If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:9:27

Video transcript

chuckles rule can only be applied to mono cyclic compounds however there are some polycyclic compounds that seem to have some form of aromatic stability and one of those compounds is naphthalene so over here on the Left we have the dot structure for naphthalene naphthalene is a is a white solid that is traditionally the component of mothballs and so has a very distinctive smell to it now nataline is aromatic however it's not as stable as benzene but we could think about it as two fused benzene like rings and so if I if I were to analyze that flame molecule using our our rules our criteria for aromaticity I could look at each carbon and naphthalene and I could see that each carbon has a double bond to it all right so each carbon is sp2 hybridized and therefore each carbon has a p orbital right so an unhybridized p orbital and so there are a total of 10 carbons in naphthalene and so if I go over here to the drawing on the right each of those carbons has ap orbital so I could draw in the P orbitals on each one of my carbons in here like that now these P orbitals are right next to each other which means they can overlap and so if you think about the criteria for a compound to be aromatic this this this would sort of meet that first criteria there right because you can see that you have overlap of these P orbitals and in this case we have delocalization of electrons across two rings now when we think about the second criteria which was who cools rule in terms of the number of pi electrons our compound has let's go ahead and analyze naphthalene even though technically we can't use Google's rule but if we look at it we can see there are two four six eight and ten PI electrons so naphthalene has 10 pi electrons if you think about Google's rule 4n plus 2 if n is equal to 2 4 times 2 plus 2 is equal to 10 pi electrons and so 10 PI electrons is is a huckel number and so it looks like that flame fulfills a two criteria even though again technically we can't we can't apply who cools Rule to polycyclic compounds but those 10 PI electrons are fully delocalized throughout both rings and one way to show that would be using resonance structures right so if we were to draw a resonance structure for naphthalene I could take these electrons and move them in here and then these electrons will go over here and then these would go over there so that's that would give me another resonance structure so if I go ahead and draw the results of the movement of those electrons right I would now have my PI electrons over here like this so it's like a benzene a benzene like ring on the left and then on the right we still had these pi electrons in here like that now in this case I've shown I've shown these these PI electrons right here I've shown them on the left side but in reality those PI electrons right are above and below our single bond right in terms of the probability of finding those electrons and so I don't have to draw it the way I did it here I could draw I could draw I could draw it like this so let me go ahead and put this is going to be equivalent to this structure so these aren't different resonance structures I'm just drawing a different way of representing that resonance structure over here so I could show those electrons in blue over here on this side like that so let me go ahead and highlight those electrons so electrons in blue are right here and these two drawings are equivalent right after I put in my other electrons right there so the the dot structures are just an imperfect way of representing the molecule right so I could show those PI electrons on the left I could show them on the right it's really the same thing once I draw this picture I'm now able to draw another resonance structure all right so I can draw another resonance structure from this one right here I could move these electrons over here move these electrons over here and then finally move these electrons over here and so when I go ahead and draw the resulting dot structure now I would have see these pi electrons are still here and then going around my ring it would look like this so there are a total of three resonance structures that you can draw for naphthalene again showing the delocalization of those 10 pi electrons and showing you a little bit about why naphthalene does exhibit some aromatic stability it's not quite as aromatic as benzene all Benzies all benzenes bonds have the exact same length but naphthalene is shown to be is shown to have some aromatic stability Natalie this is simplest example of what's called a polycyclic aromatic hydrocarbon and there are several several different examples of poly Saiki of polycyclic aromatic hydrocarbons another example would be something like anther scene and so there there are many many examples of ring systems that contain that contain fused benzene like rings throughout the system but instead of focusing on those I wanted to do another example which is an isomer of naphthalene and it's called a saline and azulene is a is a beautiful beauty carbon which is extremely rare in organic chemistry to have a hydrocarbon that's blue it also has some other interesting properties it has an increased dipole moments associated with the molecule there's also increased electron density on the five membered ring so over here on the Left we have a saline and here's the five membered ring over here on the left it turns out there are more electrons on the five membered ring than we would expect giving it a larger dipole moment and if we think about a possible resonance structure for azulene we can we can figure out why so if I took these pi electrons right here and move them in here that would push these electrons off onto this carbon so if I go ahead and draw the resulting a resonance structure right I would have I would have an ion alright that looks like this so if I think about my formal charges all right if I think about these electrons in blue right here all right there's going to go off onto that top carbon so there's that top carbon is going to get a lone pair of electrons which gives that top carbon a negative 1 formal charge all right so as a negative formal charge on that carbon and then if I think about I think about this carbon over here this carbon lost a bond all right and so that's going to end up with a positive charge so we have a carbo cation right here like that and if I if I analyze this resonance structure it has formal charges in it but if I if I look over on the right I can see I can see on the right there's a seven membered ring on the right and if I look at it I can see there are six PI electrons and then right here I have a carbo-cation and so this is one of the examples we did in the last video right so every carbon is sp2 hybridized and so we have overlapping P orbitals and we have a total of six pi electrons and so this seven membered ring is aromatic if I look over here on the left I can see that I have a five membered ring and I have some PI electrons right here and the PI electrons right here as we saw an example of naphthalene are are actually being shared by both rings so I can I can pretend like those electrons are right here on my ring okay and then this ring also has electrons like that with a negative 1 formal charge and again in the last video we saw that this ion is aromatic and it's total of six PI electrons right so go ahead and highlight those so these these and these are all PI electrons when you think about resonance structures and so six PI electrons and all the carbons turn out to be sp2 hybridized again look at the previous video for much more detailed explanation as to why these two ions are aromatic and so since these ions are aromatic they have some aromatic stability and this resonance structure over here on the right is a much greater contributor into the to the overall picture of the molecule and the negative charge is delocalized throughout throughout this five membered ring over here and the positive charge is delocalized or spread out throughout this seven membered ring and that is what gives azulene its larger dipole moment right so there's a larger dipole moment and as lean than expected because of the fact that this would give us two aromatic rings right which which confers of course extra stability and so once again this is this this ion down here was the cyclopentadienyl anion and then this cation over here was the cyclo help tetragonal cation from the previous video the redistribution of these electrons allows azulene to absorb in the orange region right which is difficult for for most organic molecules because it's a longer wavelength and that allows it to reflect in the blue region which is again rare especially for a hydrocarbon and the fact that it's blue is where this part of the name comes comes in there like a zoo or as in blue so these are this does just two examples of some ring systems that also exhibit some form of aromatic stability