Comparing E2, E1, Sn2, Sn1 reactions
Comparing E2 E1 Sn2 Sn1 reactions. Created by Sal Khan.
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- I was under the impression that bromobenzene was unreactive for SN1 because the carbocation was unfavored due to the rings geometry and the instability of the formed orbital, and for SN2 because it would imply an inversion of the groups which, again, is impossible due to the ring. Yet here you imply it's possible. Just wondering how?(3 votes)
- he doesn't mention anything to do with bromobenzene here. Bromobenzene is a six carbon ring, with three double (pi) bonds, and a bromo group attached. I imagine that all those pi bonds make the situation completely different to what we have here, which is just a cycloalkane.(17 votes)
- I agree with you that both products are formed. Generally, in organic chemistry, we are asked which product is favored. Are you saying that both products are formed in equal amounts (50/50) or is one product going to occur in a higher yield than the other? If so, which one?(3 votes)
- SN2 is favored. It is likely that the Br on C1 is in equatorial position for stability reason. In order for E2 to happen, Br and the adjacent H to be eliminated must both be anti-periplanar (both in axial position).(6 votes)
- Could someone please explain the difference of all 4 reactions in terms of mechanism? I couldn't really understand.(1 vote)
- SN1 and E1 — the leaving group leaves first.
SN2 and E2 — the leaving group leaves last.
SN1 and SN2 — the X:⁻ attacks a carbon atom.
E1 and E2 — the X:⁻ attacks a β hydrogen atom.(11 votes)
- at3:28sal says that methoxide is a srtonger base than hydroxide. how?(4 votes)
- The H atoms are slightly less electronegative than C, The electronegative O atom pulls electron density from the C, which in turn gets some electron density from the H atoms. Since there is greater electron density on the O in methoxide ion, methoxide ion is a stronger base.(4 votes)
- What are the reasons the reaction is not SN1 or E1?(2 votes)
- They can't occur in a polar aprotic solvent, i believe.(5 votes)
- So, in the event that I have this as a test question, which of these reactions (the SN2 or E2) would most likely take place? Would the E2 reaction be considered to have a more stable product?(4 votes)
- what are alpa and beta carbons?(2 votes)
- An α carbon is the carbon atom next to a functional group.
In E and SN reactions, the functional group is the leaving group, so the α carbon is the carbon atom bearing the leaving group.
A β carbon is the carbon atom next to an α carbon.(5 votes)
- So how do you know how strong an acid/base is going to be? [primarily bases, I have no idea] I've been going by the general 'list of strong bases', but this isn't helping at all...so I can't distinguish between E2 and E1!!(2 votes)
- Usually you are not expected to know the exact strength of a base, just its relative strength to other bases. The key to doing this is figuring out the stability of the conjugate acid of a base. The more stable a conjugate acid, the stronger the original base because it will have a higher tendency to gain a proton/give an electron pair. (The same is also true for strong acids and their weak conjugate bases.)Look online for a list of what makes conjugate acids and bases stable.(4 votes)
- what are the major v minor products though??(1 vote)
- In asymmetrically substituted alkenes, two different carbocations are possible. The major product is generated from the more stable carbocation, while the minor product forms from the less stable one.(4 votes)
- Do Sn2 and E2 reaction favor an unstable carbon? I still don't get why? Can anyone explain for me why a primary carbon tends to have Sn2/E2 reactions? Is it because in Sn2/E2 mechanism, nucleophile attacks at the same time the leaving group leaves, so primary carbon (which has less stereohinderance) will result in faster reaction?(1 vote)
- A simple answer is that, in these mechanisms, ease of reaction matters. It is easier to, say, remove a bromine from a primary carbon than it is to remove it from a tertiary carbon when carbocations can't be formed.
