

[hilbert2]

The equilibrium constant of any reaction, and therefore the direction the reaction proceeds to, depends on the Gibbs energies of formation of all reactants and products. Therefore the stability of the electrically neutral species also affects the position of the equilibrium. Also note that the solubility of the product ions can be a significant factor: sometimes alkyl halides can be converted to alkyl nitrates with silver nitrate, because the precipitation of insoluble silver halide salts drives the reaction forward.

 

[TeethWhitener]

2) this matters because in the case of polar protic solvents F- becomes more stable than OH-. So in this chemical reaction: R3COH+F- --> R3F+OH- in a polar protic solvent, shouldn't this reaction be considered viable since F- is more stable than OH-? However I thought this reaction is still not viable despite the increased stability of F-?

The main reason this reaction doesn't work (in either direction) has more to do with kinetics than thermodynamics. Both F^- and OH^- are terrible leaving groups.

 

[Dr Uma Sharma]

It is the overall stability of all species on the product side which will determine the spontaneity of the reaction.Ions or charged particles are always dominating species so their stability factor will matter most.

 

[Steven Hanna]

For organic chemistry we typically look at the stability of the end product to determine if the reaction will proceed.

First, this is not a good strategy to determine if a reaction is therodynamically favorable. A reaction is thermodynamically favorable if all of the products are more stable than all of the reactants. So in your example, you should compare the stability of R₃COH + F^-
to that of R₃CF + OH^-.

Now I'll try to answer your questions.

1) We usually compare the stabilities of anions, why don't we compare the stabilities of neutral species?
Answer: You need to compare the stabilities of both things: neutral species and anions.
You likely first learned this anion comparison strategy when you learned about acids and bases. If you want to compare the acidities of two things, you can just compare the stabilities of the conjugate bases. This is because for an acid base reaction you have HA + H2O ⇌ A^- + H3O⁺. If you have two of these reactions and want to subtract them to determine the difference in their free energies, the H2O's and H3O⁺'s cancel each other out. It's worth noting that it is perfectly valid to predict relative acidities by comparing the relative stabilities of two neutral HA's. You'll arrive at the same answer either way. It is more common to compare two A^-'s because chemists are familiar with the factors that stabilize/destabilize charges, e.g. inductive effects, resonance, solvation, etc.
This strategy falls apart when you try to look at a single reaction. In your example, looking at the stability of OH- vs F- ignores the differences in C-F vs. C-O bond energies. Thus, you need to look at both neutral species and charged ions. That being said, it is common to just look at anions and just assume the neutral species are approximately equal in energy, but this assumption is not always valid. This assumption is often made because ions are quite high in free energy in organic solvents, as their charges are not well screened. Thus the overall change in free energy of the reaction is assumed to be driven by factors that stabilize the charge.

2) "in the case of polar protic solvents F- becomes more stable than OH-. So in this chemical reaction: R₃COH + F^- --> R₃F+OH⁼ in a polar protic solvent, shouldn't this reaction be considered viable since F- is more stable than OH-? However I thought this reaction is still not viable despite the increased stability of F-?"
I think there's a flaw in your logic. If fluoride ion is more stable in protic solvents, this would be a reason that the reaction isn't viable. The more stable the reactants, the less favorable the reaction.

Hope this helps.