I am facing this question from quite long time (Since 2 year
) , Got different Explanations from different teacher . But none of them satisfied me .
I will reply the explanations later which I got from different teachers .
So observation is , The order of acidity is Methanol>Water>Ethanol .
What the basic knowledge says , the order should be Water>Methanol>Ethanol , Due to inductive effect .
So first Thought came to in my mind -:
there should be some solvent effect , like in their "conjugated base form" the solvation is almost equal since basically you have to solvate O
- .
Solvation of Water will be highest , since it has 2 partially positively proton (Small in size , High charge density , better solvation) this make the ΔG reaction more positive , since reactant side is getting stabilized .
Whereas in case of Methanol,Ethanol the solvation will be poorer . So Inductive effect and Solvation have just opposite effect on acidity . So In case of Methanol it most balanced , So It has Highest acidity .
But
I got confused again when I saw the data of Proton Affinity (in KJ/mol) .
Hydroxide (1635) > Methoxide (1587) > Ethoxide (1574)
So hydroxide in gaseous phase is best base , so poorest base .
Just opposite of Inductive effect . (Note: I've read that It is
assumed that alkyl group has +I effect , The results are not clear when alkyl is connected to a saturated atom )
Also Second thing which made me double confused that
In gaseous phase the the proton affinity is
Water<Methanol<Ethanol< Propanol
As predicted ,bigger alkyl have more powerful +I .
So whats going on ?? Is there any connection with Electron affinity of Oxygen ?? As Electron affinity of O
- is negative and O (neutral) is positive (at the end this is coming in my mind) ??