 So, we know that organic acids HA dissociate into H plus and A negative. A negative is what we call our conjugate base. I'm just going to write CB. Our organic acid dissociates into H plus and A negative mainly because of two reasons. First, that HA bond is not that strong and the other reason that A negative, our conjugate base, is stable after dissociation. Alright. As our conjugate base becomes more and more stable, our HA molecule becomes more and more acidic. Let's talk about our A negative stability. The idea is that we'll be comparing the stability of A negative of two molecules, molecule one and molecule two, which of these molecule is more acidic. As the stability of A negative increases, the acidity of HA increases as well. Now, because our A negative is an anion, we need to stabilize our anion as much as possible. In our previous video, we have studied about something called electronic effects. In this video, we'll be checking the stability of our anions or our conjugate base using these electronic effects because an anion can be stabilized or destabilized by electronic effect. We'll be mainly focusing on aromatic stabilization, which is way more than resonance, which is more than inductive effect. As we go about our examples today, we'll see how anions are stabilized by aromaticity, resonance and inductive effect. From here, let me show you the molecules I have for you today. This is one of the molecules like this and the second molecule I have is this. This is a connected, yes. One is cyclopentadiene, and the other one is cyclopropene. To compare the acidity of this molecule and this molecule, we need to first write the conjugate bases of these molecules. Let's start with cyclopentadiene. Before I start removing hydrogens, I need to check which of the hydrogens will be lost most easily. We have one hydrogen here, one hydrogen here, one of it here, and one of it here. The other kind of hydrogens we have is this. Why do I say other kind? I say it because hydrogens over here, here, here and here are equivalent. The effect that these hydrogen will cause after they are lost is similar. While when this hydrogen is lost, they are equal. So we have two types of hydrogens. One where the hydrogen is connected to an sp2 carbon and one where it is connected to sp3. We find out that these kind of hydrogens are the most acidic hydrogen and we make our conjugate base. Our conjugate base looks like this, where the negative charge is born on this carbon right here. We know that an sp3 carbon bearing a negative charge is highly unstable. But wait, I can see that this is a cyclic molecule. I can see a chance of delocalization of this negative charge over the entire molecules because there is pi bonds connected here. Let's try and do that. If I try to displace this negative charge over here, this makes our pi bond shift to this edge and gives us a structure like this. Right? And then I can do the same over here, make it shift right here and it will look something like this. And the same can go over here. The same thing happens again. It looks something like this. What I can see here is that the negative charge is delocalized over the entire molecules making each carbon sp2. Right? I can see that all our carbons are sp2 because of this resonance. See, here we had an sp3 carbon but as the delocalization began, this carbon also acquired sp2 hybridization. Same happened with all the carbons in this molecule. I remember if a molecule is cyclic and if all the carbons are sp2, the molecule becomes planar. A planar molecule being cyclic opens the possibility of aromaticity. Over and above this, we need one more rule to check if there is aromatic stabilization of this molecule. We call that Huckel's Rule. Let me erase some part of this screen so that it gives me more space to write. Let's do it over here. Okay. So because we have Huckel's Rule, Huckel's Rule says that the number of pi electrons, pi electrons and lone electrons should be equal to 4n plus 2 where n equal to 0, 1 or 2 and 3. Let's check how many pi electrons and lone electrons do we have in this conjugate base. We have 2 pi electrons here, 2 pi electrons here that gives us 4 pi electrons and how many lone electrons? It has negative charge so it means it has 2 lone electrons which gives us 6. I want you guys to pause the video and check if 6 satisfies the 4n plus 2 rule. If I put n equal to 1, we will get 4 into 1 plus 2 which gives us the magic number 6. So yes, this conjugate base is aromatic and as a molecule is aromatic it gains a lot of stabilization. So it's safe to say that our conjugate base is very stable. But before I come to the conclusion if cyclopectadiene or if cyclopropene is more stable, I need to check the same for the next molecule. We have cyclopropene which looks like this. Before we start writing its conjugate base, let's identify which of the carbons will lose its hydrogen most easily. We have one hydrogen here, one hydrogen here and the next type of hydrogen we have is here. I say these two are similar because both of them are connected to sp2, here as well as sp2 and here it is sp3. So let's lose the sp3 hydrogen and see how it looks. But a carbon which is sp3 is highly stable when it has a negative charge. The only way it sustains this negative charge is due to resonance where the negative charge is delocalized to this spot and the pi electrons move to this point making it look something like this. I have a negative charge here and the same happens over here again. If you remember from the last time I can see a similar pattern happening here. Right? So a cyclic molecule having all its carbon as sp2 now, here it was sp3 but now because of resonance, this carbon also becomes sp2. So it's a cyclic molecule, it has all its carbons as sp2. When I say sp2 it essentially means that the molecule is planar and because the molecule is planar there is a continuous pi electron. That means the delocalization is happening all over the molecule. The third rule we need to check is Huckel's rule. Why don't you guys pause the video and check if this molecule satisfies Huckel's rule or not. I am going to clear off this part. Huckel's rule says there should be 4n plus 2 pi electrons and lone electrons. Let's count how many pi electrons and lone electrons do we have in this conjugate base. We have 2 pi electrons in this pi bond and 2 lone electrons in this negative charge. So we have 2 plus 2 which is 4 and there is no possible value of n which can give us 4. Even if I put n equals to 0 it makes 4 into 0 plus 2 equals to 2. As this molecule does not follow Huckel's rule, this molecule is not aromatic. As this is not aromatic but it still has resonance stabilization but as we saw in the sequence we know that aromaticity provides the most stabilization followed by resonance followed by inductive effect. Our molecule cyclopentadiene and cyclopropene as this molecule followed Huckel's rule it gained aromatic stabilization whereas cyclopropene is not aromatic. The conjugate base of this molecule becomes more stable than this conjugate base. So we can say that cyclopentadiene is more acidic than cyclopropene.