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SN1, SN2, E1, E2 Mechanisms · Acedra
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SN1, SN2, E1, E2 Mechanisms
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0:00
PROFESSOR: Welcome back, class. So what we want to talk about today is being able to look at a reaction and predict if it's going to be an SN1, E1, or an SN2, E2 type of mechanism. And there's going to be these guiding principles that, when you look at them, and you look at them in a holistic way, look at all of those principles, you will be able to identify what mechanism you should invoke. And then once you've decided on the mechanism, then you can use that mechanism to predict the product
0:40
and what it should look like. So the factors that we're going to look at are-- we can enumerate these right here. We could say, the strength of the attacking species.
1:02
And then we'll take a look at the concentration-- and I'll just go like that with the brackets for concentration-- of the attacking species, and I'll preview that with att. The third thing that we want to take a look at is the leaving group ability and how that affects the mechanisms and the type of carbon the leaving group is attached to. So another way of saying that-- we could just say, hybridization states.
1:47
What is the hybridization state of the carbon that the leaving group is attached to? And that's really important. So another thing is solvent effects. So with solvent effects, that's going to affect the strength of the attacking species. So, kind of, one in five are connected. And then we have another principle that we have to watch out for, and that would be heat. And we represent heat in reactions with that delta symbol. So those are the principles that we want to discuss.
2:32
And so we are going to jump right into that and take a look at the first one, the strength of the attacking species, and why is that so important? Now, just to make sure we're all on the same page, these principles are going to tell us, when we see a reaction, what mechanism is going to occur.
3:01
So for the first factor here, the strength of the attacking species, here is a bullet point that you'll want to write down.
3:09
The stronger the nucleophile, which is the attacking species-- so attacking species are what? They could be nucleophiles, or they can be bases.
3:24
So when I'm talking about an SN2 reaction, the stronger the nucleophile-- so we can make that bold-- the stronger the nucleophile, that's going to mean, that's going to promote a faster SN2 reaction. Barely got that on the screen. Stronger the nucleophile, faster the reaction. So that's a key point to remember. So let's see here. I want to see if there's something on the PowerPoint that I want to talk about here.
4:09
So taking a look at that PowerPoint here-- I'm looking on my screen. I can't see it right on the screen that you see it. I have to look at a different screen, so that's why I'm looking away.
4:22
So here is a list on that PowerPoint of strong nucleophiles and weak nucleophiles. And what you're seeing there are, typically, strong nucleophiles are what? Negatively-charged. And then the weak nucleophiles are generally neutral. And so we're going to focus our attention on the strong nucleophiles at the moment and how they favor or promote a SN2 reaction to make them go faster. But before we look at that, when we take a look at two nucleophiles, how do we know which one is stronger?
5:05
So if we come back to the PowerPoint slide, let's take a look at that RO minus versus, let's say, the carboxylate, which would be the RCO2 minus. Do you see that on the PowerPoint slide? So we want to compare those two. But now we have to be careful-- the R group-- I'm going to translate it now into something else. The RO minus-- let's say that is going to be methoxide.
5:43
And I'm cutting that off, so I'm going to turn off the PowerPoint now. So methoxide versus-- let's just keep it generic here, like that. So that to that. When we look at these two, which one is going to be the stronger nucleophile? Well, we just use the same principles that we've learned when were looking at the strength of acids and how we use the cardinal rule to look at the stability of the conjugate base. So when we use the cardinality rule here, we can see--
6:25
let's write out the cardinal rule.
6:31
So there's the cardinal rule. And we go through it, and we compare these two. They both have the same charge. They both have the charge on the same atom. So we go to the next one. We see that this molecule doesn't have resonance. This molecule does have resonance. Hence, this guy is going to be the more stable base. That's how it's drawn. Or let's just keep the-- sorry. That's going to be the more stable nucleophile-- more stable, right? That's the more stable nucleophile.
7:12
So what does that mean from an energetics perspective? Well, if we draw it right here, just energetically right here, because the carboxylate is more stable, it's going to be right down here.
7:34
And so just by process of elimination, that would make the methoxide up here like so. So now that we've established which one is higher in energy, we can then also say, hey, this guy is higher in energy, shown right there. So it's going to be the most unstable, which will then say, hey, that's the most reactive.
8:02
That guy is the most reactive because it's higher in energy. Remember, when molecules are high in energy and are uncomfortable, they want to get more comfortable. And the way that they do that is by reacting with something because, when you react, you can get lower in energy and more stable.
8:22
So this guy is the more reactive species, which then, because it's more reactive, because it's the most unstable, it is going to be the stronger nucleophile. So that's our stronger nucleophile. So when we take a SN2 reaction, and we say which of these two nucleophiles is going to make the SN2 reaction run faster or proceed faster, we would say, the stronger nucleophile. And we've identified, from this whole process, that that is our stronger nucleophile. So this one will have a faster SN2
9:03
reaction than that nucleophile.
9:11
So let's clear out this board.
9:17
Now let's take a look at an SN1 reaction and how the nucleophilic strength affects the rate. So if we take a look at our PowerPoint again, let's take a look at some of those weak nucleophiles because weak nucleophiles favor SN1 mechanisms. So if we take a look and compare the reaction rate of a alcohol versus, let's say, an amine, so we see-- 3. There we go. There's ammonia. So we have a generic alcohol, an ammonia. And when we look at the PowerPoint slide, which one's the stronger nucleophile?
10:09
Well, the amine is going to be the stronger nucleophile, OK? When we take a look at a reaction here-- let's do a molecule that looks like this. That's somewhat correct.
10:33
Put a chlorine here, actually-- we need a good leaving group. Now, if we compare the ROH versus the amine, which one is going to go faster?
10:53
Now, if you're thinking SN2, we've learned that the stronger the nucleophile, the SN2 reaction will go faster with the stronger nucleophile. But with SN1, the rates of these two is the same-- same rate. The strength of the nucleophile does not influence SN1 reactions. And why is that the case? Because when you look at the rate law or the rate for an SN1 reaction-- remember this? So if our rate is going to equal-- and this is for SN1-- that's going to equal just the what?
