Birch reduction II
The effect of electron-withdrawing and electron-donating groups. Created by Jay.
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- Why did the attack by Na take place on para with EWG?
(In this case there might be steric hindrance but what if EWG is quite small in size?)(10 votes)
- According to the video (5:45onward) it is because the EWG stabilizes the negative charge at the ipso (the substituted carbon) and the para positions in the intermediate resonance structure. This pathway likely has lower activation energy to generate this stabilized intermediate.(3 votes)
- At6:07, how does the ipso group stabilize the para position? How does this explain why the Sodium attacks at the para position, if the para position is not more stable BEFORE the attack by sodium?(5 votes)
- It has to do with molecular orbital theory and the shape of the LUMO (lowest unoccupied molecular orbital) of the original aromatic compound.
In benzene there are two degenerate LUMOs due to the symmetry of the molecule. When the ring is substituted, the degeneracy in these two orbitals is removed, and which of the two becomes the LUMO depends on the nature of the substituent.
When the substituent is electron-withdrawing, the energy of the orbital with a lobe at the ipso position is lowered (stabilized), so this orbital becomes the LUMO. The lobes at the ipso and para positions are the largest of this orbital.
When an electron is added to the LUMO, it is added to the position where the largest lobe of the LUMO occurs. In this case, the para position is more accessible.(6 votes)
- I just wonder why the ortho is being protonated first, then the meta, and why its not the other way round.(3 votes)
- The meta position can't stabilize through resonance the way the ortho position can. The double bonds on either side of the meta position hinder the resonance while the Donating group provides and avenue for resonance for the ortho position.(4 votes)
- How do you know exactly where the sodium will donate its electron to in the first step?(3 votes)
- in all the examples i found and in this video. the EDG (electron donating group) is always a -O... never an alkyl group like CH3.
my question is what happens with, for example, in ethyl benzene? does it react in a Birch Reduction like this?(2 votes)
- Alkyl groups are also EDGs. The Birch reduction has the same regiospecificity as with anisole.(3 votes)
- why ester won't donate the electrons ? like the methoxy group.(2 votes)
- The C=O group of the ester is an electron withdrawing group. It removes electrons from the ring, as we see at4:45in the video. The methoxy group is an electron donating group via resonance.(2 votes)
- Why are single arrow electron movements happening with the first 2 pi bonds but not the third? That would change the final product but how do i remember that i use single arrows for the first two bonds and then randomly stop for the third one. Please let me know if i should clarify my question(2 votes)
- Adam, I think I understand what you are asking so I'll take a crack at answering. In simplest terms, the third pi bond is not bonding carbons that are involved in the reduction. As a result, those electrons are not "pushed" in the mechanism. Of course, what I just said is not technically correct due to the delocalization of the pi electrons in the ring. To make it easier to understand, it is simpler to think of the pi bonds as being localized, i.e. isolated from each other. Did that help or did I miss the mark on your question?(2 votes)
- what is ipso substitution?(2 votes)
- the ipso position is the position at which a substituent group is attached...for example, in toluene in the ipso position is the position on the benzene ring at which the methyl group is bonded. ipso substitution is substitution at this position.(1 vote)
- It is stated in the video that it is shown in studies that ortho protination occurs first before meta protinationfor benzene rings with EDGs. Does this imply that para protination always occurs first before the ipso position (as shown by experimentation) since that is what's shown in the video?(1 vote)
- this attack in birch reduction (the choice of the carbon) at which it happens is not accidental
I think You should also show mesomeric structures before Birch reduction to indicate which positions are electron rich and which are electron defficient in presence or EDG and EWG groups
This would help a lot(1 vote)
In the previous video, we looked at the mechanism for the Birch reduction. In this video, we're going to see what happens with the Birch reduction when you have a substituted benzene ring. So if you have a benzene ring with a substituent on it, and you add sodium, liquid ammonia, and an alcohol, the substituent is going to affect which carbons are reduced. So for example, if you have an electron withdrawing group on your ring, the carbon that's bonded to your Y substituent here is reduced. If you have an electron donating group on your ring, the carbon that's bonded to that substituent is not reduced. Let's see if we can learn why by going through a mechanism for each one of these examples. And we'll start with an electron withdrawing group. And so here we have an electron withdrawing group that