A complex answer is that:
This is because 'pentavalent' transition states are formed (basically, the attacking center also latches onto the carbon from the opposite side in the transition state causing carbon to appear to have 5 bonds) This transition state requires a less hindered carbon, like a primary carbon over, say, a tertiary carbon.(4 votes)
What I want to do in this video is to try to figure out what type of reaction or reactions might occur if we have-- what is this? One, two, three, four, five It's in a cycle. This is bromocyclopentane. If we have some bromocyclopentane dissolved and our solvent is dimethylformamide. Sometimes you'll see that just written as DMF. And I've actually drawn the formula for it here, so we can think about what type of a solvent it is. And also, in our solution, we have the methoxide ion. So we also have the methoxide ion right here. So let's think about what type of reaction might occur. And just to narrow things down, we'll think about it in the context of the last four types of reactions we've looked at. So this might be an Sn2 reaction, an Sn1 reaction, an E2 reaction, or an E1 reaction. We're going to look at all the clues and figure out what's likely to occur, and then actually draw the mechanism for it occurring. Now, the first thing since they gave us the solvent and other things that are in the solvent, let's think about how those might affect the reaction. So if we look at this solvent right here, whenever you look at any of these reactions, when you look at the solvent, you just want to think about it. Is it protic or not? And protic means that it has hydrogens that can kind of be released or that their electrons could be nabbed off and these protons could just float around. And if we look over here, we do have hydrogens, but all of the hydrogens are bonded to carbon. And carbon is unlikely to just steal a hydrogen's electrons and let the hydrogen float around. Carbon is not that electronegative. If you had hydrogens bonded to an oxygen, that'd be a different question. Then you would have a protic solvent. But in this case, all the hydrogens bonded to carbons, not likely to get their electrons nabbed off and float around as free protons. So this is an aprotic solvent. Now, we've gone over this a little bit with Sn2 and Sn1, but the same idea applies. In order to have an Sn2 or an E2 reaction, you have to have either a strong nucleophile or a strong base, and the same thing could actually be both, although they're not always correlated. We've seen that before. Now, if you had a protic solvent, it would stabilize the strong base or the strong nucleophile. The protons would react with them. They would take the electrons from that strong base or that strong nucleophile. So in order to have an Sn2 or an E2, you have to have no protons flying around, so you need an aprotic solvent. So this aprotic solvent will favor Sn2 or an E2 reaction. Now, so our mind is already thinking in Sn2 or E2, let's think about the reactants themselves. So over here, we have the methoxide ion. And let's think about whether it's a strong or weak. Let's think about it first as a strong or weak nucleophile. It's actually a pretty strong nucleophile. It is a strong nucleophile. So that would put us in the direction of an Sn1. So we have two data points. I'm sorry, for an Sn2. We have two data points for Sn2 because remember, it has to just kind of go in there and be active. It's not too big of a molecule, so it's not going to be hindered. But it's also an extremely strong base, even stronger than hydroxide. So it's also an extremely strong base, which might lead us or that does imply that we're going to have an E2 reaction. Now, the last thing we need to think about is the carbon where the leaving group might leave from. And immediately, when you look at the bromocyclopentane, there's only one functional group attached to the chain, and that is the bromo group right here, right there. It is attached to this carbon. We could call that the alpha carbon, and it is a secondary carbon. This carbon right here is bonded to one, two other carbons. This alpha carbon-- let me write it this way. This alpha carbon is a secondary carbon, and that kind of makes it neutral in this mix. If it was a methyl or primary carbon, it would favor Sn2, actually. I mean methyl, the only thing you could have is an Sn2. And if it was a tertiary carbon, it would favor Sn1 or E1 because it would favor a stable carbocation. The leaving group could just leave. And if this guy was bonded to another carbon, it would be very stable. But in this situation, it's a secondary carbon bonded to two carbons. It's a little bit neutral. Any of these reactions might occur. When we look at all of the other data points, they're pointing at both Sn2 or E2. We have a strong nucleophile/base. We have an aprotic solvent. It's going to be Sn2 or an E2 reaction. So let's actually draw the reactions. Let me do the Sn2 first. So let me do it in orange. So if we were to have an Sn2 reaction, let me redraw the molecule. Let me draw the cyclopentane part. I want to make sure-- let me draw it the same way I had it drawn up there. So the pentagon is facing upwards. And then we have our bromo group right there. So we have our methoxide ion right over here. So CH3O minus. Or another way we could view it is that this oxygen has one, two, three, four, five, six, seven valence electrons with a negative charge. One of these electrons right over here, this can attack the substrate right over there, that carbon. Right when that happens, simultaneously this bromine is going to be able to nab an