11:46
What is so depen-- sorry. For an SN1 reaction, what species is responsible for the rate? It's the alkyl halide. Compare that to the rate of a SN2. That is dependent upon the alkyl halide and the attacking species, which I'm just going to represent like that.
12:19
Or since it's SN2, we can actually put in there the nucleophile. So do you see the difference in the rates? So in an SN2, the rate is dependent on the nucleophile. And so if you have a stronger nucleophile, the reaction is going to go faster. But in an SN1 reaction, there is no attacking species over here. That doesn't influence the rate at all. It's just the concentration of the alkyl halide. And so the strength of the nucleophile for a SN1 doesn't influence it.
13:06
So what we can say in conclusion for this concept of attacking species strength is that, if you have a strong nucleophile, that's going to favor SN2. And if you have a weak nucleophile, that's going to favor SN1. Now, the same principles that I've taught you here for SN1 and SN2 also apply for E2 and E1. So if you have a strong base, that's going to favor E2. And if you have a strong base-- or sorry, not strong-- a weak base, that's going to favor E1. So same principles here, OK?
14:09
Because when you look at the rate of an E2, it is only dependent-- it is dependent upon the alkyl halide and the base.
14:26
But if you look at the rate of an E1, that is only dependent upon the concentration of the alkyl halide. So it's basically the same concept here.
14:43
But what we have to be very careful of is just because you have a strong nucleophile does not mean that nucleophile can be a strong base. Let me show you that via the PowerPoint. So if we take a look here at the PowerPoint, this is showing you the strength of nucleophiles. But in this next slide, right here at the very top, you will see that it is the strength of the base. So this slide with the strength of the nucleophile is now on the bottom of this slide.
15:28
Now, I want you to make some comparisons here. On the far right of that slide, you see that R minus. It is a strong nucleophile, and it is a strong base. You see that comparison when you're looking at the slide? But now take a look at iodine, the I minus. Do you see how it is a red and a strong nucleophile but a blue, weak base? Do you see that? So the iodine is-- or iodide because it's an anion-- the iodide is a very strong nucleophile but a weak base.
16:20
And so we have to be careful about that because, when we talk about electron-rich attacking species, you could fall into the trap of thinking that, oh, it's electron-rich, so it's automatically a strong nucleophile and strong base-- not necessarily. Now, if we take a look at the iodide, for example, and just reason through it. Why is it a strong nucleophile? Well, because it has negative charge, and it is very polarizable, which means that the electrons in the electron cloud can move around.
17:04
And polarizability is a key feature for nucleophilic attack. But why would iodide be a weak base? Well, think about it. What happens if we treat this as a base? It's going to want to abstract a proton from an acid, right? And so that would just mechanistically come in, abstract that proton like so, and generate what?
17:38
That would generate hydroiodic acid.
17:44
That's an acid. What's this pKa? Around negative 10 or 11. So we know if-- with iodide, when it behaves as a base, it's going to generate an acid, which then-- it's an acid. Acids want to donate their protons. So iodide is a weak base because it generates a very, very strong acid. So it's just not going to do it.
18:17
And you guys didn't see any of that because of the PowerPoint slide. So let me rehash that. There we go. So let's start over.
18:30
So why is iodide a weak base? Because when it abstracts a proton from an acid, it generates hydroiodic acid. So you guys saw all of that part over there, but you missed this part, OK? So, in summary, why is, when you take a look at our iodide, our bromide, and our chloride, why are they weak bases? Because they generate strong acids, if they act as a base. But why are they strong nucleophiles? Because of their negative charge and their polarizability, all right?
19:21
Another thing that you have to be aware of is, when you look at that PowerPoint slide, do you see that the OR minus is a strong base and a strong nucleophile? All right, so there are some exceptions to that.
19:47
Erase all this-- all right. So when we look at that PowerPoint slide, we see RO minus. And according to that PowerPoint, that is a strong base and nucleophile. That's what we see on the PowerPoint slide, right? But this R group right there is going to influence if it's both of these. So, for example, this OR minus could represent ethoxide, like that. And according to our chart, I look up on the PowerPoint and it says, RO minus is a strong base and a strong nucleophile.
20:38
And for this particular molecule, that is true. So we'll just go strong base, strong nucleophile-- like that. But if we added some steric bulk to that R group-- so compare it to something that looks like this, like that-- tert-butoxide. And if we are using our PowerPoint as our guide, we can get a little misinformation here because this does, in fact, represent an RO minus. But this one is only a strong base. And the reason why that is only a strong base
21:27
is because of this steric bulk right here. That steric bulk is going to prevent the tert-butoxide from being a strong nucleophile. So you're going to encounter strong, bulky bases, meaning they're strong because they have a negative charge. But because they're bulky, they can only act as a base. That cannot act as a nucleophile.
22:04
So that right there is the first factor-- the strength of the attacking species. All right, so that principle right there-- there's a lot to it. The second factor that I want to take a look at is the concentration, right? And this one is pretty straightforward.
22:33
So recall when we were looking at the rates. And let's look at the rate of an SN2. What's the rate dependent upon? The concentration of the alkyl halide and the concentration of the attacking species. Then, in this case, it's going to be a nucleophile. But when you compare that to the rate of an SN1, that's only dependent upon the concentration of the alkyl halide.
23:12
So what we see here that, when we increase-- let's say we double the concentration of the nucleophile. It's going to double the rate. But if we double the concentration of the nucleophile in an SN1 mechanism, it doesn't affect the rate at all. So what we can say here is, when we look at a reaction, a concentrated attacking species is going to favor SN2. So if our concentration of our nucleophile is concentrated, that's going to favor SN2.
24:04
But if our nucleophile concentration is low or dilute, it's going to favor SN1. And so the same principles apply when that attacking species is a base. So if our base, being really strong, if it has high concentration, that's going to favor a E2 mechanism. And if our base right here-- a weak base-- is low in concentration, that's going to favor a E1 mechanism. And it's all-- it comes back down to these rate laws right here, OK? So when you see a reaction-- let's do something like this.
