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Studying for a test? Prepare with these 3 lessons on Resonance and acid-base chemistry.
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Voiceover: Let's look at a few of the patterns for drawing resonance structures, and the first pattern we're gonna look at, is a lone pair of electrons next to a pi bond. And so, here's a lone pair of electrons; I'm gonna highlight it in magenta, that lone pair of electrons is located on this carbon, let me go ahead and put this carbon in green, here. And I'm saying, there's a negative-one formal charge on that carbon in green, and so that carbon in green is also bonded to a hydrogen, so once again, you need to be very familiar with assigning formal charges. So we have a lone pair of electrons next to a pi bond, because over here, we have a double-bond between the carbon and the oxygen, one of those bonds is a sigma bond, and one of those bonds is a pi bond, so I'm just gonna say that these are the pi electrons. So our goal in drawing a resonance structure is to de-localize that negative-one formal charge, so spread out some electron density. And so, we could take the electrons in magenta, and move them into here, to form a double-bond, between the carbon in green and this carbon right here, and that'd be too many bonds to the carbon in yellow, so the electrons in blue have to come off, onto this top oxygen here. So we go ahead, and draw in our brackets, and we put our double-headed resonance arrow, and we draw the other resonance structure, so we have our ring, like that, and then we have, now, a double-bond between those two carbons, and then this top oxygen here, now has only one bond to it. The oxygen used to have two lone pairs of electrons, now it has three, because it just picked up a pair of electrons from that pi bond. So let's go ahead, and follow the electrons. The electrons in magenta moved in here, to form our pi bond, like that, and the electrons in the pi bond, in blue, moved off, onto this oxygen, so I'm saying that they are those electrons. That gives the top oxygen a negative-one formal charge, and so we have our two resonance structures for the enalate anion. We know that both resonance structures contribute to the overall hybrid, and if you think about which one contributes more, for the example on the left, we have had a negative-one formal charge on the carbon in green, so that's a carb anion; and for the resonance structure on the right, we had a negative one formal charge on the oxygen, so that's an oxyanion. Oxygen is more electronegative than carbon, which means it's more likely to support a negative-one formal charge, and so the resonance structure on the right contributes more to the overall hybrid for an enalate anion. All right, let's do another pattern, a lone pair of electrons next to a positive charge, this time. So, let's look at nitromethane, and we could look at this lone pair of electrons here, on this oxygen, and that lone pair of electrons is next to a positive charge; this nitrogen has a plus one formal charge on it. And, so, let's think about drawing the resonance structure, so our goal is to de-localize charge, to spread charge out. We could take the electrons in magenta, and move them into here, to form a double-bond between the nitrogen and the oxygen, but that's too many bonds to this nitrogen; that would give us five bonds to that nitrogen, which we know doesn't happen, because of nitrogen's position on the periodic table. So, that means that the electrons in this pi bond here, are gonna come off, onto the oxygen so these electrons in blue, come off, onto this oxygen, and we draw our other resonance structure for nitromethane, so we have a CH three. We now have a double-bond between nitrogen and this oxygen; this oxygen used to have three lone pairs of electrons, but the electrons in magenta moved in here, to form this bond, and so that means we have only two lone pairs left, on this oxygen. For the oxygen on the bottom-right, there's only one bond now, between the nitrogen and the oxygen, because the electrons in blue moved off, onto this oxygen, and that means this oxygen has two more lone pairs of electrons. And so, when we go ahead and put in our resonance bracket here, you always need to think about assigning formal charge, so what happened to the charge? Well, this oxygen now, has a negative-one formal charge, and this nitrogen still has a plus-one formal charge, so we've de-localized that negative charge; it's actually over both of those oxygens. And notice that the overall charge for nitromethane is zero, for both resonance structures. So we have one positive charge and one negative charge on the left, so that gives us zero; and we have one positive charge and one negative charge on the right, so that gives us zero: So conservation of charge. All right, let's do another example