Monosaccharides - Oxidative Cleavage - Online Tutor, Practice Problems & Exam Prep

Periodic acid cleaves vicinal diols in sugars, leading to oxidative cleavage. Aldoses yield formic acid upon oxidation, while ketoses produce carbon dioxide. Internal alcohols also generate formic acid, whereas terminal alcohols result in formaldehyde. Understanding these cleavage patterns is crucial for predicting products when reacting monosaccharides with periodic acid. The mechanism involves forming a cyclic periodic ester, followed by breaking sigma bonds to create carbonyls, illustrating the relationship between oxidation and cleavage in carbohydrate chemistry.

Monosaccharides - Oxidative Cleavage

Hey everyone. In this video, we're going to talk about how you can use periodic acid to cleave or break apart a monosaccharide completely. Let's take a look. So guys, in general, periodic acid has the ability to cleave vicinal diols, and this is not a reaction that's unique to sugars. It's actually a reaction that occurs any time that you expose periodic acid, which I'll show you in a second, to a vicinal diol. It's always going to split the diol in half, leaving oxygens on both sides. Okay. Now what do we know about sugars? Sugars have multiple diols or multiple alcohols. And when I say diols here, I'm specifically saying vicinal diols. So, sugars have multiple alcohols that are next to each other that could potentially be cleaved. Okay? So what we're going to learn here is not a new reaction completely. What it really is, is it's an application of another reaction that already exists. That other reaction says that if you take a diol and you expose it to periodic acid, it's going to cleave. So now why don't we do that with sugars? And it turns out that the mechanism we're going to use is even identical to the general oxidative cleavage of diols with periodic acid. This This is just reinforcing what I'm saying, how this is not a brand new reaction. It's just an application of another reaction to sugars. Cool. So guys, let's start off with the basics. Vicinal diol. Remember that a vicinal diol is an alcohol, or it's a diol that the two alcohols are next to each other. They're not geminal. Geminal would be on the same carbon. Vicinal, they're next to each other. Okay. So you need a vicinal diol to make this work. We need to expose to some form of periodic acid. Now here, I've gone ahead and drawn out the periodic acid for you. This is what it should look like if you're given this form of periodic acid, which is the most common form. But guys, it turns out that lots of different professors and textbooks for some reason like to overcomplicate periodic acid and they like to draw it a bunch of different ways. So here what I did is I listed out all the different ways that I've seen it written and that I want you to think of as synonyms for periodic acid. So please don't, like I said, don't overcomplicate it, just think of them all as forms of periodic acid. So you could see, let's just go in order, you could see periodic acid and water. Same thing. You could see instead of 4 oxygens, 3 oxygens. That's called iodic acid, but that also functions very similarly. You could see it as IO₄⁻ which is the anion, so that's called periodate. And then finally, you could even see it with 6 oxygens, which is actually just another form of periodic acid. So again, don't worry too much about it. Just think if you see a lot of oxygens around an iodine, this is a form of periodic acid and this is going to be some form of cleavage if diols are present. Okay. So let's say that you expose the periodic acid to the diol, what you're going to wind up getting is this cyclic structure. Okay. Now you don't need to know the mechanism for this part but we are going to quickly go over the mechanism for the cyclic part because this part of the mechanism is shared with the general reaction of periodic acid cleaving diols. Okay? So first of all, let me just name this for you. This, structure here is called a cyclic periodic ester. Okay. The cyclic periodic ester is one of the intermediate steps of oxidative cleavage with periodic acid. Okay. And at this point, we formed our cyclic ester and now all we need to do is break apart the sigma bond that's holding the two carbons together. The sigma bond I'm talking about is this one. This is the one that we're going to try to break apart because that's what cleavage means. We're going to break carbons apart. So the mechanism for this is pretty straightforward. All it is is that you take the electrons that used to be connecting those carbons together and you make a carbonyl out of them. So you make a carbonyl out of one of them. Now if you can go either way. I could go to the left or to the right. It's going to be a cyclic mechanism and it's all concerted, meaning it all happens at the same time. Now if I make that double bond O, I'm going to have to break a bond. Right? So the way that that bond is going to break is that we're going to then take electrons and put them on the eye. Okay. So because that oxygen needs to break a bond in order to not be positively charged, so we're going to take those electrons away, put them on the eye, and now what we have to do is we have to basically keep the formal charge of the eye the same, so we're going to have to break a bond on the eye and we're going to take these electrons and make a carbonyl down here. So as you can see, we're just redistributing the same electrons that were already there and what we're going to get at the end is now two carbonyls with this bond breaking. Right? That bond is gone and now we're going to get two carbonyls. So let's draw what that would look like. So what this would look like is on the left side, I would have o double bond. Whatever that group was on the left, I'm just gonna keep it as a stick because I don't know what that is. And then on the right side, I'm gonna have an H. Now this H is this H right here. Okay. And I mean just to be super clear, this stick is this stick right here. Cool. And then on the other side, I would just get the same exact thing just flipped around. What I would get is then I have that stick there and this H here. That's that and oops. Almost perfect. There we go. So look what we've accomplished and then you get your iodic acid as a byproduct. Okay. So notice what we're accomplishing. We're cleaving because we're taking two carbons and we're separating them and we're also oxidizing. That's why this is called oxidative cleavage because I start off with alcohols and I'm ending up here with aldehydes. So guys, that's the general mechanism. Your professor may or may not want you to know that part of the reaction, but now in the next video what I'm going to do is I'm going to show you what you absolutely need to know, the four cleavage patterns for oxidative cleavage. Okay? So let's go to that video.

