Phenol Acidity

Hey guys, in this set of videos, I want to talk about predicting acidity for phenols.
Guys, phenols are alcohols, but they're substantially more acidic than a regular alcohol due to the resonance effect. Remember that you had these effects in your acid/base chapter way back in the day that told you when something was going to be a better acid. It said that if you're able to stabilize the conjugate base, then your acid would be more acidic.
Well, think about phenol. Phenol, after it gives up its proton becomes phenoxide. Phenoxide is a negative charge. Negative charges aren't that happy. But phenoxide can resonate. Notice that it can make resonance structures inside the ring, so we'd be able to resonate the negative charge to here, here and here.
What that means is that normally the pKa of an alcohol is about 16, same as water pretty much, but the pKa of phenol is closer to 10. That's because this resonance stabilized conjugate base that you can make. It's more stable and therefore the phenol is going to be more willing to give up its proton.
It turns out that we need to also understand how electron withdrawing groups and electron donating groups play into this situation as well. In this example, when I have D, that just stands for ED group, an electron donating group. When I have W, that's an electron withdrawing group.
In general, we can say that if you're pushing electrons into the ring, do you think that's going to make it more acidic or less acidic, if you're a donating group? That's going to make it less acidic because of the fact that you're destabilizing the conjugate base. The conjugate base already has a full negative charge. Do you think it wants more electrons being jammed up into that benzene? No. Whereas, what do you think about electron withdrawing groups? Absolutely. That's going to make it more acidic. The more electron withdrawing groups we have, the better. That's going to pull more electron density out of the ring and it's going to stabilize that conjugate base. Got it?
Really quick let's just do a quick example already. Go ahead and look at these four phenols and tell me which one you think is going to be the most acidic phenol.

So this question is asking us to identify the most acidic phenol, which is going to be as simple as identifying withdrawing groups and donating current. Remember that withdrawing groups are going to increase the acidity. So, we're just looking for the strongest withdrawing group to be there. So, let's go ahead and I'm going to start in reverse order I'll start at d and then we'll work our way backwards. So, look at, let's just identify what our substituents are. So, I'm dealing with an R group, I'm dealing with a nitrogen, I'm dealing with an oxygen and a chlorine okay, cool? So, let's identify each of these. So, ch3 is that a withdrawing group or a donating group? Well, since that's an R group, that's going to be a donating group, right? So, R it's not the strongest donating but it's weakly donating, cool? Moving on, we have to identify all of them, nitrogen it's got a lone pair, okay? So, we know that for sure, pretty much any atom that has a lone pair is going to be donating, right? there's one exception of that but almost all of them are donating, so it turns out that normally nitrogen would be strong donating group but since it has this carbonyl next to it it's just a moderate donating group, cool? Moderate donating, that's getting worse. Now, this next one is O with a lone pair, this is going to be a strong donating group, So, so far donating group and what I mean by donating guys again it's electron donating group. So, these are all in the EDG category, our donating group is good for acidity? No, chlorine, chlorine is a halogen. Remember, halogen is kind of the black sheep of the family, different than the rest, halogen is going to be a withdrawing group, okay? It's it's weakly withdrawing, weakly withdrawing or weakly EWG. So, which one's going to win? well, I know that for sure I've got three donating group so it just sucks because that's going to make it less stable but I do have one group that's going to pull a little bit of electrons out and stabilize a little bit more, so this is going to be an easy choice, my most acidic phenol is definitely a. Alright, cool. Guys. So, let's flip to the next page.

