E2 Elimination Reactions. Created by Sal Khan.
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- How do you know which hydrogen atom the methoxy anion attacks?(21 votes)
- it can attack both, one will just be a minor contributor and one will be a major contributor. To figure that out draw both reactions and determine which structure has more R groups connected to the alkene the one with the more R groups is the most stable making it the major contributor.(21 votes)
- At6:40+ does the Chloride bond with the Sodium at any point?(6 votes)
- no, remember that in solution the sodium and chloride would exist as ions anyways because of the ionic bonding(10 votes)
- What's stopping the methoxide ion from attacking the carbon attached to the chloro group, since the methoxide is a nucleophile and the nucleophilicity of the substituted carbon is more than than the attacked hydrogen?(3 votes)
- Yep, it's good to keep in mind that most bases can also act as equally good nucleophiles; methoxide included. Sal is showing the E2 mechanism just to demonstrative, but in reality it's equally as likely for the methoxide to act as a nucleophile and perform an SN2 reaction too.
Using a strong nucleophile/base we have the possibility for either an E2 or SN2 reaction. We can further narrow down the reaction mechanism by the type of substrate. 1° will promote SN2, 2° will promote both E2 and SN2 almost equally, and 3° will promote E2.
In this problem we have a strong nucleophile/base with a 2° substrate so we can expect nearly equal amount of E2 and SN2 products.
Hope that helps.(3 votes)
- why don't Cl- and Na+ bond togethere?(2 votes)
- They each have a complete octet. Neither wants to share electrons with the other.
But the opposite charges attract each other. This is called ionic bonding.(5 votes)
- Hi! I was told in my class that the reaction will only occur if the hydrogen forming the alkene product and the leaving group is anti-coplanar. How do you figure out if this is indeed the case. Should I just use Newman projections and wedge-dash diagrams? Is there a better (and easier) way to do this? Also, why is this the case?(3 votes)
- You can use either Newman or wedge-dash projections, but I find wedge-dash easier to use.
The anti-coplanar arrangement is necessary for all the orbitals to line up properly to form the alkene.(3 votes)
- Why is the Chlorine drawn with a negative charge if it has a full octet?(2 votes)
- why is it that Elimination is more probable to happen with a tertiary halogenoalcane but a SN is more likely to happen with a primary halogenoalcane? Ty.(2 votes)
- The nucleophile can easily attack the back side of a primary halogenoalkane in an SN reaction.
In a tertiary halogenoalkane, the bulky alkyl groups prevent backside attack of the carbon bearing the alkyl group (steric hindrance). So the nucleophile acts as a base and does the next best thing — it attacks the hydrogen on the β carbon in an E2 elimination.(4 votes)
- What's stopping the methoxide ion from attacking the carbon attatched to the chloro group, since the methoxide is a nucleophile and the nucleophilicity of the substituted carbon is more than than the attacked hydrogen?(2 votes)
- Nothing is stopping it. But the methoxide ion is also a very strong base, and it is easier to attack the H atoms than the C atom, so elimination is the major reaction. However, you always get some substitution as well.(3 votes)
Let's try to come up with the reaction of when we have this molecule right here reacting with sodium methoxide in a methanol solution, or with methanol as the solvent. Just so we get a little practice with naming, let's see, this is one, two, three, four carbons. So it has but- as a prefix and no double bond or triple bonds, so it's butane. And we have a chloro group here. So if we start numbering at the side closest to it, one, two. So it's 2-chlorobutane. Let's think about what might happen here. The first thing-- let me just redraw the molecule right there. The first thing you need to realize is this sodium methoxide is a salt. When it's not dissolved, it's made up of a positive sodium cation. Let me draw them right here. It's a positive sodium cation and a negative methoxide anion. Let me draw the methoxide part right here. It normally would have just two pairs, but now we have three pairs here. The oxygen has an extra electron. Actually, let me draw another electron as the extra electron. This'll be useful for our mechanism. It could be any of them. The oxygen has an extra electron. It has a negative charge. In solid form, when they're not dissolved, they had formed an ionic bond and they form a crystal-like structure. It's a salt, but when you dissolve it in something, in a solution, in this case, we have methanol as the solution, they will disassociate from each other. And what you have right here is this methoxide right here. This is a very, very, very strong base. This right here is a strong base. And if you use the Lewis definition of a base, that means it really, really, really