Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: orgoclear on March 29, 2010, 05:47:56 AM
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A quite silly doubt though
"can phenol give iodoform reaction ?"
I was thinking the answer to be NO till quite recently when i saw a book giving the answer as YES
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Strange. I can't see how this reaction would work. Do you have a reference for that book?
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It involves conversion of phenol to its highly unstable keto form. which then gives iodoform. (that was the given mechanism). But I highly doubt the conversion of phenol to keto form
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Hm. It's certainly on the low, low end of the equilibrium. It seems like the iodoform reaction could be reasonably irreversible, in which case you could pull the overall conversion through through LeChatlier's.
Just thinking
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Even if we suppose the conversion from enol to keto form, I still can't see where the carbon for the iodoform molecule is coming from. The haloform reaction requires a methyl ketone.. something that can be enolized three times so that we can add a halide three times. Once cleaved we have our bromoform or iodoform depending on which halide we used.
So phenol, even in it's keto form, only has two enolizable protons on either side of the ketone... how can we get a third one?
Are you sure we're not just putting an iodine on the ring (like an electrophilic aromatic substitution) rather than doing a full iodoform reaction??
I'm confused ???
Would it be possible to scan the textbook's mechanism?
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Even if we suppose the conversion from enol to keto form, I still can't see where the carbon for the iodoform molecule is coming from. The haloform reaction requires a methyl ketone.. something that can be enolized three times so that we can add a halide three times.
Good point.
I have a feeling the book is wrong, but I can't be too sure. I think it would be helpful if someone could actually perform this in the lab and post the results. I'm just out of high school and have no access to a lab. Can someone help us here?
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Definition between iodoform test being positive, and the reaction producing iodoform is an important distinction. Iodoform is yellow, and so is insoluble triiodophenol, which may be being formed in this reaction....
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Definition between iodoform test being positive, and the reaction producing iodoform is an important distinction. Iodoform is yellow, and so is insoluble triiodophenol, which may be being formed in this reaction....
True, but the OP is referring to the mechanism of the reaction as well. The book says that the result of the iodoform test is positive because of the conversion of phenol to its keto form, and the eventual formation of iodoform.
But as stewie griffin rightly pointed out, how is iodoform formed? There is no free Carbon.
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Even if we suppose the conversion from enol to keto form, I still can't see where the carbon for the iodoform molecule is coming from. The haloform reaction requires a methyl ketone.. something that can be enolized three times so that we can add a halide three times. Once cleaved we have our bromoform or iodoform depending on which halide we used.
So phenol, even in it's keto form, only has two enolizable protons on either side of the ketone... how can we get a third one?
Are you sure we're not just putting an iodine on the ring (like an electrophilic aromatic substitution) rather than doing a full iodoform reaction??
I'm confused ???
Would it be possible to scan the textbook's mechanism?
the scanners busted..
But the mechanism was after keto form was obtained, two I's were substituted in place of the acidic H's
Then some sort of cleavage took place. and then CHI3 was formed
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But the mechanism was after keto form was obtained, two I's were substituted in place of the acidic H's
Then some sort of cleavage took place. and then CHI3 was formed
Loss of aromaticity under the iodoform reaction conditions is quite difficult, and on top of that bond cleavage takes place. I mean... bond cleavage is taking place in a molecule that is found in traces in the reaction mixture, and this is a test the author expects phenol to 'answer'... omg :o
Can you tell me the name of the book? I might have it.
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What year is the book?
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I still don't believe this is true/correct. Even though your scanner is broken, could you draw out the answer in chemdraw and upload it here?
I have been searching this around google today and came across some qualitative tests to determine between ethanol and phenol. The answer was iodoform test. Ethanol can be oxidized by I2 to the aldehyde (so it's now a carbonyl with an alpha methyl group), and then the reaction proceeds like a normal methyl ketone. Phenol can't do the same.
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I've been drawing out some reaction mechanisms, and none of them seem non-contrived. The most plausible seems to be a hydroxide attack on C-3 after initial iodination, which results in ring opening and formation of a diiodomethylketone anion.
Still looks pretty improbable, though.
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The most plausible seems to be a hydroxide attack on C-3 after initial iodination
Can you please explain this part in detail? How and why does the 'initial' iodination take place?
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Maybe what happens is NOT that iodoform forms at all, but that the highly electron-rich phenoxide anion reacts with iodine in all three ortho and para positions, decolouring the reagent, as in a POSITIVE iodoform reaction. A FALSE positive.
However, there is no way to form iodoform in such reaction...
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Maybe what happens is NOT that iodoform forms at all, but that the highly electron-rich phenoxide anion reacts with iodine in all three ortho and para positions, decolouring the reagent, as in a POSITIVE iodoform reaction. A FALSE positive.
However, there is no way to form iodoform in such reaction...
That is what I have been getting at. I still believe this to be the case.
Or that someone did april fools a few days early.
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Yes ok I agree then that it's a false positive. Sorry 408.. I didn't fully understand what you were originally trying to say about giving a "positive" test a few posts ago, but now I understand you.
But then I guess there's still a semantic argument to be made and the book needs to do a much better job at explaining itself. The iodoform is a qualitative test where a positive is appearance of a yellowish solid. But it's also only truly positive (IMHO) if this solid produced is actually iodoform. In this phenol example, we're not making iodoform, but instead the triiodophenol. That's fine and dandy if triiodophenol just happens to be a yellowish solid, BUT it should NOT be considered a positive iodoform test. At the very best it should be considered a false positive, and a full explanation should follow to explain why the false positive is occurring.
What book is this? I think it's done you a disservice.
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The iodoform is a qualitative test where a positive is appearance of a yellowish solid. But it's also only truly positive (IMHO) if this solid produced is actually iodoform. In this phenol example, we're not making iodoform, but instead the triiodophenol. That's fine and dandy if triiodophenol just happens to be a yellowish solid, BUT it should NOT be considered a positive iodoform test. At the very best it should be considered a false positive, and a full explanation should follow to explain why the false positive is occurring.
What book is this? I think it's done you a disservice.
I completely agree.
http://www.iodine.com/triiodophenol.htm clearly states that triiodophenol is a white powder. So, we are not even getting close to iodoform.
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p-iodophenol though is a 'tan solid', so that's at least getting closer. One of the list of possible (mono/di/tri)iodophenols may be yellow
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yup that just might be what it wants to say. but not at all very clearly.
The book's just into its first edition and there are bound to be mistakes (i guess)
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p-iodophenol though is a 'tan solid', so that's at least getting closer. One of the list of possible (mono/di/tri)iodophenols may be yellow
But we have an activated ring here. So, the reaction, IMHO, will not stop with mono/di iodination. Iodine is a weakly deactivating group, isn't it?
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yes, weakly deactivating. Fairly sterically large as well, but I'm not sure if that will come into play here.