[ultrashogun]

So the process basically is that the base removes a proton from position alpha, forming the enolate, then the enolate attacks Br, adds it to the double bond and the lone pair on O reforms the pi bond to C. We get back the ketone.

I completely understand the mechanism of how this process repeats itself two more times, due to the alpha proton becoming more acidic with each added Br.

What I dont however understand is this, after the first Br is added I have a carbonyle with a Br in alpha, should this actually give an extremely fast E2 reaction with hydroxide as the nucleophile?

What prevents this from happening until I get the haloform at the end?

[agrobert]

Think about electronegativity.  Draw acetone and label the delta (-) and delta (+) centers.  Now do this for the alpha bromo carbonyl.  The alpha carbon is not as electropositive as it would be for good E2 conditions, the alpha carbon experiences electron pull from the carbonyl and the Br.  This is why the conjugate base formed after deprotonation of the alpha carbon is favorable.  If you did eliminate the Br you would form the enolate again which would go back to attacking the Br.  This reaction is driven to equilibrium by decreasing pKa of alpha protons by the addition of Br.

[ultrashogun]

Ohh crap, Im really sorry but its not an E2 reaction but a Sn2 reaction that happens at the alpha carbon!! Thanks anyway for your answer.

[agrobert]

Same answer for S_N2