Talk:Phenol red

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I removed the structure picture Image:Phenol_Red.gif since it is incorrect. The correct structure can be found here. AxelBoldt 19:49, 30 September 2006 (UTC)

I've drawn a new structure here: commons:image:Phenol_red_2.png. In good faith, I asked that the commons file which the original was linked to be replaced with my version.

Your structure seems like there is a keto-enol tautomerization of sorts. Is the NIH version more accurate? I googled the structure and most of the hits were like the one listed here: http://www.usca.edu/chemistry/spectra/phenlred.htm. --Rifleman 82 20:22, 30 September 2006 (UTC)

It exists in two structures, depending on pH. Here are the two structures: [1] . Regarding your picture: I don't think it's needed anymore, since commons has this: image:Phenolrot.png. We need a picture of the structure under acidic conditions though, when the cycle closes. AxelBoldt 23:56, 30 September 2006 (UTC)

(Copied from user talk:AxelBoldt:)

I've updated the picture at commons which was inaccurate. You can take a look at commons:Image:Phenol red 2.png, commons:Image:Phenol red.png, and commons:Commons:Help_desk#Image:Phenol red.png. One of the admins there have moved from 2 to the phenol red.png file. It's not false; I believe it's just the server taking a while to refresh. Do assume good faith. --Rifleman 82 00:00, 1 October 2006 (UTC)

Sorry, my fault, I should have cleaned my cache first. The picture is fine now. AxelBoldt 00:26, 1 October 2006 (UTC)

It's alright. I've added the acid base equilibria as you've suggested on the article talk page. Comments? --Rifleman 82 01:10, 1 October 2006 (UTC)

Yes, I have a couple:

  • I assume that at neutral pH most of it is in the cyclic form, no? Maybe we should have our main structure picture in the upper right then also be the cyclic form. That's what PubChem does.
  • The pKa constant shows that it is a weak acid. Which proton does it lose, the one by the sulfur? Is my understanding correct that the pKa constant does not have anything to do with the interconversion between the two structures?
  • The two structures, for low pH and high pH, should be presented in opposite order to match the table above with the yellow and red colors.
  • I'm not so happy with the arrows indicating how the structure rearranges, because it's labeled "under basic conditions". The rearrangement doesn't happen under basic conditions. Also, why is there a proton leaving at the top? Doesn't the proton drop off the hydroxide by the sulfur?
  • I think it would be nicer if we had separate pictures for the two structures, presented them underneath the colored boxes in the proper order, and then explained in a new paragraph how the structures interconvert, using your arrow picture.

AxelBoldt 02:16, 1 October 2006 (UTC)

I tried to search online, but I couldn't find much information. In my opinion, the neutral molecule will be an equilibrium between (II) in my diagram and the acyclic -SO3H form at the Chembox. I think of it as an extended keto-enol tautomerization. Why are you convinced that it is the cyclic form? While pubchem shows the cyclic keto-structure, I found a few other structures online which showed the enol form.

I'm not very sure what you mean by your second para. Usually, where there are more than one acidic protons, each pKa can be identified. However, if there is a keto-enol tautomerization like I think there is, that will be a bit complicated.

I've made the amendment for the yellow/red to be reversed: Image:Phenol red - acid base equilibria - 2.gif.

I don't understand what you mean about the rearrangement in the original diagram. It is not a rearrangement. Rather, the sulfate O- is acting as a nucleophile, making a conjugate addition to the extended alkene system finally ending up with the C=O being protonated. I am drawing the arrows in the same convention as electron pushing - the arrow represents the movement of an electron pair. --Rifleman 82 02:53, 1 October 2006 (UTC)

I believe that the substance is predominantly in the cyclic yellow form at neutral pH because "yellow pH" 6.6 is so much closer to neutral than the "red pH" 8.0. But I could be completely wrong and I have no other evidence.

I'll try to explain my second point again: at very high pH, the substance is in acyclic form and negatively charged, since it has lost a proton from the sulfur's hydroxide, as shown in your picture above, to the right. As you decrease the pH, you go to the uncharged acyclic structure shown in our chemobox. If you decrease the pH further, the structure will switch to the cyclic, uncharged one given in your picture to the left. Is this understanding correct? Or do you think that gaining of the proton and switching to the cyclic structure is essentially one and the same process, described by the pKa=8? If that were so, then the substance would essentially never exist in the neutral acyclic form shown in our chemobox and it would make more sense to display the cyclic form.

(Oh, and one minor technical thing: in your picture, the minus at the O is very small and could be mistaken for a dot (i.e. unpaired electron) by the non-expert.) AxelBoldt 16:09, 1 October 2006 (UTC)

Easy question first. I've set chemsketch to follow the guidelines at Wikipedia:WikiProject Chemistry/Structure drawing workgroup. It's a superscript minus at the same font size as the character. Unless you have other software to draw it... there's not much I can do. If we expect the reader to follow the arrow pushing diagram, I would expect them to understand it as an O-.

I think of it as the two canonical structures for acetone - 2-propanone and 2-hydroxy-1-propene. Both exist in equilibria. On SciFinder, I see both structures listed, but the cyclic form for the neutral compound is more common. That being the case, I'll make the change in the next day or two. --Rifleman 82 17:31, 1 October 2006 (UTC)

Thanks, I've put your above picture in the article since I believe it fits better with the flow. I also found some data on solubility in the Merck Index (which shows only the cyclic form and doesn't even mention the rearrangement, so we're clearly better than they are). AxelBoldt 17:38, 2 October 2006 (UTC)

Ah... I was going to tidy up the diagram a bit first, and repost it without the " - 2" in the filename. If you notice, some of the bonds are not perfectly straight. Nevermind. I'll do that later when I fulfill the 4 day requirement for commons to upload over this file. That too, for the cyclic neutral molecule. --Rifleman 82 17:53, 2 October 2006 (UTC)