Resonance structures and hybridization

Voiceover: Let's look at the amide, or "amid" functional group, and let's start by assigning a steric number to this nitrogen. So, the steric number is equal to the number of sigma bonds; so here's a sigma bond, here's a sigma bond, and here's a sigma bond; so three sigma bonds. Plus number of lone pairs of electrons, so there's one lone pair of electrons on that nitrogen, so I'll go ahead and highlight them there. So three plus one gives us a total of four for the steric number, which means four hybrid orbitals, which implies SP three hybridization for that nitrogen, and from earlier videos, you know that SP three hybridization means, a trigonal, pyramidal, geometry for that nitrogen. And so, that's one way of looking at this function group, and that lone pair of electrons being localized to that nitrogen; however, now that we know resonance structures, we know that, that lone pair of electrons is not localized to that nitrogen; it's de-localized in resonance. So we could take that lone pair of electrons, and move it in here, to form a pi bond, so that would force us to push some pi electrons off, onto this oxygen, so let's go ahead and draw the other resonance structure. So this top oxygen would now have three lone pairs of electrons around it, a negative one formal charge, and there'd be a double-bond between this carbon and this nitrogen, so let's go ahead and draw in everything. This nitrogen how has a plus one formal charge, and we can go ahead and complete our resonance bracket here. So let's follow those electrons along, the electrons in magenta, the lone pair of electrons moved in here to form our pi bond, and the pi electrons over here in blue, came off onto the oxygen, to give the oxygen a negative one formal charge. All right, let's now calculate a steric number for this nitrogen, in our second resonance structure. So, steric number is equal to number of sigma bonds: So here's a sigma bond, here's a sigma bond, and our double-bond; we know one of them is a sigma bond, and one of them is a pi bond. So we have a total of three sigma bonds around our nitrogen, zero lone pairs of electrons, so three plus zero gives us a steric number of three; that implies three hybrid orbitals, which means SP two hybridization, and a trigonal planar geometry around that nitrogen, so a planar geometry. And, here we've shown our electrons being de-localized, so the lone pair of electrons are de-localized due to resonance, and so, experimental studies have shown that the amide function group is planar, so these atoms actually are planar here, which means that the electrons in magenta are not localized to that nitrogen; they are actually de-localized. And so, that implies this nitrogen is SP two hybridized, and has a P orbital, and that allow that lone pair of electrons in magenta to be de-localized. And so, here's a situation where drawing a resonance structure helps clue us into what's actually happening: That lone pair is participating in resonance, which makes this nitrogen SP two hybridized, so it has a P orbital. All right, let's look at this example down here, and let's look first at this left side of the molecule, and so we can see our amide functional group, and if I look at the lone pair of electrons on this nitrogen, we've just talked about the fact that this lone pair of electrons is actually de-localized, so this lone pair is participating in resonance. And so, that affects the geometry, and how you think about the hybridization of this nitrogen, here. So, the electrons in magenta are de-localized because they participate in resonance, and if I think about, let's make this a different color here; let's make these electrons in here blue, so the electrons in blue on the amine, these electrons have nowhere to go; they can't participate in resonance; that lone pair of electrons in blue is localized to that nitrogen. And so, this is why you can think about an amide being different from an amine, in terms of functional group, and in terms of how they react and how they behave. If we look at another example, so this molecule right here, and we assume the lone pair of electrons on that nitrogen is localized to that nitrogen, let's go ahead and calculate the steric number. So the steric number'd be equal to sigma bonds, so that's one, two, and three; so three sigma bonds. Plus lone pairs of electrons, there's one lone pair of electrons on that nitrogen, so three plus one is four; so four hybrid orbitals, which implies SP three hybridization on that nitrogen. But we know that, that lone pair of electrons is not localized to that nitrogen; that lone pair of electrons is de-localized; it participates in resonance, because we have this pattern here. All right, so this pattern of a lone pair of electrons, in blue, next to a pi bond, which I will make magenta, and so we could draw a resonance structure. So I could take the electrons in blue, move them into here, too many bonds to this carbon, so I push the electrons in magenta off, onto this carbon. So we draw the resonance structure, so I have my ring here, the nitrogen's bonded to a hydrogen, the electrons in blue moved in to form a pi bond, and the electrons in magenta, moved off, onto this carbon right here, to give that carbon a negative one formal charge. All right, let's go ahead and calculate a steric number for the nitrogen here, which gets a plus one formal charge. All right, so the steric number will be equal to number of sigma bonds: So there's one sigma bond, here's another