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

Ruff degradation is a chain-shortening reaction that utilizes decarboxylation to remove carbon atoms from sugars. Starting with an aldehyde, it is oxidized to a carboxylic acid using bromine water, followed by a radical mechanism involving hydrogen peroxide and iron sulfate to facilitate decarboxylation. This process results in the loss of one carbon as CO₂ and transforms the C2 stereocenter into an achiral aldehyde, yielding a specific degradation product, such as D-arabinose from D-mannose.

Monosaccharides - Ruff Degradation

Hey guys. In this video, I want to talk about a reaction that's kind of opposite to Kiliani Fischer chain lengthening and that is the Ruff degradation, which is a chain shortening reaction. Let's look into it. So guys, we know that Kiliani Fischer exists to lengthen chains. We would use these cyanohydrins and then reduce them and eventually we'd get a new aldehyde group at the top. But it turns out that reactions have also been designed to take carbons away. And one of the ways that we learn how to remove carbons in Organic Chemistry 2 is through a reaction called decarboxylation. Do you guys remember decarboxylation? It was a reaction where a carboxylic acid turns into CO₂ gas and you lose a carbon in the process. Well, genius, in a genius move they decided to apply that to sugars. And they said, hey if we can turn the aldehyde from a sugar into a carboxylic acid then we could probably decarboxylate it somehow, right? And that would shorten the chain and that's exactly what we're going to learn how to do today. So every time you do one of the Ruff degradation cycles, you're going to lose 1 carbon and you're going to lose it because 1 CO₂ molecule is flying off into the atmosphere. Okay? Now unlike Kiliani Fischer where remember how Kiliani Fischer would make 2 different epimers because you were adding a new chiral center and you didn't know which direction the OH would go. But in a chain shortening reaction, we're actually having fewer chiral centers, so that means that we're going to have a stereo specific product. We're not going to have to worry about a mixture of epimers like we do with Kiliani Fischer, okay. Now, just as a quick disclaimer, the C2 stereocenter is the one that's lost in every cycle. So what we're talking about is this guy right here. Notice that right now I picked D Mannose as my original monosaccharide, which means that the OH is faced this way. But afterwards that information is going to be lost because it's going to turn into an aldehyde. Notice that right now it's chiral, but after my reaction it's going to be achiral. So there's going to be no more stereo specific information at that position. Okay? Cool. So now I know you guys are ready to get into the reagents. I've been talking a lot. So what are the reagents used for a Ruff degradation? Well guys, the first one you already know, it's bromine water. So remember that I said scientists were thinking hey, there's got to be a way that we can do a decarboxylation. Guys, the easiest way that we know to turn an aldehyde into a carboxylic acid into an aldonic acid is just to use weak oxidation with bromine water. We've done this before, you don't need to know the mechanism, but you do need to know that this will oxidize to a carboxylic acid. Cool? So that's the easy part. We already know that from before. Now the part that's a little bit more tricky is that this is actually not the type of carboxylic acid that is easy to decarboxylate. Do you guys remember on this you can go ahead and look this up if you type in decarboxylation into the search bar, CLUTCH, you'll see my whole video on this reaction. But from memory, do you guys happen to remember which types of carboxylic acids were the easy ones to decarboxylate? It was the beta carbonyl, or the beta keto, carboxylic acid. So remember that it would always help that, if this is your alpha carbon, you want to have like a carbonyl next to it and that would make the whole mechanism go quickly and you'd be able to easily decarboxylate it. Okay? But we don't have that. In fact, we have no other carbonyls. So technically this shouldn't really decarboxylate that easily, and that's why we're going to need very special reagents to do the next step. The next step is actually not going to proceed through the same decar mechanism you learned in the past. It's going to proceed through a new mechanism that's actually mostly unknown. All we really know is that it's a radical mechanism because it uses hydrogen peroxide, which is a radical initiator and then an iron sulfate complex. These two things together, what they're going to do is not only are they going to use radicals to decarboxylate, but they're also going to oxidize. Okay. So they're going to do all that. They're going to use radicals to decarboxylate, take it off and then to oxidize the final alcohol that's left. So what you need to know here is not really the mechanism, but what happens and how it works. So remember that in decarboxylation you cleave off whatever carboxylic acid you have and the C and 2 O's become CO₂. So that's where this is going to go, it's going to become CO₂ gas. Okay? Now the three sugar, I mean sorry not 3 sugars, these three hydroxyl groups are the same as the 3 hydroxyl groups over there. The only difference is that now my C2 position right here like I told you guys is going to become an aldehyde. So it's going to basically turn into a double bond. I have it drawn to the left over here. Here, I have it drawn to the right. It doesn't matter because it's trigonal planar. So there's free rotation around that bond, it doesn't matter which direction you draw it in, they're the same thing. Okay? And again, I'm not going to show you the whole mechanism, but you should just know that this is what this step does. The radical decarboxylation step takes off the aldehyde and oxidizes that C2 alcohol into an aldehyde. Okay? And then notice guys what we're left over with is only the last 3 hydroxyls in the same place. Everything else above those last 3 hydroxyls got chopped off or changed, okay. And that is how we get D arabinose in this case, which is the degradation product of D mannose. Cool, awesome guys. So hopefully that made sense, let's go ahead and move on to a practice problem.

Which aldohexoses produce the same Ruff Degradation product

Circle the Aldohexoses that would produce the same rough degradation product. If none would share a degradation product then just write Na. So guys before we can determine which ones are going to share a product, we need to do the same transformation to all of them. So do you guys remember how to do this? It's pretty easy. All we're going to do is take off the top carbonyl. So let's just take that off and pretend it doesn't exist anymore. Okay. If it even if it makes it easier for you scratch them out because they really don't matter anymore, okay.

