Ranking Acidity - Online Tutor, Practice Problems & Exam Prep

Understanding acidity involves several factors, including the stability of the conjugate base, which influences the acid's ability to donate protons. Key factors affecting acidity include the element effect, where electronegativity and atomic size play crucial roles; inductive effects from electronegative atoms not directly attached to the acidic hydrogen; resonance effects that stabilize conjugate bases through charge delocalization; hybridization effects related to s character; and steric effects that consider the size of substituents. These principles guide the comparison of acids and their strengths effectively.

Why we need factors affecting acidity and when to use them.

Alright guys, so at this point you're already pretty good at determining the equilibrium for an acid-base reaction if you have the pKa information. So I already taught you that you would compare the pKa of the acid and the conjugate acid. You would see which one is stronger and weaker, and that would determine your equilibrium. But what if you don't have pKa information? So what if you have an acid-base equilibrium question and there are 2 compounds that you just don't know the pKa values for? Or there are other situations too where pKa information might not be that helpful. And that's when we're going to use the factors affecting acidity. Okay? So for the next few pages, what I'm going to do is I'm going to introduce 5 factors that, even without knowing pKa values, we can still tell which one is going to be more acidic and less acidic based on these factors. So let's go ahead and get started. As I mentioned, there are 5 major factors, and we're really going to use these factors in 2 different situations. Okay? The first situation is that pKa information is unavailable for a molecule. So what that means is that maybe you didn't memorize it or maybe your professor didn't give it to you or maybe you did memorize it, and you just forgot it. Hey, if you're at a test and you don't remember it, it's unavailable. Alright? So then we might want to use factors affecting acidity. A second instance would be if the pKa's of 2 molecules are too similar to make a determination of highest acidity. So imagine if you're comparing 2 carboxylic acids. Well, they're both going to have a pKa of around 5, so how do I tell which one is more acidic and which one is less acidic? Well, with pKa information, you wouldn't really be able to do that. So we're going to need to look even more in-depth into these acids. Okay? So whenever we're analyzing these 5 different factors, what we're going to do is instead of looking at the acid, we're actually going to look at something else. That's going to be the stability of the conjugate base. Okay? How does this work? Okay. Well, the reason we look at the stability of the conjugate base is because that's going to tell us how willing the molecule is to give away a proton. So the more stable the conjugate, the more willing the acid is going to be to donate a proton. How does that make sense? Well, remember the conjugate base is what the acid becomes after it reacts. Okay? If the conjugate base is very, very stable, then it's going to say, hey, I'm fine giving up a proton as an acid. Because if I give up a proton, I'm just going to be really nice stable conjugate base. Awesome. But what if the conjugate base sucks? What if it's just like the worst conjugate of life? It's not going to want to exist very much. Okay? So instead, it's going to say, hey, I'd rather stay as the acid and have the proton on myself. So basically, the dissociation constant, the likelihood of me giving up a proton, is going to increase as my conjugate base becomes more and more stable. Does that make sense? And that's what these five effects have to do with. They are going to either increase the stability or decrease the stability of the conjugate base. Alright?

Understanding the Element Effect.

Go ahead and look at the first and easiest one and that's the element effect. The element effect determines how loosely or strongly a particular element bonds with hydrogen. We can use these effects to compare different protonated elements to each other. For example, a perfect example of using the element effect would be having a nitrogen attached to an H and a sulfur attached to the H. Do we know the pKa of the nitrogen? Actually, yes. I taught you that the pKa of NH3 should be around 38. What about SH2? I actually didn't teach you that one. So then in this case, if I were to ask you which one is the most, the strongest acid, you would have really no clue on how to tell me because I never gave you the pKa of SH2. So that's why we have to use these factors.

It turns out that the element effect is going to consist of two trends, and the first one is electronegativity. Electronegativity just says that the stronger the electronegativity, the more willing the molecule will be to accept a lone pair as the conjugate. Here's the electronegativity trend, let's say HF, which is an H attached to the most electronegative atom, and then we have also CH4 over here. After each of these molecules gives up a proton, H⁺, it's going to turn into F^- for the fluorine. This is the conjugate base. Now let's look at the conjugate base for the carbon, which would look like this, CH3-. Which of these conjugate bases is the most stable? It would be the fluorine since it is the most electronegative, so it's more comfortable having electrons on it compared to carbon. Does that first trend make sense so far?

Now let's look at our periodic table here. We were comparing nitrogen and sulfur. Just with the electronegativity trend, the sulfur would be the better acid, as even though you don't know the pKa of the sulfur, you know that the sulfur is more electronegative.

However, there's another effect that we need to know which is the size trend. What the size trend says is that the bigger or the squishier the atom is, the more willing it will be to accept a lone pair. Imagine F^- versus a much bigger atom, I^-. Because iodine is much bigger and more diffused, it can distribute the electrons over a larger space, making it a more stable conjugate base. Therefore, the stronger acid is HI, not HF. Hopefully, that makes sense to you guys in terms of the element effect. Just keep in mind that the element effect only has to do with hydrogens that are directly attached to different atoms. Now, take a break and try to solve this one and predict which one is going to be more acidic.

