E2 E1 Sn2 Sn1 reactions example 3
E2 E1 Sn2 Sn1 reactions: Example 3. Created by Sal Khan.
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- Sal, the carbon that underwent the Sn1 reaction was chiral, so wouldn't that mean you got a racemic mixture of R and S product?(25 votes)
- I believe you are right, with Sn1 reactions if you have a chiral center you will get racemic mixtures because both enantiomers will form.(38 votes)
- I thought protic meant that the solvent is able to hydrogen bond.(7 votes)
- protic solvents can form good hyrogen bonds
they can also give out protons(7 votes)
- Why there are E1 products? I thought if the base/Nu is not bulky, it will mostly undergo SN1 because of steric effect(5 votes)
- you get a mixture of both kinds of products when doing sn1 and el; usually e1 will form the majority products if there are three or four alkyl substituent groups around the c-c doubel bond(5 votes)
- Which type of reaction (Sn1 or E1) is favored though?(5 votes)
- When there is a carbocation outside of the cyclopentane wouldn't it become a cyclohexane since it would be stabler?(2 votes)
- Whenever a carbocation is formed, there is always the possibility of rearrangement. You could get a hydride shift, an alkyl shift, or a ring expansion. The energy difference between 5- and 6-membered rings is is not large. In fact, there are cases in which 6-membered rings contract to 5-membered rings.
In this carefully chosen example, the initially formed carbocation is already 3°, so the ion has little incentive to rearrange. Only if a migration gives a more stable carbocation will the rearrangement occur,(7 votes)
- What are other examples besides oxygen that can be used as good protic solvents?(4 votes)
- Water and alcohols typically are good protic solvents (methanol, ethanol, isopropanol, t-butanol).(3 votes)
- I thought the SN1 would not form at all because the carbon with the leaving group on it is tertiary, so the methanol nucleophile couldn't fit in there. They both form? Are there more E1 than SN1 products?(3 votes)
- SN1 would definitely occur because the tertiary carbocation exhibits sp2 hybridization and therefore has a trigonal PLANER geometry. This means there is plenty of "open space" above or below the plane for the methanol to attack. SN*2* reactions are inhibited by steric hindrance associated with tertiary carbons.(4 votes)
- For the Sn1 reaction from the above example, I would like to ask, after the Iodide leave, why the Oxygen will bond to the tertiary carbon, then the Hydrogen leave. If I were asked to do this question, I would write in this way. Firstly, the Iodide leave, then the Oxygen hag the electron from Hydrogen to form a oxygen anion. The Hydrogen bond to the oxygen anion then leaves. Finally, the oxygen bond to the tertiary carbon to form a covalent bond. Am I right? Thank you very much.(2 votes)
- So i think you are debating whether the oxygen loses the H+ then bonds to the carbocation, or if the oxygen bonds to the carbodcation then loses the H+...I think its neither and both....meaning that it happens simultaneously. I am not sure though, someone please correct me if I'm wrong.(3 votes)
- Shouldn't I put OCH3 on a dash in Sn1 product? I guess it's away from us because CH3 is up out the paper?(2 votes)
- Yes, the way he drew the molecule, the bond to OCH₃ should be dashed.(3 votes)
- What about steriochemistry on the Sn2 reaction? Would the new product still form on a wedge?(2 votes)
- I believe you mean the stereochemistry on the Sn1 reaction. And yes, I believe so. Since the new group joins on the opposite side of the leaving group, it -should- be a wedge. However since it is Sn1, there should be a racemic mix, so I believe it could be a dash as well. Hopefully someone a bit more knowledgeable can help clear that up! :)(2 votes)
Let's think about what type of reaction might occur if we have this molecule right over here. I won't go through the trouble of naming it. It would take up too much time in this video. But it's dissolved in methanol. When we talk about what type of reactions, we're going to pick between Sn2, Sn1, E2 and E1 reactions. Now, maybe the first place to start or the place I like to start is to just look at the solvent itself. And when we're trying to decide what type of reaction will occur, the important thing to think about is, is the solvent protic or aprotic? And if you look at this solvent right here, this is methanol. It is protic. And in case you don't remember what protic means, it means that there are protons flying around in the solvent, that they can kind of go loose and jump around from one molecule to another. And the reason why I know that methanol is protic is because you have hydrogen bonded to a very electronegative atom in oxygen. So every now and then, in one of the methanol molecules, the oxygen can steal hydrogen's electron. And then the hydrogen itself, without the electron, the hydrogen proton will be flying around because it doesn't have a neutron. So this is a protic solvent. Now, you might say, well does anything with a hydrogen, would that be protic? And the