Monosaccharides - Wohl Degradation - Online Tutor, Practice Problems & Exam Prep

Wald degradation is a chain-shortening reaction involving aldehydes, where cyanohydrins are formed and then removed to eliminate one carbon. Unlike Kiliani-Fischer synthesis, this process yields a single epimer product, retaining stereochemistry. The mechanism begins with D-glucose converting to an oxime via hydroxylamine, followed by a Beckman rearrangement to form a cyanohydrin. A strong base then facilitates the elimination of HCN, resulting in a monosaccharide like D-arabinose. This reaction exemplifies alpha elimination, emphasizing the importance of carbonyl chemistry in carbohydrate transformations.

Monosaccharides - Wohl Degradation

Hey guys. Now we're going to discuss a reaction that is essentially the opposite of Kiliani Fischer chain lengthening and that is the Wald degradation, which is a chain shortening reaction. Let's go ahead and look into it. So guys, as I mentioned with Kiliani Fischer, aldose aldehydes are susceptible to the same exact carbonyl reactions that we would have learned in our aldehydes and ketones section, and scientists realized a long time ago that they could strategically take advantage of that aldehyde group to do the same types of reactions that we would have done in carbonyl chemistry, but for purposes of cyanohydrins. You can use cyanohydrins to add carbons in Kiliani Fischer, right? But I also told you that cyanohydrins are reversibly added. So it is possible to actually take them off after you've put them on. So opposite to Kiliani Fischer, in this type of reaction which is called Wald degradation, the aldehydes are transformed into cyanohydrins to then be reversibly removed in base. So the idea is that we're making cyanohydrins again, but for the opposite purpose. This time we want to put them on so that we can take them off and remove a carbon. Okay? Now the chain is going to be shortened by 1 carbon at every cycle of Wald degradation. And theoretically, you could keep shortening the chain until you run out of alcohol groups. So just like Kiliani Fischer, you could just keep shortening that chain. Okay? Now one big difference between Kiliani Fischer and Wald degradation is that because this is a chain shortening reaction and not lengthening, you're not going to create multiple epimers. In fact, the Wald degradation is always going to result in a single epimer product. So notice that there are no squiggly lines here, we know the exact stereochemistry of every alcohol because all the stereochemistry is retained, which I'll show you in a second. Okay? Now the reason this happens is because in Wald degradation, the C2 stereo center is lost at every cycle. Now what am I talking about? I'm talking about second carbon, that would be this guy right here, okay. Notice that right now that is an OH that's facing towards the right direction, so that is an R configuration, right? But what we're going to do is we're actually going to, at the end of this reaction, turn that carbon into an aldehyde. So is it still going to be chiral after the fact, after you've done the Wald degradation? No, it used to be chiral and now it's going to be achiral. So what that means is that it's possible for multiple sugars to produce the same product because, for example, if I reacted the Wald degradation with an OH that was faced towards the right on the C2 position, and then I also did a Wald degradation with another sugar that was the same exact sugar except that the OH was faced towards the left. So it's at the opposite C2 epimer. If you reacted with both of those epimers, you'd get the same exact product because, regardless of which one you started with, it's gonna turn into an aldehyde. Okay? Now I know that I've been just talking a lot, now I'm going to go into the mechanism but we are going to practice this concept later. So don't worry, we're going to come back and I'm going to show you guys how multiple sugars could lead to the same product using the Wald degradation. So what I want to do now is really just go into the reagents because I've been talking too much. So first of all, let's start off with D-glucose. D-glucose, the very first step is that we're going to turn it into an amine derivative. Now what we're going to do here guys is we're going to take advantage of the fact that in carbonyl chemistry, what did we learn about ketones and aldehydes? That if you react them with primary amines, let's say NH2R, right? So it's a primary amine. What are you going to get as a product? This is a review. In review of the section of carbonyl chemistry called imines, you could search for it in your search bar if you want. What you're going to find is that this becomes a double bond N and a single bond R. Okay? And this is what we called an imine, right? But then we also talked about in that section how if you have a Z group instead of the R group, so instead of NHR, imagine that it's NH2Z, meaning it stands for something electronegative, something like O, okay? Then the product instead of being NR, it's going to be NZ, okay? And guys that Z is going to be the hydroxyl group in this case. So specifically the primary amine derivative that we're using is called hydroxylamine, and when you react hydroxylamine which is having that O with aldehyde, what you're going to wind up getting is an imine derivative. So this is called an imine derivative. And specifically, the imine derivative is called an oxime. Why is it called an oxime? That's the specific name that we give to the imine derivative of hydroxylamine and all this is reviewed in your imine derivative section of Klutch. So if you want to just type that in to go over these reactions again, that's fine. But I'm just letting you know this is all you need to know that you go from the first step is you go from your saccharide and then you make an oxime, which is really just a type of imine. Great. Now we have our oxime, this is all review. The next step is that we actually want to rearrange that oxime or that imine derivative into a cyanohydrin. Now this reaction is a reaction that I don't expect you to know as well because it's a much it's a much less emphasized reaction in organic chemistry and it's called the Beckman rearrangement. Now guys, the Beckman rearrangement is not unique to the Wald degradation. There are other reactions in organic chemistry that use the Beckman reagent rearrangement, but it's just one that we don't focus on as much as we do it means. So I'm not going to make you I'm not going to hold you accountable to know this reaction very well. In fact, what I want you to do more than anything is just memorize what happens in this step. And what happens in this step is that we use an anhydride. This is the formula for anhydride. Remember anhydrides? We're going to use an anhydride to dehydrate that oxime and rearrange it into a nitrile or a cyano group. Okay. So this is basically a rearrangement that I'm not going to hold you accountable for this mechanism. Your professor won't either, not in this section. You don't need to draw the whole rearrangement, you just need to know that that oxime is going to rearrange to a cyanohydrin, Okay. Now here's the cool part, we learned that cyanohydrins add reversibly. So that means that it's possible to also kick them off if you use base And that's going to be our 3rd step. Our 3rd step is going t