Using factors affecting acidity to rank acids

Hey everyone. So here it says without using pKa values, which of the following pairs is more acidic? Let's look at the first option. Here we have ammonia versus hydrosulfuric acid. Alright. We have H connected to nitrogen, and H connected to sulfur. Here we're going to try to figure out what type of technique we could use to determine which one is more acidic. Alright. In this case, we can rely on the element effect. If we look at our periodic table, we can see that nitrogen is in Group 5A. Sulfur is in Group 6A. But Sulfur is not in the same row as Nitrogen; it's one lower.

If we look at their electronegativities, although sulfur is more to the right than nitrogen, sulfur is not more electronegative. Sulfur, being lowered down, decreases your electronegative values. Nitrogen is just slightly more electronegative. But here the deciding factor in terms of the element effect is that Sulfur is larger in size. Remember, the trend is as we head towards the right side of the periodic table, our atomic radius or atomic size is going to decrease. But as you go down a group, it's going to increase. Sulfur being larger than nitrogen means that sulfur would be more acidic. Its bonds will be longer, and it'd be easier for us to remove an H⁺ from the sulfur atom. So here, SH2 would be more acidic than NH3. Again, it's based on our Element Effect or atomic size.

If we want to verify this, I know the question says without using pKa values, but we've decided on what our answer is. If we looked up the pKa values, you would see that NH3 has a pKa value of about 38. And Hydrosulfuric acid here has a pKa of around 10. It has a lower pKa value, which means it's more acidic. So here, Hydrosulfuric acid will be the more acidic of the pair, and it's based on the element effect or atomic size of sulfur versus nitrogen. Sulfur being larger means it's more acidic overall. Alright. Let's just use that one for this particular pair and continue with the other ones to see what the pairs will be for those.

Using factors affecting acidity to rank acids

Alright, guys. So for this one, what we're looking at are the atoms that are directly attached to the hydrogen. So I would have a carbon here that I'm looking at, and I would have an oxygen here that I'm looking at because both of those are directly attached. And for these, both of them are in the same row. What that means is that carbon and oxygen have the same size, so the size effect isn't going to matter at all. Okay? But what is going to matter is electronegativity. And remember that electronegativity gets higher as we go to the right, so that means that water is going to have a higher, I mean, going to have a higher acidity than my methane (CH₄). And remember, I said you can't use pKa values, you can just try to use the element effect. But if you wanted to double check, you could think about the pKa values. Remember the pKa of water is 16 and the pKa of CH₄ would be around 50. Okay? So obviously water is going to be a much better acid. Alright? So let's move on to the next question.

Using factors affecting acidity to rank acids

Alright. So for this one, what we had to look at is what are the atoms that are directly attached to the hydrogens. Okay? So in this case, I have an oxygen, but there are no hydrogens attached to it, so I have to look at the carbons next to it. And the carbons are what have the hydrogens on them. Okay? So I would have CH versus NH. Those are my 2 bonds that I am looking at. So then I would look at where those are in the periodic table. They happen to be right next to each other. Okay? So the size effect once again isn't really going to come into play, but the electronegativity is. This one is less electronegative and this one is more electronegative. So that means that the NH is going to be the one that is the strongest acid. Okay? So that means that it would be this one. And if you use pKa values, which again I told you not to use, but if you just wanted to confirm it, you would see that the pKa of this one is around 50 since it's just sp3CH and the pKa of this one is about 38. Alright. So now we're going to move on to our last one.

Using factors affecting acidity to rank acids

Alright. So for this one, I'm going to take myself out of the screen so I don't cover it up. And, this one we already solved, but I just wanted to point out that the reason is because the HI has a larger size. So it's going to be more acidic than HF, even though F is more electronegative. Alright? Now, some of you might be wondering, why do I have to look at the conjugate base? Can't I just look at the acid? Because so far, I haven't been drawing any conjugate bases. I've just been looking at acids. Yes, you can. As long as you understand the reason why it's more stable is that the conjugate base is more stable. That’s why it's a better acid. Okay? So, hopefully, that made sense. Let's go on to our next factor that affects acidity.

Understanding the Inductive Effect.

Let's talk about the second factor that affects acidity, and that is inductive effects. So inductive effects describe the stabilizing properties that electronegative—write that down, electronegative—atoms that are not connected directly to acidic hydrogen have on the overall acidity. Now notice what I did to this word NOT CONNECTED, this phrase, I made it bold, I made it underlined, and I made it all caps. So do you think that's kind of important? Yeah. Right? It's very important. I need you to realize that this is kind of different from element effects. Element effects had to do with the atoms that are directly attached to the H. Okay? Whereas inductive effects have to do with atoms that are not attached to the H. That means they're on other parts of the molecule. They could be 2 carbons away, but they're still going to affect the acidity of the hydrogen. Okay? So that's the difference between element and inductive effects.

