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Mechanism of an E1 elimination reaction. Created by Jay.
Video transcript
Let's take a look at the general reaction for the dehydration of an alcohol to form an alkene. So over here on the left, we have our alcohol. The carbon that's bonded to the OH is our alpha carbon. And then the carbon that's bonded to that carbon is our beta carbon. In this mechanism a beta hydrogen is required, so we'll take a look at that in a few minutes. You also need sulfuric acid and you need to heat up this reaction. So phosphoric acid can also be used. And sometimes phosphoric acid does a better job of getting your desired product. Over here on the right, you can see that we have lost our OH. It left in the form of water, in terms of the dehydration. And we form a double bond between our alpha and our beta carbons like that. So if you are a flash card person, you would put these two things on one side of your flash card, and then your product on the back of your flash card. Let's take a look at the mechanism. The steps of how this reaction occurs. So we're going to start with our alcohol. So let's go ahead and redraw our alcohol here. And put in that beta hydrogen. And we have our OH right here with two lone pairs of electrons. So put in our two lone pairs of electrons on our oxygen like that. I know that sulfuric acid is present. Sulfuric acid is a strong Bronsted Lowry acid. It donates protons in solution. So there are protons floating around here in solution. So I'm going to go ahead and draw a proton H plus like that. The alcohol portion of the molecule is going to function as a Bronsted Lowry base. It's going to accept that proton. So one of these lone pairs of electrons is going to form a bond with this proton. So I'm going to say it's this lone pair of electrons. So that's going to attack this proton, form a new covalent bond. So let's go ahead and draw the product of that acid based reaction. So let's go ahead and put in our beta hydrogen here. And this carbon right here. So what would we get? Well, one lone pair of electrons did nothing. So that lone pair of electrons is still there on the oxygen. This bottom lone pair formed a new covalent bond between that oxygen and that proton. And then we still had one hydrogen attached to that oxygen. So it would look like that. When you calculate formal charge on this oxygen, we talked about that earlier video, you're going to get a plus 1 formal charge. So that oxygen is positively charged. And oxygen doesn't like to be positively charged, it's very electronegative. So that oxygen right now wants some electrons. Well the best place that oxygen can get some electrons are from this bond between carbon and oxygen. So in the next step of your mechanism, these two electrons in this bond are going to kick off onto this oxygen atom. So let's go ahead and draw the result of that. So what would happen? So we have this hydrogen is still here. All right let me get that bond a little straighter. And then we just broke the bond between that carbon and that oxygen. So now water has separated here. So we have oxygen bonded to two hydrogens. That oxygen had one lone pair of electrons. And then it gained another lone pair of electrons from that bond that it used to be bonded to that carbon with. So we now have water floating around like that. Now, this gives a new formal charge on this carbon atom on the right here. And so this carbon is going to end up with a plus 1 formal charge. So that carbon is not happy. It needs electrons. It does not have an octet of electrons around it. So in the last step here, we need to think about how to form a double bond. And it depends on which textbook you are looking in. Most textbooks would say that this water can now function as a Bronsted Lowry base. And it's going to take this proton, just that proton, not the electron that proton brought to the dot structure. So if I go ahead and put my electrons in here on this bond, remember one of those electrons was hydrogen. But it's only the nucleus of that hydrogen atom that that's going to be picked up by that water molecule. Leaving those two electrons behind. And those two electrons are going to move in here to form your double bonds. Let's go ahead and draw the products. Let's see if we can go up here. So that is going to form the double bond between your two carbon atoms like that. And then you can have acid base reactions with water and HSO4 minus and H2SO4, and all that kind of stuff. So you have now formed our alkene. Let's go back and let's try to follow those electrons a little bit better. Let's see, these two electrons that we started with right here. These are the ones that ended up forming this bond. And so we'll put those electrons in blue. Let's see, the next electrons that we're trying to follow, let's follow these electrons right here. So the electrons in this bond, those are the ones that kicked off onto the oxygen right here. So those are these electrons right here. And then the last thing we can do, we could color coordinate these two electrons, right? These two electrons are the ones that moved in to form your alkene in your mechanism. So there's the mechanism for the dehydration of alcohols to form alkenes. So notice that we have a positively charged carbon in this mechanism, which we called in an earlier video a carbocation. So this is a carbocation mechanism. And in a future video we'll talk much more in depth about carbocations and possible rearrangements. Right now we're just concerned with the general mechanism. Let's see if we can write a general mechanism for an actual reaction. So let's take a look at cyclohexanol. So if we draw cyclohexanol like that. And if we add sulfuric acid to it, we can dehydrate the molecule. We can lose water. So let's think about running the mechanism for this. Well we know that the carbon bonded to our OH is our alpha carbon. And we know that the carbon next to our alpha carbon is the beta carbon. So this could be a beta carbon and this could be a beta carbon down here. So we know that each of those beta carbons have two hydrogens on them, right. There's