answer is no. If you have a bunch of hydrogens bonded to just carbons, that is not protic. Carbon is not so electronegative that it could steal a hydrogen's electron and have the hydrogens floating around. So a big giveaway is hydrogen bonded to a very electronegative atom like oxygen. So this is protic. And when we think about protic out of all of the reactions we studied, that favors-- well, even a better way to think about it is it disfavors. So it tells us that it's unlikely to have an Sn2 or an E2 reaction. And the logic there is an Sn2 reaction needs a strong nucleophile. An E2 reaction needs a strong base. Now, if you have protons flying around, the nucleophile or the base is likely to react with the proton. It would not be likely to react with the substrate itself. So a protic solution, you're unlikely to have an Sn2 or E2. What you are likely to have is an Sn1 or an E1 reaction. Both of these need the leaving group to leave on its own, and actually, having protons around might help to stabilize the leaving group to some degree. So it makes Sn2, E2 unlikely, Sn1, E1 a little more likely. So far, these are our good candidates. Now, the next thing to think about is to just look at the leaving group itself, or see if there is even a leaving group. And over here, everything we see on this molecule is either a carbon or a hydrogen, except for this iodine right here. And we know that iodide is a good leaving group. Well, a good leaving group, it does not make it any less likely that you'd have Sn2 or E2. Both of those can do well with a good leaving group. But it's a necessary requirement for an Sn1 or an E1 reaction. Remember, an Sn1 and E1, in both of them, the first step is that the leaving group leaves on its own. That is the rate-determining step. So that's a requirement for Sn1 or E1, so it still looks likely. We haven't seen anything that would make us think that we wouldn't have an Sn1 or E1. This is a good leaving group. Let me write it here. Good leaving group. Now, the last thing that we can think about right here is the carbon that we might be leaving from. So far, everything is pointing in the Sn1 and E1 direction, and kind of the final thing is when this leaving group leaves, it's going to form a carbocation from the carbon that it's bonded to right now. And in order for that carbocation to be reasonably stable, at minimum it should be a secondary carbocation bonded to at least two carbons, but ideally, it would be bonded to three carbons. It would be a tertiary carbon. Now, the carbon that the leaving group is bonded to is a tertiary carbon. It's bonded to one, two, three carbons. so it is a tertiary carbon. It can actually be a stable carbocation. It's a tertiary carbon, which it once again favors Sn1 and E1. So all of the clues here tell us that Sn1 and E1 are going to happen. And actually, they'll both happen. So let's think about the mechanism. In the very first step, the leaving group leaves in either one of these reactions. So if we look at the iodine, it already has seven valence electrons. One, two, three, four, five, six, seven. And in the first step, the rate-determining step of the Sn1 or E1 reaction, the iodine's going to nab an electron off the carbon. I'll do that electron in green right over there. That's going to get nabbed onto iodine to make iodide, and then that carbon is going to lose an electron and become a carbocation. So after that very first step, we have something that looks like this. So that's our molecule. Just so it's clear what we did, this arrow I'll do all the way over here, so it's clear that this is our next step. And what's happened here? Our tertiary carbon has lost its electron. So now this carbon right here has a positive charge. It's a tertiary carbocation. And now the iodine has become iodide. It has left the molecule. So it had its original seven valence electrons: one, two, three, four, five, six, seven. It nabbed one more electron from the carbon and now it is-- I wanted to do that in green. It nabbed one more electron from the carbon, now it is the iodide anion. So this step right here is common to both Sn1 and E1 reaction. The leaving group has to leave. Now, after this, they start to diverge. In an Sn1, the leaving group essentially gets substituted with a weak nucleophile. In an E1, a weak base strips off one of the beta hydrogens and forms an alkene. So let's do them separately. So over here, I'm going to do the Sn1. And on the right-hand side, I will do the E1 reaction. So let me start over here. So the Sn1 is starting over here at this step. I'll just redo this step over here. So this has a positive charge. That has a positive charge here. The iodide has left. I don't have to draw all its valence electrons anymore. And what's going to happen next? We're going to get substituted with the weak base, and the weak base here is actually the methanol. The weak base here is the methanol. So let me draw some methanol here. It's got two unbonded pairs of electrons and one of them, it's a weak base. It was willing to give an electron. It has a partial negative charge over here