The way that works is that whenever a charge can be delocalized over more than one atom, that conjugate base or that charge is going to be more stable. So remember what's going on here. Remember that all of these acids, they're always going to give up a proton, and they're going to get a lone pair in return, and they're going to turn negative, and that's called the conjugate base. If I can spread out that negative charge over multiple atoms, that's going to make my molecule more stable. Alright? So the way that that works is that, by the way, delocalizing all that means is spreading out. Okay? So when I delocalize a charge, that means instead of it being in one place, I let it be in 2 or 3 places. I spread it out over an entire molecule.

Using electron clouds to understand the inductive effect.

So let's go ahead and look at this example here. Here I have 2 acids. Okay. So we're going to go ahead and draw some electron clouds in order to figure this out. But first of all, I just want to ask you guys, if you didn't know about this trend, could you solve this question with pKa's? Which one is the stronger acid? Could you solve it with pKa's? Actually, no, you couldn't because we said the pKa of alcohol is roughly 16. So this one is 16 and this is also an alcohol at 16. So according to my pKa rules, I cannot tell the difference of which one's more acidic and which one's less acidic. Alright? So this is one of those examples where my pKa's are too similar to tell the difference, so I'm going to need to use a factor affecting acidity.

Now check this out. Both of these have the same exact element effect because in both cases, I have an oxygen attached directly to the H. So is the element effect going to be different at all? No. They're the same. The electronegativity is the same and the size is the same because they're both oxygen. Okay? But what is different about these is that one of these has 3 fluorines really far away and then one of these has 2 fluorines really close. So let's see how that's going to affect it.

This is the part that I was talking about: electronegative things that are on other parts of the molecule will affect the H. So what I want to do is give away an H. Give away an H and what we're going to get now is the conjugate bases. Let's look at the conjugate bases. The conjugate bases look like this. Basically, a negative charge in the O here and a negative charge in the O here. The only difference is that when I draw, like if you want if I want to draw like an electron cloud of where those electrons are, for the O on the left-hand side, it would just look like this. Pretty much all those electrons would reside just on that O, meaning that these are localized.

But then check this one out over here. This one has these 2 very electronegative atoms right next to it. So what that means is that instead of all the electrons being around the O, some of them are going to get spread out over these fluorines. Okay? So what that means is that I'm going to have less of a charge around my O and I'm going to have a little bit more of a charge in other parts of the molecule. In fact, if I were to draw this again, I'd probably make this part around the O even a little bit smaller. It would probably look more like this, like more spread out like that.

Okay? Now you're never going to be asked to draw this. This is just my example. I'm just trying to get you guys to see how one of the conjugate bases is going to look different. But if you had to guess which of these conjugate bases is more stable, it would be the one on the right. It would be this one because this one is delocalized or spread out. Okay? So what that means is that if one of these acids, if both of these acids had the same opportunity to give up a proton, the one that would say, oh, me first me first would be the one on the right. Why? Because that's going to be the one that forms the more stable conjugate base right here. Whereas the other conjugate base, that one kind of sucks because it's all just in one place. Does that make sense, guys? Cool.

The 3 factors that determine the strength of inductive effects.

So now what I want to talk about is what are the things that increase or decrease inductive effects. And there are actually three things that we want to look out for. Okay? So first of all, the strength of the electronegative entities. Okay. So the reason I'm using the word entities, I know that's like a weird word, is because it's not always going to be an atom. Sometimes it could be like a part of the molecule. Sometimes it could be like many atoms together. Okay? But anyway, the strength, that means that Fluorine is actually the best thing that you can use to have an inductive effect because Fluorine is the most electronegative atom. So that means if I were to rank the halogens in order, it would be Fluorine, then Chlorine, then Bromine, and then Iodine. K? Now check this out. This actually runs completely opposite to the element effect. Remember that in the element effect, the best halogen was iodine. Right? Because with the iodine, remember, it was really big and squishy and it could have a lot of electrons in it. But in this case, remember that the H is never directly attached to this halogen in inductive effects. In inductive effects, all I care about is which one is the most electronegative to take the most electrons away. Okay? So what that means is that actually, it's going to be the opposite. In this case, Fluorine is always going to be the best thing that you can have. Okay? Then the next thing is the number. The number just means the more the better. Okay? So if you have, like let's say that you have 3 Fluorines on one of them and 2 Fluorines on the other, the 3 Fluorines would win. Okay? And then finally, the proximity and that means the closer the better. The reason that proximity is important is because if your electronegative things are too far away, they're just not going to have an effect on it at all. And actually, that's what happened up far away, they're just not going to have an effect on it at all. And actually that's what happened in this conjugate base. This conjugate base had 3 fluorines, so you might have thought that it was going to actually be better. But it's so far away from my oxygen that it's not going to be able to have any effect on that electron cloud. Does that make sense? And the general rule is that if you are 3 carbons away or more, so 3 carbons or more or 3 atoms or more, then you would have no effect. Okay? So what that means is this would be 1, 2, 3. Anything after that is not really going to have an effect on the oxygen because it's just too far away. Alright. And I just realized that I was writing that off the page, so I'm just going to move that up a little bit so you guys can see that I said 3 atoms or more would equal no effect in terms of distance. Okay? So what that means is I want the strongest things, fluorines, more of them, as many as possible, and the closer the better. And that's going to be what makes my acid more acidic because it's going to stabilize the conjugate base more. Cool? Once again, a few of you guys might be asking me, but Johnny, can I just look at the acids instead of the conjugate? Yes, you can. But I need you to understand why it's more stable. And the reason has to do with these electron clouds. Okay?

