In this video, we're going to go over SN1, SN2, E1, and E2 reactions. So the first thing you need to know is if you have a methyl substrate, for example like methyl bromide, the reaction will proceed using an SN2 mechanism. It really doesn't matter what type of solvent we're using or if we're using a base, it's going to be an SN2 reaction.
Now, if we have, let's say, a primary substrate, like ethyl bromide, in this case, it's going to be an SN2 reaction, unless you decide to use a bulky base. in which case it's going to be an E2 reaction. Now let's say if you use a strong unhindered base like hydroxide.
It's a strong base, but it's not bulky. The predominant mechanism will be SN2. However, if the substrate is sterically hindered, then you could get the E2 reaction. SN2 reactions work with substrates that are not sterically hindered, but if it is, then the E2 can win in competition against the SN2 reaction.
So if you have a sterically hindered base and a primary substrate, it's going to favor the E2 reaction. But if the substrate is not sterically hindered, then it's going to be SN2, let's say if you use a strong base like hydroxide or even methoxide. Now, ethoxide would also fit this category. So we see ET, that's ethyl, CH3CH2.
One good example of a bulky base would be tert-butoxide. And so if you react it with a primary alkyl halide, you're going to get an E2 reaction. Now if we have a secondary alkyl halide, that's when things can get interesting.
So if you use an aprotic solvent such as acetone or DMF or DMSO with a very good nucleophile like iodide, cyanide, or a sulfur with a negative charge, this is going to favor the SN2 reaction. Now if you use an aprotic solvent like water, or methanol or even ethanol where you have hydrogen bonds typically this will favor the SN1 and the E1 mechanism but this can vary too sometimes it could favor an SN2 reaction so you got to pay special attention to this category but the protic solvent favors the SN1 and the E1 mechanism over the SN2 mechanism Now what mechanism will occur if we use a secondary alkyl halide with a strong base? In this case, you can get both the SN2 mechanism and the E2 mechanism.
However, the E2 mechanism will usually be the major product. So if you have a bulky or sterically hindered secondary alkyl halide, it's going to favor the E2 reaction over the SN2 reaction. Using a bulky base will also favor E2 over SN2. However, if you have, let's say, hydroxide and a secondary alcohol halide that's not sterically hindered, then you can get a good mixture of the SN2 reaction and the E2 reaction.
Typically, the E2 would slightly be, you'll get the E2 in a better yield over the SN2, but you'll still get a good mix of both of them. So I'm going to put the E2 on top here because this is going to be the major product in this category. But know that the SN2 reaction still exists in this category.
It's simply the minor product. Now, if we have a tertiary alkyl halide, we're not going to get the SN2 reaction because it's too sterically hindered. And so we're going to get the SN1 and the E1 reaction regardless of the solvent that we choose to use.
Protic solvents favor the SN1 reaction. A-protic solvents favor the SN2 reaction. And heat favors E1 over SN1, by the way. So if you wish to increase the E1 yield over the SN1 yield, you need to raise its temperature.
Using a strong base with a tertiary alkyl halide will favor the E2 reaction. You're not going to get the SN2 reaction. So this table is not perfect.
There are exceptions. But this is simply a general guideline to help you predict which mechanism will be the major pathway. in a substitution or elimination reaction.
So let's go ahead and work on some examples. So let's start with this one, 2-bromobutane and let's react it with sodium cyanide and acetone. So which mechanism will this reaction proceed with?
Is it going to be SN1, SN2, E1, or E2? And then once we know the mechanism, let's predict the major product and show the mechanism to get to that product. So we have a secondary alkyl halide. The carbon that bears the bromine atom, that carbon, is attached to two other carbon atoms.
So we have a secondary alkyl halide. Now the cyanide ion, it's the nucleophile in this reaction. And so the nucleophile is going to attack the carbon from the back, also known as the backside attack.
And it's going to expel the leaving group. Because it attacks from the back, we're going to get the inverted product. So the SN2 reaction proceeds with inversion of configuration. And so here on the left side we have the R isomer.
This is R 2-bromobutane and now we have the S isomer. Now let's move on to our next example. Let's say we have tert-butyl chloride and let's react it with water. Go ahead and predict the major product for this reaction. So first, we need to identify the mechanism.
This carbon is tertiary, so we have a tertiary alkyl halide. And we have a protic solvent, H2O. So the dominant mechanism will be the SN1 and the E1 mechanism. So let's draw the products for the SN1 mechanism and for the E1 mechanism.