Guys, it turns out that not all positions are created equal. It turns out that some positions are going to have more effect on acidity than others. In fact, the meta position is going to have a much lesser effect on acidity than the ortho and para positions, meaning that whatever type of group is on there, it just matters less if it's on the meta and it matters more if it's on an ortho or para.
Why is that? Think about the resonance structures that occur when you make a phenoxide. Remember that I told you guys that where would that negative charge resonate to? It would resonate to the top, to the side, and to the side. Notice what are those positions called respective to the O? Those are the ortho, para positions. That means the negative charge will rest directly on ortho, para positions and it will never rest on a meta position.
With that logic, what that means is that we know that a donator is going to make it less acidic because it's going to destabilize the negative. And we know that a withdrawing group is going to make it more acidic. But, if those same exact groups are in the meta position – so these are meta and these are ortho, I'm looking at ortho versus meta. If I have a donating group in the meta position, it is going to be a little less acidic, but only slightly less acidic than normal.
Why? Because this donating effect doesn't really matter that much because it's on the meta position, the negative charge never goes there. Same thing goes with the withdrawing group. The withdrawing group, you think this is great. It's going to make it really acidic, but it's only slightly more acidic if you rest on the meta position. Why? Because, once again, the meta position doesn't really matter because the negative charge never actually sits on it.
In terms of acidity, that brings us to the following acronym. If you're looking for an acidic phenol, what you're going to be looking for is a WOP. What a WOP stands for is a withdrawing group in the ortho and para positions. You don't care about the meta positions. The meta positions aren't helpful for us. We're looking mostly at those having withdrawing groups in the ortho and para positions.
That being said, why don't we take a stab at this question and tell me which one you think is going to be the most acidic phenol. 

Alright guys, so notice that my groups are identical for all four phenols, the only difference is the position that they're in, okay? So, what I have for the first one is that I have an ortho a meta and a para, okay? These are all withdrawing groups very strong withdrawing groups, okay? What I have for the second one is an ortho a meta and a meta, what I have for the third one is an ortho a para and a meta and then finally what I have for the fourth one is an ortho para and ortho. So, which one's going to win, which one's going to have the low stabilizing effects on that phenoxide, and guys it's obviously going to be d because, d is the WOP, d has withdrawing groups on the ortho and para. Now, does that mean the other ones aren't acidic? No, they are very acidic they're doing great but this one's a little bit better because all the groups happen to be in those ortho and para positions. So, it's going to be able to stabilize the negative charge in every position that it goes, got it? With the withdrawing group, makes sense? Cool guys, let's turn the page.

Alright guys. So, which one is the most acidic phenol in this case? Well, actually guys I'm just going to tell you right now, the answer was a, did you pick a? I'm sensing that a lot of you did not pick a, but that's because maybe you forgot about withdrawing and donating groups, what kind of group is NMe2? Well, this is going to be in amine derivative or an aniline derivative because you've got a lone pair right there that means this is a strong donating group, right? Electron donating group. So, what did we learn about electron donating groups? do they make a molecule or phenol more acidic or less acidic? they make it less acidic, okay? So, we found out that no, if we're going to have a withdrawing group then we want it to be in the ortho and para position, okay? But, if all you're given is donating groups then you want to minimize the effect that they have because donating groups actually destabilize your anion. So, then you want it to be in the meta position because the meta position is going to have the smallest effect. So, when you don't have any withdrawing groups then you go to donating groups and you look for them in the meta position, okay? So, in this case, what happens to donating groups. Now, that immediately is going to lose because, I don't want two of them they're already bad as they as bad as they are, I don't want two things that are destabilizing my phenol. So, that's terrible and then I have ortho, then I have on donating groups that's ortho versus donating groups that's meta and the donate group that's meta is going to win because the drawing group that's ortho is going to be significantly destabilized when my negative charge goes there but my donating group, that's meta, will be fine because the negative charge never actually hits that carbon. So, it's not going to be as destabilized and it's going to continue to remain acidic. So, see how we have to think kind of backwards here. Since we weren't given any withdrawing groups you have to go with the donating group that has the lowest effect on the phenol, all alright, so the answer is a. So, guys here's another one it's going to be tricky, try your hardest and then I'll tell you the answer.

Alright, so to find the most acidic phenol again we're looking for WOP, we're looking for withdrawing in the ortho/para, do I have withdrawing groups or donating groups? So, guys iodine is a withdrawing group, okay? So, thankfully that means that we're, this is going to be more simple, we're just going to look for the one that has the most withdrawing groups in ortho and para positions. So, this one is o, o this one is m,m, this one is p, m, and this one is m, o, so the best one is going to be a once again, because a has all of my withdrawing groups in the ortho positions, which stabilizes the radical, not radical, the anion. Awesome guys. So, I hope that made sense, let's move on to the next topic.