25:01
And we have a--
25:04
let's do hydroxide like this, OK? What's our product going to be? Well, it depends on the mechanism. But that's not what I want to talk about. What I want to talk about is, when you see a reaction, you're always going to assume that the attacking species concentration is high, that the concentration of that attacking species is concentrated. That's the assumption you go by every single time. The only way that you're going to know the concentration of that attacking species,
25:44
if it's dilute or not, is if I put and spell it out. If I go, like, in parentheses, and go, "dilute."
25:55
Now you know that because this attacking species is dilute, the mechanism may, because it's dilute, favor a SN1 or an E1, OK? Now, I use the word may be a SN1 and an E1 is because we cannot look at just one of these principles and make a decision on the mechanism. I cannot look at this reaction and say, oh, it has a dilute attacking species, so it's either SN1 or E1. No, I can't make that judgment call because I'm ignoring all the other principles that I had listed on the board.
26:35
So we have to look at all of the principles and come to a conclusion. But this right here, that dilute, is one bit of information that suggests that the reaction should be SN1 or E1 because it's dilute. So that one is a pretty straightforward principle.
27:01
The next one, or factor three, is the leaving group ability.
27:11
So let's clean up this board here.
27:16
So factor three is the leaving group ability. So if we take a look at these two alkyl halides, and we're comparing those, which one is the better leaving group? Because alkyl halide, alkyl halide, so we know those are leaving groups, OK? And so what's that going to generate here? That's going to generate a secondary carbocation plus our chloride. Same for this guy down here-- secondary carbocation plus our bromide right there. Which one is the better leaving group?
28:02
In this particular case, we could just use a size argument, like, the chloride is smaller than the bromide, and we've learned that the larger the volume that the atom has or the negative charge to move around in, it's going to be lower in energy and more stable. So we can say from this analysis that the bromine right here is the better leaving group because, after it has left, it's going to be more stable than the chloride. And so if these two atoms could talk,
28:37
the bromide is like, yeah, I can leave. It doesn't bother me at all, whereas the chloride is like, I can leave, but it's going to hurt a little bit, and I'm going to be uncomfortable in this way. So leaving group ability-- in order to do these four mechanisms, you have to have a leaving group. Now, for the SN2, E2, SN1, and E1, every single one of those mechanisms needs a good leaving group. But now, I want to take a look at, with respect to leaving groups, which mechanism
29:20
does it affect the most? So the leaving group ability of this bromine right here-- would it affect the rate of an SN2 reaction more or an SN1 mechanism? So if we take a look at the mechanism in a SN2 fashion, what's the step? In an SN2, it comes in, breaks off in a concerted fashion, right? Very simple, straightforward-- SN2-- that. But in a SN1, what's the first step in the mechanism? Leaving group leaves, generate a carbocation, then the-- we'll just use water, in this case.
30:23
The attacking species will then come in and attack and generate OH, like this. And then we would have to do a proton transfer with some more of that water to generate our alcohol, like so. So looking at these mechanisms here, this one is an SN2. This is an SN1.
30:54
We're having the same alkyl halide, same leaving group. But this bromine right here-- the leaving group ability is going to affect the SN1 reactions more than the SN2 because, what's the first step in the SN1 mechanism? The leaving group has to leave first. In an SN2, the leaving group does, in fact, leave, but it's kind of leaving because the oxygen atom here is kind of kicking it off.
31:39
And so what we can say here is that, the better the leaving group-- where do we want to write this?
31:49
The better the leaving group is going to influence and affect SN1 and E1 mechanisms far greater than the leaving group ability will affect SN2 an E2 reactions. So it is true that all four mechanisms need a good leaving group. It's just a better leaving group is going to favor our SN1 and E1s more so, OK? It affects it more.
32:28
Now, if we take a look at a alcohol--
32:38
let's take a look at it at an alcohol.
32:44
Can we say to the alcohol, hey, I want you to leave and generate a species that looks like this? Can that happen? I hope you're all saying no. Alcohols are horrible, horrible leaving groups. But what if you do really, really need to get rid of the alcohol? What if you do need to get it to say, hey, I need to leave? Well, you can convert alcohols into good leaving groups by treating it with a strong acid like hydrobromic acid. Look at what happens here. We have some lone pairs on that alcohol.
33:32
Make sure-- OK. That can come in, grab that proton, break it like so. And then we're going to have what?
33:43
Attach the water there. And now the oxygen atom is positively-charged. And we still have our bromide floating around. And now look at what can happen. That molecule right there looks like water. It is water that's attached to this carbon. So water is a good leaving group. So we can have the bromide come in an SN2 fashion, attack, kicking off the leaving group to give us our product. And I know this board is getting messy, so our product is going to be down here.
34:24
Br plus water just left, like that.
34:31
So we can, in fact, convert bad leaving groups into good leaving groups, if we need to. Also, I want to highlight, what kind of molecule did we make? We made a alkyl halide. So put that in your notes where you keep track of reactions and what products you form. And so you could phrase the question in your notebook, how do I make alkyl halides? And you would then answer your own question by saying, you take an alcohol and treat it with a strong acid, and you get your alkaline, right?
35:20
OK, so that is factor three-- leaving group ability. And if you can remember, the leaving groups affect SN1 and E1 better, right?
35:46
So let's take a look at factor four, which is basically the hybridization states--
35:57
hybridization states of the carbon that the leaving group is attached to.
36:05
So I'm going to start drawing a molecule, and I want you guys to take a moment and figure out how many good leaving groups are in this molecule, all right?
36:25
So let's see here.
36:38
Yeah, we'll do it like that. How many good leaving groups are in this molecule?
36:47
Pause for a second, and answer that question honestly. Now, if you said, hey, there's two leaving groups, then we're missing something because, in previous lecture videos, I said, the leaving group, which we see two halogens here-- so you are thinking that there are two good leaving groups. But there's another layer of understanding. Those good leaving groups have to be attached to a SP3 hybridized carbon. And so we see the hybridization state of that carbon is SP2. And that one is SP3.