SN1 stands for the first-order nucleophilic substitution reaction. And so we're going to replace the chlorine atom. with an OH group.
In an E1 reaction or elimination reaction, we're going to remove a hydrogen and the leaving group, forming a double bond or an alkene. Now the first step for the SN1 reaction is that the leaving group leaves, and so we're going to get a tertiary carbocation. And then the water molecule is going to add to this tertiary carbocation. And so we're going to get this auxonium species. And then we need to use another water molecule to take off a hydrogen.
So the end result is an alcohol. And so this is the SN1 product for this reaction. Now let's draw the E1 product for that reaction. So the first step in the S1 reaction and the E1 reaction is the same, and that is ionization.
The leaving group is going to leave, and we're going to get the same tertiary carbocation. Now for the E1 mechanism, this is where the path changes compared to the S1 mechanism. For the E1 reaction, the water is not going to act as a nucleophile, attacking the carbocation. Instead, it's going to act as a base, going for the hydrogen, and then the carbon-hydrogen bond will break. Those two electrons will be used to form the pi bond, giving us an alkene.
So that's the E1 mechanism. Now let's try this example. Let's react this alkyl halide with methanol. Go ahead and predict the products of this reaction. So on the left we have a tertiary alkyl halide and we have a protic solvent.
When the solvent acts as a nucleophile, it's called a solvolysis reaction. When you see that it's typically an SM1E1 reaction. So let's focus on the SM1 product. The S1 reaction is a substitution reaction, so we're going to substitute the leaving group with this group, and we're going to get rid of the hydrogen.
So we're going to get an ether. However, notice that this carbon is chiral, and so we're going to get a mixture of stereoisomers. So the OCH3 group, it can be in the front, it could be on the wedge, or it can be on a dash.
So these are the SM1 products for this reaction. Now let's talk about why we get a mixture of stereoisomers, why we get this racemic mixture. Well, we know the first step is ionization.
So this will produce a tertiary carbocatine. Now the methanol molecule is going to act as a nucleophile. And notice that it can approach the carbocatine from either side. It can go from the back, leading to the inverted product, or it can go to the front, which is retention.
So let's say the bromine atom is in the front, or on the wedge. So if it attacks from the back, we're going to get this, the inverted product. If it attacks from the front, we're going to get retention.
Now, of course, there's one more step. We need to take off the hydrogen. But will we get an equal amount of the inverted product and the retention product? Or will one stereoisomer have a greater yield over the other? Now the bromide ion is still in the vicinity of the carbocation.
And so when the methanol molecule approaches from the front, it's going to be repelled. by the negative charge of the bromide ion. Keep in mind that the oxygen atom has a partial negative charge and so it's repelled by the bromide ion.
Therefore, we're going to get a lower yield of the retention product than the inverted product. So the backside attack is still more favorable. So it could be 60% of the inverted product and 40% of the retention product or it could be a 70-30 ratio. Either case, you get slightly more.
of the inverted isomer as opposed to the retention stereoisomer. Now for the last step of the mechanism, all we need to do is use another methanol molecule to remove the hydrogen and so this will give us our ether product. So one is going to be on the wedge and the other will be on a dash. Now let's focus on the E1 reaction. Feel free to pause the video and write a mechanism for the E1 reaction.
So the first step is ionization and this will give us our tertiary carbocation. Now in the next step for the E1 reaction, methanol is going to act as a base instead of a nucleophile. So it's going to abstract one hydrogen away from the carbocation. So the hydrogen has to be one carbon away from the carbocation.
It can take the white hydrogen, it can take the blue hydrogen, or it can take the green hydrogen. So if it takes the green hydrogen, we can get this product. Now if it goes for the blue hydrogen, we can get this product.
And if it goes for the white hydrogen, we can get this product. And so we get a mixture of E1 products. for this reaction.
However, not all of these alkenes are equally stable. So this particular alkene is a disubstituted alkene. There's two carbon atoms that are attached to the two double bonded carbon atoms.
This alkene is trisubstituted and the same is true for this one. It has three R groups and so this is the least stable alkene. So therefore, the major product for the E1 reaction will be these two alkenes.
They will both form in good yield. this will be the minor E1 product. Now let's say if we have 2-bromo-3-methylbutane and let's react it with sodium methoxide and methanol. What is the major product in this reaction? Go ahead and try it.
So we have a secondary alcohol halide and methoxide is a strong base. It's not a sterically hindered base, but it is a strong base. And so whenever you have a secondary alcohol halide with a strong base, the major product is typically the E2 product.
