All right, So here's our first reaction with periodic acid in a sugar. What we see is that we're actually starting off with a key toes here, and we're reacting with periodic acid in water, which is just a form of periodic acid to react with. This is perfect. We know that what we're going to get is oxidative cleavage. Um, and actually, you could predict ahead of time how maney molecules you're going to get. Because how maney carbons do we have total 12345 with oxidative cleavage of mono sack rides. You're never gonna You should be cleaving at every single spot. Okay, If it's a mono sack right, that's in a straight chain, you should be cleaving in every single spot, meaning that every carbon should be its own molecule. Okay, cool. So let's go ahead. And just maybe we can start from top to bottom and you guys can tell me which oxidation product would you expect at each carbon? Let's start off with one. What? Oxidation products. And I'm gonna actually going to use letters and then we'll put it all together. So at carbon A. What would we expect to be the oxidation product. Well, I see it's a terminal alcohol. So I would expect that this is gonna be formaldehyde, and I'm going to draw it on its side so that it looks a little bit more like it did from originally. The O's going this way. I'm gonna draw it this way as well. Maybe it will make it easier to visualize a great job. You guys all said formaldehyde. Good job. Okay, so let's go toe. Go to be So then be Is this carbon? What are we going to get from that key tone? Remember, that key tone reacts with periodic acid to give what? Co two. So let's go ahead and write that in Cedo 10 no, I have one mole of co two. This is getting cool. Let's start off. Let's go down to see. So what would we get with? See what I see with C. Is that is this an internal alcohol or a terminal alcohol? Its internal? Because I have stuff on both sides, Right? So that means that I'm going to get one mole of formic acid. Good job. So I'm gonna go ahead and I'll just draw it like this so that we can keep the O. H and H and the same orientation. Okay. Remember formic acid, you're gonna get your O h and your age and a carbon you'll because it's a carb oxalic acid. Now let's go down to D so d what would I get? Well, guys, this is another internal alcohol because I've got alcohol. I got carbon on the top, Carbon on the bottom. This is not the end of the chain. So this is going to be another form it acid. And then finally, e what am I gonna get for E is guys, this is just another way to draw a terminal alcohol. It's the same exact thing Is that just drawn differently. So should give me the same thing. It should give me a formaldehyde, which we're gonna draw like this, like this. And guys, there you have it. That's actually my product. But let's go ahead and just put it together a little bit more nicely. Um, for the final answer. So the way that we should probably write the final answer is that it's going to be co two plus on to formic acids plus two formal decides and that is our answer. Okay, That wasn't so bad. Right? So now you guys know how to react with periodic acid. This is actually a really tricky example. So any other mon assess like straight chain mono sack, right? Should be relatively easy for you guys to react with. Okay, So, great job. Let's want to the next video.

guys, this is, Ah, hard problem. Let me go ahead and read it and then try. Toe will try to work through it together in aqueous base, De Glucose has the ability to memorize into small amounts of demand, a pira nose and then on also rearrange into de facto Farinos Fisher glycol. Sedation can then transform these sack rights into, oh, Glick asides. So the two part question says a predict the structure of the black aside products after treatment with acid in methanol. And then be, How could the treatment of those like asides with periodic acid distinguish if EP, memorization or rearrangement is more favorite cool? So that was a mouthful, guys. So this is what it's basically saying in the first little sentence. It's saying that de glucose through two separate processes can either become that thing or this thing. Okay, now I have videos on both of these processes on EP, memorization and rearrangement. So if you're really curious about how to get these exact cyclic sugars, then please go ahead and watch those videos. But because this is beyond the scope of this question, I'm not going to draw out the exact way that we got to these sugars in this question, OK, because we're really trying to focus on is what happens after these form. Okay, Just so just what I'm basically saying is, take my word for it. That D glucose can provide these two cyclic sugars A for a nose and a PIRA knows now where the question really starts, as it says. Okay, predict the structure of the OGE like asides that would form after each of these is treated with methanol and acid. So this part is actually pretty easy because you guys should have learned about rogue like oxidation already. Or also Fisher like oxidation. And you guys should be familiar with which alcohol's react when you react within with which alcohols in a sugar react with alcohol and asset. Do you guys remember Notice that all of the O's are missing their ages so that we're gonna have to predict together if it remains an alcohol or if it gets a metal group on it? So what do you guys think? Here I have potential alcohol's or by potential oxygen's that could now have a methyl group on them. Ch three are all of them going to get sage three are none of them. What number do you guys remember? Well, if you go back to your old like oxidation video, we talked about how Onley one of these can react. And that's because of the intermediate, the intermediate. That's very important. Can Onley form on the an American carbon? So that means that my first product for both of these should have h is remaining in all the places that are not an americ. And then the an America oxygen should get a ch three. And this is what we call an oak like aside and you can see that it's named correctly. It says that now it's d man Oh, piranha side, Which means that you have now an O R group on the an American position. Okay, now I'll get out of the way. Uh, here, just so you guys can see what I'm gonna do here. We can see that it's the same thing. Um, I would just put h is everywhere. That's not an americ. And this is my animated position here. I'm now gonna put a ch oops a c h three here. Cool. So not terrible. Let's scroll up. And we actually already got points for the first part of the question. You know, if this if this was your professors question, at least we're halfway there right now. The next question says, What if you reacted both of these with periodic acid, Right? So let's go ahead and write that down periodic acid for both of these, Could you use periodic acid to prove which one is more favored? Okay, And what we're hinting at here is