first circle, the two mono sack rags that would produce identical al derek acids upon treatment with nitric acid and then label the products of these oxidation as either optically active, which is another name for Cairo or optically inactive, which is another name for a Cairo. Okay, cool, guys. So let's go ahead and start with the very first one, which is I have these four model sack rides. Two of them are just to give me the same exact Al Derek acid product. How do we determine that? Well, guys, the first thing we should probably do is just go ahead and, um, replace all the groups, all the Alba hide groups and all of the alcohol groups at the bottom with carb oxalic acids. Now, for the sake of time, let's just go ahead and try to just draw over what's already here instead of drawing them out all over again. So I'm just gonna put like, ohh ohh ohh! And then down here, we also know this is gonna turn into car. Looks like acid. So let me just try to do that the best that I can. Okay. Cool. You guys see what I'm doing I'm just kind of making up my own carb oxalic acid over here. Cool. Okay, so now we see that the top in the bottoms of these should be the same. The function loop should both be carbon selig acids. Now, how would we know which of these actually, which of these are the same? Okay, well, guys, the way that you would know is by flipping them and seeing if, by turning one over on its head can you achieve any of the other mono sack rides that air here Now, guys, this is very important that the Onley type of rotation that you're doing is that your turning it on its head or like, 180 degree turn What? I also refer to an oracle one as a deejay. Spin so much It's like a record and you're spinning it. You cannot do this. You cannot flip it this way into the page. You cannot flip it upside down this way. Those are rotations that are not allowed. According to Fisher projections, the Onley rotation that you could ever do with official projection because of the unique orientation of the group's is a deejay, spin or flipping it on its head like this. Okay, so if you think about now that I've given you that those instructions, which of these by turning it around this way would equal to the others. Okay, so let's let's look for some easy examples that we can rule out. So, like, for example, I'm looking at De Manos de manos. Once I do this, it's actually gonna be the same exact compound, right? Because I got to over here and two over here, and then they're gonna switch swap places, and it's just gonna be itself. So that one definitely can't be the same as any other mono sack rides here. But what I'm saying is, by turning this 1 108 degrees, it's not gonna look like any of these other ones. I'm gonna cross out D Mantle's Let's look at Algal Actos. Elga, Lactose has those two in the middle. That air right now coming at me, right? I'm looking at these two guys right here. Okay, If I were to turn it around, what I would then get and let me just actually show you is I would get them then facing the other direction. Ohh Ohh Ohh Ohh! Does this make sense so far? What I'm saying is that if you were to turn this around, you would then get those two middle ones face that way and then bracketed by the 20 H is on the top of the bottom coming this way. So that is a legitimate rotation. Does that look like any of the mono sack rights here? No. And the reason that I kind of focused on this one to rule it out is because none of the other mono sack rights have the two middle O. H is going in the same direction. In fact, notice the ones that I'm left with, D glucose has them going in opposite directions. And LG Youlus also has been going in opposite directions. So I kind of looked at l galactus and I was like, That looks like it doesn't belong because it's the only one that they're both faced in one direction, right? So we can go ahead and rule it out cool. So that leaves me presumably with my answers, because two of these have to be the same. But let's go ahead and just make sure let's prove it Okay, So if we were to take D glucose and turn it around, what would we get? So what I would get is basically, let's draw our four lines. 1234 I'm even gonna draw the car. Looks like acids weaken. Do a full flip. So just keep in mind, guys, that this car box like acid here is actually this one because I'm turning it around like this. Okay, that's what I keep meaning When I say deejay spin, I'm saying you spinning it around, so let's just draw out exactly where everything would be. So this guy here, what's start? Because the fact that we're starting at the green card looks like acid in working down. That should be green and working up on the other one. So that means that, um let's say this is now the one carbon, This is now the two carbon. This is now the three, the four, the five and the six. Because I'm I'm working down this way. Okay, So now this is my one. My two should