- Alcohols and phenols questions
- Alcohol nomenclature
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- Biological oxidation of alcohols
- Oxidation of alcohols
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- Aromatic stability of benzene
- Aromatic heterocycles
Created by Jay.
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- Was a lot helpful....just a request. Will you also emphasize parts important for the MCAT would give a great deal to students, thanks(30 votes)
- 4:21why is it that only one of the electron pairs can participate in resonance? Also how are you supposed to know if electrons are localized to the atom or participate in the hybrid resonance?(2 votes)
- you will have to analyze where the electrons are in a hybridized orbital or an unhybridized orbital. resonance structures are very helpful(1 vote)
- I'm confused how steric number relates to knowing if something is aromaticity -- it's not really explained. I don't get why the point about steric number is brought up as a way to tell if it's participating in resonance, when in the 2nd example we ignore the steps we took in the 1st example.(2 votes)
- So if we conclude that thymine is aromatic as demonstrated at11:12, then is it correct to conclude that the "native form" or most commonly encountered form of thymine has 2 positively charged N and 2 negatively charged O? Moreover, what prevents the thymine from becoming a dienol at this stage?(1 vote)
- As I understand it, the molecule isn't alternating between the three structures shown, rather, it exists as one structure which is an "average" or somewhere in between all possible resonance structures. So in that form, there are no formal charges, only partial charges.(2 votes)
- Would it ever be possible to talk about the reactions that Pyridine can undergo, such as Electrophilic substitution or Electrophilic addition?(1 vote)
- @9:16Isn't the first resonance structure of thymine more likely than the one that was drawn in the video?(1 vote)
In previous videos, we've already seen that benzene is aromatic because it fits the following criteria. Benzene contains a ring of continuously overlapping p orbitals. So each of the six carbons in benzene has a double bond to it. So each of those six carbons is SP2 hybridized, which means that each of those carbons has a free p orbital. And because benzene is a planar molecule, those p orbitals can overlap side by side and allow for a delocalization of the pi electrons in benzene. So if we count the number of pi electrons in benzene, we can see there are two, four, and six. So six pi electrons fits Huckel's rule, which is the second criterion, which says that the ring has to have 4n plus 2 pi electrons. In this case, n is equal to 1. So 4 times 1 plus 2 gives us 6. So six pi electrons for benzene. Just to remind you, n can be equal to 0, 1, 2, 3, or any other whole number. And that's called Huckel's rule. So you need Huckel's rule pi electrons in the ring for a compound to be aromatic. If we look at the pyridine molecule, pyridine is an analog to benzene. The only difference is that pyridine has a nitrogen in the ring instead of one of these carbons right here. So we say that pyridine is a heterocycle. A heterocycle is a cyclic compound that contains a heteroatom in the ring. A heteroatom is any atom other than carbon. So something like nitrogen, oxygen, or sulfur. And heterocycles can be aromatic too. So let's go ahead and analyze pyridine in a little bit more detail here. So here's the dot structure for pyridine. And we'll start by looking at the carbons on pyridine. So pyridine has five carbons. Each of those carbons has a double bond to it. So each of those carbons is SP2 hybridized, meaning there's a free p orbital on each of those five carbons. I'm just going to sketch in those p orbitals on those five carbons like that. Let's analyze the nitrogen, and let's figure out the hybridization of this nitrogen atom. The best way to do it is to figure out the steric number of this nitrogen atom. And so the steric number is equal to the number of sigma bonds plus the number of lone pairs of electrons. So watch an earlier video if you'd like to see how to do steric number in more detail. So we could say this is the sigma bond. We could say this is a sigma bond. And then we could say there's one pair of electrons on that nitrogen. So the steric number is equal to the number of sigma bonds, which is 2, plus the number of lone pairs of electrons, which is 1. And so the steric number is equal to 3. So this nitrogen must have three hybrid orbitals. And therefore, it's SP2 hybridized. So it has three SP2 hybridized orbitals. And therefore, one p orbital. All right, so an unhybridized p orbital. So this nitrogen is also SP2 hybridized, and so we can go ahead and sketch in the p orbital like that. And if we look at the number pi electrons in pyridine, there's two, four, and six pi electrons. So that fulfills Huckel's rule. So there are six pi electrons. And we can see that the pyridine molecule is a ring of continuously overlapping p orbitals. These p orbitals can overlap side by side. And those six pi electrons can be delocalized throughout the ring. And so since this meets both of the criterion, pyridine is an aromatic molecule. So it has some extra stability associated with it. Now this lone pair of electrons on this nitrogen, that lone pair of electrons occupies an SP2 hybridized orbital, so we said this nitrogen is SP2 hybridized, which means it has three SP2 hybrid orbitals. So one of those SP2 hybridized orbitals formed a bond with this carbon over here. One of them formed a bond with this carbon over here. And the last SP2 hybrid orbital actually contains that lone pair of electrons. So the lone pair of electrons on a nitrogen does not participate in resonance. That lone pair of electrons is localized to that nitrogen. And so anytime you see a situation like pyridine where you have a nitrogen with a lone pair of electrons and some electrons already participating in resonance-- so those would be the electrons in magenta here-- the electrons in magenta participate in resonance so the electrons in blue cannot participate in resonance. They are localized to this nitrogen atom. OK. So we've seen that pyridine is aromatic. Let's go ahead and do an example that's similar to pyridine. This is pyrimidine. So let's see if we can analyze the pyrimidine molecule the same way that we analyze the pyridine. So once again, if I start with my carbons here, each of these carbons is connected to a double bond. So I have four carbons. And therefore, each carbon is SP2 hybridized. I can go ahead and sketch in a p orbital on each of my SP2 