[g_orbital]

What is the mechanism of bromination of beta-keto-ester (as shown in the image file attached)?
Is a tautomerization needed for this reaction? Does it involve a bridged bromonium ion?
I'll emphasize that the reaction is done without acid.

Thank you!

[Albert]

In my opinion, the key factor is to find a protecting group for the ester, that makes the other alpha-position the best one for a base. I don't know if benzyl alcohol could be the answer.

[lemonoman]

I think that you're gonna have a bromination on the carbon in between the two carbonyl groups.  But then, assuming you have the conditions for enolization, you're going to have a reverse-Dieckmann to break open the ring, a ring flip (by rotation), and then a nucleophilic attack to close the ring again.

Maybe? lol

[g_orbital]

Thank you for your kind reply! However, there are some problems in your suggestions:

1. There is no any protecting group in this reaction (taken from an article).
2. There is no acid or base (hence, Claisen/Dieckmann reactions seem implausible,  in my opinion).

Are acid conditions crucial for tautomerization? Furthermore: Does the regioselectivty of the bromination of the specific terminal-alpha-position relate to the fact that Bromine is a bulky atom? Maybe there's a steric interference that prevents the bromine entering between the two carbonyls?

Please tell me what you think...

Thanks again!

[Albert]

No base? Sounds impossible (but, nevertheless, it could be, of course).

However, generally speaking, the hydrogen in between the carbonyl group and the carboxyl group/ester is really more acidic than the other one in alpha-position.

However, I agree with you in the sense that steric interference probably plays a key role here.

Are you sure you wrote all the information present in the article?

[g_orbital]

The experimental procedure is attached herewith.
I've noticed three facts that may be a clue:
1) The beta-keto-ester reactant is dissolved in diethylether.
2) Sodium carbonate is used here but only in the product workup. Is it reasonable that this is the base? (I think not).
3) The reaction conditions are 0-25 celsius degrees under nitrogen. Perhaps there's a kinetic control here that direct the reaction to the less-stable product? (maybe under high temprature there's a thermodynamic control?)

Thank you very much indeed!

[Albert]

I'd say sodium carbonate can be a sufficiently strong base for this reaction.
Temperature is always something of paramount importance. Maybe temperature and steric interference direct the bromination to that position.

However, I would like to know movies' opinion.

[g_orbital]

I'm so desperate... How can it be that sodium carbonate would act as a base after the reaction mixture (which includes only reagent + bromine + diethyleter) is poured into sodium carbonate only after 90 minutes? I suppose that sodium carbonate is used simply for workup - it neutralizes HBr as byproduct of the reaction.

What about the possibility that the mechanism for this reaction is radical maybe the conditions mentioned give a clue on that...

Movies (and others) - I'd really like to know your opinion.

Thanks in advance!

[movies]

Heh, I've been watching this thread and hoping that someone else would figure this out because I didn't know the answer.

I did a little digging and came up with a few papers that are on point here:

Mayo, F. R. J. Am. Chem. Soc. 1937, 59, 1655-1657.
Mayo, F. R. Chem. Rev. 1940, 40, 351-412.

You might not like the answer, and there isn't a whole lot of experimental evidence, as Mayo admits.  There is enough to suggest some steps though.

First off, the bromine ends up between the two carbonyls, everyone seems to agree on that.  The next part is a little tricky though.  Mayo did some experiments where he showed that this initial product doesn't rearrange in a vacuum, but it rearranges rapidly in the presence of air.  His hypothesis is that the initally formed bromo-beta-ketoester is dehalogenated by either oxygen or hydrogen peroxide.  That forms a radical between the two carbonyls which can then rearrange to the other alpha position and then react with a bromine atom carrying on the chain.  (Scheme attached).  The process may also be affected by the presence of small amounts of HBr, but the role of this reagent isn't clear, it might not even be necessary.

The argument for why you get the rearranged product has to do with the stability of the radical intermediates.  The first on that you form is tertiary and therefore relatively low in energy.  The rearranged radical is secondary and therefore higher in energy and more reactive.  However, the two brominated compounds are relatively similar in energy, so the reaction is driven by the fact that it is harder to go from the product to the secondary radical than it is to go from the starting material to the tertiary radical.

This is specific to beta-ketoester compounds, the rearrangement of simple alpha-bromo-ketones is different.  See: Tetrahedron Lett. 1978, 19, 1659-1662.

All in all very interesting!  Definitely not obvious though.