[AlbertoA]

When you have a ketone like propan-2-one wich alpha hydrogen will be more acidic, the one who is at the therminal C or in 3-C?

By the way if you use LDA, you'll have the kinetic product deprotonating the therminal C?
and deprotonating the 3-C would give the thermodinamic product?
just a guess, I think it's wrong

[voidSetup]

Propan-2-one (acetone) is symmetrical, the alpha hydrogens should have the same reactivity. Do you mean butan-2-one?

[AlbertoA]

I didn´t see that i wrote that! I'm so stupid!
but yes, I meant butan-2-one

[Exploring]

The alpha acid is the one adjacent to the carbonyle but not the terminal one, but the substituted one (the secondary, not the tertiary)
R-CH2-CO-CH3
alpha carbons have more acidic Hs than terminals

[Nosterius]

Exploring: both are alpha hydrogen atoms. "Alpha" simply means that it is directly on the first carbon atom(s) adjacent to the carbonyl group. Butanone has a total of 5 alpha hydrogen atoms.

I guess you found this information on the wiki page: http://en.wikipedia.org/wiki/Alpha_and_beta_carbon, but I find it missleading.

[voidSetup]

Typically a primary carbanion is more stable than a secondary or tertiary one which would make the alpha hydrogen on the terminal carbon more acidic. However, when a hydrogen that is alpha to a carbonyl is removed, an enol is formed so I think you look at the stability of the alkene that is formed. With an aldehyde you only have one option, ketone there is two.  Look up enol stability, there is another situation involving a hydrogen alpha to two carbonyls that is stabilized by resonance and hydrogen bonding.

[Nosterius]

You should check in the Carey Sundberg part B. In the 4th edition, the first chapter is dedicated to the alkylation of nucleophilic carbon intermediates, which turns out to be mostly about enolates. A key sentence is: "Ideal conditions for kinetic control of enolate formation are those in which deprotonation is rapid, quantitative and irreversible".

Using LDA (a bulky strong base) on butanone you should be able to deprotonate quantitatively the least hindered carbon atom, which turns out to be the terminal methyl group. However, this is rarely perfect and some deprotonation of the most hindered carbon atom occurs at the same time.

However, you need to use a slight excess of strong base for this to be true. In the case where the ratio of LDA: ketone would be slightly below 1.0, you would end up with a mixture of the kinetic enolate and some leftover ketone. The kinetic enolate could then deprotonate the leftover ketone to give a mixture of kinetic and thermodynamic enolates (remember, the deprotonation is rarely 100% selective) + the resulting ketone (since you protonated the enolate during the process). This newly formed enolate could then undergo the same reaction, until you reach a point of equilibrium where the ratio of the two differen enolates is determined by their respective stability (thermodynamic enolate).

[orgopete]

Just to add my two cents here. Since carbon is a better electron donor than hydrogen, I prefer to argue that deprotonation of the methyl occurs because it is more acidic not less hindered. Replacing hydrogen with carbon will increase electron donation and decrease acidity. However, since alkene stability favors the more substituted, equilibration favors the more substituted. Hence kinetic is the removal of the methyl hydrogens, but the thermodynamic is the CH2 hydrogens.