because oxygen is electronegative, but it doesn't have a full negative charge, so it's not a strong nucleophile. But it can donate an electron to this carbocation, and that's what is going to happen. It will donate an electron to this carbocation. And then after that happens, it will look like this. That's our original molecule. Now this magenta electron has been donated to the carbocation. The other end of it is this blue electron right here on the oxygen. It is now bonded. That is our oxygen. Here's that other pair of electrons on that oxygen, and it is bonded to a hydrogen and a methyl group. And then the last step of this is another weak base might be able to come and nab off the hydrogen proton right there. Oh, I want to be very clear here. The oxygen was neutral. The methanol here is neutral. It is giving away an electron to the carbocation. The carbocation had a positive charge because it had lost it originally. Now it gets an electron back. It becomes neutral. The methanol, on the other hand, was neutral, gives away an electron, so now it becomes-- it now is positive. So now you might have another methanol. You might have another methanol molecule sitting out here someplace that might also nab the proton off of this positive ion. So this one right here, it would nab it or it would bond with it. It would give the electron to the hydrogen proton, really. The hydrogen's electron gets nabbed by the oxygen, and so then that becomes neutral. So in the final step, it'll all look like this. We have that over here. The methanol that had originally bonded has lost its hydrogen, so it looks like this. We just have the oxygen and the CH3 there. It is now neutral because it gained an electron when that hydrogen proton was nabbed. So if you wanted to draw it, it has actually those two extra electrons, just like that. And if you want to draw this last methanol, it's now a positive cation, so it looks like this. So it's OH, CH3, H, and then it has unbonded pair right there, and now this has a positive charge. So that was the Sn1 reaction. Now, the other reaction that's going to occur is the E1. Once again, our first step-- nope, I didn't want to do that. Our first step looks like that. So that is our first step. Let me get everything straight. So the leaving group had left. So in each situation, the leaving group had left. Our iodide is up here. And in an E1 reaction, you don't get substituted. What happens is one of the beta carbons gets a hydrogen swiped off of it by a weak base. Now, let's think about what a beta carbon is. The alpha carbon is the carbocation carbon. That's right over there. That's the alpha carbon. Beta carbons are one carbon away. So this is a beta carbon, this is a beta carbon, and then that is a beta carbon. Now, this carbon over here is not bonded to any hydrogens. It's only bonded to other carbons. So that one cannot lose any hydrogens. And then we have to pick between this carbon that's bonded to three hydrogens and this carbon that's actually bonded to two hydrogens. I didn't draw it before, but it's implicitly there. It's bonded to two hydrogens. Now Zaitsev's rule tells us that the dominating product is going to be produced when the carbon that has less hydrogens loses a hydrogen. So out of this, this carbon has two hydrogens, this one has three. So the one that's going to lose the hydrogen to the base, or more likely to lose the hydrogen to the base, is the one that has two hydrogens, not three, the fewer hydrogens. So our base in this case is once again the methanol, acted as a nucleophile in the Sn1, acted as a weak nucleophile in Sn1. Now it'll act as a weak base. So we have methanol right over here. That's a hydrogen. CH3 has some electrons right over here. It has a partial negative charge. It will give one of the electrons to the hydrogen, just to the hydrogen proton. The hydrogen's not going to take its electron with it. That electron is then going to be given to the carbocation to make it neutral, and it will form a double bond. So after that happens, we are left with something that looks like this. This was our original molecule. Now, this carbon right here in yellow, it has lost its hydrogen. This hydrogen back here is still there. I could draw it if I like. I don't need to. It just makes the thing messy. But it has now formed a double bond with the primary carbon. It has now formed a double bond. That electron that had bonded with the hydrogen was now given to the carbocation. It has a double bond with that. And then you have your methanol has now turned into a positive ion. It has now turned into a positive ion: CH3. And now it has this bond. It has this bond with the hydrogen. I'll even make that electron that it gave away in magenta. And then it has that extra lone pair of electrons. So, in this circumstance, we looked at all of the clues. All of the clues were against Sn2 and E2. They favored Sn1 and E1. So if you were to actually make this, if you were to actually try to see what happens is this reaction, you would get products of both Sn1 and E1 reactions, these products and these products over here.