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Current time:0:00Total duration:10:31

Video transcript

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 right 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 hydrogen's on there so since this carbon right here has four single bonds to it it is sp3 hybridized which means that that carbon does not have ap orbital and so cyclo pentadiene is not aromatic it violates the first criteria does not contain a ring of continuously overlapping P orbitals since it violates the first criteria we can go ahead and say that cyclo pentadiene is non aromatic so you don't even need to need to worry about the second criteria right 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 and actually has a pKa of approximately 16 for one of these two one of these two protons here that I that I drew in yellow and so there must be some sort of stability associated with the conjugate base an order for cyclo pentadiene to be so acidic and so if we think about a base coming along so I'm giving 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 right we have of 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 of the fact that cyclopentadiene is so acidic remember the more stable the conjugate is the more acidic the compound and so in this case the the extra stability is associated with the fact that this ion is aromatic and it might not look like that because if we if we look at this top carbon right this top carbon still looks like it's sp3 hybridized because the carb anion however that lone pair of electrons can participate in resonance right 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 right if we draw one of the possible resonance structures for this so I'm gonna go ahead and draw and show that these electrons have now moved into here all right 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 or if you 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 right this lone pair of electrons in here in magenta is now occupying AP orbital because it's participating in resonance and so there are actually many more resonance structures you can draw and 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 resonance and therefore that lone pair of electrons is actually delocalized and occupying AP orbital in the ring and so now we have a ring of continuously overlapping P orbitals right so over here on the left we'd already said that these that these carbons are sp2 hybridized and if you think about these resonance structures you could 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 alright so if I go ahead and sketch in the fact that this is our anion right with a negative 1 formal charge when I'm looking for pi electrons right I'm looking for PI electrons in this molecule so I will 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 ap orbital and those are actually PI electrons now so we have a total of six pi electrons right we have six pi electrons for this anion and we've now seen that all of these carbons right all these carbons in here are sp2 hybridized when you draw all of your resonance structures and so each one of those carbons has ap orbital and so I can sketch in a p orbital on this diagram alright so we have 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 right to D localize 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 a 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 right 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 right the ones below the center line and two anti-bonding molecular orbitals like that I need to fill my molecular orbitals with my six pi electrons and so I go ahead and put in my six pi electrons like that and you can see that I have I have filled my bonding molecular orbitals analogous to having a full outer shell and I have Google's rule I have 4n plus 2 pi electrons right and once again we can see that using using 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 right contains a ring of continuously overlapping P orbitals it also has 4n plus 2 pi electrons in the ring and so this conjugate base this conjugate base up here is stable because it's an aromatic anion which is the reason for such a low PKA value for cyclo pentadiene so even though cyclopentadiene itself is non aromatic right this ion over here turns out to be it turns out to be aromatic which explains the stability let's analyze one more one more ion here so here we have the cyclohexanol cation and if I'm looking for my my PI electrons right 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 right everything that has a double bond these are all sp2 hybridized and then my carbo cation right here this carbon is also sp2 hybridized so all seven carbons are sp2 hybridized which means that each one of those carbons has ap orbital and so a little bit difficult to sketch in here but you can see that each one of my seven carbons has ap orbital which makes it which makes it relatively easy for those those orbitals to overlap in my ring to D localize those pi electrons and so I satisfy my first criteria a ring of continuously overlapping P orbitals here so a total of 7 P orbitals so 7 atomic orbitals and once again those 7 atomic orbitals give me 7 molecular orbitals which I can represent using my frost circle so I go over here to my frost circle I draw my dividing line between my bonding and my anti-bonding molecular orbitals and now since I have a seven membered ring I have to attempt to sketch in a seven sided Fink a seven sided polygon into my frost circle and so this is probably one of the trickiest one of the trickiest ones to ah here and so let's see if we can do something that approximates a seven sided figure inside of my circle here so something something like this all right gives me a seven sided figure and once again we care about the points of intersection right 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 anti bonding molecular orbitals like that and we're going to fill six pi electrons in our diagram so once again electrons fill the lowest orbitals first and so we can take care of our six pi electrons like that and you can see that we've filled our all of our bonding molecular orbitals that's the extra stability that goes along with with an eye on being aromatic here right so n is equal to one alright so four times one gives me four plus the two gives me a total of six PI electrons and so both criteria have been fulfilled right we have a ring of continuously overlapping P orbitals which is over here alright so so that's this is our over here on the left is our first criteria alright so this is our ring and then we also have 4n plus 2 pi electrons and so the cycle will have to try anal cation is aromatic so this is an aromatic aromatic cation and we can expect some extra stability associated with it