sigma bond, and then, in our double-bond, one of them is a sigma, and one of them is a pi; so I'm saying that's our sigma bond. So our steric number is equal to three, plus the number of lone pairs of electrons, now zero. So that's a steric number of three, which implies three hybrid orbitals, which says SP two hybridization. And since we know that, that lone pair is de-localized, it's going to occupy a P orbital, and so therefore this nitrogen is SP two hybridized, because we know SP two hybridization has three SP two hybrid orbitals, and one P orbital. So that lone pair in blue is actually de-localized; it's occupying a P orbital, and so let's go ahead and draw that, down here, so let's say this is the nitrogen, and you're looking at it, at a bit of an angle. If that nitrogen is SP two hybridized, that nitrogen has a P orbital, so we can go ahead and draw in a P orbital, on that nitrogen. And so, the electrons in blue, since those electrons are de-localized, those electrons are going to occupy that P orbital; those electrons can participate in resonance. Alright, so let's go ahead and draw a hydrogen here, just to finish that off, and I could have kept on going with resonance structures, so I actually forgot to put in these electrons, right here, because I could have kept on going; I could take these electrons in magenta, move them into here, and then push these electrons off, onto that carbon, and I could keep on going. And so there are many more resonance structures that you could draw; I'm not going to do so for time purposes; here I'm just trying to show how resonance affects the hybridization, or how you should think about the hybridization, when you're drawing these imperfect dot structures. Let's go ahead and finish our picture of the molecule, because we know that these four carbons here, all have a double-bond to them. And so those four carbons are therefore SP two hybridized, and if those carbons are SP two hybridized, each one of those carbons has a P orbital, so I can draw in a P orbital on all of my carbons, here. And you could think about those electrons, let me go ahead an highlight them, the electrons in red here, so these pi electrons, so those pi electrons are participating in resonance, and so we have a total of six pi electrons participating in resonance: those four in red that I just highlighted here, and then these electrons here, in blue, which occupy this P orbital. And so this is going to be extremely important, when you talk about aromatic compounds later in the year. So, for right now, just try to identify the skill of de-localized electrons versus localized; in this case the electrons on the nitrogen are de-localized; they participate in resonance. Let's do one more example, so one final example here. All right, and let's go ahead and start by calculating the steric number of this nitrogen, so how I've drawn it here. So, steric number is equal to number of sigma bonds. So here's a sigma bond; I have a double-bond, one of them is a sigma bond. So I have two sigma bonds, plus lone pairs of electrons; there's one pair of lone electrons here, on this nitrogen, so two plus one gives me a steric number of three; that implies three hybrid orbitals, or SP two hybridization. So if that nitrogen is SP two hybridized, that nitrogen's going to have a P orbital, so let me go ahead and sketch in a P orbital here, on this nitrogen, and let's think about the other carbons too, so let's think about these carbons, I should say. So I have one, two, three, four, five carbons; all these carbons have a double bond to them, so all of those carbons are SP two hybridized, so I can draw in a P orbital on all of those carbons. So all of those carbons have three SP two hybrid orbitals, and one un-hybridized P orbital. So, it's the same idea with this nitrogen here: if this nitrogen is SP two hybridized, it has three SP two hybrid orbitals, and one un-hybridized P orbital. I already drew in the un-hybridized P orbital; let's go ahead and put in those SP two hybrid orbitals, and I'll use magenta. So, one of those SP two hybrid orbitals is forming this sigma bond over here; another one's forming this sigma bond back here, and then a third SP two hybrid orbital must be here. And that's actually where that lone pair of electrons is, so this lone pair of electrons here in blue, is localized to this SP two hybrid orbital, so this is actually, let me go ahead, and draw an arrow here. So the lone pair of electrons is actually localized to that SP two hybrid orbital. So that lone pair of electrons can't participate in resonance, because that nitrogen already has a pi bond. So, if I go over to here, these pi electrons are using that P orbital, and since those pi electrons in green are using this P orbital, over here on the right, so this P orbital over here; that means that those electrons in blue can't use it, and so, therefore, those electrons in blue are actually localized to that nitrogen. Which might not be what you first thought because we talked about the pattern, of course, of the electrons in blue here, lone pair of electrons next to a pi bond, so you might think we could draw a resonance structure; however, the reason we can't, is once again, those electrons in green are actually participating in resonance, and so they're already using that P orbital. All right, so this is kind of an exception to what we talked about before, where we said that there's a lone pair of electrons next to a pi bond, that lone pair is de-localized, so this is the one thing to think about.