Monosaccharides - Oxidative Cleavage

So, this is the part that can get a little bit tricky if you don't have everything mapped out already. It turns out that different types of alcohols and carbonyls on sugars respond differently to periodic acid. It's not like a one size fits all, that it's always an aldehyde. It's actually not. You have to kind of do a little bit of memorization here. So, what I've done is I'm trying to organize it for you in the easiest way possible so that you're going to just be able to look at this chart and know exactly what to do. Just to repeat, you can't just guess what the product will be. The product is very specific depending on what type of alcohol or carbonyl you're starting with.

So, what I'd like to start with is basically the difference between an aldose and a ketose. Okay, remember that most sugars always come in either the form of an aldose or a ketose. Remember, an aldose at the top is going to have an aldehyde. Remember, a ketose at the top is going to have a ketone, and that's what I've just drawn here. Okay. Well, that oxygen that's either in the aldose or the ketose will react with periodic acid. Now, this isn't the same exact reaction as the vicinal diols reaction because this is not an alcohol. It's a carbonyl, but it still is going to oxidize that carbonyl. Okay? So how would we oxidize these carbonyls?

Well, if you're starting off with an aldehyde, what you're gonna wind up getting at the end is, and I'm going to say this a lot, formic acid. Now, formic acid is the common name for this molecule. You could also call it methanoic acid, which would be the IUPAC name, but most commonly it's called formic acid and that's just the simplest carboxylic acid possible. It's a carboxylic acid that only has one carbon. So if you have an aldehyde at a terminal end, well, aldehydes are always terminal, you're going to get formic acid as your oxidation product.

Now if you start off with a ketone, for a ketose at the top, then you're not going to get formic acid, you're actually going to get CO₂. So I'm just going to put here CO₂. We all know that's carbon dioxide. So you can see that in both cases, we're oxidizing. We're adding more bonds to O, but the exact products are a little bit different.

So, everyone got that so far? We've got formic acid. We've got CO₂. But now we have to look at the alcohols. Those are the carbonyls, but what about the alcohols? Well, if you have an alcohol, that is an internal alcohol. What do I mean by internal? There are things on both sides, so it has something on the top and something on the bottom. What you're going to wind up getting from that is also formic acid. You're also going to get one equivalent of formic acid for every internal alcohol that you have. Okay? Cool.

And let's say you have a terminal alcohol. Now by the way, this terminal alcohol I drew it as if it was the bottom, but we know that terminal alcohols could also exist on the top because maybe you have a ketone on the top, so then you have an alcohol there. That's fine. Whenever you have a terminal alcohol, your product is going to be this, which is the condensed formula for it is CH₂O, which is also known as formaldehyde. Oops, let's try that again. Formaldehyde, and we know that formaldehyde is the simplest aldehyde. Okay. So, you're either going to get the smallest aldehyde or the smallest carboxylic acid, or CO₂, carbon dioxide. Okay?

So, that's what you need to know, and if you know these four cleavage patterns, then you should be able to take any molecule, react any sugar, sorry, any monosaccharide, react it with periodic acid and predict exactly what products you're going to get. Okay. So why don't we go ahead and do an example of this together to get our practice in figuring out what the products would look like.

Predict the products of the following oxidative cleavage

Alright. So here's our first reaction with periodic acid and a sugar. What we see is that we're actually starting off with a ketose here and we're reacting with periodic acid and water, which is just a form of periodic acid to react with, so this is perfect. We know that what we're going to get is oxidative cleavage, and actually you could predict ahead of time how many molecules you're going to get because how many carbons do we have total? 1, 2, 3, 4, 5. With oxidative cleavage of monosaccharides, you're never going to be cleaving at every single spot. If it's a monosaccharide that's in a straight chain, you should be cleaving at 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 1. What oxidation product and I'm going to actually 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 going to be formaldehyde and I'm going to draw it on its side so that it looks a little bit more like it did originally. The O is going this way. I'm going to draw it this way as well. Maybe it will make it easier to visualize. Great job. You guys all said formaldehyde. Good job. Okay. So let's go to b. So then b is this carbon. What are we going to get from that ketone? Remember that ketone reacts with periodic acid to give what? CO₂ . So let's go ahead and write that in. So now we have 1 mole of CO₂ . This is getting cool. Let's start off. Let's go down to C. So what would we get with C? Well, what I see with C is that is this an internal alcohol or a terminal alcohol? It's internal because that has stuff on both sides, right? So that means that I'm going to get 1 mole of formic acid.