electron from that same carbon. And then we are going to be left with-- the bromine now becomes the bromide anion. It had one, two, three, four, five, six, seven valence electrons. One, two, three, four, five, six, seven. Now it nabbed one more electron, making it bromide. Now it has a negative charge. And if we were to draw the chain, it would look like this. Well, we could draw it on this, and I might as well draw it on this side, just so it's attacking from the other side. This is the chiral substrate, so we don't have to be too particular about how we draw the connections to the carbon. We're not actually even showing anything popping in or out. But we would have the methoxide ion, where now it's bonded, so it's no longer an ion, so it's OCH3, just like that. It has bonded to this carbon. Obviously, implicitly this carbon had another hydrogen that we are not showing. Just that quickly, that was the Sn2 reaction. That is the mechanism. Now let's think about what the E2 reaction is. To do the E2 properly, to give it justice, we're going to have to draw some of the hydrogens. So on the E2 reaction, let me draw that in blue. Let me draw the cyclopentane part. Let me draw it big. Actually, over here, it's less important to draw it too big. So let me draw the pentagon. The pentagon just like that. That is the bromine, three, four, five, six, and then it has a seventh valence electron right over here. This is the alpha carbon. That right there is the alpha carbon. And then there are two beta carbons. There are two beta carbons right over there and there. They each have two hydrogens on them. They each have two hydrogens. I know it's becoming a little hard to read. They each have two hydrogens on them. And in an E2 reaction, the strong base will react-- let me make it a little cleaner than that. Let me get rid of the beta. The beta makes it's a little dirty. OK, so they each have two hydrogens on them. Now in an E2 reaction, the strong base-- over here, the methoxide ion was acting as a strong nucleophile. In E2, it's going to act as a strong base. It's going to nab off a hydrogen off of one of the beta carbons. And you might want to say, OK, which one? Let's look at Zaitsev's rule. It doesn't matter. These are symmetric. They are both bonded to two other carbons. They both are bonded to the same number of hydrogens. It doesn't matter. It's actually going to be random which one, and you actually won't be able tell the difference because it's symmetric. So let's just draw it like this. Let me draw the methoxide ion. One, two, three, four-- or anion, maybe I should say-- five, six. And then it has one bond to the CH3. It has a negative charge, very, very, very strong base. It can go over here and nab the hydrogen and leave hydrogen's electron behind. Maybe I'll take a color. This electron can be given to the hydrogen so that it forms a bond with it. Hydrogen's electron-- let me do this in a suitably different color. Hydrogen's electron that is sitting right over there can now be given to the alpha carbon. It can now be given to the alpha carbon to form a double bond. And now that the alpha carbon is getting that electron, now the bromo group can leave. It's a decent leaving group. And that was another thing that we should think about in our equation. But a good leaving group actually favors all of the reactions: Sn2, E2, Sn1, E1. And so the carbon's getting the electron, and then the bromine can then take this carbon's electron. And just in one step that's what's distinctive about the E2 and the Sn2 reactions. All of the reactions are involved in the rate-determining step and there really is only one step. Just like that, after that happens, what we're left with is the methoxide anion takes the hydrogen, so it becomes methanol. Let me draw that. So it becomes methanol. So it had one, two, three, four and then five, that's this one right there, but then this guy goes and bonds with the hydrogen. This guy goes and bonds with the hydrogen, just like that, and hydrogen leaves its electron behind. Let me draw the cyclopentane part now. And so the cyclopentane looked like this before, if I just focus on the ring. Now, this guy was bonded to a hydrogen. He was bonded to this hydrogen over here, but now that electron is going to be used to form a bond with this alpha carbon right over here. Let me draw the alpha carbon. And the alpha carbon is right over there. Obviously, implicitly at every one of these edges, we have a carbon. But now, a double bond is going to form with that alpha carbon. We could just draw it like that, a double bond. Obviously, there's another carbon here. I could write up another carbon over there. And now this double bond will form. And now the bromide has left. It's taken an electron with it from that carbon now that the carbon doesn't need it. It was already starting to hog it because it's so electronegative. So that's bromine. It takes that orange electron. Now, it is bromide. And we're done. And so just to go back to the original question here, which reaction is likely to occur or which mechanism? It's actually both Sn2 and E2. You would see a mix of both of these occurring because you have all of the environmental factors that would enable both. And so you would have both of these mechanisms. Here's the-- let me separate them out. Here's the Sn2 reaction. You would have the Sn2 reaction occurring in your whatever, your vial, or your pot, or whatever you're making all of this stuff occur in, and you would also have your E2 reaction. So you would see some of all of these, some of all of those products and these products right over there.