Using Inductive Effect to determine acidity

Now what I'm going to do is I'm going to give you guys this problem that is actually really easy, but I'm still going to give it to you guys just as a free response. Go ahead and try to solve which one would be the more acidic, the right or the left. So go ahead and pause the video and tell me. So the answer is obviously this one would be more acidic. Okay? Because this one actually meets all three criteria. It has some electronegative elements that are not attached. It actually has 4 fluorines. It has a lot of them, 4, and they're all really close. They're all exactly on the alcohol. So this is an example of just a really easy question that you could get.

Now some of you guys might also be asking this, but Johnny, how about if my professor gives me one but not the other and if he switches them up? For example, what if on one of them, one of the acids has a bromine but then the other one has 2 bromines but then the other one has one fluorine. How do I know which one is better? And my answer to that is that professors are never that mean. So what they're going to do is because the only way to figure that out would actually be to do math and to actually do empirical calculations. So you're never going to have to worry about, oh, this one has more but this one is closer, stuff like that. Unless it's really obvious, like the one I gave you up there was obvious because it was way too far away. But if it was like you can't really tell the difference, you wouldn't get that kind of question. It's always going to be just these three things pulled together. So for example, one of them, let's say this was an alcohol, one of them could be like having 1 chlorine and then the other alcohol could be like having 1 fluorine. And then you just have to pick the difference between those 2. Okay. It wouldn't be like that you're having to pick the difference between 1 fluorine and 2 chlorines because that would be way too hard. Alright. So hopefully that makes sense to you guys. Let's go ahead and move on.

Understanding Resonance Effects

The next three factors aren't quite as complex, so I went ahead and grouped them into a category just called others. So let's go ahead and start with that third factor that affects acidity and that would be resonance effects. Alright? So resonance effects is actually just really easy. It's just whenever the donation of a proton leads to the formation of a possible resonance structure, okay, that conjugate base will be more stable. And the reason that the conjugate base will be more stable is because it can resonate. Remember that resonance structures enable for a charge to be in multiple places. That would make sense that it's delocalizing the charge. Alright? And that means that if the conjugate base is more stable, then the molecule will be a better acid. So check this out. For this example, I've given you two OH bonds. So I have OH and I have OH. So we're not going to use pKas to figure this out. It says which of the following pairs of acids would have the lower? But I'm not asking you to remember it, I mean even though you do, you should remember it, but I just want to use the factors affecting acidity to figure this out. So for both of these, do they have different element effects? Remember the element effect is the atom that's directly attached to the H. So in this case, they both have the same element effect. So I'm just going to write that here. Same element effect. Okay? And the reason is because they both have an oxygen attached to an H. That doesn't change. Alright? So the size, the electronegativity is exactly the same, but it turns out that one of these is a much better acid. In fact, one of these is called carboxylic acid and the other one is just an alcohol. So what is it about the carboxylic acid that makes it so much better as an acid? And it's the fact that let's look at the conjugate bases. The conjugate base for my carboxylic acid looks like this. O negative. Okay? The conjugate base, so I'm just going to put here as CB conjugate base. Alright? The conjugate base for my alcohol looks like this. Alright? So which of these is gonna be the one that's more stable? Well, both of them have a negative charge on the O, but notice that this O is stuck. This one, that negative charge isn't going to be able to go anywhere, so it's completely localized. Whereas on the other one, this one is actually going to be able to resonate using the two arrow rule that I taught you guys in resonance structures. So then I would make a bond there and I would break a bond there and I would actually get a new resonance structure that looks like this. And sorry it's a little bit crowded. I would get a new resonance structure that looks like that. So what that means is that I'm able to distribute this negative charge over those three atoms. Isn't that interesting? What that means is that one of them is going to be way more stable than the other and that is why carboxylic acid has a pKa of 5 whereas alcohol has a pKa of 16. That's a huge difference. Really, if it weren't for the carbonyl, that carboxylic acid would have a pKa of 16. But the carbonyl changes it so it can resonate. So now the pKa is basically like a trillion times better or like in terms of it's like a trillion times more acidic. Almost a trillion. Alright? So it's literally a way better acid. So that's called the resonance effect. Anytime you can make resonance structures, it comes into play.

