O'Chem keto-enol tautomerization

[brake4country]

Homework Statement

I have a general question about keto-enol mechanisms. In a carbonyl compound, the carbonyl carbon is susceptible to nucleophilic attack. This is clearly different from what happens in base/acid catalyzed keto-enol tautomers. How can nucleophilic attack be avoided in a base solution in order to promote the keto-enol form?

Homework Equations

N/A

The Attempt at a Solution

When I looked at the examples from my book on keto-enol (a base such as OH- involved), I couldn't help but wonder why that OH- in solution didn't act as a nucleophile instead of abstracting the alpha hydrogen. Why is this and what am I missing here? Does this perhaps have something to do with the acidity of the alpha hydrogens?

 

[Ygggdrasil]

What product would form if you had a hydroxide react with the carbonyl carbon? What reactions would that product undergo?

 

[CrazyNinja]

There a lot of complex factors involved in this.

As @Ygggdrasil said, there is no point in the -OH group attacking the carbonyl group because it would simply reform the carbonyl group due to instability. In reality, the correct statement should be that - " The equilibrium constant for this reaction is quite low".

On the other hand, plucking the α-H out could lead to new possibilities. Let's denote the action of plucking of α-H by -OH as Reaction 1 (in short (1)). As most reactions in organic chemistry, this is in equilibrium. Now let's say you have another nucleophilic center in the reaction mixture ( another -CO- group ). The product formed in (1) can now act as a nucleophile for that group. Let this reaction be Reaction 2 (2).

The entire point of this is - The continuous consumption of the product in (1) as reactant in (2) drives the reaction (1) forward (Le-Chatlier's Principle). As a result, this reaction is favorable and we get a corresponding product. If (2) was not there, then (1) would again be in unstable equilibrium, leaving no scope for removal of α-H. Conversely, if the products of (2) are stable, then (1) would be preferred over other possibilities (like attack on carbonyl group).