hybridized carbons like that. When I study the nitrogens in pyrimidine, I could see it's the exact same situation that we had in pyridine. So I can see that there's a sigma bond here, a sigma bond here, and a lone pair of electrons here like that. And so I can see that this nitrogen is SP2 hybridized. And I can see that these pi electrons here are going to be participating in resonance. So for that nitrogen, it's SP2 hybridized. It has a free p orbital. So I can go ahead and draw in the p orbital there. And I know that the lone pair of electrons in blue, since this nitrogen is SP2 hybridized, that lone pair of electrons is going to occupy an SP2 hybridized orbital. It's the exact same situation for this nitrogen. There's a sigma bond. There's a sigma bond. We have a lone pair of electrons on that nitrogen. And then we also have some electrons already participating in resonance. And so this nitrogen is also SP2 hybridized. I can go ahead and draw a p orbital on that nitrogen, which means that lone pair of electrons is not going to participate in resonance. That lone pair of electrons is going to occupy an SP2 hybridized orbital. And it's going to be out to the side like that. And so for pyrimidine. Once again, I have a total of six pi electrons. And those six pi electrons are going to be delocalized as the p orbitals overlap side by side in your ring. And so pyrimidine is also aromatic. It meets the criteria for it. And the lone pair of electrons on those nitrogens, those lone pairs are localized to those nitrogens. The pyrimidine general structure is actually very important in biochemistry. So when you study biochemistry, you'll see how important it is. And here's an example. This is the thymine molecule, which is, of course, found in DNA. And you'll always see in textbooks that thymine has a pyrimidine ring, but at first it's not so obvious that a pyrimidine ring is present in thymine. Because if I look at the nitrogens in thymine-- we'll start with this nitrogen up top here-- I can see that this nitrogen has three sigma bonds to it and one lone pair of electrons. So three sigma bonds, right? So the steric number would be equal to 3 sigma bonds plus 1 lone pair of electrons. So the steric number should be equal to 4, which implies four hybrid orbitals, which would mean that that nitrogen is SP3 hybridized. And if it's SP3 hybridized, you wouldn't have any p orbitals to participate for aromaticity. And so this must not be the case. There must be a way to see a pyrimidine ring here. And the answer is because this nitrogen is actually not SP3 hybridized. It actually has a lone pair of electrons that are delocalized and not localized to this nitrogen, meaning we can draw a resonant structure for the thymine molecule. So this lone pair of electrons right here in this nitrogen are not localized to that nitrogen as we saw in the previous dot structures. Those electrons can move in here to form a pi bond between the nitrogen and that carbon. That would, of course, push these electrons in here off onto this oxygen. So we can go ahead and draw a resonance structure. So let's go ahead and put the nitrogen in our ring. And let's go ahead and draw in the rest of our ring like that. And so that lone pair of electrons moved in to form a double bond between that nitrogen and that carbon so that's our situation now. And for the top oxygen here, it had two lone pairs of electrons, but it picked up one more lone pair of electrons giving it a negative 1 formal charge. This nitrogen is still bonded to another hydrogen. And we can go ahead and draw in the rest of the molecule as well. So there is our resonance structure for thymine. Now let's go ahead and analyze the nitrogen after we drew the resonance structure here. So if I wanted to figure out the steric number for this nitrogen now, I can see that there's a sigma bond here, there's a sigma bond here, and there's a sigma bond here. And now no lone pairs of electrons around this nitrogen. So now the steric number would be equal to 3 plus 0, which is of course equal to 3. So with a steric number of 3, we can say that this nitrogen is SP2 hybridized. And so it has a p orbital now. And the lone pair of electrons that was in the nitrogen over here-- I'm going to go ahead and put those in magenta-- they're not localized to that nitrogen. They're actually delocalized. And that lone pair of electrons can now participate in resonance. But we still don't have the exact pyrimidine structure here. And so we can draw yet another resonance structure. So we can do the exact same thing with this lone pair of electrons down here on this nitrogen. So we can look at the other nitrogen now, and we could do the exact same thing. So at first thought, it might look like that lone pair of electrons is localized to that nitrogen, but it's not. It's actually delocalized because of the resonance structure that we can draw. So pretty much the exact same thing we did before. So let's go ahead and draw in our ring here. So we have our two nitrogens in our ring. This nitrogen is bonded to a hydrogen right here. The lone pair of electrons now moves into here to form a double bond, a pi bond. And this oxygen had two lone pairs of electrons, it picked up an extra lone pair, giving it a negative 1 formal charge like that. And also we can draw in the rest of the molecule here. So double bond methyl group, we have here our negatively charged oxygen up here. So negative 1 formal charge. And over here on our nitrogen, we have our hydrogen and we have this in here. And I forgot to give a plus 1 formal charge on this nitrogen on this resonance structure right here. So there's a plus 1 formal charge on this nitrogen. Obviously, it still has a plus 1 formal charge over here. And we have a plus 1 formal charge over here for this nitrogen. But if we focus in on both nitrogens, this nitrogen is now the exact same situation as this top nitrogen. They're both actually SP2 hybridized with a steric number of 2. So they have a p orbital. Each one of them has a p orbital. And we now can see a little bit better that there actually are six pi electrons that can be delocalized throughout this ring. And now maybe it's a little bit more obvious that the thymine molecule contains the pyrimidine ring. And therefore, it is aromatic and has some extra stability associated with it. So sometimes drawing resonance structures will allow you to see the possible aromaticity or extra stability in a molecule. So this is an example of a biological aromatic heterocycle. So a molecule found in biochemistry which is obviously extremely important that we can analyze using these simple concepts of aromaticity in organic chemistry.