Understanding hybridization effects.

Now let's look at the hybridization effect. The definition of the hybridization effect is this: the higher the s character in an acid, the more stable the conjugate base becomes. We discussed s character when we talked about hybridization. Recall that 25% s character comes from sp₃ hybridization, where you have one s and three p orbitals; therefore, 25% of the hybrid is s character. The s orbital is the closest to the nucleus and is the smallest, which means the more s character present, the closer the lone pairs are held to the nucleus. This proximity makes the conjugate base more stable by holding the electrons tighter to the positive, enhancing stability.

Now, considering the acidity trend, it would emerge that sp hybridization has 50% s character because it consists of one part s and one part p. Sp₂ would have 33% s character, and sp₃ would have 25% s character. Let's examine the pKa values to understand how this trend affects acidity. The pKa for an sp hybridized CH is 25. For sp₂, characteristic of alkyne, the pKa is 44. And for sp₃, typically found in alkane, it is 50. As the s character increases, so does the acidity. Does that make sense? When the s character proportion increases, the entirety of the hybrid orbital being s leads to greater acidity.

Which of the following hydrocarbons is the most acidic?

So this is just a way to explain that pKa trend. Which of the following hydrocarbons would be the most acidic? And you guys already know that it would be this one because this one would have an H there and that one would be sp hybridized, so it would have a pKa of 25. Is that cool?

Understanding steric effects.

All right. So we get to our last one and this is called steric effects. Now I still haven't even talked to you guys about what sterics are, but that will be coming shortly. But basically, what steric effects say is that particularly with alcohols, this is really just going to deal with alcohols, the more easily solvated the conjugate base is, the more stable it will be. Alright, so what the heck does that mean? Solvated. Wow. That's a huge word. It just means how easily it will dissolve in an aqueous solution or how easily it will be mixed in an aqueous solution. All right? What that means is that we want our alcohols to not be very bulky in order to mix. If they are bulky, they're not going to have as easy of a time mixing into the solution. So the rule here that you just need to know is that the smaller the R group, the more acidic the alcohol. Oops. Acidic. Okay. So the smaller the R group, the more acidic the alcohol, going all the way down to water.

Let me show you an example. So basically, if I have HOH and then CH3OH and then let's say I just keep adding, I keep putting more and more groups. So let's say I had something like that. Okay, I have these 3 different these are not all alcohols. These 2 are alcohols and then the one on top of it is water. Okay? So just so you know, this first alcohol, that I have right here, let's say this is going to be, this would be like whatever. I'm just going to say this is just CH3OH. This first one would have a pKa of around 16. Okay? As I add R groups to it, so as I make R and R here, that pKa is going to start to go up because it's going to be a little bit less acidic. So in this one, we have a pKa of maybe around 17. Okay? It gets to the point where if you add enough R groups, the pKa can go all the way up to about 19 for terbutanol, which is one of the bulkiest alcohols. Alright? Then let's look at water. Water, instead of having an r group, it just has an H there. H is the smallest one of all. Remember I said smaller the R group? Well, H is really the smallest that it can get. Right? So that means that water actually has a pKa of 15.7, making it actually the best acid out of all of these. Why? Because it's the one with the smallest group next to the alcohol. Does that make sense? So basically, the smaller the bigger your group gets, the worse of an acid it gets. Okay? The smaller your group is, the better an acid. Okay? So, then the other trend is just the opposite of that. It would say the bigger that the R group is, the more basic the alkoxide. The alkoxide is just the name for the conjugate base. So that would just be anything that has an O negative on it. Okay?

Which of the oxides is the most basic?

Alright. So, yep, the one that would be the most basic would be this one right here. And the reason is because that's called Terbutoxide and it's the bulkiest base out of all of these that has the most R group sticking off of it, so it would be the one that is the worst solvated, so that means that it's going to be the least stable conjugate base, which means it's the most basic. Alright. So I know that was a lot of like it means this, which means this. Just remember the rules. If you're confused, just memorize these rules and you'll be fine. Alright? So now let's go ahead and do some practice.

Would the following reactions go to the right or the left? Draw the products and label ALL species. Provide the full mechanism.

Alright guys, so for the next few questions, let's do some cumulative practice based on predicting equilibrium, pKa values, and factors affecting acidity. For all these questions, make sure to try and draw them all out on your own first and then proceed.

Would the following reactions go to the right or the left? Draw the products and label ALL species. Provide the full mechanism.

