why not haloform reaction results in an α keto acid?

[adianadiadi]

During the haloform reaction, the base hydrolysis of α,α,α-trihalocarbonyl compound gives the haloform as the final product.

 

But why not the halogen groups are substituted by the OH^- groups to give a trihydroxy compound which can lose water molecule to give finally an α keto acid?

[mississippichem]

During the haloform reaction, the base hydrolysis of α,α,α-trihalocarbonyl compound gives the haloform as the final product.

 

But why not the halogen groups are substituted by the OH^- groups to give a trihydroxy compound which can lose water molecule to give finally an α keto acid?

 

Because [ce] CF_{3}^{-} [/ce] or [ce] CCl_{3}^{-}[/ce] are such a good leaving groups when next to carbonyls and because keto-enol tautomerization is very fast. Substitution by the hypohalite would be competing with the [imath] S_{N}2 [/imath] you speak of. Becuse of the carbonyl, the individual halide molecules would be poor leaving groups in this scenario anyway.

[adianadiadi]

Becuse of the carbonyl, the individual halide molecules would be poor leaving groups in this scenario anyway.

 

Could you please elaborate this point?

[Horza2002]

To help make this easier, the mechanism is below.

 

You are right that typically, a halide would be displaced by the hydroxide because they are reasonable leaving groups. However, the carbonyl groups makes the protons in the alpha position more acidic...the hydroxide can then deprotonate it to give the enolate which then reacts further with the halogen. Now, at his point (the prdouct of step 2) the prescence of the iodidine on the alpha position makes the two remaining protons more acidic because of its electron withdrawing capabilities. This means that a second proton is deprotonated faster than than a substitution reaction can occur. The last proton is then again easier to remove because the resulting anion is stabilised by the two iodides.

 

If you were to actually do this reaction however, you would normally use three equivalent of a sterically hindered base (e.g. potassium t-butoxide) to get you through steps 1-6 and then at this point add in the hydroxide. Because even though the major product will be what you expect, there will always be some side reactions...and remember that alpha-halo carbonyl compounds are more activated to substitution reactions because of the overlap of the C-X (X is a halogen) sigma^* and the C=O pi^* makes them more reactive.

Edited May 2, 2011 by Horza2002