So molecule A. Did I have any Cairo centers? Meaning, Did I have any atoms that had four different groups on them? OK, and the answer is no. Okay, the closest thing to that was this carbon right here. What that carbon has is it has a three carbon chain on one side. It hasn't O h on the other side, but then at the bottom, it has to h is. Okay, so does it have four different groups? No. Two of the groups that it has on it are exactly the same. So are there any Cairo centers? No. Okay, let's talk about tribunal centers. Are there any double bonds that conform system trans on here? There's not any double bonds at all, So no. So according to test number two, would this be Cairo, or would this be a Cairo? Okay. And the answer is that this would be a Cairo. Why? Because of the fact that it doesn't have any carol centers. So remember that Cairo centers are a source of Chire ality. So we can assume that if it doesn't have any Carol centers that it's gonna be a Cairo isn't that easy? So you just say Okay. Doesn't have Carl centers. No. Then it's gonna be a Cairo. It's as easy as that. Okay, so now I want you guys work on molecule B.

So, first of all, I want you guys to notice that B is a ring. So, technically, I could have used test one on this. Remember what test? One said Test once said test for an internal line of symmetry. So I could have used an internal line of symmetry test. But for the sake of this problem, I want to go ahead and use the second test as well. Okay, but then we'll compare the two and make sure that they agree. All right. So do I have any atoms with four different groups on them? No. The closest thing I have is this Adam right here that Adam has a chlorine. It has a hydrogen, and then it has rings on both sides. It's got this side of the ring on one side, and it's got this side of the ring on the other. Okay, the only problem is that even though these these carbon groups look kind of different, they're actually exactly the same. So both sides of the ring are perfectly symmetrical. So this is going to count as three groups. Not for why? Because if I count around, what I'm gonna find is that I have the same exact atoms going in both directions. If I count in the green direction, what I'll get is ch two ch two ch two ch two. So it's all the same thing then If I go to the blue direction, I get ch two ch two ch two ch two. As you can see, there is no difference depending on which path I take. I'm still gonna be the same atoms in the same order, so they both count is the same exact group. Ah, lot of times I like to call these are groups here. Those were both are groups, but they would both be the same. Our group Okay, so r one R two in this case are one equals R two because of the fact that they're both exactly the same. That means that I have no Carol centers. Tribunal centers only applied to double bonds. We're gonna cross that out. So that means that the answer is, I don't have any stereo genic centers and this should also be a Cairo. Does that make sense? And the reason I'm concluding this is a Cairo is because I have no Cairo centers. All right, now, Let's talk about test one. Internal line of symmetry test. Could I have used that test? Yeah. I also could have used that test. What would that tests have told me? Same thing I could have drawn my internal and a symmetry like this. Okay, I would have had an internal line of symmetry, so it would have been symmetrical. And what's a symmetrical say? It says that it's a Cairo, so I could use both tests in this case. But notice that this test works for both chains and for rings. Whereas Test one on Lee really works for rings. That's the advantage of using tests to because test to works for everything. Okay, so let's move on to the next question.

molecule. See? What did you guys say? This molecule is also a Cairo. Why? Because let's look at Cairo centers. Do I have any Cairo centers? Well, the closest thing is that one. And once again I only have three groups. I have the H, which is obviously won the O A tresses. Obviously different. But then I have once again to our groups that are the same. R one r two are identical. So that means that there's no Carol Centers. There's no tribunal centers. So this should also be a carol. See how easy this is. Okay, so let's move on to molecule D.

So is Molecule D Cairo. The answer is yes. This is actually a Cairo molecule. Why is that? Well, let's test for Cairo centers. So did we have any Cairo centers here? And the answer is yes, actually. Have more than one. I have to. Cairo centers. I have a Cairo center here and have a carol center here. How does that? How does that work? Well, because notice that have got a ch three then that means in the back here I would have an age. Okay, so I have a ch three in an age. And then what else? Well, I would have one direction of the ring. Let's say that's the green direction. And then I have another direction of the ring. Let's say that's the blue direction. Okay, so my question to you is are both sides of this ring exactly the scene, for example, if I take the green path is the same as if I take the blue and actually, no, they're completely different. Notice that the green path will get to that metal group in two carbons. Okay. Whereas the blue path of I take it around, it's going to take four carbons to get to the same place. Okay, so are both sides of this ring perfectly symmetrical with respect to that carbon? Know that carbon has a two carbon path and a four carbon path. So what that means is that if we have our one, let's say this is our one. And this is our to this is an example. Where are one does not equal are to their different from each other. Okay, so that means I have four different groups, So this would be a Cairo center. If I use the same logic, I would find this carbon over here is also a Kyrill center. So then the answer is I have to Carol centers. That's the first time we've gotten to What about tribunal centers? No, there's no double bonds. So the question here is is this gonna be Kyrill or not? OK, and the answer is yes. This will be Cairo. Okay, because basically, just say if we have one or more Cairo centers, that will be a Cairo structure. Okay, we're simplifying that right now. It turns out there's some more rules that we're gonna learn later, But for right now, I just want you guys to know if as a Cairo center, it's Cairo. I don't wanna overcomplicate it right now. Is that cool? So I have to Carl centers. So it's Kyle. I just want us to think that way. Cool. So let's move on to molecule eat.

