[xshadow]

I have some doubts about this reaction in acid catalysis

Here the text I read:



Why if I add a second equivalent of Br₂ I'll have the attack  on the other  α-carbon (the one not brominated in tjebfrist "step") and not on the same one??

When I add the second equivalent the rate determining step is the deprotonation of the H bonded to  a C_α of bromoketone by a weak base (pH acid)


Now  a C_α with:

- hydrogen but also a bromine atom has its hydrongen more acidic than another C_α without an halide bonded.

And a more acidic  hydrogen means that  the base can attack the H⁺ more easily.

So why the second bromination doesn't occour on the same C_α  of the first halogenation?


Thanks!!

[sharbeldam]

Very good question.
My guess would be is that the product of the second halogenation is less favourable because there is a + charge on the oxygen and the bromide is electronegative so it makes that product less stable and hence less likely to happen.