Phenol Acidity - Online Tutor, Practice Problems & Exam Prep

Donating vs Withdrawing Groups

In this set of videos, I want to talk about predicting acidity for phenols. 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 will be more acidic. Think about phenol, guys. Phenol, after it gives up its proton, becomes phenoxide. Phenoxide, it's a negative charge. Negative charges aren't that happy. But phenoxide can resonate. Notice that it can make resonance structures inside the ring. We'll 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, the same as water pretty much. But the pKa of phenol is closer to 10. That's because of 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 means it 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 phenyls and tell me which one you think is going to be the most acidic phenyl.

O,P-positions vs. Meta-Positions

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 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 donating group is going to make it less acidic because it's going to destabilize the negative. 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 orthos 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. With 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 WAP. What a WAP 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.

Identify the most acidic phenol

Notice that my groups are identical for all 4 phenols. The only difference is the positions that they're in. What I have for the first one is that I have an ortho, a meta, and a para. These are all withdrawing groups, very strong withdrawing groups. 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. Finally, what I have for the fourth one is an ortho, para, and ortho. Which one is going to win? Which one is going to have the most stabilizing effects on that phenoxide? It's obviously going to be D because D is the WAP. D has withdrawing groups on the ortho and para. Now, does that mean the other ones aren't acidic? No. They're very acidic. They're doing great. But this one's a little bit better because all of the groups happen to be in those ortho and para positions. It's going to be able to stabilize the negative charge in every position that it goes with the withdrawing group. Make sense? Cool guys. Let's turn the page.

Identify the most acidic phenol

Which one was the most acidic phenol in this case? Actually, I'm just going to tell you right now. The answer was A. Did you pick A? Yikes. 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? This is going to be an amine derivative or an aniline derivative because you've got a lone pair right there. That means that this is a strong donating group, electron donating group. 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. We found out that if we're going to have a withdrawing group, then we want it to be in the ortho position. Destabilize your anion. Destabilize your anion. Then you want it to be in the meta position because the meta position is going to have the smallest effect. When you don't have any withdrawing groups, then you go to donating groups and you look for them in the meta position. In this case, what I have is 2 donating groups. That one's immediately going to lose because I don't want 2 of them. They're already bad as bad as they are. I don't want 2 things that are destabilizing my phenol. That's terrible. Then I have ortho. Then I have a donating group that's ortho versus a donating group that's meta. The donating group that's meta is going to win because the donating 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. 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. The answer is A. Here's another one. It's going to be tricky. Try your hardest and then I'll tell you the answer.

Rank the following phenols in order of increasing acidity. 

• A. D < A < B < E < C
• B. D < E < A < C < B
• C. E < A < B < D < C
• D. B < E < D < A < C

Here’s what students ask on this topic:

Why are phenols more acidic than regular alcohols?

Phenols are more acidic than regular alcohols due to the resonance stabilization of their conjugate base, phenoxide. When phenol loses a proton (H⁺), it forms phenoxide, which has a negative charge. This negative charge can be delocalized over the aromatic ring through resonance, making the phenoxide ion more stable. In contrast, the conjugate base of a regular alcohol does not have this resonance stabilization, making it less stable and thus less acidic. The pKa of phenol is around 10, compared to about 16 for regular alcohols, indicating higher acidity.

How do electron withdrawing groups affect the acidity of phenols?

Electron withdrawing groups (EWGs) increase the acidity of phenols by stabilizing the conjugate base, phenoxide. EWGs pull electron density away from the aromatic ring, reducing the negative charge on the phenoxide ion. This stabilization makes it easier for phenol to lose a proton, thereby increasing its acidity. The effect is more pronounced when the EWG is in the ortho or para positions relative to the hydroxyl group, as these positions directly influence the resonance structures of the phenoxide ion.

What is the pKa of phenol and how does it compare to regular alcohols?

The pKa of phenol is around 10, which is significantly lower than the pKa of regular alcohols, which is about 16. This lower pKa value indicates that phenol is more acidic than regular alcohols. The increased acidity of phenol is due to the resonance stabilization of its conjugate base, phenoxide, which is not available in regular alcohols.

Why do ortho and para positions have a greater effect on phenol acidity than the meta position?

Ortho and para positions have a greater effect on phenol acidity because the negative charge on the phenoxide ion can resonate to these positions. This resonance stabilization directly influences the acidity of phenol. In contrast, the meta position does not participate in the resonance structures of phenoxide, making its effect on acidity minimal. Therefore, electron withdrawing or donating groups at the ortho and para positions significantly impact phenol's acidity, while those at the meta position do not.

How do electron donating groups affect the acidity of phenols?

Electron donating groups (EDGs) decrease the acidity of phenols by destabilizing the conjugate base, phenoxide. EDGs push electron density into the aromatic ring, increasing the negative charge on the phenoxide ion. This destabilization makes it harder for phenol to lose a proton, thereby decreasing its acidity. The effect is more pronounced when the EDG is in the ortho or para positions, as these positions directly influence the resonance structures of the phenoxide ion.