Let's go ahead and solve this first one. So what's the first rule that I want to follow when I'm predicting equilibrium? Well, I want to identify the base and the acid. And remember that there are sometimes when it's difficult to determine which one is the acid or the base, but in this case, it's a really easy question because I have a formal charge. And remember that formal charges are always really good indicators of acids and bases. Remember that we said if it's positive, then that's an acid and if it's negative, then that's a base. Okay? So in this case, I actually see a positive and a negative, but only one of them is important. Why? Because the positive is on a k. Remember that a k is a spectator ion. So I'm just going to cross that out. I don't care about it. What I do care about is the fact that I have a negative on an o, which means that this species or this molecule is my base, and if that's my base, that means that the other side must be or the other molecule must be my acid. Okay? So it was that easy. Sometimes it's going to be more difficult to predict the acid and the base, but in this case, since we had a full formal charge, it was easy. Now the next thing that we want to do is predict the conjugates. So sometimes the conjugates are going to be drawn for you. Sometimes you have to draw them yourself. If you have to draw them yourself, like in this case, it's not that hard. Just think about what you need to do. For example, let's start with the acid. Remember that the acid is a proton donor. So what it should do is it should make a base, a conjugate base, that just is missing a proton pair, but it's going to have an additional lone pair from the one that it's going to receive after giving away the hydrogen. So that's going to give me a negative charge. Okay? Then what's the other conjugate? Well, the other conjugate should be this O- that's going to gain an H, so I should have plus a molecule that looks like this, OH, where one of the lone pairs is donated, so this should be a neutral compound. So now I should have an alcohol on the other side. So this should be my conjugate base and this should be my conjugate acid. Cool so far? Now when we were doing these questions before and they were already drawn out, I always start off with the mechanism. Okay? But it turns out for these questions, I actually don't want to start off with the mechanism. The reason is that I don't know what direction the equilibrium is going in yet. And remember that when I'm drawing a mechanism, I want to make sure that the mechanism and the products are on different sides. So if the products, let's say, if the major products are on this side, let's say this is the product side here, that means that this is the favored side, then I would want to make sure that my mechanism is here. Okay? Since I don't know the equilibrium yet, I shouldn't draw the mechanism until I know that. Okay? So we're going to hold off on drawing the mechanism and we're going to do our last step, which is determining the equilibrium. So how do I do that? Do you guys remember? I would compare the pKa's of the acids. Okay. And that's if I know the pKa's. So in this case, do I know these pKa's? Yes, I do. This acid on the left-hand side would be 38. This would be an amine. And then my conjugate acid has a pKa of 16. Okay? So what I always like to say is that you want to make sure your number is going from lower to bigger. Okay. Because you want to make sure you're going from a stronger acid to a weaker acid. So am I going from lower to bigger? No, I'm not. I'm going to scratch out this arrow and I'm actually going to draw equilibrium arrows. The way we're going to draw them is like this. It's going to be a big equilibrium arrow going towards the left and then a small one going towards the right. You know what that means? That means that even though these two sets of products are in equilibrium with each other, there's much more on the left-hand side. Much more of it is favored to look like what's on the left side because that's the weaker base and the weaker acid. Okay? So just by drawing those equilibrium arrows like that, I've already answered the question of to the right or to the left. So now all we have to do is provide the full mechanism. So now that I know that my left is the one that is favored. Alright, so this would be my products area. This would be the favored area because this is the one that's actually making the thing that is going to exist most of the time. Okay. That means that like I was saying before, I want my mechanism to be on the other side, so my mechanism should actually be over here because what I want to show is that I'm making a reaction that goes in this direction. Okay? So if I had to draw my mechanism on the right-hand side like I have right on top of me, where would I start that mechanism? And the answer is I would start always from the same place, from the negative charge to the acidic H. So I would do this. Okay. So I have from negative to H. Now I have grabbed an H. Is that it? No, that's not it because of the fact that H only wants to have one bond, not two. So even though there isn't a bond drawn there, we know that there actually is a bond between the O and the H. Right? This is just a condensed structure. So I'm going to take the two electrons from there and jump them onto the O. That's what's going to give me my O- on the other side and my NH on the other side. Does that make sense? So if you guys said that the answer was to the left, you got this one right. Alright? So let's go ahead and move on to the next one.

Would the following reactions go to the right or the left? Draw the products and label ALL species. Provide the full mechanism.