So does Molecule e have any stereo genic centers? Yes, it does. Is it Kyra? Well, no, it's not because this molecule has no Cairo centers. Can't feel notice. Most of these atoms here don't even have four groups around them. For example, this carbon right there. It only has three groups because as a roaming ah, method and then the double bond. So you can never have a Cairo center directly on a double bond. That doesn't even make sense. You need a Tetra Hydro. Okay, But let's say we do have one Adam here that has, or a two or three one that kind of has four different groups. It might, Let's see, has a double bond on one side has a method on the other, but then it has to ages. Okay, so any time you see a ch two, you can kind of just throw it away. Don't worry about it, because there's two ages, so it can't be a Cairo center. So the answer is, I have no Cairo centers here. Now it's like a tribunal centers. Do I have any double bonds capable of forming stereo eyes, summers or Sister Trans? If I were to swap groups. And the answer is yes, I do, because I have a double bond here, okay? And that job on, if I were to switch the position of the roaming and the metal, let's draw that out and see what happens then. What that means is that I would have a metal going up, a Brahmin going down, and then I would have still the Ethel Group and the metal. Okay. Is this equal to this? Are they the exact same thing? No. It turns out they're not remember how to do NZ. We would take the highest priority on both sides. So on this side, I would say, Okay, my highest priority on the red side, my highest priority must be the bro Ming, because Brahman has a higher molecular weight. Are you cool with that on the blue side? What's my highest priority? Is it methyl or ethyl? It's gonna be Ethel. Okay, So would this be ears e? Well, the big groups are assist to each other, so this would be equal to Z. Remember that Z stands persists then over here I would have the opposite situation, will have roaming and Ethel. But they're on different sides of the fence. Okay, so if they're on different sides of the fence, this would be E. Okay. By the way, if this is confusing you if you have never really done years before, I already taught this. So go ahead and look back towards the ears. E part of, you know, a past lesson. And that might help you guys. Okay, So the point here is that I will get a different stereo a summer if I swap groups. So this does have one tribunal center. Does that make sense? Cool. So I have zero Carol centers, One tribunal center. Is this Cairo? No, it's still a Cairo. The reason is because I told us that tribunal centers by themselves are not Cairo. Okay? They're just a source off steroids summers. You can make steroids summers around them, but by themselves, they're not Kyra. So this is also a Cairo. All right, so let's go on to the last question, which is question F

So for this one, I'm gonna take myself out of the screen so my big head doesn't get in the way. All right. So did you guys find any Cairo centers here? And the answer is you shouldn't have actually, OK, because if you look at, let's look at this carbon right here. What four groups doesn't have. Well, it has a metal. It has a hydrogen in the back. Then it has this side of the ring. Let's call that one are one. And then it has this side of the ring. Let's call that one or two and you have to figure out if they're different or if they're the same. So in this case, it looks like they're the same because, you know, we basically it takes us a ch to hear a ch two here. And then if you go around, you both get to the metal group at the same time. So once again, both of these roads lead to the same exact place in the same exact way. So they're both the same. This is an example where r one equals R two. So this would have zero Cairo centers because of the fact that no atoms have four groups. Okay? Also has zero tribunal centers. So that means that overall, this is gonna be in a Cairo molecule. All right, so I hope you guys understood that part now. I wanted to also show you guys that this happens to be, what a ring. So could I have used my first test on it? Absolutely. I could have used my first test. And the first test is the internal line of symmetry test. So where is the internal line of symmetry here? Because we said that this is a Cairo, which means that it must also have an internal line of symmetry. So where is it? Well, the answer is that the line of symmetry is gonna be right down the middle like this. Okay. Where If I sliced it right down the middle with some scissors, what I would wind up getting is too perfectly equal halves. So we could have used So basically could have used either test one or test to, um to figure this one out. In this case, test one might have gone a little faster, but test to works as well. Alright. And in all these cases, the tests will always agree with each other. Okay, so hope that makes sense to you guys. Let me know if you have any questions.