[ptryon]

Familiar with the iodoform test? Then read on...

I have just tried the iodoform test using reagent grade methanol and iodine dissolved in 1moldm^-3 NaOH. When the aqueous alkaline iodine was added to the methanol a substantial quantity of precipitate was formed. If my understaning of the reaction is correct this shouldn't happen!

According to my undergrad organic chemistry textbook, the mechanism involves the following steps:

(1) the oxidation of the alcohol to a carbonyl group
(2) hydroxide ion deprotonates the carbon adjacent to the cabonyl group to form an enolate ion
(3) enolate attacks the I₂ so that one iodine atom is added to the carbon (the one that lost the H in step 2)
(4) Repeat steps 2-3- further addition of iodine atoms until CI₃ is formed which is a better leaving group than OH
(5) Nucleophilic attack of hydroxide ion at carbonyl carbon (which has been happening throughout)
(6) Negative charge on oxygen refroms c=o and C-C bond broken to eliminate CI₃^- leaving an acid
(7) The CI₃^- carbanion deprotonates the acid to form triiodomethane and an alkanoate ion
   
Step 2-7 seem to rule out the possibility of methanol participating in this reaction since there is no carbon adjacent to the carbonyl group. If it works for methanol, where does the triiodomethane come from? Can anyone help???

[Grundalizer]

SN1

[jake.n]

I would guess that by successive oxidations to the formaldehyde form and nucleophillic attack to form the iodomethanol, you could conceivably form a substantial amount of triiodomethanol or the sodium salt.  From there I don't know how you would get to triiodomethane.

[ptryon]

Hi
Thanks for your thoughts Jake. Could triiodomethanol account for the precipitate I observed? I see two problems with nucleophilic substitution of mechanism:

1. The only possible nucleophile I can think of is an iodide ion. Wouldn't it require the hetrolytic fission of the I-I sigma bond to generate the I^- ion? If so what would cause this in aqueous alkaline conditions?

2. Formaldehyde only has very bad leaving groups- 2 hydrogen atoms which would need to be kicked out as hydride ions (pKa HI = -10 cf pKa H₂ = "very large") If the iodide ion did attack the carbon, it would be kicked out again in the following step (taking us back to formaldehyde)

Any other thoughts...

[jake.n]

1) The hydroxide ion can produce iodide and hypoiodic acid.  Additionally, in polar alcohols iodine is in equilibrium with the triiodide ion which can act as a iodide donor.

2) Formaldehyde does have two leaving groups, which is why the attack of the iodide ion would result in nucleophillic addition, not substitution (see figure)

I still don't understand how one could produce iodoform from the triiodomethanol, however, the triiodomethanol may actually be the yellow precipitate you are seeing.  I can't seem to find anything about the appearance of it or its salts.

[nox]

Here's the thing: in basic aqueous solution iodine forms hypoiodite initially. The hypoiodite then rapidly disproportionates to iodide and iodate.

Iodide probably just sits around and doesn't do much, but iodate is a very powerful oxidizer, and methanol probably gets oxidized to formaldehyde and formic acid. Since you have NaOH you could form sodium formate, which *might* precipitate if you have a large excess of methanol compared to water.

Actually why don't you run some tests on the precipitate? NMR and IR should give you some clues as to whether it's an organic or inorganic precipitate. I have a hunch the precipitate is NOT iodoform, and probably something inorganic.

Also triiodomethanol would be unstable, its sodium salt much more so. An iodide would simply be kicked out to form carbonyl iodide, the iodine analogue of phosgene. This would be supremely reactive and react either with water to give carbonic acid which spontaneously decarboxylates, or react with methanol to give dimethyl carbonate. Of course neither of the two are solids, so...

[ptryon]

Thanks for your replies. This is more complicated than I first thought...

Jake: I didn't even consider nucleophilic addition. By the way did you make the diagram? If so what software did you use?

nox: Your answer is clear and insightful but makes the problem seem even more mysterious! Please let me know if you can think of any insoluble inorganic compounds that could form under these conditions and explain the mysterious precipitate. Unfortunately I work in a high school in Malaysia and don't have access to any spectrometers... I may have to request an NMR sprctrometer on next years budget!

For the time being I will do what all good chemists do and blame it on 'impurities'