Aromatic stability IV
Aromaticity of the cyclopentadienyl anion and cycloheptatrienyl cation. Created by Jay.
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- At about9:10, how do you know there is 3 bonding and 4 anti bonding instead of the other way around?(5 votes)
- When constructing molecular orbitals using a Frost circle, one of the vertices is always drawn in the centre pointing down. In an aromatic compound there will only be one arrangement where all of the p orbitals are in phase. This arrangement has maximal orbital overlap so will be the lowest energy bonding orbital (in the pi system). This single lowest-energy MO is represented by the vertex at the bottom of the Frost circle (it will always stand alone without any degenerate orbitals). Having a vertex at the bottom of the Frost circle (as opposed to a side of the polygon at the bottom) means that there are 3 bonding and 4 anti-bonding MOs.(8 votes)
- Is there ever a situation where you calculate the pi electrons at the start and when you get to the frost circle, they don't create a 'full outer shell'? Or is the Frost circle just a way to demonstrate BMO etc, but in reality you don't need it to work out if a molecule is AROMATIC?(3 votes)
- The Frost circle is just an easy visual way to work out if a molecule is aromatic without having to do the mathematical calculations.(5 votes)
- What is pKa and what does it numeric value suggest?(1 vote)
- Ka is the value of the equilibrium constant for the ionization of an acid in water. The greater the Ka, the more of the acid ionizes, and the stronger the acid. Most common weak acids have Ka values ranging from 10^-2 to 10^-14. A Ka value of 10^-2 means that an acid is stronger than one with a Ka of 10^-12.
pKa is the negative logarithm (to the base 10) of Ka. Thus the pKa values of these acids range from 2 to 14. The lower the value of pKa, the stronger the acid. So an acid with pKa = 2 is stronger than an acid with pKa = 14.(8 votes)
- (1:30) A pKa value of 16 indicates a weak acid, doesn't it?(3 votes)
- in the very first example... arent those hrydrogens on the sp2 double bonded carbons acidic... i thought that becoz sp2 orbitals are more acidic than sp3 orbitals??(2 votes)
- Quite correct! Normal sp3 bonded H atoms have a pka ~ 51, while sp2 bonded H atoms have pKa ~44, and acetylenic (sp bonded) H atoms have pKa ~25. But in cyclopentadiene the sp3 H atoms have pKa ~16. They are 10^9 times more acidic than even an alkyne H.
The reason is that the cyclopentadienide ion (C5H5-) is exceptionally stable because it is aromatic. So it is much easier to lose the sp3 bonded H (the activation energy is much less), and the H is therefore more acidic than you might expect.(4 votes)
- I've read in my textbook that Hükel's rule doesn't work with polycyclic aromatic compounds does anyone know why?(1 vote)
- For the last figure, at8:16, why does the does the top carbon have a pi orbital. Isn't it only the electrons involved with the pi bond?(1 vote)
- The top carbon is positively charged, so is sp2 hybridized. Thus, it has an unhybridized p orbital. This p orbital has the necessary orientation to be involved in resonance structures with the double bonds in the ring. Thus, through resonance, the p_ orbital on the carbocation is part of the _pi system.(3 votes)
- do frost circles only work with cyclic compounds?(1 vote)
- It's even more limited than that. They work only for planar cyclic conjugated compounds.(4 votes)
- if having a high pKa value indicates that it is a weak acid, doesnt that mean that the molecule is not willing to give up a hydrogen very easily to reach this apparently very stable conformation? so why is it constantly referred to as "so acidic" in the video?(1 vote)
- It is very acidic relative to other hydrocarbons. When Jay first talks about this1:12he says "extremely acidic for a hydrocarbon".
Typical pKa values:
alkanes pKa ~50
alkenes are pKa ~45.
Various sources list the pKa of cyclopentane as being ~45 to >60 and I can't find anything for cyclopentene.