Which aldohexoses produce the same Wohl Degradation product

Circle the aldohexoses that would produce the same wall degradation product. If none of them would share a degradation product, then just write NA. Cool. So, guys, remember that what we're basically discussing here is that wall degradation is going to get rid of your C2 stereochemistry. So whatever that C2 is going to be gone and all you have to look at is what's below C2, that's what's going to be retained. So since these are all aldohexoses, I would expect that after wall degradation, all these carbonyls disappear, right? So, those carbonyls are gone. They turn into CN groups or HCN. And then, all of these oxygens become carbonyls. Let's turn them into double bonds right now. This is going to be a little bit like makeshift but it's going to work. Just put double bonds. Double bond, double bond, double bond, double bond. Okay?

Now keep in mind, guys, what is the stereochemistry of this carbon now once it has a double bond? Trigonal planar. So, it doesn't matter what side the O double bond O is facing. It doesn't matter because it's trigonal planar anyway. So those are the same thing. There's free rotation around that single bond. Does that make sense? So, it doesn't matter. It could be the same if it's towards the right or towards the left. Another thing to keep in mind, I'm ignoring the H because it's going to eventually leave. So, like this H would eventually go away. So I'm literally just saying without a mechanism let's draw carbonyls. Cool.

So, if all these are the same, all those aldehydes left, and if all those carbonyls are the same at the top, which of these 2 are identical? Do you see any 2 of these that are identical sugars? And yeah, I do. I see glucose and mannose. Okay? Glucose and mannose will become the same exact monosaccharide because the top carbonyl is going to leave, and then they're just both going to turn into a carbonyl at that position, the blue position, and then all these other groups are going in the same exact directions. Okay? So, that's actually the answer that mannose and glucose produce the same exact wall degradation product because of the fact that they're C2 epimers of each other. And C2 epimers are lost, the stereochemistry is lost every time you do a wall degradation. Cool? Cool. So that's the end of the question. Let's move on to the next video.

Here’s what students ask on this topic:

What is the Wohl degradation reaction in organic chemistry?

The Wohl degradation is a chain-shortening reaction used in carbohydrate chemistry. It involves converting an aldose (a sugar with an aldehyde group) into a cyanohydrin, which is then removed to eliminate one carbon atom. The process starts with the formation of an oxime from the aldose using hydroxylamine. This oxime undergoes a Beckman rearrangement to form a cyanohydrin. Finally, a strong base, typically methoxide, is used to eliminate HCN, resulting in a monosaccharide with one less carbon. This reaction retains the stereochemistry of the original sugar and produces a single epimer product.

How does the Wohl degradation differ from the Kiliani-Fischer synthesis?

The Wohl degradation and Kiliani-Fischer synthesis are opposite reactions in carbohydrate chemistry. The Wohl degradation is a chain-shortening reaction that removes one carbon atom from an aldose, while the Kiliani-Fischer synthesis is a chain-lengthening reaction that adds a carbon atom. In Wohl degradation, cyanohydrins are formed and then removed, resulting in a single epimer product with retained stereochemistry. In contrast, Kiliani-Fischer synthesis uses cyanohydrins to add carbons, often resulting in multiple epimers due to the creation of new chiral centers.

What are the key reagents used in the Wohl degradation reaction?

The Wohl degradation reaction involves several key reagents. First, hydroxylamine (NH₂OH) is used to convert the aldose into an oxime. Next, an anhydride is employed to facilitate the Beckman rearrangement, transforming the oxime into a cyanohydrin. Finally, a strong base, typically methoxide (CH₃O^-), is used to eliminate HCN and reform the carbonyl group, resulting in the shortened monosaccharide.

What is the role of the Beckman rearrangement in the Wohl degradation?

The Beckman rearrangement plays a crucial role in the Wohl degradation by converting the oxime into a cyanohydrin. This step involves the use of an anhydride to dehydrate the oxime, facilitating its rearrangement into a nitrile or cyano group. This transformation is essential for the subsequent elimination step, where the cyanohydrin is removed using a strong base, ultimately shortening the carbon chain of the monosaccharide.

Why does the Wohl degradation result in a single epimer product?

The Wohl degradation results in a single epimer product because the reaction retains the stereochemistry of the original sugar. During the process, the C2 stereocenter is lost as it is converted into an aldehyde, making the resulting product achiral at that position. This means that regardless of the initial configuration of the C2 hydroxyl group, the final product will be the same, leading to a single epimer product with consistent stereochemistry.