[xshadow]

Hi

I have a question:
In an energy profile is the ketone( reagent) higher or lower (in energy) than the cyanohydrine product ??

I know that the equilibrium of this reaction favours the product...
So is it correct to say that in the energy  profile the cyanohydrin is at the bottom?

Moreover  " I " convert π-bond (less stabilization) in a C-C σ-bond ....so I'll get more stabilization in this process



Thanks

[mjc123]

I know that the equilibrium of this reaction favours the product...
So is it correct to say that in the energy  profile the cyanohydrin is at the bottom?

If the equilibrium favours the product, that means that the free energy (G) is lower for the product.
Since addition to the ketone means reducing the number of molecules, I would expect ΔS to be negative.
Therefore, from ΔG = ΔH - TΔS, ΔH must be negative too.
Bond energy considerations are doubtful. You break C-H and C=O, and form C-C, C-O and O-H. Using figures from tables, which are averages and not specific to this situation, I get ΔH = -7 kJ/mol, which is small enough for me to doubt whether it is really positive or negative. Besides, this ignores interactions with the solvent.