So the acidity of cyclopentadiene about 10^29 times greater than for a typical alkene!(2 votes)
- how can we say that a carbon cation is sp2 hybridised. pls explain.(1 vote)
- It has 2 electrons each in s and 2 p. Try drawing the electronic configuration of Carbon. It'll look something like, 1s² 2s² 2px² 2py² 2pz0. Or use VSEPR theory formula. It's pretty straightforward.(1 vote)
Here we have the cyclopentadiene molecule. And let's analyze it to see if it fits the criteria to be aromatic. And so we're going to start with the first criteria. So does it contain a ring of continuously overlapping p orbitals. Well, if we analyze these carbons here, this carbon is double bonded, so it's sp2 hybridized. And therefore has a free p orbital. Same with this carbon. And same with these other two, here. So those four carbons are sp2 hybridized. If I look at this carbon, however, it's sp3 hybridized. That's a little bit easier to see if I go ahead and draw in some hydrogens on there. So since this carbon right here has four single bonds to it, it is sp3 hybridized, which means that carbon does not have a p orbital. And so cyclopentadiene is not aromatic. It violates the first criteria. It does not contain a ring of continuously overlapping p orbitals. Since it violates the first criteria we can go ahead and say that cyclopentadiene is non-aromatic. So you don't even need to worry about the second criteria. It doesn't fulfill the first criteria for something to be aromatic. However, cyclopentadiene has an interesting property. It's extremely acidic for a hydrocarbon. It actually has a pka of approximately 16 for one of these two protons here that I drew in yellow. And so, there must be some sort of stability associated with the conjugate base in order for cyclopentadiene to be so acidic. And so, if we think about a base coming along, so maybe a lone pair of electrons, a negative charge, just some generic base. It's going to take this proton right here, leaving these two electrons behind on that top carbon. So if we go ahead and draw the structure of the conjugate base, we have our pi electrons here. And now we have a lone pair of electrons on our top carbon, which makes this negatively charged. So here's the conjugate base. And there must be some sort of stabilization associated with this conjugate base because of the fact that cyclopentadiene is so acidic. Remember, the more stable the conjugate base, the more acidic the compound. And so, in this case, the extra stability is associated with the fact that this ion is aromatic. And it might not look like that, because if we look at this top carbon, this top carbon still looks like it's sp3 hybridized because it's a carbanion. However, that lone pair of electrons can participate in resonance. So if we take this lone pair of electrons and we move them in here, and that would kick these electrons in here off onto this carbon. If we draw one of the possible resonance structures for this. I'm going to go ahead and draw and show that these electrons have now moved into here. These electrons are now off on this carbon right here. So it's a negative 1 formal charge. And then we have these pi electrons over here like that. Now, if we analyze this top carbon here, now we can see that it's actually sp2 hybridized. So in this resonance structure, it looks like it's sp2 hybridized. And that lone pair of electrons here in magenta is now occupying a p orbital because it is participating in resonance. And so there are actually many more resonance structures you could draw. We're not going to do that here in this video. Here, I'm just trying to point out that the electrons in magenta, the lone pair of electrons in magenta participate in the resonance. And therefore, that lone pair of electrons is actually delocalized and occupying a p orbital in the ring. And so now we have a ring of continuously overlapping p orbitals. So over here on the left, we'd already said that these carbons are sp2 hybridized. And if you think about these resonance structures, you can have all those carbons in that ring are now sp2 hybridized. And so you fulfill the first criteria. You have a ring of continuously overlapping p orbitals. Let's analyze this anion a little bit more using our frost circles. So if I go ahead and sketch in the fact that this is our anion, with a negative of formal charge. When I'm looking for pi electrons in this molecule I will use this color here. So here are 2 pi electrons, and then 4 pi electrons. And then we've seen that this lone pair participates in resonance. So this lone pair of electrons occupies a p orbital. And those are actually pi electrons now. So we have a total of 6 pi electrons. We have 6 pi electrons for this anion. And we've now seen that all of these carbons in here are sp2 