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Current time:0:00Total duration:9:22

Video transcript

Let's think about what'll happen if we have this molecule. Let's name it. We have one, two, three, four, five carbons. No double bond. Five tells us pent. It's pentane, and it has two groups on the number three carbon, one, two, three. It doesn't matter which side we start counting from. We have a bromo group, and we have an ethyl group, two carbons right there. On the three carbon, we have three bromo, three ethyl pentane right here. So we have 3-bromo 3-ethyl pentane dissolved in a solvent, in this right here. It's an alcohol and it has two carbons right there. Meth eth, so it is ethanol. This right there is ethanol. Let's think about what might happen if we have 3-bromo 3-ethyl pentane dissolved in some ethanol. Now ethanol already has a hydrogen. It's not super eager to get another proton, although it does have a partial negative charge. It is polar. Oxygen is very electronegative. It has a partial negative charge, so maybe it might be willing to take on another proton, but doesn't want to do so very badly. It's actually a weak base. Ethanol right here is a weak base. It's not strong enough to just go nabbing hydrogens off of carbons, like we saw in an E2 reaction. It's just going to sit passively here and maybe wait for something to happen. What might happen? Well, we have this bromo group right here. We have this bromine and the bromide anion is actually a pretty good leaving group. It's a fairly large molecule. It's able to keep that charge because it's spread out over a large electronic cloud, and it's connected to a tertiary carbon. This carbon right here. This carbon right here is connected to one, two, three carbons. So if it were to lose its electron, that electron right there, it would be-- it might not like to do it-- but it would be reasonably stable. It's within the realm of possibilities. It could occur. Maybe in this first step since bromine is a good leaving group, and this carbon can be stable as a carbocation, and bromine is already more electronegative-- it's already hogging this electron-- maybe it takes it all together. Let me draw. Neutral bromine has one, two, three, four, five, six, seven valence electrons. Maybe it swipes this electron from the carbon, and now it'll have eight valence electrons and become bromide. What happens now? And of course, the ethanol did nothing. It's a weak base. It wasn't strong enough to react with this just yet. What is happening now? This is going to be the slow reaction. What I said was that this isn't going to happen super fast but it could happen. This is actually the rate-determining step. What happens after that? Let me just paste everything again so this is our set up to begin with. But now that this little reaction occurred, what will it look like? The bromine has left so let me clear that out. We clear out the bromine. It actually took an electron with it so it's bromide. Let me draw it like this. I'll do it in blue. This is the bromine. The bromine is right over here. It had one, two, three, four, five, six, seven valence electrons. It swiped this magenta electron from the carbon, now it has eight valence electrons. It has a negative charge. The carbon lost an electron, so it has a positive charge and it's somewhat stable because it's a tertiary carbocation. Now let's think about what's happening. And I want to point out one thing. In this first step of a reaction, only one of the reactants was involved. This rate-determining, the slow step of reaction, if this doesn't occur nothing else will. But now that this does occur everything else will happen quickly. In our rate-determining step, we only had one of the reactants involved. It's analogous to the SN1 reaction but what we're going to see here is that we're actually eliminating. We're going to call this an E1 reaction. We're going to see that in a second. Actually, elimination is already occurred. The bromide has already left so hopefully you see why this is called an E1 reaction. It's elimination. E for elimination and the rate-determining step only involves one of the reactants right here. It didn't involve in this case the weak base. Now that the bromide has left, let's think about whether this weak base, this ethanol, can actually do anything. It does have a partial negative charge over here. It does have a partial negative charge and on these ends it has partial positive charges, so it is somewhat attracted to hydrogen, or to protons I should say, to positive charges. But not so much that it can swipe it off of things that aren't reasonably acidic. Now that this guy's a carbocation, this entire molecule actually now becomes pretty acidic, which means it wants to give away protons. Another way you could view it is it wants to take electrons, depending on whether you want to use the Bronsted-Lowry definition of acid, or the Lewis definition. Either way, it wants to give away a proton. It could be this one. It could be that one. It has excess positive charge. It wants to get rid of its excess positive charge. So it's reasonably acidic, enough so that it can react with this weak base. What you have now is the situation, where on this partial negative charge of this oxygen-- let me pick a nice color here-- let's say this purple electron right here, it can be donated, or it will swipe the hydrogen proton. Then hydrogen's electron will be taken by the larger molecule. In fact, it'll be attracted to the carbocation. So it will go to the carbocation just like that. Now in that situation, what occurs? What's our final product? Let me draw it here. This part of the reaction is going to happen fast. The rate-determining step happened slow. The leaving group had to leave. The carbocation had to form. That's not going to happen super fast but once that forms, it's not that stable and then this thing will happen. This is fast. Let me paste everything again. So now we already had the bromide. It had left. Now the hydrogen is gone. The hydrogen from that carbon right there is gone. This electron is still on this carbon but the electron that was with this hydrogen is now on what was the carbocation. That electron right here is now over here, and now this bond right over here, is this bond. We formed an alkene and now, what was an ethanol took a hydrogen proton and now becomes a positive cation. Let me draw that. So this electron ends up being given. It's no longer with the ethanol. It gets given to this hydrogen right here. That hydrogen right there. And now they have formed a new bond and since this oxygen gave away an electron, it now has a positive charge. And all along, the bromide anion had left in the previous step. The bromide anion is floating around with its eight valence electrons, one, two, three, four, five, six, seven, and then it has this one right over here. That makes it negative. Then our reaction is done. We had a weak base and a good leaving group, a tertiary carbon, and the leaving group left. We only had one of the reactants involved. It was eliminated. And then once it was eliminated, then the weak base was then able to take a hydrogen off of this molecule, and that allowed this molecule to become an alkene, formed a double bond. This is called, and I already told you, an E1 reaction. E for elimination, in this case of the halide. One, because the rate-determining step only involved one of the molecules. It did not involve the weak base. We'll talk more about this, and especially different circumstances where you might have the different types of E1 reactions you could see, which hydrogen is going to be picked off, and all the things like that. And we're going to see with E1, E2, SN1, and SN2, what kind of environments or reactants need to be there for each one of those to occur in different circumstances.