wants to give away this electron to something else. If you use the Bronsted-Lowry definition, this means that it really, really, really wants to take a proton off of something else. In this situation, that is exactly what it will do. I'll actually give you the most likely reaction to occur here, and we'll talk about other reactions, and why this is the most likely reaction in future videos. So it wants to nab a proton. It is a strong base. It wants to give away this electron. Let's say that it gives away this electron to this hydrogen right over here. Now, this hydrogen already had an electron. If it's getting an electron from the methoxide anion, the methoxide base, then it can give away its electron to the rest of the molecule. It can give away this electron to the rest of the molecule. Now, carbon won't need the electron. Carbon doesn't want to have a negative charge. Maybe simultaneously that electron goes to that carbon right over there. But once again, this carbon doesn't want it. It already has four bonds, but what we see is we have this chloro here. This is a highly electronegative group. The chlorine is very electronegative, so the whole time, the chlorine was already tugging on this electron right here. Now, all of a sudden, it's all happening simultaneously, when an electron becomes available to this carbon, this carbon doesn't need this electron anymore. The chlorine already wanted it, so now the chlorine can take the electron. And just like that, in exactly one step, let's think about what happened. If all of these simultaneous reactions occurred, what are we left with? Let me redraw my two chlorobutanes. I'm going to have to change it now. So now, the chlorine has disappeared. So we clear it. That's disappeared. The chlorine has now left. This chlorine is now up here. It has left. It had this electron right over there, which is right over there. The other electron it was paired with that was forming a bond with is now also on the chlorine. It becomes a chloride anion. Let me draw the rest of the valence electrons. One, two, three, four, five, six, and it has the seventh one right here. One, two, three, four, five, six, seven, eight, so it now has a negative charge. This electron up here-- let me clear this part out as well. Now, this electron right here, this magenta electron, this is now given to this carbon. Let me draw it here. That magenta electron is now given to that carbon. And if we look at the other end, the one that it was paired with, that it was bonded with, it still is bonded with it, so that green electron is now still on that carbon and now they are bonded. Now, they're forming a double bond. This all of the sudden has become an alkene and now the methoxide took the hydrogen. Let me redraw the methoxide. OCH3, the oxygen had one, two, three, four, five. It had six. Actually, it had seven valence electrons, one with the carbon and then all of these six unpaired ones. Neutral oxygen has six. But now it gave one of them away to a hydrogen. It gave that green electron there to the hydrogen-- I'll make this hydrogen the same color-- to this hydrogen right over here, so now this hydrogen is now bonded with it. So what are we left with? We have a chloride anion, so this is chloride. We now have methanol. This was a strong base. Now it has become its conjugate acid. So now we have methanol, which is the same as the solvent, so it's now mixed in. And now we're left off with one, two, three, four, still four. We still have four carbons, but now it's an alkene. We have a double bond, so we could call this but-2-ene, or sometimes called 2-butene. Let's think about what happened here. It all happened simultaneously. Both reactants were present. There was actually only one step so this is the rate-determining step. If we would try to name it, it would probably have a 2 in it someplace. And something got eliminated. The chloride, or I guess you could call it the chloro group, got eliminated. It left. It was a leaving group in this situation. So this was eliminated, and this type of reaction where something is eliminated and both of the reactants are participating in the rate-determining step, and we only had one step here so that was the rate-determining step, is called an E2 reaction. And once again, E stands for elimination. And the 2 stands for that both reactants were involved in the rate-determining step. We had one reactant and two reactants. They were both involved with the rate-determining step. E2, just like Sn2. It's obviously not the same reaction, but in Sn2, we had substitution. The leaving group left, nucleophile came. It was substitution using a nucleophile so that's the S and the n. And then the 2 was we had both reactants. It was all happening simultaneously. I'll leave you there. We're going to go into more detail on the different types of each reactions, which ones are favored, and then we'll talk a little bit about E1. You could maybe guess what that's going to be based on our experience with Sn2 and Sn1.