Monosaccharides - Strong Oxidation (Aldaric Acid) - Video Tutorials & Practice Problems

Monosaccharides - Strong Oxidation (Aldaric Acid)

Hey guys. So in this video, I want to discuss a pretty easy reaction called strong oxidation of monosaccharides. So guys we've already learned that bromine and water mixed together when exposed to a monosaccharide will make something called an aldonic acid, where you're going to turn the aldehyde at the top into a carboxylic 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 can do a stronger type of oxidation than this because that was called weak oxidation, remember? So if you're saying, well why is it weak? Well, it's weak in relation to this oxidation that we're about to discuss. This oxidation not only oxidizes the top aldehyde to a carboxylic acid, like bromine water did, but, and this is the important part, it's also going to oxidize the bottom alcohol into another carboxylic acid. So the product is going to be what we call a diacid. Okay? A diacid is just one of the names that's given for a dicarboxylic acid. You can call it a diacid.

So when exposed to nitric acid, which is the reagent of strong oxidation, monosaccharides can be oxidized into diacid derivatives called aldaric acids. Now, those names are a little bit confusing because they're so close together. Aldonic acid versus aldaric 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 flashcard so that you can say okay, this aldonic acid has one carboxylic acid and the aldaric acid has two. Okay? Now, let's just go over the reagents very quickly. The reagents guys are very simple. It's just going to be some kind of aqueous nitric acid and a lot of times you'll find that in the problems or in the textbook they'll add heat. But if there was no heat, you would still know this is going to be a strong oxidation. The heat isn't 100% essential that it needs to be written out. Even though reaction conditions typically will be hot, but you know that sometimes different problems can be written in different ways. So just letting you know that it always revolves around nitric acid.

Now one thing that's interesting about creating these aldaric acids is that now we've potentially created a symmetrical molecule because notice that the top functional group is identical to the bottom functional group. So what that means is that due to the symmetry of those carboxylic acids, different monosaccharides can lead to the same aldaric acid product. So what I'm basically saying is that you could be starting off with two different disaccharides, but because of the fact that now both the top and the bottom are the same functional group, they could lead to the same product because they could be identical once you flip them. Okay? And I'm going to 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 monosaccharides making the same, aldonic, I'm sorry, aldaric acid product. 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?

Hey everyone, let's take a look at the following example question. Here it says, first circle the two monosaccharides that will produce identical aldaric acids upon treatment with nitric acid. Then, label the products of these oxidations as optically active (meaning chiral) or optically inactive (meaning achiral). Alright. So let's do the first part. We know that treatment of these monosaccharides with nitric acid is going to oxidize not only their aldehydes but also the primary alcohols on the bottom. So these aldehydes, we're just going to say they're all going to become OH groups. So let's work on showing that. So all of these will become OH groups, and then all of these down here will become carboxylic acids. So we'll have a carboxylic acid, carboxylic acid, same here, same here. And then down here, we'll have our carboxylic acid. So the goal is products or diacids that are created from oxidation with nitric acid. We have to look to see which ones would be identical. Here, it's not as easy to see, but so we had all of these monosaccharides. And if we take a look, we have L-gulose, which would undergo this oxidation to create this structure. And we have D-glucose, which is our first structure, it undergoes oxidation as well. It may not look like it, but they're actually identical. So, let's just imagine that we took this and we flipped it. So, the top part comes down here and this bottom part gets rotated up here. Doing that gives us what type of image? Well, if we flip it that way, let's draw what we would get. We get this carboxylic acid up here. And then, let's draw the rest of it. And then we get down here. So I'm flipping it this way. Alright. So when I flip it this way, these OHs now, they're gonna be over here because I'm flipping it that way. Right? And this H would be here, and this OH would be down here. And flipping it that way, we would say that these hydrogens after we've flipped it, they would be up here, and then this OH would be down here, and this H would be here. If you look at that, what does that resemble? This structure here resembles this structure right here. So with this oxidation, we would say that the products, the oxidized products of the glucose and L-gulose will be identical. So that's the first part of this question. We couldn't do that with either of the monosaccharides in the middle in the form of L-galactose or D-mannose. Right. So we're going to say here for this first part, it's the D-glucose plus the L-gulose.

Now, looking at our dicarboxylic acids that we've created, our dark carboxylic acids. Let's answer this next part. It says, then label the products of these oxidations as optically active, meaning chiral, or optically inactive. Now, to be optically inactive, where are the scenarios that help to create something as being optically inactive and therefore achiral? Well, first, they'd have to have either no chiral centers, or they'd have to be a meso compound. The first one is out of the question because all of these have chiral centers. All four of these monosaccharides even in their oxidized forms would still have chiral centers. So we're not looking at that. What we're looking for is which one represents a meso compound. Remember, a meso compound has an internal plane of symmetry. The way we can check this is just cut these diacids down the middle, and look, do we have some type of internal plane of symmetry? Does the top portion match the bottom portion? If we take a look, let's look at this top portion here. This top portion doesn't look like the bottom portion. So let's make this one 1 and this one 1, and this 2 and this is 2, and this is 3, and this is 3. Okay. So the ones are the same but that's not in reference to a chiral center. 2 and 2. Alright. So we would say that yes, both of them have an OH on the right, so they are modeling each other. But then, if we look at these as threes, they don't match up. It's not looking in a mirror. This OH is here, so there'd have to be an OH here. So there is no internal plane of symmetry and therefore, this is not a meso compound, meaning that it will be chiral overall and optically active. Let's go to the next one. Here's our mirror here. Let's look in the mirror. So we do 1, 2, 3, 1, 2, 3. Looking in this mirror, there's an OH here, and for 2, there's an OH here. For 3, there's an OH here and there's an OH here. So this thing is looking at itself in a mirror internally and it's seeing itself back, its own reflection. So there is an internal plane of symmetry within this, within this oxidized form of L-galactose. So in its diacid oxidized form, we would say that it's optically inactive because it is a meso compound. For the next one, 1, 2, 3, 1, 2, 3. Looking into the mirror, for 2, there's an OH on the right, but for this 2, it's on the left. So it doesn't match up automatically. It can't be a meso compound, so it's optically active. And then finally, this last one, 1, 2, 3. Looking in the mirror for carbon 2, it sees an OH on the left. There's an OH on the left. Okay. For carbon 3, this one has its OH on the right. Oh, but this one does not. So there is no internal plane of symmetry where the top part matches perfectly with the bottom part, so this could not be a meso compound, and therefore it's optically active. So going back to the beginning of this question, we would say that the two monosaccharides that form identical aldaric acids upon oxidation with nitric acid would be D-glucose and L-gulose. And we would say that all of these diacids, their oxidized forms are optically active except for the diacid form of L-galactose. If we cut it right in the middle, we would say that the top part and the bottom part match. There's an internal plane of symmetry at play making this a meso compound and therefore optically inactive. Right? So just keep this in mind when you are faced with questions like this, take a similar approach to determine if things are identical and whether they are chiral or achiral in nature. Doing this and applying these concepts will help you with questions further on.