So this one was pretty tricky because this is not your typical Brønsted-Lowry acid-base reaction. Remember that in Brønsted-Lowry, we exchange a proton. So what that means is that on one side the proton is in one place and on the other side the proton is in another. Okay? And these are in equilibrium with each other and this is what we've been doing over and over again. Well, in this case, I can't use pK_a values because this is not a Brønsted-Lowry reaction. This is actually a Lewis-based Lewis acid reaction. And remember that I told you guys that Lewis is the more general definition and it encompasses all reactions. So basically, anytime you have a nucleophile and an electrophile reacting together, that is a Lewis acid-Lewis base. And that's what this is. So really, this is going to be more like the types of reactions that we're going to see later in organic chemistry where we're just going to see nucleophiles and electrophiles reacting together and the difference is that instead of getting equilibrium, because equilibrium means like you're exchanging one thing for another, all we're going to get is products here. Okay? So what that means is that in this question I'm not asking you guys, does it go to the right or the left? It's just going to go it always goes to the right. Your challenge here is to figure out what does the product look like. Okay? And like I said, this is very similar to the type of reactions that you're going to see later on in organic chemistry where we're just going to say predict the product. Okay? So now that I've said all of that, how can we identify the acid and the base? Okay? And honestly, this one just requires a little bit of memorizing. I already taught you guys something really special about BH3. Do you guys remember? Remember that boron has an empty p orbital. Okay? And remember there was one other atom that had an empty p orbital. Do you remember that? Also, aluminum. Okay. So if you ever see aluminum, that's another one. And this means that it's going to be an amazing Lewis acid because it's really good to accept it's easy for it to accept electrons. Okay? So this is actually going to be a super strong acid. It's just not a good proton donor. Instead, it's just an electron pair acceptor. Okay? So that's my Lewis acid, so obviously the double bond must be my base. But how can it be a base? Well, remember, double bonds have extra electrons. Okay? So remember that I told you guys that double bonds are good nucleophiles. Okay? So it actually makes sense that this would be my Lewis base. Okay? And if we were using general nucleophile-electrophile terms, this would be my nucleophile because it's good at giving away electrons and this would be my electrophile because it's good at accepting electrons. Okay? Cool. So now let's go ahead and draw the arrow. Like I said, usually, we would hold off on drawing the mechanism until we know which side is favored. But I already told you the right side is favored because this is just going to be a reaction. It's just going to give me a single product. Okay? So where should the arrow start from? It should start from the negatively charged thing, which is the nucleophile. So what I should do is I should take the electrons from my double bond and put them straight into my p orbital. That means that now I'm making a new bond between the boron and one of the carbons on the double bond. Does that make sense? Now another thing to keep in mind is how many hydrogens did each carbon on the double bond have before? This one had 1 and this one also had 1. Are you guys cool with that? Because that would make the octet for both of those. Now what I want to do is draw what the product I'm going to get is that it's still attached to an H up here, but now it's also attached, it has a new single bond and then obviously H3. Okay? Is that cool so far? Now the bottom one still has only this is a single bond in between, not a double bond. And now this only has one H. So are there any formal charges that I need to add to this molecule? Actually, there's 2. Okay? You should have added a formal charge positive on this carbon. Why? Because that carbon now has only 3 bonds instead of 4 because the double bond broke and nothing came in to replace it. So, that would be what we call a carbocation. And then the boron likes to have 3 bonds because it's in the 3rd column or 3rd group, but now it has 4, so the boron should get a negative charge. Okay? And that's actually going to be our final product for now. Okay? For now, because we just learned acids and bases. But just so you guys know, this is the first step in a reaction that we're going to learn in chapter 8 that I'm just going to tell you, it's called hydroboration, it's like a really big word. Okay? But this is actually the first step in a reaction that we're going to learn later because like I told you, later we're going to be dealing with nucleophiles and electrophiles a lot and you're going to see stuff like this all the time. Okay? So hopefully, that wasn't too confusing. I was just trying to throw in some Brønsted-Lowry and also some Lewis. Okay? But just keep in mind that the difference with Lewis is that it always goes to the right, so let me write this down, just so you guys can know. Lewis is going to always go to the right. Okay? So that's something about it that you just need to know. It's always going to go to the right non-equilibrium. Okay? And it's always going to just make a single product. Excellent. And then also, you can't really use pK_a values for it because you're not donating protons. Remember the definition of pK_a was how likely you are to donate a proton. So since we're not donating protons, you can't use pK_as here, but thankfully you don't have to because, like I said, you can just always say it's going to go to the right. Alright, guys, let's move on to the last question.

Would the following reactions go to the right or the left? Draw the products and label ALL species. Provide the full mechanism.