37:30
So this is not a good leaving group because it's attached to SN2. So in that particular molecule, there's only one good leaving group.
37:46
So let's draw another molecule. Let's take a look at something that looks like this.
38:04
And let's treat it with some water in an SN1 fashion here. How many of those leaving groups would work for this reaction? How many of these groups here would react with the water? Well, we first have to establish what we just learned over here. Is every single carbon that's the leaving groups are attached to-- are they SP3? Yes, they are. They're all SP3 hybridized carbons. So we have one, two, three good leaving groups. But when we invoke a SN1 mechanism,
38:46
there's only going to be two plausible choices. And we have to remember that if we choose to attack here or use this as a leaving group in an SN1 fashion, what's the first thing that happens in an SN1 mechanism? The leaving group has to leave. And if that leaving group leaves, what are we going to generate? We will generate-- let's draw it-- erase this for a second. We would generate a primary carbocation. We do not generate primary carbon cations. It's too unstable-- does not work.
39:29
But can we generate a secondary carbocation or a tertiary carbocation? Yes, we can. So what we have to be careful of when we are looking at a molecule like this--
39:46
not only do we have to identify that the alkyl halide is attached to SP3 hybridized carbons, but we have to identify that, hey, this alkyl halide right here is a primary. That one is a secondary. And that one is a tertiary alkyl halide. And so that primary alkyl halide-- if it undergoes an SN1, it would become a primary carbocation. That's no-go. That would become a secondary carbocation and a tertiary. And those are OK because those carbon cations are stable enough that they
40:27
can form those intermediates. So what we can say for a SN1-- we have primary alkyl halides, secondary, and tertiary.
40:43
Which ones are not an option? Cannot do that one, right? What about a SN2 mechanism? So in an SN2 mechanism, we have to use a strong nucleophile, which looks like that. We invoke SN2.
41:10
How many of these alkyl halides would react with the hydroxide in an SN2 fashion?
41:20
Hopefully, you're seeing, based off of what we've talked about previously, that the tertiary is off the table-- can't do it. And why is that? Because of the mechanism. The hydroxide here comes in-- backside attack. And if it tried to do backside attack here on that tertiary alkyl halide, the sterics here would prevent that from happening. So you cannot have a SN2 mechanism on a tertiary alkyl halide due to sterics. But can that mechanism work for a secondary and a primary? Yes, it can.
42:08
Now, let's take a look at the same molecule here. It's getting a little bit messy, so let's clean it up.
42:22
What if we used the hydroxide right there, as well? And let's invoke an E2.
42:35
What if we have a E2 mechanism? We have primary, secondary, and tertiary.
42:46
Can it react with all three of these alkyl halides? How do E2s work? Yes, it can. So the E2 mechanism is a little bit more promiscuous that it can react with both of them, right? However, if you invoke a strong, bulky base, like tert-butoxide right there, those are going to favor primary and secondary over this tertiary alkyl halide due to a steric argument, all right? So this E2 right here-- you have to really look at the base to see what kind of base
43:38
it is-- if it's a small, strong base or a strong, bulky base. But if we just left-- if we just looked at this particular scenario, a strong, small base without steric bolt-- it will do all three, OK?
44:03
Let's see here. Ah, we've got one more. Yes. Let's do E1. So that would have to be a weak base. So E1-- we would have primary, secondary, and tertiary. Cut off the tertiary. Let's take it down here. So E1-- primary, secondary, and tertiary. Which ones of these-- do all three alkyl halides work for an E1?
44:44
No, it doesn't, because what's the first step in a E1 mechanism? The leaving group has to leave. And so it follows the same trend as the SN1. Primary is off the table. Secondary and tertiary are. Those can work. And it all comes back to the mechanism. The leaving group leaves first, generating the carbocation, so we have to assess the stability of those carbocations. OK? Now, primary carbocations don't form. That's what I've been saying. But there's something that--
45:26
there's some-- I don't want to call it really an exception. It's just another layer of understanding. But students, at first glance, would be like, hey, that doesn't work when, in fact, it does. And so I want to show you two examples that are very common here. And that is when you have a benzene ring, and you have a good leaving group or a halogen. Can this-- so what is the hybridization state of that carbon. That is SP3. So the halogen can, in fact, leave.
46:11
And when you look at this carbon there, as well, is that a primary, secondary, or tertiary alkyl halide? And it is, in fact, a primary alkyl halide. But look at this. It does, in fact, leave. It can generate that. And it does kind of look like a primary carbocation, but it's not. And that's because it's attached to this benzene ring. Look at what this benzene ring can do. It can have a resonance structure that helps to stabilize what looks like a primary carbocation. Well, we call this--
47:01
we don't call it a primary carbocation. We call it a benzyl cation or a benzylic cation. Benzyl benzylic-- interchangeable in this particular case. Another example that you need to be aware of is something that looks like this.
47:36
Can that bromine just leave? Yes, it can. But, first, we identify that, hey that is still SP3 when we look at it. And then that looks like its primary, and so we're thinking that it can't happen, the leaving group can't leave. But it does because what do we generate? We generate something that looks like that. And so what is that guy called? See, this is an allylic.
48:20
And allylic cations can, in fact, form. So why is that so stable? Why can that form? You guys can already see what's going to happen. We have resonance.
48:43
So that carbocation is resonance-stabilized. And so we have a benzylic cation and an allylic cation that, at first glance, they look like they're primary, but they're not. They have a different name, and they are stable due to resonance. So what we can take a look at is the relative stabilities of some typical cations. I just want to make a list for you so you can see their stabilities.
49:24
So we have a cation that looks like this. That's called a vinylic cation, and that's very unstable. And then we have our methyl cation that's a little bit more stable. And then we have--
49:45
let's see here-- our primary carbocation that is representing a secondary cation. And those secondary carbocations right there are approximately just as stable as our allylic, OK? And also approximately like the benzylic. And I'm going to cut that off. Darn it. I'll draw it right here like that. You can see that. That's good.
