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Sn1 Sn2 E1 E2 reactions: Primary and tertiary alkyl halides

Predicting whether an alkyl halide will undergo an Sn1, Sn2, E1, and/or E2 reaction based on the sterics of the alkyl halide and the nucleophile.  Created by Jay.
Video transcript
So we've talked about Sn1 reactions, Sn2, E1, and E2. Now we have to figure out when those reactions occur. And the trick is to first look at the structure of your alkyl halide. And for primary alkyl halides, not all these reactions occur. For example, Sn1 reactions and E1 reactions do not occur with a primary alkyl halide. Because Sn1 and E1 reactions require formation of a carbocadon, and a primary carbocadon is not very stable at all. So you only have to choose between and Sn2 and E2 reactions if you have a primary alkyl halide. And Sn2 reactions are inherently favored. Let's look at an example. So we'll do bromoethane here. So if we react bromoethane with sodium hydroxide, our task is to figure out the products. And the first thing to do is to identify your nucleophile. So the hydroxide anion would be my nucleophile. And how would we classify the hydroxide anion? Well, a negative 1 formal charge, so it's probably a strong a nucleophile. So just like we saw in the last video. So strong nucleophile would make you think of an Sn2 reaction, because Sn2 mechanisms requires strong nucleophile. Let's think about hydroxide as a base. So we know that as we know that this hydroxide anion would be a relatively strong base. So strong bases would make you think of the E2 mechanism. So we have Sn2 versus E2 here. And Sn2 reactions actually happen faster for primary alkyl halides. So Sn2 into is inherently favored. If we think about what we would get as our product of this proceeded via an Sn2 mechanism, right, so an Sn2 mechanism. Nucleophilic attacks by our hydroxide and at the same time, these electrons kick off here to the bromine, so we're going to form ethanol as the major product here. So Sn2 is going to be the major product. You might get some E2 product as a minor product. All right, let's do another one here. Let's do the same reactants. So bromo ethane. And let's react that with potassium tert-butoxide. OK so potassium tert-butoxide. Is going to look like this. K plus. And then we have our alkoxide anion over here on the left. So first identify your nucleophile, right? Well, of course it is going to be our alkoxide over here. So is that nucleophile going to be strong or weak? In the last video, we said that alkoxide anions were strong nucleophile. That's usually the case. However, this is a very bulky nucleophile, so these methyl groups here decrease the nuclear felicity of it. There's increased steric hindrance. So that's going to make this a relatively weak nucleophile since it's a sterically hindered nucleophile. And so that's going to decrease the likelihood that it's Sn2. Decrease the likelihood it's Sn2 because Sn2 reactions require strong nucleophiles, nucleophiles are unhindered. Sterically unhindered. So it decreases the likelihood it's Sn2. So we think about the basicity of our alkoxide anion there. We know that it is a strong base. And the strong bases would favor an elimination reaction. So an E2 reaction. A concerted mechanism. So we have decreased likelihood it's Sn2 because this sterically hindered base. So we have a weak nucleophile. But it's a strong base. This is going to react via an E2 mechanism. It's going to react to via an E2 mechanism here. And we're going to form our alkene like that. So we're going to take a proton, the bromine's going to leave, we're going for form our alkene, and then we are done. So if you see a primary alkyl halide, think Sn2 versus E2. Sn2 is inherently favored unless you have a week nucleophile and a strong base, in which case the strong base would take over and it would react to via an E2 mechanism. So let's look at tertiary alkyl halides now. So primary alkyl halides are the easiest one. All you to do is determine between Sn2 and E2. Tertiary alkyl halides are the next easiest. So tertiary alkyl halides, so once again possibilities. Sn1, Sn2, E1, and E2. Well, Sn2 reactions are not likely with a tertiary alkyl halide. These will not happen. And the reason has to do with the mechanism. Sn2 mechanisms require decreased steric hindrance. A tertiary alkyl halide has a lot of steric hindrance that would get in the way of an Sn2 mechanism. So Sn2 mechanisms will not occur with a tertiary alkyl halide. However, you can have Sn1, E1, or E2. And elimination reactions are inherently favored when you're talking about a tertiary alkyl halide. so if you have a strong base that's present, you're going to get an elimination reaction. Let's look at our tertiary alkyl halide here. So tert-butyl chloride reacting with potassium hydroxide, like that. So once again identify your possible nucleophile. That would be my hydroxide anion right here. So I think about, is hydroxide a strong or weak nucleophile? I know that it is strong, which might make me think, hey, maybe Sn2. I know Sn2 isn't possible here. So I move on to the strength of the base. And I know that hydroxide is a relatively strong base, which makes me think of an E2 mechanism. I know that elimination is inherently favored here. So if you have a strong base present, you're going to get elimination. So this is going to occur via an E2 mechanism. And the fact that it's sterically hindered doesn't really bother an E2 mechanism. Let's just run through the E2 mechanism real fast for you here. It is a proton right here. Hydroxide is going to function as a base. It's going to take that proton. These electrons are going to move in here. At the same time these electrons kick off onto the chlorine. So we're going to form this alkene as our product. So the fact that it's sterically hindered doesn't really bother the E2 mechanism. Once again, if a strong base is present, elimination will occur. So that one's E2. let's look at another reaction here. We'll use tert-butyl chloride again. So we'll use tert-butyl chloride right here. Like that. And we're going to react tert-butyl chloride with water this time. So water's going to be our possible nucleophile. So those lone pair of electrons. So this is our solvolysis reaction here. And how would I classify water? So I need to do my nucleophilic classification here. Well, the lone pairs of electrons mean that water can act as a nucleophile. There's not a formal charge or anything. So it's a relatively weak nucleophile. A relatively weak nucleophile makes me think of a possible Sn1 mechanism. And what about the strength of water as a base? Well, we know that water can act as a base, but it doesn't react. it's not very good base. It's a weak base. Which makes me think of a possible E1 mechanism. So I have I thought of Sn1 and E1 mechanism, which both require formation of a carbocadon. So in your first step, kick these electrons off onto the chlorine, form a carbocadon like that. A tertiary carbocadon, so it's relatively stable. And now we have two possibilities. So we have two possibilities here. We could react via an Sn1 mechanism, or we could react via E1 mechanism. So in an Sn1 mechanism, the lone pair of electrons on the oxygen atom would now attack your carbocadon. After you lose a proton, you would form tert-butanol here. So that's the possible Sn1 product. The possible E1 product, water's going to function as a weak base. Take a proton, and you would end up with your alkene here. Like that. So, you're going to get two products. Sn1 product and E1 product. So we've seen for tertiary alkyl halide, you could get E2, you could get Sn1, you could get E1. And so there are three possibilities. Remember to first look at the structure of alkyl halide, then look at your nucleophile, classify it as a strong or weak nucleophile. Then think about is being a base, either strong or weak base. And with those clues you can usually figure out the products of your reaction. Also an E1 reaction here is favored at high temperatures. So if there's high temperatures, increased temperatures, an E1 one mechanism is favored. So you could get more of an E1 product if you just increase the temperature of your reaction. In the next video, we'll do secondary alkyl halides. And you'll see there, it's a little bit harder than primary and tertiary.