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Current time:0:00Total duration:10:01

Video transcript

last video we observed that benzene exhibits aromatic stabilization and when chemists first made cyclooctatetraene which this molecule right here they assumed it would react like benzene because it looks like it has alternating single double bonds and it has a ring and so they just assumed that it would behave like benzene it turns out cyclooctadiene does not react like benzene so benzene did not react the way cyclohexene did up here so cyclohexene will give us a mix of enantiomers when the bromine adds across double bond benzene does not do that cyclooctatetraene does as a matter of fact so it will give you it will give you a mixture of enantiomers and the bromine will add across one of those double bonds it turns out cyclooctadiene isn't even conjugated so it looks like it has alternating single and double bonds in this dot structure but it behaves like isolated double bonds like four different isolated double bonds and it's not aromatic so let's see if we can we can analyze the reasons for for this reaction here and so we're going to just real quickly review the criteria to determine if a compound is aromatic or not so a compound or ion is aromatic if it contains a ring of continuously overlapping P orbitals and also has 4n plus 2 where n is an integer pi electrons in the ring which is hookahs rule so here is my cyclooctatetraene molecule and if I count my PI electrons right I can see there are a total of eight PI electrons in this molecule so eight pi electrons and each carbon in cyclooctadiene is sp2 hybridized so each carbon has a free p orbital now the fact that cyclooctatetraene has eight atoms in the rings means that there's a little bit of angle strain in this molecule so one possible conformation for the molecule to adopt is what's called a tub confirmation so you can see like these two carbons can swing up on either side to adopt a tub confirmation now if each carbon and the top conformation is sp2 hybridized right I could draw a p orbital on each carbon in the tub confirmation here so when I do that all right I would get this as a picture now the problem is its difficult for the P orbitals to overlap right so it might be easy for these guys to overlap right here but because of the tub confirmation it'd be hard to get overlap of these orbitals and so it turns out that this is the reason why the molecule acts like it's not conjugated it's because it does have P orbitals but they can't really overlap in the tub confirmation and so this violates the first criteria for a mountain for a compound to be aromatic and therefore we say that cyclooctadiene is non aromatic so it's non aromatic because it violates the first criteria does not have a ring of continuously overlapping P orbitals the or the P orbitals don't really overlap very well in the tub confirmation what if cyclooctadiene adopted a planar conformation so we're just going to pretend like cyclo octet serine adopts a planar conformation down here and once again each carbon is sp2 hybridized so each carbon is going to get ap orbital like that and so we have a total of eight carbons right so eight p orbitals so eight atomic orbitals so let me go ahead and write that here so in a total of eight atomic orbitals for cyclooctadiene and according to mo theory those eight atomic orbitals are going to give us eight molecular orbitals right so the atomic orbitals are going to cease to exist and give us eight molecular orbitals drawing those eight molecular orbitals would be way too much for this video so we're not going to actually draw them but we are going to show where they are in terms of energy using our frost circle and so this is what we saw in the last video so how to draw a frost circle I'm going to go ahead and put the line to the center of my frost circle to help me draw the polygon in here okay so what kind of a polygon do I draw well I have an eight an eight membered ring so I'm going to draw an eight sided polygon and here I'm going to start from the from the bottom here so I'm going to attempt to draw an eight sided figure in here so it'd be something like this now remember when you are drawing your frost circle and your inscribing your polygon the important thing is where the polygon intersects with your circle and every point where the polygon intersects with your circle represents a molecular orbital and so I can see that I would have a total of eight molecular orbitals right because I have eight points of intersection between my polygons and between my circle and again the nice thing about a frost circle is it shows you the relative energies of your eight molecular orbitals all right so I would have three bonding molecular orbitals which would be the ones down here all right so these three points of intersection will give you the relative energies and so I have three bonding molecular orbitals and then up here at the top right I also have three molecular orbitals but these are my anti bonding like in orbitals nose or higher energy and my two points of intersection that are right on the center line here represent two non bonding molecular orbitals like that so when I fill my my molecular orbitals again it's analogous to electron configurations I have a total of eight pi electrons and I need to worry about for a planar cyclooctadiene molecule and so I can go ahead and start to fill in my eight my eight pi electrons like that so that takes care of six of them and I have two more and since this is analogous to electron configurations I'm going to follow hoons rule and not pair up my electrons in an orbital here so that represents my eight my eight PI electrons and since I have unpaired electrons right I have two unpaired electrons that predicts a very unstable molecule if it were to adopt a planar conformation and if I think about in terms of who cools rule right I know it doesn't follow who cools rule huckel's rule is 4n plus 2 where n is an integer and that comes from the 2 comes from the fact that I have this orbital down here right and I do have 4n right here but I don't have 4 n where n is an integer for these two electrons up here and so this is where this this is where it breaks down and so I have a total of eight PI electrons which is not which is not follow who Kools rule and so because the number of electrons is incorrect this molecule is definitely not going to adopt a planar conformation and so cyclooctadiene has a tub confirmation and not planar it is not aromatic it is considered to be non aromatic because of the violation of the first criteria all right but it is possible to react cyclooctatetraene it's possible to oxidize it and so let's see what happens when we do that so if we take cyclooctadiene and we oxidize it so it's going to lose some electrons so I'm going to say that these pi electrons are going to stay and we're going to lose the pi electrons on the left so if I take away a bond from these two carbons right that used to have that double bond there they're going to be positively charged and so I could draw a resonance structure for this so I could move these electrons over here and so if I if I go ahead and show that resonance structure right then this carbon would have a still has a plus one charge and the other positive charge moves over here to this carbon like that and you could continue drawing resonance structures for this molecule and I'm not going to do that I just want to show you that the positive charges are spread out throughout the entire ion here and so one way to represent that would be to just show the electrons are spread out throughout this entire ion and the whole thing has a two plus charge like that and so when I when I analyze this dye cation that I got from cyclooctadiene I realized that all of the carbons are sp2 hybridized right so if I look at these right my carbo cations right those are sp2 hybridized everything with a double bond on it is sp2 hybridized and so I have eight sp2 hybridized carbons and so each one of those sp2 hybridized carbons has ap orbital and and I also have six pi electrons right I have two four and six pi electrons so six pi electrons follows who cools rule and so this ion turns out to be planar all right so it actually does look like this up here and so there's opportunity for those P orbitals to overlap side by side right and so this ion actually fulfills the first criteria right it contains a ring of continuously overlapping P orbitals and when you analyze the second criteria right it has six PI electrons and and so it would fill the bonding molecular orbitals right and no longer has these two electrons up here so as 6 PI lecture it fulfills fulfills huckel's rule right so there's a total of six pi electrons for this dye cation and so this ion is said to be aromatic right so this is an aromatic ion it fulfills both of the criteria for it to be aromatic so it's extra stable and that's the reason why it turns out to be planar because it has eight carbons like cyclooctadiene and so because of that number it it it has some strain that it has to overcome right so it would like to turn into a tub to overcome some that strain but the fact that it's a planar dye cation means that there must be some sort of extra stability in that ion that's forcing it to be planar and that extra stability is due to the fact that it is aromatic and so this ion has been proved to be planar so that's that's this video just an overview of cyclooctadiene why it itself is non aromatic but if you turn it into into a cation it can be aromatic which is further proof for the concept of aromatic stabilization