face the opposite direction. So right now it's face. So the right I should now face it to the left. My three is face to the right. I should face it to the left. My four is face the left. I should face it to the right. My five is face the right. I should face it to the left. And there you go. That's our structure. Does this structure match any of these other structures? Doesn't look the same. This is the same as L Gulas guys. So notice that these d glucose and l Gulas were completely different Mono sack rights to begin with. But after we oxidized them with nitric acid, they both yielded the same exact product. Okay, What? This is what I'm trying to say is that both d glucose would make this and l Gulas is also gonna make this. So when I said earlier that sometimes it will lead to the formation of two different ones can lead to the formation of the same compound. This is what I'm talking about. And these are the types of practice problems that you're likely to see in terms of nitric acid because they're gonna gonna want to test you on how well you understand that concept. Cool. The last thing, guys, because we're not done. It says then after you're done, label all the products of these oxidation as optically active or optically inactive. Okay, Now, for this part of the exercise, we don't need to actually draw all the products. What we can do is just work with what we already drew here. Okay. Oops, that's that's still that. Okay, but what I'm saying is you can just look at it this way. So, guys, why would something be optically active? Because it's Cairo, Okay? And it's Cairo because it's a symmetrical. Has carol centers etcetera? Why would something be optically inactive? Well, there's a few different ways because we said it's a Cairo. A Cairo could mean it either. Has no Kyrill centers at all. You guys remember this? Just know Carol Centers. It's a Cairo, right? But another reason that it's a Cairo could be because it's a me so compound. Do you remember what a missile compound is? Missile compound is a compound that has two or more Cairo centers but is perfectly symmetrical. Okay, so two or more Kyle centers, but is perfectly symmetrical, has a plane of symmetry. Um, and that those who basically the two reasons Okay, so I know for sure that the first reason zero Cairo centers would not happen here. Because all of my structures have many, many Carol centers. Carl centers everywhere. Basically, all of these guys Air Carl Centers. Okay, so for sure, it's not going to be because of no Carl Centers. But after oxidation, do any of these become miso compounds and basically what we're looking for to be me? So is that it has carol centers. Plus, it has a plane of symmetry. It looks like I'm saying something else, another type of acronym, but it really just means plane of symmetry. Okay, so do any of these achieve a plane of symmetry after I have oxidized them? Let's just look, let's just cut them down the middle. Is there a plane of symmetry on this guy? Absolutely not. What I see is that the top looks different from the bottom. The wages are facing in different directions. Is there a plane of symmetry on this guy? Yes. Notice that the top of the sugar looks identical to the bottom of the sugar. So it's as label these. I'm gonna have to label this as an active Well, I should put optically in cool. We know the other one was active kids. Carol, how about this one? Does this one have a plane of symmetry? Could be a little tricky, You could think. Oh, but the top of the bottom or flip mirrors of each other. But that's not a plane of symmetry means. Okay. Plane of symmetry means that they have to be identical on both sides. So this would not have a plane of symmetry. And then finally, what about D and I Actually, I dropped my pen happened sometimes to her life. Um, do we have a plan of symmetry there? No, because once again, like the d glucose, there's a lot of just differences in the directions of the O. H. Is so then for all of these other ones, we would put optically active right, which means that it's Cairo and guys, I don't just make this stuff up. The reason that I'm testing you both on or that I'm teaching you both about the Monta soccer is making the same product as well as touching on the fact of optical activity versus optical in activity. That's because many professors will test on this idea of strong oxidation in this way, because strong oxidation is a really easy reaction. So your question is not likely to be like draw the product predictive product. It's likely to be something like this. Where is the product of strong oxidation? Optically active? And I have to think about it. I have to think. Well, what are we doing? There's a plane of symmetry, etcetera. So that's why I included all of that in this practice problem. Okay, awesome guys. So I hope that you understand strong oxidation with nitric acid. Better let's move onto the next video.