And the SN2 reaction will give you the minor product. But we're going to draw both. So let's start with the SN2 reaction. So the methoxide will attack this carbon kicking out the bromine atom.
Now we don't have to worry about the stereochemistry because it wasn't specified at this carbon even though we could get a mixture of stereoisomers because this compound can be R or S we don't know but we're not going to worry about stereochemistry. The SN2 reaction will give you the inverter products so let's say if this was in the front then the OCH3 group would have to be in the back. Or, if this was in the back, then the OCH3 group would have to be in the front. So since the stereochemistry wasn't specified, we're not going to specify it here either. So this is going to be the SN2 product, and this is going to be the minor product in this reaction.
It can happen, but... it's not going to be in good yield. Now for the E2 reaction, no rearrangements can occur.
The base can take off a hydrogen one carbon away from the carbon that bears the bromide atom. So it can take the white hydrogen or the blue hydrogen. Methoxide will preferentially take the blue hydrogen over the white hydrogen. And so we're going to get this particular E2 reaction. So this is going to be the major E2 product.
Now, it could take the white hydrogen, put in a double bond here, and this will be the minor E2 product. Now, you need to know that internal alkenes are generally more stable than terminal alkenes. So this alkene here is more stable than the one on the right. And the reason being is it's more substituted. Here we have a tri-substituted alkene.
On the right, there's only one carbon attached to these two double-bonded carbon atoms. And so this is a mono-substituted alkene. Tri-substituted alkenes are more stable than mono-substituted alkenes.
And so that's why this is the major product. It's simply the more stable alkene. The most stable alkene is known as the Zetsev product, whereas the less stable alkene is known as the Hoffman product.
And so this is the answer for this problem. So the major mechanism for this reaction is the E2 mechanism, and this is the major product overall. Keep in mind, the SN2 reaction can occur for this particular problem.
but it will lead to a minor product. Now let's say if we have the same alkyl halide, the same substrate, but this time we're going to use a sterically hindered base, such as terbutoxide and terbutanol. What's going to happen?
So we have a protoxyl solvent, but we also have a strong base. And so If you're using the triad that I mentioned in the beginning, you need to use the fourth column because the strong base takes priority over the protic solvent. Because this has a negative charge, it's far more reactive than terbunol. So we have a sterically hindered base and also a relatively sterically hindered alkyl halide due to this methyl group on the top. So whenever you have a bulky base and a sterically hindered substrate, if both the substrate and the base are sterically hindered, this will give you the Hoffman product and so you're going to get the less stable alkene so because terpitoxide is so bulky it wants to go for the most accessible hydrogen and the hydrogen atoms on a primary carbon are more accessible than the hydrogen atom on a tertiary carbon terpitoxide doesn't want to have to struggle against these two methyl groups to get to that hydrogen and so it's not gonna Form the Zeta product in good yield it can still work, but it's going to be the minor product instead It's going to go for the most accessible hydrogen atom so we're going to get the Hoffman product as the major product for this reaction Now we can still get the Zeta product, but it's going to be the minor product in this case.
So the E2 reaction is the major mechanism in this example. And so this is the major product. Now I want to compare these two reactions together. 2-bromopentane and 2-fluoropentane.
So let's start with 2-bromopentane. I'm going to use methoxide as a strong base dissolved in methanol. What is the major product for this reaction? So in this case, methoxide is going to go for the hydrogen that's one carbon away from the carbon-bromine bond. And so we're going to get the Zeta product as the major product.
You can still get the Hoffman, but that's going to be the minor product. Now, if we use an alkyl fluoride, something different is going to happen. The methoxide is going to go for the primary hydrogen as opposed to the secondary hydrogen.
And so we're going to get the Hoffman product as the major product. The question is why? Why do we get the Hoffman product as the major product if fluorine is the leaving group? Whereas if bromine or chlorine or iodine is the leaving group, we're going to get the Zeta product.
For one thing, chlorine, bromine, and iodine, they're good leaving groups. And so, when you have a good leaving group, it's going to favor the Zeta product. Fluorine, on the other hand, it's a bad leaving group. And so you're going to get the Hoffman product. Now the reason for this is when you have a good leaving group, the transition state resembles an alkene.
And so because of that, you need to find the most stable alkene. So the major product will have or will be associated with the most stable alkene. Because that's going to give you the most stable transition state.
In the case of an alkyl fluoride. Because it's a bad leaving group, when the methoxide ion approaches to attack the hydrogen, the fluorine is still sticking around. It doesn't want to leave. And so you have this buildup of negative charge on this molecule.
