Organic Chemistry

Learn the toughest concepts covered in Organic Chemistry with step-by-step video tutorials and practice problems by world-class tutors.

24. Carbohydrates

Monosaccharides - Strong Oxidation (Aldaric Acid)

There is a stronger oxidation out there that exists besides bromine/water, which turns monosaccharides into aldaric acid, but oxidizing both the aldehyde at the top and the bottom alcohol forming a diacid.



Play a video:
Was this helpful?
Hey, guys. So in this video, I want to discuss a pretty easy reaction called strong oxidation of mono sack rides. So, guys, we've already learned that, bro Ming and water mixed together when exposed to mono sack ride will make something called an al Danek acid where you're going to turn the Alba height at the top into a car box. Look, acid at the top. You guys should remember this pretty easy reaction. I already taught it to you. Okay, But it turns out that we could do a stronger type of oxidation than this because that was called week oxidation, remember? So I'd be saying, Well, why is it weak? Well, it's week in relation to this oxidation that we're about to discuss this oxidation, not Onley oxidizes the top alga hide, tow a car. Books like acid, like roaming water did, but and this is the important part. It's also gonna oxidize the bottom alcohol into another car box like acid. So the product is going to be what we call a die acid. Okay, A die acid is just one of the names that's given for a dye carb oxalic acid. You can call it a die acid. So when exposed to nitric acid, which is the re agent of strong oxidation, mono sack rides can be oxidized into di acid derivatives called Al Dora Gas. It's now those names are a little bit confusing because they're so close together. Al Danek Acid versus al dark acid I don't have a perfect way for you to memorize it, so just keep in mind that it's tricky and you might want to put that on a flash card so that you can say Okay, this is what the Al Danek acid has. The one car books, one car books, like acid in the Eldar ic acid has to. Okay, now, let's just go over the re agents very quickly. The re agents, guys air. Very simple. It's just gonna be some kind of acquis nitric acid. And a lot of times you'll find that in the problems or in the textbook, they'll add heat. Um, but if there was no heat, you would still know this is gonna be a strong oxidation that he isn't like 100% essential that it needs to be written out, even though reaction conditions typically will be hot. But you know that sometimes different problems could be written in different ways, so just letting you know that it's always revolves around nitric acid. Now, one thing that's interesting about creating these all dark acids is that now we've potentially created a simple, symmetrical molecule because noticed that the top functional group is identical to the bottom functional group. So what that means is that due to the symmetry of those carb oxalic acids, different mono sack rides can lead to the same Al Derek acid product. So what I'm basically saying is that you could be starting off with two different dice, aka rides. But because of the fact that now both the top and the bottom of the same um are the same functional group, they could actually lead to the same products because they could be identical once you flip them. Okay, and I'm gonna show you an example, because I know it might be hard to visualize now, but I'm just letting you know that this is a thing that sometimes you could get two different mono sack rides making the same Aldo Nick, I'm sorry. All dark acid products. Okay, so let's go on to the next video where I'm going to do a practice problem with you guys.


Which monosaccharides would produce aldaric acid?

Play a video:
Was this helpful?
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.

Note: We are in the process of re-doing the above video. In the meantime, focus on the concepts of oxidation and optical activitity, while ignoring the structures of D-Glucose & L-Gulose being similar.:)