50:37
And then you barely see that less than sign right there. But then we can have our tertiary carbocations are typically the most stable. So this is just a general way of looking at the stabilities of carbocations and how you would rank them. Now, as we progress in the course, we can see a little-- over in this area, a little switching occurring depending on the situation. And we'll see those when we talk about them. But, for right now, this is a good way
51:18
of looking at the stabilities of these cations.
51:28
OK, let's take a look at factor five, which is solvent effects.
51:57
So I'm just going to go straight to the answer, and then we'll talk about the reason why that is the answer, OK? So we have two types of solvents that we need to be aware of in organic chemistry. And the first one is polar aprotic solvents. And those are going to favor SN2 and E2 mechanisms, all right? And then you have something called polar protic solvents, and those are going to favor S1 and E1 mechanisms.
52:41
Now, why is this the case? So that's the answer.
52:48
But why? Well, let's go to the PowerPoint because that's going to show us some things right here.
53:01
So these slides-- it looks like I skipped this slide right here. So I want to highlight this one real quick, and this one is all about the leaving group ability. So you've been taught that the halogens are good leaving groups. That is true. But you also have other types of leaving groups shown on the slide called sulfonate anions. And those are good leaving groups. And we have a variety of them. We have a mesylate, a tosylate, and a triflate. And you can see how they look.
53:38
So if I wanted to draw a mesylate leaving group, it would look like this-- O attached to a sulfur double-bonded O, double bonded O, like that. So that would be a mesylate. Now, mesylates and all of these sulfonates right here behave the same way. If you want the bromine to leave, it leaves. If you want the mesylate to leave, it leaves-- same thing. It just looks a little bit different. And then when you look at the tosylate, the tosylate would look like this--
54:25
six-membered ring and a methyl group, right? So there's our tosylate. It would behave the same exact way. But when we draw our tosylates, drawing this big piece out every single time can be a little pain in the butt. So instead of drawing it all out, we could just go like this-- OTS.
54:52
And if we wanted to abbreviate our mesylate, it would just simply be O-mesylate. And you did see that, so let me redraw it like that.
55:06
How would you do the triflate? You would just go, OTF, OK? So those are just new leaving groups that you need to be aware of. And they play a significant role in organic chemistry.
55:23
So this slide right here is just showing you good leaving groups.
55:34
OK, now back to factor five, which is solvent effects here. There's two. We already talked about the two types of solvents. But I want you to pay close attention to the product solvent on the left. Do you notice how close the solvent is around that anion? Now, let's back up for a second and evaluate the difference between a protic solvent and a aprotic solvent. That's really important to understand. So a protic solvent or a polar protic solvent-- let's do it that way--
56:24
polar protic solvents are solvents much like alcohols-- so ethanol-- or water.
56:36
When we take a look at these molecules, we know, from general chemistry, that both of these molecules are polar. So that checks that word right there. They're polar molecules, and they're solvents. Now the protic right here is referring to protons. Now, the proton of interest has to be one that can hydrogen bond. How do we know if a hydrogen bond can form? it's if the hydrogen is attached to a fluorine, oxygen, or nitrogen atom. So this hydrogen right there can hydrogen bond.
57:21
And because we have a molecule that has a hydrogen that's attached to an oxygen, fluorine, or nitrogen, that can hydrogen bond, and so that makes it a protic solvent. So the protic is referring to a proton that can hydrogen bond. When you compare polar aprotic--
57:48
for example, acetone, which is also shown on the slide there on the right. That's acetone. It looks like this-- we can see that molecule is, in fact, polar. But aprotic means that it does not have a hydrogen on the molecule that can hydrogen bond. Yes, it does have those three hydrogens on those carbons. Yes, it has hydrogens. But it does not have hydrogens that can hydrogen bond. That's what we're looking for. So hydrogen-bonding capability-- protic.
58:28
Hydrogen that cannot participate in hydrogen bonding-- aprotic. So this is a polar aprotic solvent. These are polar protic solvents. And because of this difference in the ability to hydrogen bond affects how they interact with anions. And we can see back to our PowerPoint slide there that the molecule on the left with water as the polar protic solvent can get really, really close to that anion, and it can hydrogen bond. Well, let me take that back. You can see that those partially positive
59:13
hydrogens on the water-- and why are they partially positive? Because of the electronegativity and the polar molecule. So we have a water molecule. If we draw it right here like this, do you see how we have a-- because it's polarized, we have one side of the molecule that is partially negative-- negative-- and the other side is partially positive. So do you see how the partially positive side of the water molecule is interacting with that anion? And that interaction is really, really strong.
1:00:11
All right.
1:00:14
But, in contrast, do you see that the acetone, or the polar aprotic solvent in the molecule on the right, do you see how its proximity-- or do you see how the solvent molecules, which are the acetone molecules, how close they are to the anion? They're not as close as the water molecules are. And one reason that that's the case is because of the sterics of those methyl groups. It doesn't allow the partial positive of the acetone-- so when we look at acetone, there's
1:00:58
a partial positive and a partial negative part. And because of these methyl groups right here, those methyl groups prevent this partial positive piece of the molecule to interact very strongly with the anion. So just look at the distance when you look at that PowerPoint slide. Look at the blue partial positive on the group on the left. Do you see how close those partial positives are to the anion? And compare that to the structure on the right. You see those partial positives are further away.
1:01:39
So there's less of an interaction between the solvent and the anion for the polar aprotic solvent. And so this has an effect on the ability of those anions to react.
1:01:59
So let's ask a question here and tie this all together. So I'm going to leave that PowerPoint slide up.