two hydrogens at each of these carbons here. So any of those hydrogens could be the one, could be the beta hydrogen that participates in this mechanism. So let's think about what happens in this mechanism. So in the first step, we can do a shortened version. We know that there were two lone pairs of electrons on our oxygen. And we know there are protons floating around in solution. So we could even draw it real fast. Put our two lone pairs on that oxygen. And we know that there are protons floating around in solutions. Let me get my two lone pairs in there. And here we have an H plus, so we can go ahead and protonate our alcohol. So we can go ahead and draw the results of that. So now we have oxygen bonded to two hydrogens. And now has only one lone pair of electrons on it. And this will get a positive 1 the formal charge. OK. Oxygen, it doesn't like that positive formal charge. It wants to try to get rid of that formal charge. So these electrons are going to kick off onto your oxygen like that. So water leaves, it's the dehydration step. And we now have a carbocation at this carbon right here. So this is a carbocation. And I could take either one of those beta hydrogens, they're, identical so I'm just going to take one of the ones from right here. And you could think about your water molecule coming along here and acting as a Bronsted Lowry base, it's going to accept a proton. So this is going to take this hydrogen. And that means these electrons are going to move in here to get rid of that positive 1 formal charge in that carbon, and to form your alkene. So your final product would be cyclohexene. So there's a quick mechanism for you. All right let's look at one more example of what we call an E1 elimination. This time we're going to use an alkyl halide. So let's take a look at an alkyl halide and let's see if we figure out the mechanism for another E1 elimination. We'll talk about where the term E1 comes from after we look at this example. So let's look this as our alkyl halide. So we'll make that bromine. Put in my lone pairs of electrons on bromine. And we're going to react this alkyl halide with ethanol. So I'll go ahead and put ethanol in here. And we're going to heat up, as well. OK so first we have to talk about this alkyl halide. This is called a tertiary alkyl halide. So this is tertiary. So why is this a tertiary alkyl halide? Well the carbon that is directly bonded to our halogen is that carbon. And that carbon is bonded to one, two, three other carbons. So that's where the tertiary part comes in. So a tertiary alkyl halide reacting with ethanol and heat via an E1 mechanism. The first step is the leaving group is going to leave to form a carbocation. In the previous example, water was are leaving group. In this example, our halogen is going to be our leaving group. So the two electrons in the bonds between this carbon and this bromine, these two electrons right here, are going to kick off onto the bromine. And our first step. So let's go ahead and let's just run through the mechanism here. So we're going to get that bromine. It used to have three lone pairs of electrons around it. Now has four lone pairs of electrons. That gives it a negative 1 formal charge. And the methyl group is still here. We just lost a bond to this carbon, the one that we had in blue on the left. So this carbon gets a plus 1 formal charge. So that's the first step in the mechanism. Formation of a carbocation. Bromine is relatively stable as a leaving group, because it now has now has electron configuration of a noble gas. So it's stable as an anion. So now we have this situation. We know that this is my alpha carbon. This is the one that had the halogen on it. So we know that, you know that one of these guys is my beta carbon. So this one or this one. So once again, it doesn't really matter which one I choose. I'm just going to take the hydrogen off of this carbon. And our ethanol molecule is going to act as a base now. So if I go ahead and draw my ethanol molecule. So completely analogous to water in the previous example. It is going to take that proton. So a lone pair of electrons in ethanol is going to take just that proton. And leaving these two electrons behind, we're just going to move in to form a double bond to get rid of that positive charge on my carbocation. And that now gives me my product. So let's go ahead and get some more room down here. I can draw the product of my elimination reaction. Right, so this was a-- let's do that one again. So I know that I had a methyl group at the top, right here. And I can see my double bond is going to form between my alpha and my beta carbon. So my double bond is going to go right in here. And this isn't the only product for this reaction. But so far this is all we know that can form. In future videos, we'll talk about what else can happen in this reaction. So we can form an alkene starting with an alkyl halide. And now let's talk about where the term E1 comes from. Elimination. So the elimination, or your leaving group, in the first example is water. In this case it's your halogen. And you're forming a double bond, that's the elimination part. The one comes in, it's unimolecular, meaning the rate of this reaction depends on the concentration of only one of your reactants. So if you go back and look at our two reactants here, our two reactants are methanol and our alkyl halides. And the rate of this reaction depends only on the concentration of your tertiary alkyl halide. And the reason for that is the formation of your carbocation. This first step here is the rate determining step. So this first step we drew here is that is the rate determining step. That is the slow step. The overall rate of reaction determines on that rate determining step. So it's completely dependent on the original concentration of your alkyl halide. But it's not dependent upon the concentration of the alcohol, because the alcohol can't do anything, the ethanol can't do anything until your carbocation forms. So since the rate depends only on the concentration of your alkyl halide, it's unimolecular, that's where the one comes from in an E1 elimination.