- Alcohols and phenols questions
- Alcohol nomenclature
- Properties of alcohols
- Biological oxidation of alcohols
- Oxidation of alcohols
- Oxidation of alcohols (examples)
- Protection of alcohols
- Preparation of mesylates and tosylates
- SN1 and SN2 reactions of alcohols
- Biological redox reactions of alcohols and phenols
- Aromatic stability of benzene
- Aromatic heterocycles
Created by Jay.
Want to join the conversation?
- So, can someone please explain to me what exactly "n" is and what determines its value? When ever my professor does these problems and apparently the maker of this video there seems to be a bit of mistery when it comes to figuring out what "n" is. In the video he says "If n is 1 then use Huckel's rule." Well, where did 1 come from? What is he counting? To me its almost like you figure out how many pi electrons there are and then solve for "n"! But that can't be right!! Thank you...(6 votes)
- n is any positive integer. You are right in saying you figure out how many pi electrons there are and then solve for n. You just want the number of pi electrons you have to follow the rule. For example, if you had 7 pi electrons and plugged that into Huckel's rule you would get n = 1.25. Since 1.25 isn't an integer the compound isn't aromatic. If n is a positive integer then that satisfies one requirement of being aromatic.(15 votes)
- Does anybody know if molecular orbital theory is on the MCAT?(1 vote)
- I realize its too late but for other people, yes it is remember guys its the bonding/antibonding thing(2 votes)
- Why makes the middle bonding pi electrons have extremely low energies and the middle antibonding elections have such a higher potential energy?(1 vote)
In this series of videos, we're going to look at aromaticity or aromatic stabilization. We've already seen that bromine will add a cross double bond of a simple alkene like cyclohexane to give us a mixture of enantiomers for our products. If we try the same reaction with benzene, we're not going to get anything for our product. So there's no reaction. And so benzene is more stable than cyclohexane. At first you might think that the stability is due to the fact that benzene is conjugated, but numerous other experiments have shown that it is even more stable than we would expect. And that extra stability is called aromaticity or aromatic stabilization. And so benzene is an aromatic molecule. Let's look at the criteria to determine if a compound is aromatic. All right. So a compound is aromatic if it contains a ring of continuously overlapping p orbitals. And so if the molecule is planar, that's what allows the p orbitals to overlap. It also has to have 4n plus 2 pi electrons in the ring where n is equal to 0, 1, 2, or any other positive integer. And this is called Huckel's rule. So let's go ahead and analyze benzene in a little bit more detail. So if I look at the dot structure, I can see that benzene has two pi electrons there, two here, and two more here for a total of six pi electrons. If I look at the carbons of benzene, I can see that each carbon has a double bond to it. So each carbon is sp2 hybridized. And if each carbon is sp2 hybridized, that means that each carbon has a free p orbital. So let me go ahead and sketch in the unhybridized free p orbital on each of the six carbons of benzene. Now since benzene is a planar molecule, that's going to allow those p orbitals to overlap side by side. So you get some overlap side by side of those p orbitals. And so benzene contains a ring of continuously overlapping p orbitals. So p orbitals are considered to be atomic orbitals. And so there are a total of six atomic orbitals in benzene. According to MO theory, those six atomic orbitals are going to cease to exist. And we will get six molecular orbitals instead. And so benzene has six molecular orbitals. Drawing out these molecular orbitals would be a little bit too complicated for this video. So checkout your textbook for some nice diagrams of the six molecular orbitals of benzene. However, it is important to understand those six molecular orbitals in terms of their relative energy levels. And the simplest way to do that is to draw a frost circle. And so here I have a circle already drawn. And inside the circle we're going to inscribe a polygon. And since benzene is a six-membered ring, we're going to inscribe a hexagon in our frost circle. I'm going to go ahead and draw a center line through the circle just to help out with the drawing here. And when you're inscribing your polygon in your frost circle, you always start at the bottom. So we're going to start down here. So we're going to inscribe a hexagon. So let's see if we can put a hexagon in here. So we have a six sided figure in our frost circle. The key point about a frost circle is everywhere your polygon intersects with your circle, that represents the energy level of a molecular orbital. And so this intersection right here, this intersection here, and then all the way around. And so we have our six molecular orbitals and we have the relative energy levels of those six molecular orbitals. So let me go ahead and draw them over here. So we have three molecular orbitals which are above the center line. And those are higher in energy. And we know that those are called antibonding molecular orbitals. So these are antibonding molecular orbitals which are the highest in energy. But look down here, there are three molecular orbitals which are below the center line and those are our bonding molecular orbitals. So those are lower in energy. And if we had some molecular orbitals that were on the center line, those would be non-bonding molecular orbitals. We're going to go ahead and fill our molecular orbitals with our pi electrons. Let me go back over here. And remember that benzene has six pi electrons. And so filling molecular orbitals is analogous to electron configurations. You're going to fill the lowest molecular orbital first. And each orbital can hold two electrons like electron configurations. And so we're going to go ahead and put two electrons into the lowest bonding molecular orbital. So I have four more pi electrons to worry about. So four more pi electrons and I go ahead and put those in. And I have filled the bonding molecular orbitals of benzene. So I have represented all six pi electrons. If I think about Huckel's rule, 4n plus 2, I have six pi electrons. So if n is equal to 1, Huckel's rule is satisfied because I would do 4 times 1 plus 2. And so I would get a total of six pi electrons. And so six pi electrons follows Huckel's rule. If we look at the frost circle, and we look at the molecular orbitals, we can understand Huckel's rule a little bit better visually. So if I think about these two electrons down here, you could think about that's where the two comes from in Huckel's rule. And if I think about these four electrons up here, that would be four electrons times our positive integer of one. So 4 times 1 plus 2 gives us 6 pi electrons. And we have filled the bonding molecular orbitals of benzene which confers the extra stability that we call aromaticity or aromatic stabilization. And so benzene is aromatic. It follows our different criteria. In the next few videos, we're going to look at several other examples of aromatic compounds and ions.