1:02:17
Which is the stronger attacking species, bromide versus chloride? Do you remember at the beginning of this video, we talked about how this right here is more stable, so that makes the chloride more reactive because it's less stable, so it's the stronger attacking species? Remember that analysis? And what principles did we invoke to come to that conclusion? It was a size argument, right? OK. But now the strength of these attacking species
1:02:55
is also going to be dependent upon the solvent because what we're seeing on the PowerPoint slide is that a polar protic solvent is going to solvate an anion very, very well. You see that there's more water molecules around the anion, and they're closer to the anion, so it stabilizes-- or it solvates the anion more. And when an anion is solvated more heavily, it becomes more stable. More stable means lower in energy. And when it's lower in energy it becomes less reactive,
1:03:39
whereas a polar aprotic solvent-- the one that's shown on the right, the image on the right in the PowerPoint-- does not interact with the anions as well. And for our purposes, we are going to just make the assumption or statement that polar aprotic solvents don't solvate anions very well, OK? So that's something that you write down. That's really important to remember. Polar aprotic solvents do not solvate anions very well, whereas polar aprotic solvents do, in fact, stabilize anions very well.
1:04:26
So if we take a look and ask the question, which is the stronger attacking species, well, it's all dependent on the solvent. So if we take these two species and put them in a solvent called DMF, which stands for dimethylformamide-- and that is a polar aprotic solvent.
1:05:02
And polar aprotic solvents do not solvate anions very well. So when we take a look at these two, we're going to just make the assumption that there's no solvent present.
1:05:16
And thinking that there's no solvent present when, in fact, it really is, but we're just ignoring it because polar aprotic solvents don't solvate anions very well, so they don't offer any stability to the anion. So all we're going to do now is look at these two and assess which one's the most stable. And we're just going to use what principles we've learned up to this point. And in this particular case, remember when we can-- let's get rid of this--
1:05:52
carbon nitrogen oxygen, fluorine chlorine bromine iodine. Remember that we have electronegativity arguments and the size argument. When we're going down a column like this, we're going to look at size as a factor and not electronegativity to explain stability. So because we're going down, the bromine-- or sorry-- the bromide is larger. We're ignoring the aprotic solvent because it doesn't solvate the anions very well. We can say, based off of size, that this is going to be the more stable anion.
1:06:39
So that makes that unstable in comparison to this, which will make the chloride the stronger attacking species.
1:06:50
All right?
1:06:53
That would be the stronger attacking species in a polar aprotic solvent. Now, let's change it up.
1:07:06
What's going to happen to the strength of these attacking species when we change the solvent to polar protic? So let's put in ethanol. What's going to be the stronger attacking species? Hmm. Now, let's think about the principles here. That's polar protic, so those do solvate anions very, very well. And we have two anions right here, so they're going to be solvated very well. Now, it comes down to which anion is going to be solvated better. And the size here does influence that, once again.
1:07:52
The smaller anion is going to have more localized charge due to its size. Because it's smaller, the negative charge is more localized. And when that happens, the solvent is going to interact more strongly with a smaller anion. So this anion is going to be solvated more than the larger anion. So what is that going to do to the energy between these two? If this is more solvated, it's going to lower it's energy more so than the bromide. And so that's going to make the bromide the least stable
1:08:39
in this situation. And so the bromide is going to be the least reactive of the two, so it's going to be the stronger attacking species.
1:08:52
Do you see how the solvent has flipped those? So in aprotic, this guy is going to be the stronger attacking species. No. Wait wait wait. Let's pause for a second.
1:09:08
Let's not mess this up. So polar protic-- this is going to be solvated more, so that makes this the stronger attacking species. But in a polar aprotic solvent, we're going to pretend that there's no solvent around because aprotic solvents don't solvate anions very well. And so when you compare the stability of these two, that one's going to be more stable because of its size. So that makes that guy the stronger attacking species in a polar aprotic solvent, OK?
1:09:50
It all has to do with how the solvent interacts with the anion.
1:09:59
So let's draw some tables here and some charts and just summarize all of this, OK?
1:10:13
So let's take a look at nucleophilic strength we go down, just like that.
1:10:30
And I want to take a look at the nucleophilic strength when we're looking at polar protic solvents.
1:10:40
So what we're showing here is that, in polar protic solvents, the iodine is going to be the stronger nucleophile than the fluoride. So what we're doing is, energetically-- we have one, two, three, four. What we're going to see right here is that the iodide is at the very top, and the fluoride right there is at the very bottom, OK? The fluoride is going to be lower in energy, more stable. Why? Because we're in polar protic solvents. The smaller atom is going to be solvated more strongly,
1:11:31
and that stabilizes it. And so the larger atom is solvated less, making it higher in energy, and it's more reactive. So nucleophilic strength, as you go down a column in polar protic solvents, the larger atom is going to be the stronger attacking species or the stronger nucleophile. Now, what if we go across?
1:12:06
What if we go-- let's do a different-colored marker here.
1:12:14
I'll draw another energy diagram right here. We'll have four energy levels, like so. What if we're going to go across the periodic table? What is going to be the stronger nucleophile? Now, this is an interesting thing is that it goes in this direction but for nucleophilic strength. So that's saying that the carbon, which is going to be a carbon anion, is going to be higher in energy than the fluoride.
1:12:59
Now, notice that this is still in polar protic solvents, OK?
1:13:06
So why do we have a trend in this direction? Why is the carbon anion a stronger nucleophile than the fluoride anion? And the principle is that when you go across the same period here, the sizes of the atoms don't change that much. So the actual volume of the carbon anion versus the fluoride anion-- about the same size. So the solvent is going to solvate these anions, basically, to the same degree. So that's a wash. How the solvent interacts with these anions
1:13:53
doesn't matter anymore when trying to determine which one is the stronger nucleophile. So we have to go to another principle. And now, we're just going to look at the stabilities of these two species. And we can't use a size argument because they're the same size. So what can we do? We can look at electronegativities. And we know the fluoride or the fluorine atom is more electronegative than the carbon. And so the carbon can tolerate a negative charge better,
1:14:27
so it's going to be lower in energy, less reactive, so making the carbon the stronger nucleophile, OK? You see that? Now, let's take and change up the solvents, now. Let's change the solvents to polar aprotic-- polar aprotic solvents. What's going to happen now?
1:15:03
Do the trends change? And they do, in fact, change.