that periodic acid is probably going to give different products for each of these. It's probably going to give something for the first one, something for the second one. So if we can find out what that difference is, then we could prove is a more common making the pirate the making the pirate pira nose ring, or is be more common making the foreign nose ring. Get what I'm saying. One of these is a six member drain, and one of these is a five member drink, and what we're trying to figure out is which one is more popular? Is it better to go to the six member or better to go to the five member. So what we're gonna do with periodic acid, we're gonna look at the products for each and then those products are going to tell us which one is the most favorite. Cool. So let's look at the cyclic sugar. So the first one, the piranha side, How many places could be Cleve on this sugar? Okay. Now, by the way, some of the rules that we talked about with straight chain mono sack rides don't apply to cyclic, um, like asides. And the reason is because remember that when you had a straight chain, every single position, Haddon Oh, that could react. Okay, but thes aarggh like aside these air cyclic like aside. So that means some of these oxygen's won't be able to react with periodic acid. So we're not going to get the same exact we're not going to get. If you have six carbons, you're not gonna get six products necessarily because some of the O's can't react. We have to look specifically on Lee for which oxygen's are which alcohols are Dial's visceral dial's to each other. And what I see is that this is a dialogue, all right, that there are visceral dial's there next to each other. So that means that I could break here. Does that make sense? That's a visceral denial. I see another visceral Doyal here between these two. So that means I could also break here, and that's actually it. None of these other ones air Dial's because notice on, I'll just use a red X this oxygen. This alcohol here doesn't have any alcohol. Divisional to it. This is not even an alcohol, so it doesn't count. And this is not even an alcohol. So this doesn't count. Does that make sense? So I can Onley Cleve in two places. So that means that if we were just to look at this through geometry, imagine you have something a rubber band, right? Zehava rubber band. And you're going to cut it in two places. How many pieces they're gonna get? Total. It sounds like a trick question, but it's actually I'm just This is just like a physical products here. Like if you take a rubber band and you cut it in two places, you're gonna get to pieces, right? You're gonna get a big piece on a small piece. That's what we're gonna expect over there. So we're gonna expect to products. Let's look at our for and ring are for a nose ring. So how many of these alcohol's count as visceral Dial's? Well, this is definitely one right Those air next to each other so I could cut their What else would count as a visceral denial? Well, um, this'll I mean, so this is not doesn't have any oxygen's around it. This does not count. This, um, does not count because it doesn't have any oxygen's around it. This is not an alcohol, so looks like we're kind of stuck like we don't have that many different places that we could that we could cut. We could only cut in one place. So I imagine you took that rubber band, and you cut it just in one place. How many pieces would you expect to get one? But it would now be a chain. It would now be a string instead of a loop. Right. So we're expecting Is that for the first one? I'm gonna get to pieces for the second one. I'm going to get just one piece now. What would those pieces look like? Well, let's go ahead and draw out what it would look like. So for the first one, do we have any fragments that we actually can match to our four cleavage patterns that we talked about? Remember the four cleavage patterns that you're separating certain things? Do any of these fragments match one of those cleavage patterns? Yep. The one that I'm looking at here is Let me use a different color. This guy right here. Do you see how that's actually an internal alcohol? Because it's an alcohol that has stuff on both sides, and it's getting cut on both sides. So that means that this fragment right here is going to give me formic acid. Okay, How did I know that it was formic acid? Because, guys, that's just a cleavage pattern that we remember that cleavage that when you have an internal alcohol, you get formic acid. Does that make sense? So far, we're gonna get formic acid plus the rest of the chain. Okay, Now for this question, whatever the rest of that chain looks like, it doesn't really help me. It's not applicable, so don't worry about it. Just say that we're going to get that little piece of formic acid plus the rest of the chain. All the other carbons does that make sense? Cool. Now let's look at this one for the five. For the five. How maney do we see any cleavage patterns that are the same or that match up to what we memorized in our page before? No, actually, because no, none of these alcohols are getting chopped on both sides like you would normally expect. Also, none of them are terminal that they can get chopped on both sides. So what that means is, guys that this is actually just going to form one big change, one chain. Okay, that's it. It's not gonna form anything else. So how could you use periodic acid to determine which one is being produced? Well, if, after periodic acid you end up getting formic acid as a byproduct, then you know that you have piranha side. So piranha side after periodic acid after cleavage will give one mole of formic acid. Okay. Whereas if you were to apply the same periodic acid to a five member ring like this, you're just going to get one chain plus no formic acid. Okay, Now, what do we know about formic acid? guys, it's acidic. So you could actually you could probably test out the acidity of the solution to figure out which one is being is being formed. Is it the first one, or is it the second one? Okay, now it turns out, guys, that it didn't ask you to determine which one is more popular or which one is more favored. It just said, How could you use periodic acid to determine which one is more favored or to determine which one is forming? We could look for formic acid. If you're finding formic acid, that means that you're getting the six member drink, and now just you guys know it Turns out that the formic asked. The six member ID ring is much more favored than the five member ID ring. So if we were to actually do this in real life, we would find that periodic acid. When you, when applied to D glucose and efforts allowed toe memories or rearrange, you would actually get formic acid, you get the presence of formic acid because it formed that six member drink. Cool, So tricky question guys. But it was a cool application of oxidative cleavage. Let's go ahead and move to the next video