# Are compounds containing active methylene groups positive for the haloform test?

Are compounds containing active methylene groups (e.g. 1,3-diketones) positive for the haloform test? Support your answer with a suitable explanation.

I was stuck on a certain question wherein a diketone was formed and I had to tell whether or not it shows the haloform test. It is clearly mentioned in my class notes that active methylene compounds don't show the haloform test; but in this problem the solution says that the diketone shows positive haloform just because it has an active methylene group. Please help!

Here's a possible mechanism that I came up with. Don't know if its correct. Sorry that it's handwritten; I don't know how to type chemistry.

• At least the wikipedia article (en.wikipedia.org/wiki/Haloform_reaction) states that diketones are suitable substrates for the haloform reaction. Therefore I see no reason why the haloform test shouldn't work. In fact I would suspect it to work even better due to the higher acidity in the alpha position. – logical x 2 Oct 15 '17 at 17:31
• @deusexmachina, as you say, the iodination at the acidic position is likely to proceed fine. The issue I see is that the positive result of the iodoform test (at least) is given by the formation of $\ce{CHI3}$, both the colour and the odour of which can be identified. With 1,3-dicarbonyls, the final step results in the expulsion of a diiodoenolate, as opposed to $\ce{CI3-}$, which is required for the positive result. I am busy tonight, so anybody is welcome to take this and write up an answer. – orthocresol Oct 15 '17 at 17:46
• Please use the built-in image uploader; images included that way are not subject to deletion on imgur. – Martin - マーチン Oct 19 '17 at 5:59
• Hmm, seems I didn’t think far ahead enough in my last comment. Your mechanism has some points for nitpicking, but I agree that the dihaloenolate expelled should give a positive result. – orthocresol Oct 19 '17 at 7:56
• @orthocresol please point them out. I think the addition of $Br_2$ is ambiguous, but it gives me the product I want! – YourAverageEuler Oct 19 '17 at 9:12

As orthocresol already commented, apart from some nitpicks, the general direction is correct.

The only issue really is the addition of bromine, which does not occur via a [2+2]-cyclo-addition (2nd from below in scheme).

Ordinarily you would expect addition the addition of bromine to a π-bond to occur first via a polarisation complex and then via a bromonium ion (bottom of scheme).[1] In the case of the enolate, the π-bond is already readily polarised, and sufficiently asymmetric that a direct addition will occur.[2] The following depicts the HOMO -1 calculated at the DF-BP86/def2-SVP level of theory, the HOMO is an (delocalised) in-plane lone pair of the carbonyl-oxygen atoms in the C2v conformation (1st row). In the Cs conformation the π-bond is the HOMO, it is also the low-energy conformer (2nd row). (On mobile the rows might differ.)

Please note that the following scheme is just a proposed mechanism, and as such probably a reasonable simplification; the actual one might depend on more factors like solvent, temperature, pressure, etc., and might have more, or even less steps, other intermediates (ion pairs), hydrogen abstraction might be via water/base bridges, etc. For educational purposes I think this is sufficient though.

1. Thanks to orthocresol pointing it out in the comments.

Enolate reaction with bromine (as opposed to ordinary alkene) is traditionally taught to not go via a bromonium ion (img), presumably because the electronic character of the two doubly bonded carbons is very different. But I have not consulted any literature, this is just based off undergrad teaching + textbooks, so...

Since I was unlucky to find any specific literature on the topic (I did not search very thoroughly), I have tried optimising a bromonium analogue at the DF-BP86/def2-SVP level of theory. I have been unable to obtain a stationary point (no local minimum and no transition state). Which is not unexpected given the HOMO/ HOMO -1 is strongly polarised.