1:15:12
So when we are looking down a group in a polar aprotic solvent, which species is going to be the stronger nucleophile? Well, it's going to be the fluoride. It's a complete switch, all right? So that would mean that we have to switch these. Our fluoride is now there. Iodide is right there. Now, why is there a switch now? Well, because we are in polar aprotic solvents. They, the aprotic solvents, don't solvate anions. So we ignore that there's a solvent present for determining
1:15:59
which one's the stronger nucleophile and just go based off of what we've been taught in the past. Do we use electronegativity arguments when we go down a group? No. We use size. And we can see that, based off size, a negative charge on a larger iodide is going to be more stable, lower in energy than a negative charge on a smaller fluoride. And so because fluoride is higher in energy, the least stable going down this group, that's going to make that the most reactive,
1:16:38
and it's going to make it the stronger nucleophile in this series. Now, does this arrow change? Does the nucleophilicity change there? And that arrow does not change because, once again, solvents are not going to solvate anions very well because that's how polar aprotic solvents work. And we have the size argument here in which the carbon anion and the fluorine anion are the relative same volume. So we just have to go off of the principles that we've learned before.
1:17:23
We can't use a size argument in determining if the carbon is more stable or the fluoride because they're the same size. But what can we use to invoke the stability? We can say electronegativity. The fluoride is more electronegative, so it can stabilize the negative charge better, so it's going to be lower in energy, making the carbon anion the stronger nucleophile, all right?
1:17:57
So I think I've already written this down somewhere, but I'll just do it again just to cover our bases here. So when we have polar aprotic solvents, that's going to favor what? SN2 and E2. Polar protic solvents is going to favor SN1 and E1. Now, another principle that I want to share with you to help put this all together is the fact that, when you have a attacking species or a anion, depending on the-- OK, let's put it this way. This anion is going to have solvents around it, OK?
1:18:49
And in order for that anion to act as a nucleophile, it has to basically shed this solvent. It's not going to be able to shed all of it, but it needs to be able to get rid of some of it, of that solvent, in order for that negative charge to come in and attack the electrophile. We've got to get rid of that solvent. So if you have an attacking species-- so if we compare a bromide versus this bromide-- and I'm going to represent solvent with this green marker.
1:19:30
So I'm going to invoke a lot of solvent on that bromide and just a little solvent around that one. So you see they're still both bromide, but this bromide right here is going to react faster and better. And it's going to be a stronger nucleophile because it has less solvent around it, whereas this right here is going to be a weaker nucleophile in this particular solvent system because it has more solvent. And in a SN2 and an E2 mechanism, what's happening?
1:20:16
It is coming and attacking in a concerted fashion, OK? So we want this attacking species to be-- the word is naked. It doesn't want solvent around it because more solvent around it is going to limit its ability to do the SN2 attack. OK? So that's another reason why polar aprotic solvents are going to favor SN2 because there's going to be less solvent around the attacking species, which allows it to attack better. And we've already learned that the strength of the attacking
1:21:02
species influences if it's going to be an SN2- or an SN1-type mechanism. Stronger attacking species promote SN2 and E2. On the flip side, why do polar protic solvents favor SN1? Well, that has to come back to this idea of leaving group.
1:21:33
So if we have this compared to this, but in this one, we have polar protic solvent and this one we have polar aprotic solvent, and the solvents-- we're just worried about the anion. So when this anion leaves like this, and this bromine leaves like that, what difference does it look like? Well, this is polar protic, so that means it's going to have a lot of solvent, and this one is going to have less. So that guy leaves. That guy leaves. But which one's going to want to leave more earnestly?
1:22:20
Well, in this case, they're both the same leaving group. But you can see that, hey, in polar protic solvents, the bromine is going to want to leave-- it will leave easier than in polar aprotic solvents because, in polar protic solvents, when it leaves, it becomes an anion. And when it's solvated by the solvent, that makes that bromide lower in energy. This bromide isn't solvated very well because of the polar aprotic solvent. So that makes this guy higher in energy than that guy.
1:22:59
And so when you compare these two scenarios, which one wants to leave better? This one because it gets solvated more strongly, and it's more stable. Isn't that cool? OK, now the next factor that we need to talk about is heat and what that does.
1:23:29
OK, so let's take a look at a reaction here.
1:23:36
Now, I see we have a secondary alkyl halide, and I want to treat that with a sodium hydroxide, so a strong attacking species here. Now, we could go in two ways here. We could go in this path that's going to give us this product right here, like that. And what type of mechanism is that right there? So let's be a little bit more particular. Let's put that as a wedge right there. And so that's going to make that a dash. Hopefully, you can see that as a SN2 mechanism. And so we have a substitution.
1:24:22
So SN2 for substitution. But we can also have a E2 mechanism right here in which we could generate the alkene plus the chloride plus what? The hydroxide obstructed a proton, turning it into water.
1:24:47
Now, when we take this species here, treat it with the hydroxide, we're going to have competition here. And that's what this whole lecture is about is trying to figure out what is it going to do.
1:25:01
But when you tack into this principle here of adding heat-- adding heat, I have said, it's going to favor the elimination product. So why is it going to favor the elimination product? Well, it comes back to Gibbs free energy, when you take a look at the equation here like that.
1:25:32
So when we apply heat-- well, let's bag heat for a moment. Let's take a step back. When you look at the substitution versus the elimination, which one has higher entropy?
1:25:48
This has-- it goes from two species to two species versus two to three. So this molecule-- or the elimination product has more entropy. When we look at this equation right here, when we increase entropy, that's going to be a favorable process, right?
1:26:16
But now when you tack in heat, what are you doing? You are multiplying the temperature by the entropy. And then, in elimination, we have more entropy. And now you're going to multiply it. And so that's going to make this term right here that much more negative, which is going to give us a more negative delta G, which is a more favorable reaction. So heat favors elimination because of this term right there.
1:26:58
OK, so you have to understand here that when we're going to go through all these principles, we use heat as a tiebreaker. And I'm going to show you what I mean by that when we go through a problem. So just remember, heat is used as a tiebreaker, and we represent heat with the delta symbol. So if we did this reaction right here, in this particular case, we could see, basically, a mixture of products. We could have our alkene and an alcohol. But if I change and added some heat to it,
1:27:42
then, clearly, it's going to be our alkene.
