[Nescafe]

Hello,

Is it possible for a compound to have -LogP and does this mean that it is incredibly polar?

LogP = [compound in octane]/[Compound in water]

So yes mathematically if it is more polar in water then it will lead to a -ve logP, can anyone confirm if I am right please?

Thanks,

Nescafe.

[kriggy]

Yes, it is posible.
if the concentration is equal in both phases (octano/water = 1) then logP = O. If its higher in water then its negative. So IMO it doesnt have to be "incredibly polar"