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:12:09

Video transcript

in the last video we use the criteria for aromaticity to see that heterocycles can be aromatic too in this video we're going to look at more aromatic heterocycle specifically five membered rings and so we'll start with pure all right down here so pure all molecule as you can see five atoms in the ring and if we take a look at the carbons in the ring we can see that those carbons all have a double bond to them therefore each of those carbons is sp2 hybridized meaning each of the carbons has a free p orbital so i can go ahead and draw a free p orbital on each of those four carbons like that in terms of the nitrogen on the ring I need to know the hybridization state of this nitrogen so the best way to do that is to calculate the steric number so we know the steric number is equal to the number of Sigma bonds plus number of lone pairs of electrons and so I can see that here is a sigma bond here is a sigma bond and here is a sigma bond so three sigma bonds plus lone pairs of electrons there's one lone pair of electrons on that nitrogen there so the steric number should be equal to four which would imply an sp3 hybridization state for puerile so we know that's not the case because P R all is an aromatic molecule an sp3 hybridized nitrogen would mean no P orbitals at that nitrogen which would violate the first criterion for this compound to be aromatic and so there must be some way to get that nitrogen to be sp2 hybridized and of course we saw how to do that in the end of the last video this lone pair of electrons on this nitrogen is actually not localized to this nitrogen we can take this lone pair of electrons and move them in here so that lone pair of electrons can participate in resonance so if those lone pairs of electrons move into there to form a PI bond that would kick these electrons off onto this carbon so the resonance structure will have nitrogen with a PI bond here now and a lone pair of electrons on this carbon which would give this carbon a negative 1 formal charge we still have we still have a PI bond over here like that and this nitrogen now would have a plus 1 formal charge now when we analyze the hybridization state of this nitrogen we can see that once again we're going for Sigma bonds so one single bond to Sigma bonds three sigma bonds so three single bonds this time no lone pairs of electrons because that lone pair of electrons is now delocalized in resonance and so three plus zeros of course three meaning that this nitrogen is now sp2 hybridized since that nitrogen is sp2 hybridized it has a free p orbital so we can go ahead and draw the p orbital on that nitrogen and so you could think about you could think about in terms of dot structure right so these two these two electrons over here alright these two electrons are actually delocalized and participate in resonance so that lone pair of electrons you could think about as occupying AP orbital here and they're actually delocalized right so we have all these all these pi electrons do you localize throughout our ring and so let's go ahead and check the the criteria for aromaticity so p row does contain a ring of continuously overlapping P orbitals and it does have 4n plus 2 pi electrons in that ring so let's go ahead and highlight those right so we had these pi electrons so that's to these pi electrons so that's four and then these pi electrons here and magenta are actually delocalized in the ring so that gives us six pi electrons so if n is equal to one four times one plus two gives me six pi electrons so pyrrole has six pi electrons and also has a ring of continuously overlapping P orbitals so we can say that it is aromatic let's go ahead and look at another molecule here so similar to it this is emitters all and so for emitters all once again we have the same sort of situation that we had for P R all with this nitrogen right here so at first it looks like that nitrogen might be sp3 hybridized but we can draw a resonance structure for it so we can take these electrons in here and move them in here which you kick these electrons off onto that top nitrogen so let's go ahead and draw the resonance structure for that so we would have some PI electrons here we would have a double bond here we would now have a plus 1 formal charge on this nitrogen and we would have a negative 1 formal charge on this top nitrogen right here like that so once again we've seen that this nitrogen in blue it's actually SP two hybridized right so let me go ahead and highlight it in blue over here on the right this nitrogen is actually sp2 hybridized so it's the exact same situation that we saw in pure all which at first it looks like those electrons might be localized to that nitrogen those electrons are actually not those electrons are delocalized in the ring because of the possible resonance structure and so we've determined that that nitrogen in blue that's bonded to that hydrogen is sp2 hybridized so over here I can go ahead and identify that nitrogen in blue that's bonded to the hydrogen it's sp2 hybridized therefore it has ap orbital like that so I can go ahead and draw the p orbital on that nitrogen if i look at the carbons if i look at the carbons in this molecule alright so we'll go ahead and highlight these carbons in red here so these three carbons of my original dot structure we can see they have a double bond to them so those are all sp2 hybridized so I can go ahead and draw in the p orbital on those carbons as well and then once again going back to the original dot structure and this time looking at the nitrogen so I'm go ahead and draw this one in magenta so this nitrogen right here if I look at this first dot structure it's just like the example in pyridine that we saw in the last video right so it's actually sp2 hybridized and since it's sp2 hybridized we could think about these electrons in here and magenta as participating resonance and the electrons in blue here as being localized to that nitrogen atom localized to an sp2 hybridized orbital and so I can go ahead and identify this