So the tricky part about this question is that even though this is a Brønsted-Lowry acid-base reaction, so I should get an exchange of protons, I'm not going to have pKa information for all the reagents that I need. Okay? So what that means is that in the event we are missing pKa values, we're going to have to use the factors affecting acidity. So I told you guys I'm mixing things up. So really the only way to get this was to know the factors affecting acidity or to look up the pKa value and I hope that you did the first one because I want you to use those factors. So the first thing, as always, let's figure out the acid and the base. It's not that obvious, but once I dissociate the spectator ion, remember, spectator ions dissociate, so I would get this to become Na+ and then I would get NH- like that. So that means that this is my base and that means this must be my acid. So now in the same way that we did before, we're going to draw our conjugates. These conjugates have to be basically whatever that thing would do, whatever that reagent would do, we have to draw the opposite. So the acid is going to donate a proton, so I would expect that my conjugate base is going to be an S- because it's going to be an S that has donated a proton and gained a lone pair. And then what I'm going to have is a nitrogen that used to have one H, but now it's also going to have another H. Okay? And that second H is going to come from the fact that we grabbed an H from the S. So this is going to be my conjugate acid. Alright, so now let's go ahead and try to figure out what the equilibrium would be. So I would say okay, what's the pKa of my acid? That's the hard part. I don't know. Do I know the pKa of my conjugate acid? Yes, this is 38. But I have no clue what the pKa is of the first one because that's not one of the ones that I told you to memorize. Okay? So what this means is that instead of using pKas, this is going to be an opportunity for us to figure out which of the five factors of acidity, factors affecting acidity, I could use to predict this. So what do you guys think? Which factor could I use to predict which one is going to be the stronger acid? Well, the answer is the element effect. So we're going to use the element effect and the reason is because you use that factor when the elements attached to the H is different. So notice that I have a sulfur attached to the H on the one side, and then I have a nitrogen attached to the H on the other side. That means that I can use the element factor to figure out which of these is going to be the more stable conjugate base or what you could just say is which one is going to be the stronger acid. Element effect basically said that as you go to the right of the periodic table and as you go down, your acids are going to get stronger. Okay? So nitrogen is around here somewhere. Where would sulfur be in relation to that? Obviously, this is a really basic periodic table. Sulfur is actually to the right and down. It's in the next row, and it's also in the next column over. So which of these should be the better acid? And the answer is that it has to be sulfur because sulfur is not only more electronegative, but it's also going to be bigger. Okay? So the answer here is that this is my stronger side, and if this is my stronger side, then which direction is this going to go? Is it going to go to the right or to the left? It's going to go to the right because I need my stronger to go into my weaker. Does that make sense? Cool. So now you guys just saw how even if I don't have pKa information, I can still determine the equilibrium using these factors. Okay? So now all I have to do is I have to draw the mechanism, and the mechanism would start on which side? What do you guys think? Should I draw it on the left side or the right side? I should draw it on the left side. Why? Because these are my products. Right over my head, these are my products because these are the ones that are favored. So what that means is that my mechanism should start on the other side because you always want your mechanism giving you products. So where would I start the arrow from? The negative charge. And I would grab the H. Am I done? No, I'm not done. I would have to take the electrons from the SH bond and give them to the S. And now I'm done. Alright. So hopefully that made sense to you guys. This is the strategy that we use for equilibrium questions. Like I said, this is really the highest form of question that you will get in this chapter. So I hope that you guys understood it, got some practice. Let me know if you have any questions.

Here’s what students ask on this topic:

What are the factors affecting acidity in organic compounds?

The factors affecting acidity in organic compounds include:

• Element Effect: Electronegativity and atomic size of the atom bonded to the acidic hydrogen.
• Inductive Effect: Stabilizing influence of electronegative atoms not directly attached to the acidic hydrogen.
• Resonance Effect: Stabilization of the conjugate base through charge delocalization.
• Hybridization Effect: Higher s character in the hybrid orbitals increases acidity.
• Steric Effect: Smaller substituents around the acidic hydrogen increase acidity by allowing better solvation of the conjugate base.

How does electronegativity affect acidity?

Electronegativity affects acidity by influencing the stability of the conjugate base. A more electronegative atom bonded to the acidic hydrogen will stabilize the negative charge on the conjugate base better, making the acid more willing to donate a proton. For example, in the comparison between HF and CH₄, the conjugate base F^- is more stable than CH₃^- because fluorine is more electronegative than carbon, making HF a stronger acid than CH₄.

What is the inductive effect and how does it influence acidity?

The inductive effect refers to the stabilizing influence of electronegative atoms that are not directly attached to the acidic hydrogen. These atoms pull electron density through sigma bonds, stabilizing the conjugate base. The strength of the inductive effect depends on the electronegativity of the atoms, their number, and their proximity to the acidic hydrogen. For example, a molecule with multiple fluorine atoms close to the acidic hydrogen will be more acidic due to the strong inductive effect of fluorine.

How do resonance effects stabilize conjugate bases?

Resonance effects stabilize conjugate bases by delocalizing the negative charge over multiple atoms. This delocalization spreads out the charge, reducing the energy and increasing the stability of the conjugate base. For instance, in carboxylic acids, the negative charge on the conjugate base can resonate between two oxygen atoms, making the conjugate base more stable and the acid stronger compared to alcohols, where the negative charge is localized on a single oxygen atom.

What is the hybridization effect on acidity?

The hybridization effect on acidity is related to the s character of the hybrid orbitals. Higher s character means the electrons are held closer to the nucleus, stabilizing the conjugate base. For example, sp hybridized carbons (50% s character) are more acidic than sp² (33% s character) and sp³ (25% s character) hybridized carbons. This is why alkynes (sp) are more acidic than alkenes (sp²) and alkanes (sp³).