Why is this section important? Because not all acid-base reactions are that easy.
Concept: Why we need factors affecting acidity and when to use them.4m
Alright guys, so at this point you're already pretty good at determining the equilibrium for an acid-base reaction if you have pKa information. So I really 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 will determine your equilibrium. But what if you don't have pKa information? So what if have an acid-base equilibria question and the two compounds you just don't know the pKa’s for, or there are other situations too where pKa information might not be that hopeful. And that’s when we’re going to use the factors affecting acidity.
So for the next few pages what I am going to do is I’m going to introduce five factors that even without known 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.
So as I mentioned, there’s five major factors and we’re really going to use these factors in two different situations. The first situation is that pKa information is unavailable for a molecule. So what that means is that mean you didn't memorize it or maybe your professor didn’t give it to you or maybe you did memorize if and you just forgot it. Hey, if you're at a test and you don't remember it, it’s unavailable, right? So then you might want to use factors affecting acidity.
A second instance would be if the pKa’s of two molecules are too similar to make a determination of highest acidity. So imagine if you're comparing two carboxylic acids. Well, they're both going to have a pKa of around five so how do I tell which one is more acidic and which one’s 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.
So whenever we’re analyzing these five different factors, what we're going to do is instead looking at acid we're actually going to look at something else. And that's going to be the stability of the conjugate base. How does this work? 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 giveaway 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. 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 this really nice stable conjugate base.
But what if the constantly sucks? How about if it’s just like the worst conjugate of life. It's not going to want to exist very much. 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 that. They are going to either increase the stability or decrease the stability of the conjugate base.
There are two situations in particular that making predicting equilibrium challenging:
No matter what, we know that the stronger acid will have the more stable conjugate base. Remember, reactivity and stability have that inverse relationship we talked about.
This effect describes the way different atoms donate protons. For example C-H vs. N-H.
Concept: Understanding the Element Effect.7m
Let's 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, yeah. I taught you guys that the pKa of NH3 should be around 38. What about SH2? Did I teach you that one? I actually didn't teach you.
Then in this case, if I were to ask you guys which one is the strongest acid, you would have really no clue on how to tell me because I never gave you the pKa of SH2. That's why we have to use these factors.
It turns out that the element effect is going to consist of two trends. 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. I'm just going to say EN. What that basically says – let's just put it this way, we have let's say, HF, which is an H attached to the most electronegative atom. And then let me say that we have also CH with also an H3 over here. I'm just giving you – that's a little bit off the screen. Can you see that that's just an H on the side? Cool. Like right there.
Anyway, we've got HF and we've got CH4. What I'm wondering is after each of these molecules gives up a proton, what is it going to look like. Well, after this gives up one proton, H+, what it's going to look like is this, F-. Does that make sense? Because you always exchange a proton for a lone pair, so it's going to turn into F-. This is the conjugate base.
Now let's look at the conjugate base for the carbon. The carbon when it gives up its proton, by the way, I'm drawing this arrow wrong. This is not a mechanism arrow; this is just me showing what happens next. What that would look like is like this CH3-.
These are my two conjugate bases. My question to you is which of these conjugate bases is the most stable. Which of these is the happiest having that lone pair or that negative charge on it? Obviously, the answer must be that it would be the fluorine.
Why? Because fluorine is the most electronegative, so it's the one that's happiest having electrons on it, whereas carbon is really not electronegative, so carbon sucks. This carbon is really, really bad. That's like a terrible happy face, sad face. This carbon is really, really bad. This fluorine is kind of happy. It's okay because it's very electronegative, so it doesn't mind having a lone pair.
So does that first trend make sense so far?
Now let's look at our periodic table here. We were comparing nitrogen and sulfur. Nitrogen at the periodic table was here. Sulfur is here. Just with the electronegativity trend, which of these do you think is going to be the better acid? The nitrogen or the sulfur?
Hopefully, you said the sulfur because even though you don't know the pKa of the sulfur, you know that the sulfur is more electronegative. Is that cool so far? So it's going to be happier with the negative charge.