nitrogen in magenta over here so I'm saying this nitrogen is in magenta I can say that it is sp2 hybridized it's the exam situation as pyridine so I can go ahead and draw a orbital on it and that lone pair of electrons in blue that's on that nitrogen that lone pair of electrons occupies and go ahead and put them in blue that lone pair of electrons is going to occupy an sp2 hybridized orbital on that nitrogen so those electrons are not involved in resonance those electrons are localized to that nitrogen and so when we go ahead and think about the criteria for aromaticity alright our first criteria are alright we have a we have overlapping P orbitals here and so I can see that that is the case right we have overlapping P orbitals everything is sp2 hybridized in our ring and of course we also need 4n plus 2 pi electrons and so let me go ahead and label those in magenta so my PI electrons right going back to my original dot structure here's 2 right here's 4 and then this lone pair on our nitrogen right those are actually those are actually PI electrons we think about this resonance structure here some what I'm going to go ahead and label them in magenta again so there actually are six PI electrons in the emitters all molecule and those pi electrons are delocalized around this ring around these overlapping P orbitals and so the emitters all molecule is aromatic as well so the emitters all molecule is actually extremely important in biochemistry so let's take a look at a famous molecule that contains the emit assault ring and this molecule is called histamine which anyone who has allergies has heard of histamine and you can see the emitters all ring over here and the left on the histamine molecule so if you want to understand biochemistry it's very useful to understand these concepts found in organic chemistry and so histamine would be an example of a biological aromatic heterocycle this portion of the molecule the emitters all ring is aromatic it satisfies the two criteria as we have seen so let's do let's do one more example of an aromatic heterocycle here this time let's look at an example that has sulfur in it so this molecule is called thiophene and this entire fiat and and this is really the sulfur analog to the Piro Piro molecule that we studied so let's start by analyzing this sulfur in terms of its steric number all right so I'm going to look at this sulfur right here I'm going to calculate the steric number of that sulfur so number of Sigma bonds to that atom so here's a sigma bond and here's a sigma bond so we have steric number is equal to number of Sigma bonds plus number of lone pairs of electrons so we have two lone pairs of electrons around that sulfur like that so that would be 2 plus 2 which is equal to 4 which implies that that sulfur is sp3 hybridized but that doesn't work for our concept of aromaticity because in order for something to be aromatic our atom needs to be sp2 hybridized so it has a free p orbital just like just like these carbons right here so once again these carbons all have a double bond to them so those carbons all have AP orbital so I can go ahead and sketch in those those P orbitals on those carbons like that so this the sulfur looks like it's sp3 hybridized but of course it's going to be a similar example to the pure all molecule I can show a resonance structure for this die open molecule I can take one of these lone pairs of electrons I'm going to say it's the lone pair on the right here and I could show them moving in to form a PI bond between the sulfur and this carbon we should of course kick these electrons off onto that carbon and so when I go ahead and draw this resonance structure alright so now there's a PI bond between the sulfur and this carbon lone pair of electrons moved out onto this carbon giving that carbon a negative 1 formal charge there was still a lone pair of electron I should say over here and there was still a lone pair of electrons left on this sulfur so now we can also give this sulphur a +1 formal charge and we can analyze it in terms of its steric number so once again number of Sigma bonds so this is a sigma bond this is a sigma bond and so the steric number will be equal to 2 plus how many lone pairs of electrons now there's only 1 so 2 plus 1 is equal to 3 so now we can see that it's actually sp2 hybridized so it has three sp2 hybrid orbitals and one of those sp2 hybrid orbitals is going to contain that lone pair of electrons in blue there so that lone pair of electrons and again it's sp2 hybridized which means has two other sp2 hybrid orbitals and those hybrid orbitals are forming bonds with those carbons there since it's sp2 hybridized it also has an unhybridized p orbital p orbital so I can go ahead and draw in that unhybridized p orbital and so I could think about one of those lone pairs of electrons on my original dot structure occupying a P orbitals let's go back here and look at look at the thiophene original dot structure I drew I could think about one of these lone pair of electrons I'm going to mark it in magenta as occupying ap orbital so I can go ahead and put it in there like that and I can think about the other lone pair of electrons on that dot structure as occupying an sp2 hybridized orbital out to the side so whenever you see this sort of situation think about where those lone pairs of electrons actually are and so I also have these as being PI electrons and these as being PI electrons so 2 4 6 a total of six pi electrons which is who cools number D localized throughout our ring throughout our overlapping P orbitals and so we can say that the thiophene molecule is aromatic it satisfies both of the criteria for this and so there are of course analogs to thiophene using oxygen and that's of course a similar situation and you can draw other resonance structures but I just wanted to draw this one to show that one of those lone pairs is actually delocalized and involved in resonance and one of those lone pairs is actually localized to the sulfur atom and localized in an sp2 hybridized orbital so this is how to analyze aromatic heterocycles