And so the transition state resembles more like a carbanion instead of an alkene. And what you need to know is that primary carbanions are more stable than secondary carbanions. And as a result, it's better to put the negative charge on a primary carbon as opposed to a secondary carbon.
And this is why the Hoffman product is the major product. Now I want you to compare these two reaction mechanisms. Let's say we have 2-bromo-3-methyl butane.
And we're going to react it with water. Now let's compare this to 2-bromo-butane. And we're going to react this molecule with the acetate ion, which is going to be dissolved in acetic acid.
So go ahead and determine the substitution mechanisms that these two reactions will proceed by. Ignore the elimination mechanism. Now let's focus on the reaction on the bottom. So we have a secondary alkyl halide and a protic solvent. So if we use the chart, this would indicate that...
it should proceed by the SN1 mechanism. However, it turns out that this reaction actually goes by the SN2 mechanism, and for one reason, acetate is a decent nucleophile. It has a negative charge, so it's more nucleophilic than H2O.
Here we have a secondary alkyl halide and a protic solvent, and it turns out that this mechanism favors the SM1, or this reaction favors the SM1 mechanism. Now let's talk about why. So as we said before, we have a better nucleophile, and so that favors the SN2 reaction. The nucleophile has a negative charge, whereas this nucleophile is neutral, even though both solvents are protic.
Now the second reason why this one favors an SN2 reaction is because that substrate is not sterically hindered. Here we have a sterically hindered secondary alkyl halide due to this methyl group. An hysterically hindered alkyl halide actually favors the SN1 reaction because once we get the secondary carbocation, it could rearrange and form a more stable tertiary carbocation. So whenever you're dealing with secondary alkyl halides, you need to be careful because sometimes it can be SN1, sometimes it can be SN2.
Now we know it's not going to be an E2 reaction because we don't have a strong base, so we can ignore that option. But when you're between these two mechanisms, ask yourself, what type of secondary alcohol halide do I have? Is this sterically hindered?
If it is, it's going to favor the SN1 mechanism. If it's not, it favors the SN2 mechanism. Is there a protic solvent or is there an aprotic solvent? If there's an aprotic solvent, it favors the SN2 reaction. If there's a protic solvent, typically, it favors the SN1 mechanism.
But in this case, this exception... We have a good nucleophile in a product solvent, so that good nucleophile favors the SN2 mechanism. And so you have to look at every detail and weigh all your options carefully.
Now let's go over the mechanism of those two reactions. So let's start with this one, where we're going to react it with H2O. And we said that this is going to be the SN1 mechanism. So the first step, the leaving group leaves. And so we're going to get a secondary carbocation.
Now, whenever you have a secondary carbocation next to a tertiary carbon, you're going to get a hydride shift. And so this is going to give us a more stable tertiary carbocation intermediate. And so water is going to attack the carbocation, giving us this intermediate. And then in the next step, we're going to use another water molecule to take off a hydrogen atom. And so our final answer is a tertiary alcohol.
Now this alcohol is not chiral because it has two methyl groups, and so we don't have to worry about stereoisomers. This is going to be the major SN1 product. Now let's go over the other example where we had two bromobutane.
reacting with the acetate ion. So we said that this reaction will proceed by the SN2 mechanism, and so the oxygen of the acetate ion is going to attack from the back, expel in the leaving group. And so we're going to get inversion of configuration.
So this is going to be our product. Now here's the interesting question for you. Let's say we have this substrate and let's react it with methanol.
What's going to be the major product in this reaction? And what type of mechanism do we have? Is it the SN2 mechanism, the E2 mechanism, SN1, E1? What's going on here?
So the carbon that has the bromine atom is a primary carbon. So we have a primary alkyl halide, which usually favors SN2 reactions. And we have a protic solvent, so are we going to get an SN2 reaction?
Notice that this primary alkyl halide is very hindered. Next to the primary carbon, we have a quaternary carbon. And so, due to the fact that this primary carbon is sterically hindered due to the adjacent carbon, it's safe to say that this reaction will not proceed by the SN2 mechanism.
It's too sterically hindered. So what's going to happen? Can it proceed by the SM1 mechanism?
It can, however... We can't just kick out the Br because if we do we're going to get a primary carbocation and primary carbocations are not stable. They're very difficult to form and so that's not going to happen. Instead we're going to get a concerted reaction.