is an ester on our benzene ring, and we're going to start with our sodium, which we know has one valence electron. I'm going to go ahead and just do the mechanism, and we'll talk about details as to why it happens that way when I'm done here. So this lone electron on sodium is going to be donated to this carbon, and we have a bond here, which consists of two electrons. One of those electrons is also going to move out onto that carbon, and one of those electrons is going to move into here. With our other bond, a similar idea. One of the electrons is going to move into here, and one of them is going to move off onto this carbon so let's go ahead and show the movement of all of those electrons. So once again, we have our ester group up here. We added an electron to this carbon from the sodium. We got another electron from that bond so that gives this carbon a negative 1 formal charge. And over here we formed a pi bond. There was already a pi over here. And then, of course, we get one electron on that carbon. So this is our radical anion, and if you had a hard time following those electrons, please go back and watch the previous video and take some time to study the mechanism really well. So in the next step in the mechanism, we know that's the protonation step. So we have an alcohol, it comes along here, and we're going to protonate our anion. So these electrons here pick up a proton and kick those electrons off onto the oxygen so let's go ahead and draw what we have now. So pi electrons here. Up here we have our electron withdrawing group, our ester. And we had an electron on this carbon. There was already a hydrogen bonded to this carbon, and those electrons picked up a proton and so we now have two hydrogens bonded to that carbon. In the third step up our mechanism, we started off with an electron and then we went to a protonation and so we know the third step is electron again. And so we could think about another sodium coming along with its valence electron, and, of course, it's going to donate its valence electron to the carbon that already has an electron on it, right? So to this carbon right here. So let's go ahead and draw the result of that. So I have my ring, I have my pi electrons in my ring, I have my hydrogens already bonded to it. Up here I have my ester group bonded to my ring. I had one electron on this carbon and the sodium just donated another one so that gives me a negative 1 formal charge, and so this is our anion right here. And the final step for our mechanism, we know that's another protonation step so I'll go ahead and show my anion picking up a proton from this alcohol here. And so we can go ahead and show and draw the final product. So I have my ring, I have these pi electrons, I have these hydrogens on this carbon, and I have my ester group coming off of my ring, and I added on a proton to that carbon so that's the final product. And you can see, of course, that the carbon that is bonded to our electron withdrawing group was reduced. It gained a proton here. And the question is why? Why do we get this as our product? And a way to think about that is to go back here to this structure. So this anion step. And I can draw our a structure for this anion because of the presence of that electron withdrawing group. So let me go ahead and put in my resonance brackets and my resonance arrow here. And this is negative charge-- I'm going to go ahead and use red here-- so these electrons right here can participate in resonance due to the presence of our electron withdrawing group, and we know that this oxygen has two lone pairs of electrons on it so if these electrons moved in here to form a bond, that would push these pi electrons off onto the oxygen. So let's go ahead and draw our resonance structures. So we have our ring, we have our pi electrons. Now, the carbon on our ring is double bonded to the carbon out here, and that carbon is bonded to an oxygen, that oxygen now has three lone pairs of electrons, which gives it a negative 1 formal charge. And we still have this part over here on our molecule and these hydrogens of course. And so let me show those electrons in red there. So those electrons in red moved in here to form a pi bond, and so there's resonance stabilization of electron density at the ipso position. So the ipso position is the position that's right next to our substituent. So it's the carbon on our ring that's bonded to our substituent. And because of the presence of our electron withdrawing group, electron density at that position is stabilized. And actually if you go back here to the radical anion, these electrons are actually delocalized. And so the presence of that electron withdrawing group stabilizes electron density at the ipso position and also the para position, and so that's why you see the reduction happening at those carbons. So now that we have an idea about why we get this as the product when you get an electron withdrawing group present on your ring, let's go ahead and do one for an electron donating group and we'll try to compare. So the electron donating group here is a methoxy group. So here's our methoxy group, which we know the oxygen has two lone pairs of electrons, and it's those lone pairs of electrons, it's one of