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Video transcript
In this video, we're going to see how induction and resonance affect the activating strength of substituents on a benzene ring, but before we get to substituents let's real quickly review the mechanism for electrophilic aromatic substitution involving benzene. And so the benzene ring is going to function as a nucleophile so all the pi electrons in the benzene ring are negatively charged, of course, and they're attracted to positively charged things, like an electrophile. And so the benzene ring is going to function as a nucleophile, and some of these pi electrons are going to form a bond with our electrophile. So we get nucleophile attacking the electrophile, and the electrophile is going to add on to our ring. And I'm just going to say the electrophile adds on to this carbon right here, which means that this carbon has a plus 1 formal charge, and that will represent our positively charged sigma complex. So if you could imagine a substituent on our benzene ring that somehow increased the electron density in that ring, that would make that benzene ring even more nucleophilic, and that increased electron density would help to stabilize the positively-charged sigma complex, which means the sigma complex is more likely to form. And so a substituent that increases electron density we could call that an electron donating group, and an electron donating group would activate the ring towards electrophilic aromatic substitution, which means that the overall rate of the reaction will be faster than that compared to benzene. And so we call an electron donating group an activator. So let me go ahead and write an activator here. If we thought about the opposite situation, if we thought about a substituent on the ring that overall decreased the electron density in that ring, the ring would not be as nucleophilic, and you could think about the ring as being a little more positively charged and so that would, of course, destabilized the positively-charged sigma complex. And so a substituent that overall decreased the electron density in the ring, and we could say it's an electron withdrawing group because it's withdrawing electron density from the ring. And that would, of course, deactivate the ring towards electrophilic aromatic substitution, and so we would call this a deactivator, and so the reaction would be slower than that of benzene by itself. Let's see how the concepts of electron donating groups and electron withdrawing groups affect activating strength. And we'll start with strong activators here. So first we can look at the phenyl molecule, and we can think about this carbon on our ring in a sigma bond to this oxygen right here. We know oxygen is more electronegative than carbon, and so the oxygen can withdraw some electron density from the ring by in inductive effect, right? So oxygen being more electronegative, it can pull the electrons through that sigma bond closer to itself. And so it's withdrawing some electron density from the ring because of electronegativity, and so we call this induction or the inductive effect. So there's some induction in this molecule. Now, since it's withdrawing some electron density, you might expect the OH group to be a deactivator, but that's not what we observe. We observe the OH to be a strong activator. And so there must be another effect here to counteract this inductive effect, and, of course, that effect is resonance. So let me go ahead and write resonance right over here. So we can draw several resonance structures for the phenyl molecule, but if you think about this lone pair of electrons on this oxygen, it's right next to our benzene ring. And so this lone pair of electrons can participate in resonance and move in here to form a pi bond, which would push these electrons in here off onto this carbon so we can go ahead and draw a pi bond between our oxygen and our carbon now. There's still a lone pair of electrons on that oxygen, it's still bonded to a hydrogen so it has a plus 1 formal charge. And so we have these pi electrons here and we have these electrons move out on to this carbon, which gives that carbon a negative 1 formal charge. And we could keep drawing more resonance structures, but I'm going to stop there because the point that I'm trying to make is that we get to some donation of electron density to the ring through a pi bond. So let me go ahead and highlight our pi bond here. So these electrons move in here to form a pi bond between that carbon and that oxygen, and so we get overlap of P orbitals between this carbon and this oxygen. So let me go ahead and sketch that really fast. So we have a carbon and we have an oxygen, and we're going to get overlap of P orbitals. Since carbon and oxygen are on the same period on the periodic table, their P orbitals are pretty much the same size, and that means that you get good overlap and therefore, the oxygen is able to donate some electron density to that ring there. So the lone pair of electrons on the oxygen is conjugated into the pi system of that ring, and so that's overall, an electron donating effect, right? You're increasing the electron density in the ring, and so the resonance effect says that the OH group is an electron donating group, which would, of course, make it a strong activator. And that's, of course, what we observe experimentally. And so we can say that the resonance effect beats the inductive effect when you're talking about a strong activator here, so an atom that has a lone pair of electrons next to your benzene ring. Now, the same idea holds true for amylin down here. So once again, thinking about nitrogen compared it to this carbon, that nitrogen is more electronegative and so it's going to withdraw some electron density from the ring via the inductive effect through that sigma bond. So I can think about drawing an arrow showing the movement of electrons towards nitrogen. Now, nitrogen is not as electronegative as oxygen there so it's not really withdrawing quite as much electron density via induction. Since that nitrogen has a lone pair of electrons on it, it can also participate in resonance. And so just like the previous example, I could take this lone pair of electrons right next to the ring, move it into here to form a pi bond to push these electrons in here off onto that carbon. So I could go ahead and show one possible resonance structure here. So I could show my pi bond between my carbon and my nitrogen. The nitrogen would now be positively charged, and we would have a lone pair of electrons on this carbon, which would make that carbon negatively charged. And once again, you could draw more resonance structures, but we just don't have the time for that in this video. And so once again, this lone pair of electrons is actually conjugated into the pi system of the ring right there. And so that's increasing the electron density of the ring, which makes the ring more nucleophilic, stabilized the positively-charged sigma complex, and therefore, it's overall an electron-donating a group, which makes it an activator. And we could also think about the P orbitals of this carbon and this nitrogen so they are also, of course, in the same period on the periodic table so when you sketch in your P orbitals here you can make them pretty much the same size, and so you get good overlap of those P orbitals. Now, amylin is actually even more reactive than the phenyl example, and that's because of the nitrogen being less electronegative than oxygen. And so in the pi bond here since this nitrogen-- let me go ahead and highlight our pi electrons, let's use green this time-- so these pi electrons right here, since nitrogen is less electronegative than oxygen that means that those pi electrons are better able to be conjugated into the pi system of the ring since the nitrogen isn't pulling on them as much as the oxygen is. And so since you get a little bit more electron density donated to your ring because of the nitrogen being less electronegative that means increase electron density makes this a better electron-donating group, which activates the ring better towards electrophilic aromatic substitution. And so in the next video, we're going to use the same concepts of induction and resonance, and we're going to analyze a moderate activator, a weak activator, and we'll also look at an example of a weak deactivator that is still an ortho/para director.