Phenolphthalein, a common nonprescription laxative, is also an ac...

Question

Phenolphthalein, a common nonprescription laxative, is also an acid–base indicator that is colorless in acid and red in base. Phenolphthalein is synthesized by the acid-catalyzed reaction of phthalic anhydride with 2 equivalents of phenol.
(a) Propose a mechanism for the synthesis of phenolphthalein.

Mechanism for phenolphthalein synthesis from phthalic anhydride and phenol.

(a) Propose a mechanism for the synthesis of phenolphthalein.

Synthesis of phenolphthalein with phthalic anhydride and phenol, showing acid-catalyzed reaction.

Accepted Answer

All right. Hi everyone. So for this question, it says that time all failing is an acid based indicator. That is colorless, below ph 9.3 and blue above ph 10.5. The synthesis of time all failing involves the reaction of phallic and hydrate with two equivalents of time all or two isopropyl five meta fino in an acidic medium. So now we must write a mechanism for the reaction that we have here. This is the general scheme. And now what I want to bring to your attention is the fact that one of the car bottles of our catholic and hydrate is now missing. Right? And it's been replaced with two equivalents of fennel. So we know that that oxygen has been detached from our phallic and hydrates somehow. So now let's begin by remembering the fact that we are in an acidic medium, which means that we are going to have protons floating around in solution. Right? So on the side here I went ahead and I redrew are phallic and hydride as a starting point. And now I'm going to draw a free floating proton which are card bottle here at the top is going to go ahead and take. So now, for our first step, we're going to have a proto nated um carbonnel it. So let me just draw outer unpremeditated car bottle, that one remains untouched. And now we're going to have an O. H. With a positive charge on oxygen. Now, oxygen is not a fan of that at all. Right, So this is actually where our first equivalent of time all is going to come in. So time or I'm gonna go ahead and draw in blue. This is our group. Our method Isopropyl. There we go. So this is going to be our first equivalent of time. All right. And what's going to happen is that the para position of this benzene ring in time a relative to R O H is going to go ahead and attack the carbon carbon of the positively charged oxygen. So the para position is going to attack and the pi electrons of the carbonnel are going to go ahead and shift towards oxygen. So now, in our second step, we're going to have a neutral um O H group where are carbonnel originally was. So this is our untouched Harbottle like I mentioned before, R O H is now going to be neutral and now we are going to have one equivalent of time all. But what I should have mentioned in the first step was that because the pi electrons in the para position went ahead and attacked, that means we're going to see a rearrangement of the electrons within our team all. So it's going to happen first. Um in order to address any electron deficiencies. Um a pair of electrons from oxygen is going to shift down to become a car bottle and therefore the pi electrons Inside the ring next to this. Now car bottle is going to shift over one and that is going to restore any um electron deficiencies that we had within the ring. So now if I go ahead and I reach your time away, this is going to be our method. And now, like we mentioned before, we're going to have a positively charged card bottle and our double bonds are like so, so now what's going to happen is that we need to restore aromatic city before we can move on to the next step. So recall that we're going to have hydrogen here in the para position and this hydrogen is going to go ahead and detach itself in order to donate its pair of electrons towards the aromatic system. So we're going to see a shift in the pi electrons within the ring. And then this card bottle is going to become a neutral O. H group because it's pi electrons are going to shift upwards towards oxygen. So now if I scroll down once again, I'm just going to read your next step. The hardest part is definitely making sure that everything is consistent because you can forget parts of your molecule really quickly. So this is our untouched carbonnel. Sorry, awaits group. And now, like we mentioned before are, oh, age of time all is going to be neutral is R O. H or isopropyl group. And now aromatic city has been restored. So now what we need to happen is that we need this O. H group from our original phallic anhydride to go ahead and leave because notice how in the final product it's gone, it's been replaced by a second equivalent of time all. So in order for that to happen, we're going to take advantage once again of the fact that we are in acidic medium, we're gonna take another free floating proton. And oxygen is going to go ahead and take it. And now recall that when an O. H. Group is predominated, it's going to become Hde 20. Plus, right? So water is going to be a fantastic leaving group and it's going to go ahead and detach itself from the molecule leaving a cargo cat ion. So now, like I mentioned before, our car bottle is gone, right, It was an O. H. Group that has since been pollinated and it left. So this is going to be our time all first equivalent. There we go. And now like we said before this is going to have a positive charge. So now this is where our second equivalent of time all is going to come in. Actually, I should draw this in a different color, Let me do that. So I'm just gonna redraw Simon once again, but in a different color in green this time, this is our ice a profile or metal and now we're going to see a repeat of the process that we've talked about earlier, right? So our para position relative to R. O. H. Is going to go ahead and attack this carbo cat, I'll and I'm gonna draw a really weird arrow so that it's not cutting off any parts of the other molecules. So now we're going to see another rearrangement within the electrons of the thigh mall here to reduce any electron deficiencies. So our pair of electrons here on oxygen is going to go ahead and jump in between itself and carbon, meaning that the pi electrons of the ring are going to shift over as well. So now if I go ahead and scroll down once again, just make sure I'm not forgetting any parts of our molecules because that's important. So now I'm going to draw our first equivalent of time or and now I'm going to draw our second equivalent. That was not a good looking ring close enough. So now we're going to have our method group over here. This would be an incredible Starik hindrance by the way but we're just going to ignore that for the sake of this mechanism. So this is going to be our Isopropyl and we are going to have a positively charged car bottle. So now what's going to happen is that the hydrogen at the para position must once again donate it's um electrons so that aromatic city is restored within our second equivalent of time all and now we're almost done guys, I swear so let me just redraw everything. Okay. O. H. Group. I saw pro po and now I'm gonna do the same thing over here, is there a wage, no neutral our methods and our Isopropyl and like we mentioned before, aromatic city has been restored and we're back again with the fennel and this is going to be the end of our mechanism. Guys. We made it. You see now what we have um attached to where the carbonnel was in our catholic and hydrate. We have two equivalents of time all, which is exactly what we have in our scheme here. So we did it. Um if you made it this far, thank you so much for watching. Let me just do a quick scroll through as a summary. This has been our quick scroll through and thank you so much for watching and I hope you found this helpful.

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