But now it turns out that there's another effect that we need to know or another trend that has to do with the element effect. That one is the size trend. What the size trend says is that the bigger or the squishier that the atom is – I'll explain what squishy means in a second – the more willing it's going to be to accept a lone pair.
I'm going to give you guys now two more examples still using fluorine. Imagine that I have F- and F- looks like this. The negative charge has to do with a lone pair that accepted over here. Now what I just drew was the electron cloud. There's a lot of electrons circulating and I just drew the electron cloud.
Now let's compare that to iodine. Iodine is at the bottom down here. Iodine is a much bigger atom. The atomic number is I think that's 53, so it's really, really big. It has 53 electrons. Instead, iodine is going to look like a sun compared to fluorine. It's going to look massive. Iodine is super, super massive. Florine is little. And iodine also has a negative charge and it also has a lone pair.
So which of these do you think is going to be most affected by the lone pair? The answer is that fluorine is more electronegative than iodine. I agree with that. But also fluorine is a lot smaller. That means that it's going to feel the effect of that negative charge way more than the iodine will.
The iodine is so big, it's like I don't care. I don't mind having an extra lone pair because I'm massive. That's what I mean by squishy. It's really like big and squishy. It has a ton of electrons everywhere, whereas fluorine is kind of small. Even though it's very electronegative, it still is way smaller, so overall it's going to experience the effects of that negative charge more than the iodine would. The iodine honestly doesn't care. It could give up an electron, it could gain an extra electron. It doesn't matter.
So the answer is that with size you also become more acidic. The further down you go, the more acidic you are. If I were to compare HF versus HI and I went ahead and I gave up a proton in each of these, what I would become is F- and I would become I- as my conjugate bases.
And then which one would be the most stable conjugate base? The I-. Why? Because the I-, like I drew, is much, much bigger, so it's going to be able to distribute those electrons in a much bigger space than the fluorine will, so it's going to be more stable. That means that the stronger acid is going to be HI.
Cool. Hopefully that makes sense to you guys in terms of the element effect. Now just keep in mind that the element effect only has to with hydrogens that are directly attached to different atoms.
Let's go ahead and do these examples down here.
So for 1, I'm going to have you guys – we already did some of these, so I'm just kind of working ahead, but showing you guys using the element effect. So for 1, go ahead and take a break and try to solve this one and predict which one is going to be more acidic.
It consists of two trends:
NOTE: This effect can only be used when comparing the way different atoms are attached to hydrogen. If you are comparing O-H vs. another O-H, it won’t work!
Example: Without using pKa values, which of the following pairs is more acidic?2m
Example: Without using pKa values, which of the following pairs is more acidic?1m
Example: Without using pKa values, which of the following pairs is more acidic?2m
Example: Without using pKa values, which of the following pairs is more acidic?1m
This effect describes the way that electronegative atoms that are NOT CONNECTED to the acidic proton make the conjugate base more stable.
Concept: Understanding the Inductive Effect.2m
Let's talk about the second factor that affects acidity and that is inductive effects. Inductive effects describe the stabilizing properties that electronegative – write that down – electronegative atoms that are not connected directly to the 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, 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 two carbons away, but they're still going to affect the acidity of the hydrogen. That's the difference between element and inductive.
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 lone pair, I mean, I'm sorry – give up a proton and they're going to get a lone pair in return and they're going to turn negative. 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.
The way that that works is that – by the way, delocalizing, all that means is spreading out. When I delocalize a charge, that means instead of it just being in one place, I let it be in two or three places. I spread it out over an entire molecule.
Any EN force that helps to pull electrons away from the conjugate is called an inductive effect. If you can spread out that negative charge over multiple atoms, that base will be more stable.
Example: Using electron clouds to understand the inductive effect.4m
Concept: The 3 factors that determine the strength of inductive effects.4m
So now what I want to talk about is what are the things that increase or decrease inductive effects? And there's actually three things that we want to look out for, OK? So first of all, the strength of the electronegative entities, OK? So, the reason I'm using the word entities I know it'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, OK? But anyway, the strength that means that Fluorine is actually the best thing that you can use to have an inductive effect because Florine is the most electronegative atom, so that means if I were to rank the Halogens in order it will be Fluorine then Chlorine then Bromine and then Iodine, OK? Now check this out this is actually runs completely opposite to the element, 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 effect all I care about is which one is the most electronegative to take the most electrons away, OK? 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, OK?
