E2 mechanism: regioselectivity

- [Instructor] Let's look at the regiochemistry of the E2 mechanism. So first we'll draw our products, we'll go through the mechanism, draw the products, and then we'll talk about why this reaction is regioselective. On the left is our alkyl halides, and here is our strong base, sodium ethoxide, so Na+ and ethoxide has a negative charge. Since we're dealing with a strong base, you know we're going to do an E2 mechanism. The carbon that's directly bonded to the bromine would be the alpha carbon, and the carbons that are directly bonded to the alpha carbon are the beta carbons. So I'm going to call this carbon beta one, and this carbon beta two, and I'm going to call this carbon beta three. So sodium ethoxide is our strong base, it's going to take a proton from one of our beta carbons. Let's think about beta one first. I'm going to draw in a proton here, on the beta one carbon. And ethoxide is a strong base, so I'm going to draw in the ethoxide anion, there's a negative one formal charge on the oxygen, and the ethoxide anion is going to take this proton, and these electrons are going to move in here to form our double bond. At the same time, these electrons move off on to the bromine. So let's draw all our products. If we take a proton from the beta one carbon, so it would look like that, and let's show those electrons. These electrons in here, the magenta, moved in here to form our double bonds. It would be the same result if we took a proton from beta two. So, we just took a proton from beta one, if we took a proton from beta two we would get the same alkene. Let's move on to the beta three carbon. So let's take a proton from our beta three carbon, and let's do that over here, so here's our proton, and let's draw in our ethoxide anion right here, so I'll put in my lone pairs of electrons on the oxygen, negative one formal charge so we're going to take this proton, and then these electrons, and move them to here, and then the electrons come off on to our leaving group. They form the bromide anion. So the alkene that would form if we took a proton from our beta three carbon, let me draw it in here, so there's our double bonds, it would form that. So our electrons in magenta came in here to form our double bonds, and these'll be the two products for this reaction. It turns out that the isomer on the right is the major product. So this one is the major product. And the one on the left is the minor product. So we are talking about regiochemistry here, or think about the region of the molecule where the double bond forms. So this reaction's said to be regioselective because one of the isomers is favored. Now, let's talk about why. The major product on the right is a more stable alkene, it's more substituted. You look at the carbons for our double bonds, we have one, two, three alkyl groups. This is a trisubstituted alkene. Whereas our minor product on the left, this carbon, has only two alkyl groups bonded to it, so this one is a disubstituted alkene. And we know, from talking about the stability of alkenes, the more substituted alkene is the more stable one. And we call this the Zaitsev product. So this would be the Zaitsev product, the more substituted alkene. And this is the major product, it's the most stable one. So that's why this reaction regioselective. Use sodium ethoxide, a major product is the more stable alkene. Now let's do this reaction using a different base. So we're starting with the alkyl halides, and I actually left up the same products, but this time, we're going to go through thinking about potassium tert-butoxide as being our base. So there's a positive charge of potassium, and a negative charge on the oxygen. Potassium tert-butoxide is a sterically hindered base, so let me just add on to the drawing that I did in the previous example, where we used sodium ethoxide as the base, and I'm going to change it to make a tert-butoxide. So let me draw in here, so now we have our tert-butoxide anion functioning as a base. So if this takes a proton from the beta one or beta two positions, we form the disubstituted alkene. If we go over here, if we take a proton from the beta three position, let me add in this stuff here so now we have our tert-butoxide anion, we're going to form our trisubstituted alkene. This time, the disubstituted alkene turns out to be the major product. So this one is the major product. And the trisubstituted alkene is the minor product. And we can explain this by thinking about the fact that potassium tert-butoxide is a sterically hindered base. So on the left, we have these bulky methyl groups, and if we're taking a proton from beta one or beta two, our base can be off to the side and relatively out of the way. So it's easy for the base to take one of these protons. But on the right, as we get a little bit closer, at the beta three position, we have more steric hindrance here, so there's more steric hindrance, that prevents the base from taking the proton as easily. So that's why the trisubstituted product turns out to be the minor product, when you are using a sterically hindered base like potassium tert-butoxide. So in this case, the major product is the less substituted alkene, and we call this the Hofmann product. Let me write that in here, so this would be the Hofmann product. And this is the major product when a sterically hindered base is used. So pay close attention to what base is used in an E2 mechanism. If you are using an unhindered strong base, something like sodium ethoxide, your major product is the Zaitsev product, the more substituted product. But if we're using a sterically hindered base, something like potassium tert-butoxide, the Hofmann product, or less substituted product is the major product.