hybridized when you draw all of your resonance structures. And so each one of those carbons has a p orbital. And so I can sketch in a p orbital on this diagram. So we have five carbons in the ring. Each carbon has a p orbital. And so there are five p orbitals in this ring. And you can see that those p orbitals in the ring can overlap to delocalize those electrons. So the overlap of those p orbitals satisfies our first criteria. It contains a ring of continuously overlapping p orbitals. I have a total of five p orbitals. And I know that those p orbitals are atomic orbitals. So five atomic orbitals, according to MO theory, are going to give me five molecular orbitals. And I can represent those molecular orbitals in terms of their energy on my frost circles. I'm going to go ahead and draw this line in here. When you're doing frost circles, remember to always start at the bottom here. A five-membered ring means I'm going to try to inscribe a five-sided polygon in my circle. So I'm going to try to draw pentagon in here. So let's see if we can do it. So we'll put those lines in here like that. And then you can see that here is my pentagon inscribed in my circle. Once again, the important thing is where that polygon intersects with my circle. That represents the energy level of my molecular orbitals. And so you can see, I'm going to have three bonding molecular orbitals. The ones below the center line. And two antibonding molecular orbitals like that. I need to fill my molecular orbitals with my 6 pi electrons. And so I go ahead and put in my 6 pi electrons like that. And you can see that I have filled my bonding molecular orbitals analogous to having a full outer shell. And I have Huckel's rule. I have 4n plus 2 pi electrons. And once again, we can see that using this energy diagram here would be the two electrons. And then 4 times 1. So n is equal to 1 here. Gives me a total of 6 pi electrons. And so this ion satisfies both criteria. It contains a ring of continuously overlapping p orbitals. It also has 4n plus 2 pi electrons in the ring. And so this conjugate base up here is stable because it's an aromatic anion, which is the reason for such a low pka value for cyclopentadiene. So even though cyclopentadiene itself as non-aromatic, this ion over here turns out to be aromatic which explains the stability. Let's analyze one more ion here. So here we have the cycloheptatrienol cation. And if I'm looking for my pi electrons, I have 2, 4, and 6 pi electrons in this ion. So a total of 6 pi electrons. When I'm looking for sp2 hybridized carbons, everything that has a double bond. These are all sp2 hybridized. And then my carbocation right here at this carbon is also sp2 hybridized. So all seven carbons are sp2 hybridized, which means that each one of those carbons has a p orbital. And so a little bit difficult to sketch in here. But you can see that each one of my seven carbons has a p orbital, which makes it relatively easy for those orbitals to overlap in my ring to delocalize those pi electrons. And so I satisfy my first criteria, a ring of continuously overlapping p orbitals here. So a total of seven p orbitals. So seven atomic orbitals. And once again those seven atomic orbitals give me seven molecular orbitals, which I can represent using my frost circle. So I go over here to my frost circle and I draw my dividing line between my bonding and my antibonding molecular orbitals. And now since I have a seven-membered ring, I have to attempt to sketch in a seven-sided polygon into my frost circle. And so this is probably one of the trickiest ones to draw here. And so let's see if we can do something that approximates a seven-sided figure here inside of my circle here. So something like this gives me a seven-sided figure. And once again we care about the points of intersection because that represents the energy levels of my molecular orbitals. And so now it's easy to see that we have three bonding molecular orbitals. And four antibonding molecular orbitals like that. And we're going to fill 6 pi electrons in our diagram. So once again, electrons fill the lowest orbitals first. And so we can take care of our 6 pi electrons like that. And you can see that we've filled all of our bonding molecular orbitals. So that's the extra stability that goes along with an ion being aromatic here. So n is equal to 1. So 4 times 1 gives me 4, plus the 2, gives me total of 6 pi electrons. And so both criteria have been fulfilled. We have a ring of continuously overlapping p orbitals, which is over here. So this, over here on the left, is our first criteria. So this is our ring. And then we also have 4n plus 2 pi electrons. And so the cycloheptatrienol cation is aromatic. So this is an aromatic cation. We can expect some extra stability associated with it.