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Resonance structures for benzene and the phenoxide anion

Examples of how to draw resonance structures for molecules with aromatic rings. Created by Jay.
Video transcript
Voiceover: Here is the dot structure for benzene C686 and we can draw a resonance structure for this. We could take these electrons right here. Move them over to here. That would mean too many bonds to this carbon. We have to take these electrons and push them to here, which would mean too many bonds to this carbon and so finally, we take these pi electrons and move them over to here. We draw our resonance brackets and go ahead and draw our other resonance structure for benzene. The electrons move over to here, to here and then finally, to here. And so, let's follow those electrons. Let's make the ones on the top left here red. These electrons in red. I'm showing them move to over here and let's make these electrons over here green. The electrons in green move down to here and then finally, we'll use blue. These electrons in blue. I showed them moving over to here and remember that the actual benzene molecule is a hybrid of these two resonance structures. If you're just drawing on a sheet of paper, you could use one or the other but remember that it's actually the hybrid because our dot structures are just not perfect ways to represent molecules or ions. Those are our resonance structure for benzene. Next, let's look at the phenoxide anion. Here is the phenoxide anion down here and I'm gonna try to color code the electrons. Let me go ahead and make these electrons in here, red and let's make these right here, green and then let's make these blue. Just like we did with the benzene ring up above and I could start off for the resonance structures for the phenoxide anion by doing the other resonance structure just like we did for benzene like that but I'm gonna save that for the end and so, let's think about what we would do first. We know one of our patterns is a lone pair next to a pi bond and that's what we have here. If you think about a lone pair of electrons on this oxygen. I'll make it magenta. That lone pair is next to the pi bond. The one in red and so, we can go ahead and draw a resonance structure and we take these electrons in magenta and move then into here. That would mean too many bonds to this carbon. We take the electrons in red and we push them off onto this carbon. Let's go ahead and draw our resonance structure. We have our ring here and we have now a double bond between the oxygen and the carbon. Only two lone pairs of electrons on this oxygen now. The electrons in magenta move in here to form a pi bond and the electrons in red move off onto this carbon right here. That's gonna give that carbon a -1 formal charge. Let's go ahead and draw a -1 formal charge here. The electrons in blue have not moved and the electrons in green, I haven't showed moving yet either. We put those in there like that. All right, next, we have the exact same pattern that we did before. We have a lone pair of electrons next to a pi bond. The lone pair of electrons are the electrons in red right here. Next to a pi bond, the electrons in blue. Let's go ahead and draw another resonance structure. We could take these electrons in red, push them into here. That would mean too many bonds to this carbon. If you take these electrons in blue and push them off onto this carbon. Let's draw that resonance structure. Once again, we have our carbon double bonded to an oxygen up here. We said that these electrons were the ones in magenta and the electrons in red moving here to form a pi bond. The electrons in blue move off onto this carbon and that gives this carbon a -1 formal charge. This carbon has a -1 formal charge. This one right here the one that has the blue electrons on it. We still have our electrons in green over here and we have the exact same pattern. We have a lone pair next to a pi bond. The lone pair are the ones in blue and this time, the pi bond are the electrons in green here. We can draw yet another resonance structure. We could take the electrons in blue, move them into here. That would mean too many bonds to this carbon. To take the electrons in green and push them off onto that carbon and let's draw that resonance structure. Once again, we have our ring. We draw our ring in here. We have this double bond up here. Put in lone pairs of electrons on the oxygen and these electrons were the ones in magenta and we go around the ring. We had our electrons in red right here. The electrons in blue move into here and finally, the electrons in green move off onto this carbon. This carbon right here in green. Therefore, that carbon gets a -1 formal charge now. Thinking about it again, we once again have a lone pair of electrons next to a pi bond. We have the electrons in green. Those electrons could move into here and then that would mean too many bonds to this carbon. We can take these electrons in magenta and push them off onto the oxygen. Let's go ahead and draw our last resonance structure here. We have now a single bond to this top oxygen and three lone pairs of electrons giving that top oxygen a -1 formal charge. The electrons in magenta, let's say that those electrons are these electrons right here. Going around our ring, we had electrons in red. We had electrons in blue right here and then finally, the electrons in green move into here. We have five total resonance structures for the phenoxide anion. I can go ahead and put brackets around all five of these and since we're talking about resonance structures in the benzene ring, we can think about going back and forth between these two as a final thought here. We could take these electrons in red, push them into here, which would take these electrons In green over to here, which would take those electrons in blue over to here and then that would give us the one that we started with as well. It's also important to think about the hybrids. The hybrid has the negative charge delocalized. The negative charge is delocalized over. We could see in this resonance structure and this one, the negative charge is on the oxygen and this one, it's on this carbon. And this one is on this carbon and this one is on this carbon. The negative charge is delocalized over oxygen and three carbons when you're thinking about the resonance hybrids.