[tvs333]

I am wondering why a carbonyl such as acetone reacting with HBr (or other HX) doesn't produce a molecule with both an OH group and a Br on one carbon.  Is it because Br is not a good enough nucleophile?  Or is it because usually there is water in solution and due to the large number of water molecules the hydrated product will predominate.  Or is it because it would undergo acid-catalyzed alpha bromination?

Thank you for your help.

[sundberg]

Because both oxygen and bromine are large very electronegative atoms, i.e they repell eachother too much.

[movies]

Such bromohydrins do form to a small extent under highly acidic conditions, but it's not entropically favorable.

[tvs333]

Hi,

Thanks for your replies.  Care to expand on how you know it is entropically unfavourable?

For the repulsion explanation, I'm not sure that makes sense to me when we can have oxygen containing nucleophiles attacking the carbonyl, which are also electronegative. It's more to do with size then?

Thanks

[movies]

Whenever you take two molecules and make one molecule, then the entropy of the system decreases.  There are fewer theoretical arrangements of one atom in space than there are for two atoms in space.

[Arsenic]

if i am not mistaking the other product in this reaction is HBr , solvolysis of this product is likewise not entropically favored in water (i.e the entropy of the solvent would decrease due to a more ordered structure for H20 molecules around Br- and H+) . what  is more likely to be the driving force of this reaction is the formation of a carbonyl double bond.

[movies]

I disagree about the entropy, but if you want to take a bond energetics tack, that's okay too.  You have to account for the energy of the bonds that you are breaking as well: a hydrogen-oxygen bond and a carbon-bromine bond.

[Arsenic]

Would you be kind enough to elaborate?

[movies]

About the entropy thing or about the bond energetics?

For entropy, water is already pretty organized into hydrogen bonding networks.  Adding an additional proton isn't going to change the ordering very much unless you have a very small amount of water.  The bromide ion would be treated similarly, just with the opposite polar end of the water.  In fact, the dissociation of HBr in water is probably favorable enough to drive the equilibrium.  Think about how difficult it is to get HCl out of water or water out of HCl.  The ionic form is favored by a lot.

For the bond energetics argument you have to consider all of the bonds in each structure, so in the first structure you drew you have to account for the bond energy of the carbon-bromine bond and the oxygen-hydrogen bond with are absent in the ketone form.

[Arsenic]

your argument about the bond energetics is absolutely right  , but the part I still cannot digest is the entropy(I am not sure if your argument is correct) , could you take a look at the following URL and tell me what you think.
http://www.chemguide.co.uk/inorganic/group7/acidityhx.html
 

[movies]

That's very interesting.  I see your argument.  I haven't seen a treatment of acidity like that before.  Perhaps my entropy argument is flawed!

Unfortunately, my skills at thermodynamics are lacking and I'm not sure how to compare the two compounds in question with any certainty.  The URL you showed certainly suggests that the change in entropy for converting the bromohydrin to the acid would have a negative entropy (induce a more ordered system) but I don't know how to account for changing from an alcohol to a ketone.  I would think that removing the possibility of hydroxyl hydrogen bonding would allow for a less ordered system (positive entropy), but I don't know for sure.  Moreover, I don't know how the magnitudes of these effects would compare.

Thanks for that link, it was very interesting to read.  By the way, did you happen to work out all of the bond energetics?  I'm curious to see how they compare.

[Arsenic]

did you happen to work out all of the bond energetics?  I'm curious to see how they compare.

no, but I will tackle it when ever I find enough time.