We're going to get a methyl shift before the Br leaves to produce the unstable carbocation. So this methyl group is going to attack this carbon kicking out. the bromine atom. And so now the methyl group is here. By doing this, we form a more stable tertiary carbocation intermediate without forming a primary unstable carbocation.
So now at this point the methanol could attack the carbocation. given us this intermediate. And then in the next step, we can use another methanol molecule to remove the hydrogen atom. And so our final product is an ether, and we don't have to worry about stereochemistry. This is not a chiral center.
And so this is the answer. So this is one of those rare situations where a sterically hindered primary alkyl halide can give you the SM1 mechanism. Now let's consider this example. So once again, we have another sterically hindered secondary alkyl halide.
But this time we're going to react it with sodium ethoxide in ethanol. So go ahead and try this problem. So we have a strong base, that's ethoxide. And based on the chart, when you have a strong base like ethoxide and a secondary alkyl halide, it can predominantly give you the E2 reaction mechanism.
But it can also give you the SN2 reaction. However... We do have a secondary sterically hindered halide and so we could say that the SN2 reaction is negligible. If it does occur it's going to be in a very very low yield so we're going to ignore the SN2 reaction.
We're going to focus on the E2 reaction because that's going to be the dominant mechanism if you have a sterically hindered substrate. So notice that we can't get the Zeta product in this case. There is no hydrogen on its carbon because it's quaternary.
So we have to remove the hydrogen on the primary carbon because we have no choice. And so ethoxide will go for this hydrogen, and we can only get one product. For the E2 reaction, there are no rearrangements.
And so this is one of those rare cases where we get the less stable alkene, since we only have one choice. we have no other choices. And so that's it for this example.
What's going to happen if we mix methyl bromide with terpitoxide? Go ahead and predict the product of this reaction. Now whenever you have a methyl substrate, the only mechanism that can work is the SN2 mechanism. There are no other choices.
Now even though we have a bulky base, such as terpetoxide, which typically favors elimination reactions over substitution reactions, we can't form a double bond with one carbon atom. You need at least two carbon atoms to form a double bond. And so there's no way we can get the E2 reaction. So the SN2 reaction will occur with a 100% yield.
So what I'm going to do is draw the terpetoxide first. then the substrate after that. So, terpitoxide will attack the methyl group from the back, expelling the bromide ion. And so, our product will be an ether.
And so, this is the only product that we can get in this reaction. Now, let's move on to this example. So here we have butyl bromide.
and we're going to react it with methoxide and methanol. So based on the chart, if we have a primary alkyl halide that's not sterically hindered and we have a strong base, what is going to be the dominant mechanism in this reaction? So since we don't have a sterically hindered base, the mechanism will predominantly be SN2. And so in this case, the methoxide ion will attack the carbon that bears the bromine from the back, expelling the leaving group. And so we're going to get an ether.
Now we don't have to worry about stereochemistry because this carbon is not chiral. It has two hydrogen atoms. And so that's it for that example. Now what if we have a primary alkyl halide but with a sterically hindered base like terbutoxide?
What's going to happen in this case? Well based on the chart, this is going to favor the E2 reaction. So the base, this bulky base, it doesn't like to behave as a nucleophile.
because it's so bulky. Unless it has to in the case of methyl bromide. But it prefers to behave as a base, abstracting a proton. And so this is going to give us one butene.
Now I want to compare the last two reactions with a new reaction. And so you could see something. So here we have a primary alkyl halide.
with an unhindered strong base. And so this gave us the SN2 reaction. In the last example, we had a primary alkyl halide that was not sterically hindered, but we had a sterically hindered base. And so this gave us the E2 reaction.
Now what if we have, let's say, asterically hindered primary alkyl halide. And let's use methoxide again. Will we get an SN2 reaction or an E2 reaction?
So, in the first example, if we don't have a sterically hindered base or a sterically hindered primary alkyl halide, we get the SN2 reaction. If we have a sterically hindered base... we're going to get the E2 reaction.
If we have a primary sterically hindered substrate, we're going to get the E2 reaction as well. Because anytime you have a sterically hindered substrate, it's going to weaken the yield of the SN2 reaction. And so therefore, we're going to get this alkene.
So make sure you understand that. So the only time you get the SN2 reaction with a strong base is if you have a primary alkyl halide that is not sterically hindered. If you have a sterically hindered base or a sterically hindered primary alkyl halide, it's going to favor the E2 reaction over the SN2 reaction.
You can still get the SN2 reaction, but the E2 reaction will give you the major product.