Now this is going to be a little bit makeshift, but try to turn try to draw it in such a way so you can turn that top O now, the one that used to be at C2, turn it into an aldehyde. So I'm just going to do this double bond, double bond, double bond, double bond. Those H's are going to go away in the mechanism so we can even just scratch them out. Mechanism so we can even just scratch them out. Cool. Alright, so now we've really done the transformation. It's just in a very like crude way, but we've actually done it in a way that's very understandable carbonyl So don't worry about the direction of the carbonyl. All we really care about is for any of these 4 sugars, do they have their alcohols facing exactly the same directions so that they can be the same Aldopentose?

Do you guys see any that are exactly the same? So I do see one that looks kind of similar, which is D glucose over here related to L Glucose which is over here, okay? And if you I don't know, if you weren't paying attention maybe someone could pick these because they would think, oh but if you flip it then it's the same thing. That's wrong, don't think that way. These are different Aldopentoses guys because this is not a symmetrical molecule. We have the same group on both sides. Remember the top is going to be an aldehyde, the bottom is an

Predict the product for the following reaction

Predict the products with the following reaction. If multiple steps are indicated, then state all the intermediate structures. Cool. So guys, let's figure out what we're working with first and then actually draw it all out. So what I'm starting off with is an aldohexose called dehydose. And my first reaction appears to be bromine water, so I know this should be a weak oxidation. My second one is peroxide and an iron 3 complex or ion and water. Okay. Now actually guys, I drew it this way because I want you guys to know that, you could see that iron sulfate complex written in a lot of different ways, as long as you have some kind of elemental iron 3+ that's all you're really looking for. You're just looking for something that's iron 3+ to promote and to do this radical reaction, okay. So guys, you should know that this is going to be the radical decarboxylation. And I could even say decarboxylation oxidation. Right. Because we know that the aldehyde gets oxidized; the alcohol gets oxidized to an aldehyde at the end. So let's draw the intermediate structures because it says if multiple steps are indicated draw all of them. So the intermediate structure is going to be the carboxylic acid of the first one which is going to be this:

<math> <mo>OH</mo> <mi>C</mi> <mo>(</mo> <mi>O</mi> <mo>)</mo> <mi>OH</mi> </math> (Intermediate structure 1, oxidation of the top aldehyde to a carboxylic acid.)

Cool. And then my second and last step is going to be to get rid of it. So then what I'm going to do is I'm just going to draw this part with an aldehyde sticking off of the top. So it's going to be I'm really just drawing the bottom part first so:

<math> <mi>O</mi> <mo>OH</mo> <mo>O</mo> <mo>CH</mo>₂<mi>OH</mi>, <mi>H</mi>, <mi>H</mi> </math>

What's that what I'm drawing what's in that circle. Cool. And now we know that this position gets oxidized to an aldehyde. So let's draw that:

<math> <mi>O</mi>=<mi>CH</mi> <mo>&plus;</mo> <mi>CO</mi>₂<mi>gas</mi> </math> (This represents the formation of aldehyde and evolution of CO₂ gas.)

Cool. And that's my rough degradation guys. We're done. That was it. Awesome. So we're done with this practice problem. Let's move on to the next video.

Here’s what students ask on this topic:

What is Ruff degradation in organic chemistry?

Ruff degradation is a chain-shortening reaction used in organic chemistry to remove carbon atoms from sugars. The process starts with an aldehyde, which is oxidized to a carboxylic acid using bromine water. This is followed by a radical mechanism involving hydrogen peroxide and iron sulfate to facilitate decarboxylation. The reaction results in the loss of one carbon as CO₂ and transforms the C2 stereocenter into an achiral aldehyde, yielding a specific degradation product, such as D-arabinose from D-mannose.

What reagents are used in Ruff degradation?

The reagents used in Ruff degradation include bromine water for the initial oxidation of the aldehyde to a carboxylic acid. This is followed by hydrogen peroxide and iron sulfate, which facilitate the radical decarboxylation process. These reagents work together to remove a carbon atom as CO₂ and convert the C2 stereocenter into an achiral aldehyde.

How does Ruff degradation differ from Kiliani-Fischer synthesis?

Ruff degradation and Kiliani-Fischer synthesis are opposite reactions. Kiliani-Fischer synthesis is a chain-lengthening reaction that adds a carbon atom to a sugar, creating a new chiral center and potentially leading to epimers. In contrast, Ruff degradation is a chain-shortening reaction that removes a carbon atom as CO₂, resulting in fewer chiral centers and a stereospecific product. Ruff degradation transforms the C2 stereocenter into an achiral aldehyde.

What is the role of hydrogen peroxide in Ruff degradation?

In Ruff degradation, hydrogen peroxide acts as a radical initiator. It helps facilitate the radical decarboxylation process, which is essential for removing a carbon atom from the sugar molecule. Hydrogen peroxide, in combination with iron sulfate, generates radicals that decarboxylate the carboxylic acid, releasing CO₂ and converting the C2 alcohol into an aldehyde.

What is the final product of Ruff degradation of D-mannose?

The final product of Ruff degradation of D-mannose is D-arabinose. During the reaction, D-mannose undergoes oxidation to form a carboxylic acid, followed by radical decarboxylation, which removes a carbon atom as CO₂. This process transforms the C2 stereocenter into an achiral aldehyde, resulting in the formation of D-arabinose.