1:27:51
So we'll leave it at that. So now I want to show you how you're going to use all of these principles in order to decide what a reaction is going to do-- if it's going to be any of the four mechanisms, OK? I'm just going to double-check my notes to make sure I haven't missed anything.
1:28:18
OK, so we will do this via PowerPoint, OK? So let's turn on our PowerPoint again. And it always starts at the beginning, so we have to go through it here. OK, so here's our reaction. We have the chloropentane, and we're going to react that with sodium bromide, OK? And I'm going to show you step by step how I can look at that particular reaction and figure out what mechanism I'm going to invoke. And once I've decided on a mechanism, I will do that mechanism, and that will show me
1:28:58
what products I've obtained and which is going to be the major. So the first thing that you have to do when you see a reaction-- it's written right there-- determine whether the substrate has a suitable leaving group. And when we look at that molecule, we see that there is a chlorine attached, and so chlorines are good leaving groups, OK? We also have to make sure that the leaving group that we've identified in step one is attached to a SP3 hybridized carbon. That is critical-- SP3.
1:29:38
So that's step two. Now, the PowerPoint slide is just reminding you of the different types of alkyl halides here. So when I look at the alkyl halide right there, it is what? A secondary alkyl halide. When you have a secondary alkyl halide, all four reaction mechanisms are on the table. A secondary alkyl halide can undergo SN1, SN2, E1, and E2. But you can see on that chart, as well, what if our molecule was only a primary alkyl halide? Then our only two options are going to be a SN2 and an E2.
1:30:32
If it's a primary alkyl halide, our job is easier because we went from four possible mechanisms all the way down to two just looking at step two. But I've chosen a harder problem to illustrate this. When it's a secondary alkyl halide, all four mechanisms are on the table at this point in time. So that's step two-- making sure what type of alkyl halide you have. Now, step three, we're going to go through the principles that I've discussed-- the strength of the attacking
1:31:15
species, the concentration, the leaving group ability, and the effects that solvent has on the reaction.
1:31:31
OK, let's see right here. OK, so what do we have here? So adding up the four factors-- so we want to take a look at the strength of the attacking species. Notice how it's sodium bromide. That is the attacking species. Sodium bromide is an ionic compound, so it is going to dissociate. So what we actually have is the bromide, which is the Br minus, right? So we're going to have our Br minus right there as our attacking species. Now, when we take a look at that, is bromide a strong nucleophile?
1:32:16
I look at my little chart there, and it says, hey, Br minus is shown in red. And that's going to be a strong nucleophile. I'm keeping a mental note there. But when I look and ask, hey, is bromate a strong base, it shows that it is a weak base. So what that does for us is, when you have a difference between weak and strong, always go with the strong. And the strong-- what type of strong is it? Is it a strong base or a strong nucleophile? Well, in this particular case, it
1:32:58
is a strong nucleophile or a weak base. So we go with the strong. So it is a strong nucleophile.
1:33:09
And so what does that tell us? We're going to create a check mark here. So it's a strong base-- sorry. It is a strong nucleophile, so that favors SN2.
1:33:26
But is it a strong base? No, it's not, so I'm not checking the E2.
1:33:36
But is it a weak base? Yes, it is. And weak bases favor E1. So do you see, on that table there, I'm looking at factor of strength? And the bromide right here is telling me that it could do a SN2 and an E1 based off of what it is. Strong nucleophile means SN2. Weak base means E1. And I put those check marks. And I'm going to keep a tally mark of all these check marks. And at the end of this analysis, the column that has the most check marks is going to be the mechanism that we choose.
1:34:29
So we go to the next idea here-- concentration. You can see that the sodium bromide doesn't have parentheses that says, hey, it's dilute, so we know it's concentrated. So what does a concentrated species favor? SN2. It also favors E2. But because of the previous factor, the strength, they told us that E2 wasn't an option. So we're not going to put, for concentration, E2. We're just going to leave it as just SN2, all right? So the concentration is high, which favors SN2.
1:35:14
Now, we go to the next-- the leaving group ability right here. So does it-- the chloride there-- is it a decent leaving group? And, yes, it is. And so leaving group's ability really affects all four reaction mechanisms. So that one is an easy one. If it has a good leaving group, all four choices are a possibility. So we have all the check marks there when we're just looking at the leaving group only. Now, the other factor there is the solvent. What type of solvent do we have? Well, we have DMSO.
1:36:03
DMSO is a polar aprotic solvent. So when you look at the solvent, you have to figure out if it is polar protic or polar aprotic. I told you that DMSO is polar aprotic. So if I know that, that it's polar aprotic, what mechanisms does that favor? So a polar aprotic solid favors SN2 and E2. And you can see that I have two check marks there. And now I just sum up each column. And you can see, SN1 only has one check mark. SN2 has four. E1 and E2 both have two. So we have a clear winner.
1:36:56
The mechanism is going to be SN2. And so now that we identified what mechanism that we have, we now need to do it, draw it out. And so there's the mechanism shown there on the PowerPoint where it's looking at the SN2 mechanism. So we're going to do backside attack. It's going to invert the stereocenter and give us our one product. Now, the factor when we consider heat, that is only used to break a tie, OK? So let's say, for example, if our particular problem had
1:37:41
SN2 with three check marks and E2 with three check marks, how do we break the tie? Well, if heat was added, then it would favor the elimination. If no heat was added, then we would favor the SN2. That's how we're going to handle that. So like the slide says, we temporarily ignore the heat component to it unless there's a tie, and then we go back and look at the reaction to see if there's any indication that heat was added. When I look at that reaction scheme, I don't see a delta symbol.
1:38:25
I don't see the word heat anywhere, so I'm just going to assume that no heat was added. But that only comes into play when we have to break the tie. So, class, what we're going to do is continue working on problems like this when we come to class. So I hope this has been helpful. And if you need any additional help or have any questions, please reach out. I'm more than happy to help.
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