Predict the structure of the glycoside products

Guys, this is a hard problem. Let me go ahead and read it, and then we'll try to work through it together. In aqueous base, D-glucose has the ability to epimerize into small amounts of D-mannopyranose and also rearrange into D-fructofuranose. Fischer glycosylation can then transform these saccharides into O-glycosides. So the two-part question says, a) predict the structure of the glycoside products after treatment with acid and methanol, and b) how could the treatment of those glycosides with periodic acid distinguish if epimerization or rearrangement is more favored? Cool. So that was a mouthful.

Guys, so this is what it's basically saying in the first little sentence. It's saying that D-glucose, through two separate processes, can either become one thing or another. Okay. Now I have videos on both of these processes, on epimerization 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. Okay? Because what we're really trying to focus on is what happens after these form. Just take my word for it that D-glucose can provide these two cyclic sugars, a furanose and a pyranose.

Now where the question really starts is it says, okay, predict the structure of the O-glycosides that would form after each of these is treated with methanol and acid. This part is actually pretty easy because you should have learned about O-glycosylation already or also Fischer glycosylation, and you should be familiar with which alcohols react when you react with alcohol and acid. Do you remember? Notice that all of the Os are missing their Hs, so we're going to have to predict together if it remains as an alcohol or if it gets a methyl group on it. So what do you think? Here I have 1, 2, 3, 4, 5 potential

Here’s what students ask on this topic:

What is the role of periodic acid in the oxidative cleavage of monosaccharides?

Periodic acid (HIO₄) plays a crucial role in the oxidative cleavage of monosaccharides by cleaving vicinal diols. When periodic acid reacts with vicinal diols in sugars, it forms a cyclic periodic ester intermediate. This intermediate then undergoes cleavage, breaking the sigma bond between the two carbons and forming carbonyl compounds. The products depend on the type of alcohol or carbonyl present in the sugar: aldoses yield formic acid, ketoses produce carbon dioxide, internal alcohols generate formic acid, and terminal alcohols result in formaldehyde.

How does periodic acid cleave vicinal diols in sugars?

Periodic acid cleaves vicinal diols in sugars through a mechanism that involves the formation of a cyclic periodic ester. Initially, periodic acid reacts with the vicinal diol to form this cyclic ester. The ester then undergoes oxidative cleavage, breaking the sigma bond between the two carbons. This cleavage results in the formation of carbonyl compounds. The specific products depend on the type of diol: internal diols yield formic acid, terminal diols produce formaldehyde, and carbonyl groups in aldoses and ketoses yield formic acid and carbon dioxide, respectively.

What are the products of oxidative cleavage of aldoses and ketoses with periodic acid?

The products of oxidative cleavage of aldoses and ketoses with periodic acid differ based on the type of sugar. For aldoses, which have an aldehyde group at the terminal end, the oxidative cleavage results in the formation of formic acid (HCOOH). For ketoses, which have a ketone group, the cleavage produces carbon dioxide (CO₂). These products are formed due to the specific oxidation patterns of the carbonyl groups in these sugars when exposed to periodic acid.

What is the mechanism of oxidative cleavage of vicinal diols by periodic acid?

The mechanism of oxidative cleavage of vicinal diols by periodic acid involves several steps. First, periodic acid reacts with the vicinal diol to form a cyclic periodic ester. This intermediate then undergoes cleavage, breaking the sigma bond between the two carbons. The electrons from the broken bond are used to form carbonyl groups, resulting in the formation of aldehydes or ketones. The overall process involves the redistribution of electrons and the formation of carbonyl compounds, illustrating the oxidative nature of the cleavage.

How do internal and terminal alcohols in sugars react with periodic acid?

Internal and terminal alcohols in sugars react differently with periodic acid. Internal alcohols, which have substituents on both sides, are oxidized to form formic acid (HCOOH). Terminal alcohols, located at the end of the sugar molecule, are oxidized to produce formaldehyde (CH₂O). These reactions are part of the oxidative cleavage process, where periodic acid cleaves the vicinal diols in the sugar, leading to the formation of specific oxidation products based on the position and type of alcohol.