for a pattern that we might see. So, for this one, we have a positive charge next to a pi bond, so let's look at this carbon. So I'm saying it has a plus-one formal charge, and if it has a plus-one formal charge, it must have only three bonds, and since it's already bonded to another carbon, so it's already bonded to- Let me go ahead and label these. So the carbon in yellow there is bonded to this carbon in green, because it has a plus-one formal charge, it must have only two other bonds, and so those must be to hydrogen. So I draw in those hydrogens. So now, it make a little more sense why it's a plus-one formal charge; it be four minus three, giving us plus one. The carbon in green has a formal charge of zero, so it already has three bonds, so it needs one more, two hydrogen, and let's go ahead and make this carbon, over here, in red, already has two bonds, it has a formal charge of zero, so it needs two more hydrogens. So, once again, our pattern is a positive charge next to a pi bond, so let me go ahead and highlight these things here, so we have a positive charge, next to a pi bond, and so here, let's say this one is our pi bond like that. So, when you're drawing resonance structures, again, your goal is to de-localize that charge, and so we could spread out that positive charge by taking the electrons in blue, the pi electrons, and moving them into here. So let's draw the resonance structure. So, we now have, let's see, we would now have a double-bond between the two carbons on the right. The hydrogens haven't moved, right, so I'm gonna leave those hydrogens in there, so there's still one hydrogen on the carbon in the middle, two hydrogens on the carbon in the right, and two hydrogens on the carbon on the left. So the electrons in blue moved to here, like that, so let me go ahead and highlight those carbons. So the carbon in green, right here, and the carbon in red. So what happened to the plus-one formal charge? Well, you can see that it's actually moved to the carbon in the red; the carbon in red right here, has only three bonds, so four minus three gives us a plus-one formal charge. And, let's go ahead and finish our resonance bracket here, so I put that in, and so when you're doing this for cations, you're not gonna move a positive charge, so when you're drawing your arrows, you're showing the movement of electrons, so the arrow that I drew over here, let me go ahead a mark it in magenta. So this arrow in magenta is showing the movement of those electrons in blue, and when those electrons in blue move, that creates a plus-one formal charge on this carbon, and so don't try to move positive charges: Remember, you're always pushing electrons around. Then finally, let's do one more. So, for this situation, this is for acetone, so we have a carbon right here, double-bonded to an oxygen, and we know that there are differences in electronegativity between carbon and oxygen: Oxygen is more electronegative. So what would happen if we took those pi electrons? Let me go ahead an highlight those; I've been using blue for pi electrons, so these pi electrons right here, and we move those pi electrons off, onto the more electronegative atom, like that, so let's go ahead and draw our resonance structure. So this top oxygen would have three lone pairs of electrons: one of those lone pairs are the ones in blue, those pi electrons; that's gonna give the oxygen a negative-one formal charge, and we took a bond away from this carbon, so we took a bond away from this carbon, and that's going to give that carbon a plus-one formal charge. And so, when you think about your resonance structures, first if all, I should point out that one negative charge and one positive charge give you an overall charge of zero, so charge is conserved, and over here, of course, the charge is zero. So if you're thinking about the resonance hybrid, we know that both structures contribute to the overall hybrid, but the one on the right isn't going to contribute as much, so this one on the right is pretty minor, and that's because you have a positive and a negative charge, and the goal, of course, is to get to overall neutral. But, what's nice about drawing this resonance structure, and thinking about this resonance structure, is it's emphasizing the difference in electronegativity, so, for this one, you could just say oxygen get a partial negative, and this carbon right here, gets a partial positive. So that's one way of thinking about it, which is very helpful for reactions. But drawing this resonance structure is just another way of thinking about, emphasizing the fact that when you're thinking about the hybrid, you're thinking about a little more electron density on that oxygen. All right, so once again, do lots of practice; the more you do, the better you get at drawing resonance structures, and the more the patterns, the easier the patterns become.