Then the next thing is the number, the number just means the more the better, OK? So, if you have like let's say that you have 3 fluorenes on one of them and 2 fluorenes on the other, the 3 fluorenes would win, OK?
And then finally the proximity and that means the closer the better, The reason that proximity is important is because if you're electronegative thins or too far away there's just not going to have an effect on it all and actually that's what happened up in this conjugate base, this conjugate base had 3 Fluorenes so you might have thought that that was going to actually be better but it's so far away from my oxygen that is 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'll have no effect, OK? So, what that means is this this would be 1 2 3 anything after that is not really going to have an effect on oxygen because that's just too far away, alright? And I just realized that I was writing that off 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, OK? So, what that means is I want the strongest things fluorenes more of them as many as possible and the close of 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 guys might be asking but Johnny can't 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, OK?
Factors that increase inductive effects:
Example: Without using pKa values, which of the following pairs is more acidic?2m
If a conjugate base is able to make a resonance structure, it will be more stable.
Concept: Understanding resonance effects. Which of the following –OH groups would be more acidic and why?4m
The next three factors aren't quite as complex, so I went ahead and grouped them into a category just called Others. Let's go ahead and start with that third factor that affects acidity and that would be resonance effects.
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, that conjugate base will be more stable. 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. That means that if the conjugate base is more stable, then the molecule will be a better acid.
Check this out. For this example I've given you two OH bonds. I have OH and I have an OH, so we're not going to use pKa's 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 should remember it. But I just want to use the factors affecting acidity to figure this out.
For both of these, do they have different element effects? Remember the element effect is the atom that's directly attached to the H. In this case, they both have the same element effect, so I'm just going to write that here. Same element effect. The reason is because they both have an oxygen attached to an H. That doesn't change.
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? It's the fact that – let's look at the conjugate bases. The conjugate base for my carboxylic acid looks like this, O-. I'm just going to put here this is CB, conjugate base. The conjugate base for my alcohol looks like this.
Which of these is going to 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 on 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. Sorry, it's a little bit crowded. I would get a new resonance structure that looks like that. 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 would still have – 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 in terms of its – it's like a trillion times more acidic, almost a trillion. It's literally a way better acid.
That's called the resonance effect. Anytime you can make resonance structures, it comes into play.
Concept: Understanding hybridization effects.3m
Now let’s look at the hybridization affect. The hybridization affect the definition is this: but the higher the s character in an acid—remember that we talked about s character when we were talking about hybridization. Remember we were like saying that 25% s character would come from sp3 where you have like an s and you have 3 p's so then 25% of the entire thing is s, and then would be the s character and the reason is because the S orbital is the one that's closest to the nucleus. Remember it’s the smallest so the more s character you have, the closer the lone pairs are going to be held to the nucleus. It’s going to make the conjugate base more stable because you going to hold those electrons tighter to the positive and it's going to be a little bit more stable.
So if we were to think about the acidity trend what it would look like, and by the way, just go ahead and add a stick there if you don’t have one already. So for the acidity trend what it would be is that sp would have 50% s character because it's basically one part s and one part p. sp2 would have 33% s character. And then sp3 would have 25% s character.
Now let's look at the pKa’s underneath. I want you guys to draw in the pKa’s so that you can remember what they are and you can see how this trend is affected. So what was the pKa of an sp hybridized CH? It was 25. What was the pKa of sp2 where this is our alkyne, alkene and alkane? sp2 would be 44. And then what was sp3? It was 50. So you can see how as my s character gets higher, as my s character is increasing my acidity is also increasing. Does that make sense? So as your s character is getting bigger, the amount of the whole hybrid orbitals is s, the more acidic you get.
The higher the %s-character of the conjugate base, the more stable it will be.