those lone pairs that can contribute and increase the electron density in your ring, and so therefore, the methoxy group is an electron donating group as we saw in some of the other videos here. So we'll start the mechanism, and once again, let me just go ahead and do the mechanism and we'll talk about why it occurs when I'm done here. So if the sodium donates its valence electron to the ring and it's going to donate it to this carbon right here on our ring. For this bond, one of the electrons is going to move out onto that same carbon that the sodium donated an electron to, and then the other electron is going to move into here. So for this bond, one of the electrons is going to move into here and one of the electrons is going to move out onto this carbon. So let's go ahead and draw the result of all those electrons moving. So once again, we have our methoxy group up here. So this carbon got an electron from sodium, this carbon got an electron from the bond, and so now it has a negative 1 formal charge on it. We formed a pi bond over here, and we also have an unpaired electron over here so this is our radical anion. Next, of course, is protonation. So we can go ahead and show our alcohol over here. So our carbanion functions as a base, picks up a proton, and let's go ahead and show the result after our ring is protonated here. So this is protonation in the ortho position. So if I think about this carbon right here, protonation occurs ortho to our methoxy group so there was already a hydrogen on that ring, and we protonated so now there are two hydrogens on that ring. Pi bond over here. We still have our unpaired electron. All right, so third step of our mechanism, of course, would be another electron, which we get from sodium. So sodium comes along and donates a valence electron right here so let's go ahead and draw the result of that. So we have our ring, we have our electron donating group, we have pi electrons here and here, we had these hydrogens, and at this carbon we had one electron, we got another one from another sodium so that gives us a negative 1 formal charge and so that's our anion. So the next step would be protonation so this time protonation is occurring meta to our electron donating group. So if I think about this carbon right here. Right now, we're talking about meta protonation compared to the position of our electron donating group. And so these electrons are going to pick up a proton from our alcohol, kicks these electrons in here off, and so now we can go ahead and show the product. So once again, our methoxy group, our pi electrons, we had hydrogens on that carbon, there was already a hydrogen on this carbon, right? It picked up a proton, and so now there are two hydrogens, and so now we're done. This is our product. And so, of course, the question is why is this a little bit different from the electron withdrawing group? Why does the electron donating group work this way? So once again we can look at the carbon that's bonded to our methoxy group, our electron donating group. That carbon was not reduced. And one way of thinking about it is in this mechanism you can see we have a lot of electron density in the ortho and the meta position so we have a lot of electron density in the ortho and the meta position. And so those are the carbons that are reduced, and it makes sense that we don't reduce the carbon that's attached to our electron donating group because, of course, our electron donating group would donate some electron density to that carbon. So let me go ahead and draw a little picture here so you can understand a little bit better about what I'm talking about. If I had my electron donating group, I can move these electrons into here, and so there's a little bit more electron density. And so if we had a negative 1 formal charge or some electron density right at this carbon, these electrons here would, of course, destabilize an anion or electron density at this position, at the ipso position. And so you're not going to see a reduction at the ipso and the para position. So that's what we saw when we had the electron withdrawing group. An electron withdrawing group stabilized electron density at the ipso and the para position here, but that's not the case for an electron donating group. It would destabilize any electron density at those positions, and so that's why you think about in ortho and a meta, it's more out of the way. And so it doesn't destabilize it as much. And of course, you could think about this ortho and this meta. It doesn't really matter because of the symmetry for this example, but that's one way to think about it. Maybe it's not the best explanation, but it's a good way of thinking about it here. And at one time, there was some disagreement over the order of protonation so it turns out that you do have an ortho protonation first and then followed by a meta protonation, but some people used to think that a meta protonation was first followed by an ortho, but I think that studies have shown that ortho was first and now meta. So this just gives you a little bit of insight into what electron withdrawing groups and what electron donating groups, what effect they can have, what effect those directors can have, on the Birch reduction.