Example: Which of the following hydrocarbons is the most acidic?1m
This rule really only applies with alcohols for now.
Concept: Understanding steric effects.3m
Alright, 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 effect say is that particularly with Alcohols so this is really just going to deal with alcohols the more easily solvated the conjugate base is the more stable will be, alright? So, what the what the heck does that mean, OK? Solvated wow that's a huge word it just means how easily it will dissolve in an aqueous solution or how easy it will be mixed in an aqueous solution, alright? And 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 small the R group the more acidic the alcohol, OK? So, the smaller 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 C and then let's say just keep adding I keep putting more and more groups let's say I have like that, I have these 3 different, these are not all alcohols these two are alcohols and the one on top is water, OK? So just so you know this first alcohol that I have right here, let's say this is going to be whatever I'm just going to say this is just CH3OH this first one I have a pKa of around 16, OK? 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 it's only a little bit less acidic so then in this one have I have a pKa of maybe around 17, OK? It gets a point where if you add enough R groups the pKa can go all it's about 19 for a torbutrol 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, remembers I said the smaller the R group remember 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 bigger your group gets the worse of an acid it gets, OK? The smaller your group is the better an acid, OK? So, then the other trend is just the opposite of that it I would say the bigger 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, OK?
Particularly with alcohols, the more easily solvated the conjugate base is, the more stable it will be.
Example: Which of the oxides is the most basic?1m
Example: Practice: Intro1m
Example: Practice 1: Acid-Base Reactions6m
Example: Practice 2: Acid-Base Reactions7m
Example: Practice 3: Acid-Base Reactions6m
Rank the following in order of basicity. (1 – least basic, 3 – most basic)
Using inductive effect and resonance considerations, determine which molecule is most acidic:
Consider the molecule given below:
a) What is its name?
b) Give its approximate pKa.
c) Draw a structure that is more acidic.
d) Draw a structure that is more basic.
Circle the weakest acid:
Match the following list of pKa values to the compounds shown below: 16, 19, 25. Your answer should include pictures and a minimum of words to support your choices.
The following compound is one of the strongest acids known. Explain using detailed pictures and a minimum of words why it is such a strong acid.
Propose a change to the structure below that would significantly increase the acidity. The molecule you draw should have a different molecular formula.
For each pair of compounds below, indicate which is more acidic.
Rank order the following in terms of anion stability. Write the letter corresponding to the correct order of anion stability, ranked from most to least stable for the molecules labeled as (a) - (d).
Rank order the following in terms of anion stability. On the answer sheet on page 1 write the letter corresponding to the correct order of anion stability, ranked from most to least stable for the molecules labeled as (a) - (e).
Rank order all of the following with respect to relative acidity. The acidic H atom in question is indicated with the arrow for each molecule. On the answer sheet on page 1 write the letter corresponding to the correct order of acidity, ranked from most to least acidic for the molecules labeled as (a) - (d).
Circle the most acidic hydrogens in the following molecules. If there are two or more equivalent hydrogens (hydrogens having the same chemical environment) circle ALL of them.
Place the following compounds in order of their basicity with #1 being the most basic and #4 being the least basic.
Which is the correct order of decreasing acid strength for these species? (most acidic → least acidic)
a) I > II > IV > III
b) II > I > III > IV
c) III > IV > II > I
d) IV > III > I > II
Which of the indicated hydrogen atoms is most acidic?
What is the order of acidity of the indicated hydrogen atoms in this molecule? Rank from most acidic to least acidic?
a) b > a > c
b) c > b > a
c) c > a > b
d) b > c > a
Consider molecule A shown below. Remember that for a Lewis structure, you must show all atoms, all bonding valence electrons, all nonbonding valence electrons, and all nonzero formal charges. Using structural considerations, identify which of the two types of Hs (CH or NH) in molecule A is the more acidic one. Provide a clear, concise explanation why one is more acidic than the other.
Circle the most acidic compound from the following list and explain why.
Aluminum, silicon, phosphorus and sulfur are in the same period and are arranged in order of increasing atomic numbers. Which anion is the strongest base?
Which is an incorrect statement?