Reducing Sugars: Study with Video Lessons, Practice Problems & Examples

Reducing sugars are defined as any straight-chain monosaccharide or cyclic monosaccharide with at least one cyclic hemiacetal group. These sugars can be oxidized to form aldonic acids, as demonstrated by tests like Tollens, Benedict's, and Fehling's, which provide visual cues such as silver mirrors or brick-red precipitates. Understanding the distinction between hemiacetals and acetals is crucial, as hemiacetals can revert to straight-chain forms, allowing them to participate in oxidation reactions. This knowledge is essential for identifying reducing sugars in carbohydrate chemistry.

Reducing Sugars

Hey fam, in this video I want to do a comprehensive review of one of the most commonly tested topics in carbohydrate chemistry and that is the topic of reducing sugars. Let's go ahead and get started. So guys, whereas a lot of different textbooks or maybe even videos online would start at the definition of reducing sugars for this topic, I think it's going to be easier for you to understand if I first relate the reactions we're looking at to reactions that we've already learned, so that you understand the reactions first and then I'm going to go into the definition of reducing sugars a little bit more, okay. So just kind of hold on tight, it's all going to make sense at the end. I'm just going to mix things up a little bit. Alright guys, so what do we already know? What we already learned is that a reaction called bromine water, but it's not called bromine water, bromine water is the reagent and it's called weak oxidation and what it can do is it can provide high yields of altoic acid. So remember that we actually had the same exact molecules displayed when we were learning about weak oxidation and what we learned is that if you start off with an aldose and you react it with bromine in water, what you're going to get through a mechanism that you don't really need to know is the formation of an aldonic acid. So you already know this, this reaction is called weak oxidation, wonderful, but this reaction has a limitation. The limitation is that even though it works really well, it does not undergo a visible transformation. Now that actually isn't a big deal in modern days because we have NMR and a lot of different analytical techniques that we can use to verify if a reaction has completed or not, but thinking back to the 1800s when these reactions were first developed, it was kind of nice to have reagents that would turn red or turn blue or green once they reacted. So the reactions we're going to learn about are reactions that accomplish the same exact transformation while giving us visual cues. And those reactions are going to be the Tollens' test, the Benedict's test, and the Fehling's test. And these tests are going to react with what we call reducing sugars, which I'm going to define very soon, into aldonic acid, but they're not just going to do it the way that bromine water did it, they're going to do it while also providing specific visual indications that the reaction took place, okay? So the reaction proceeds forward by oxidizing. The test will oxidize any sugar capable of forming a straight chain aldose or ketose and this is going to be our definition of reducing sugars for right now. So when I said that it reacts with reducing sugars, we're going to say the reducing sugar is any sugar that's capable of forming a straight chain aldose or ketose. So D-Mannose, is that a reducing sugar? Yeah, because it's a straight chain aldose, right? So what we're saying is that these reactions, Tollens', Benedict's, Fehling's, they're all going to do the same thing. They're going to turn that D-Mannose into they're going to oxidize it and turn it into a carboxylic acid and then they're going to give us specific visual indications. Okay. So let's go really a little bit more into what the reagents are for each of these that you can know what to expect when you look when you find this on the exam let's say. So let's start off with Tollens', which is the most common of the 3. So the Tollens' test has been around for a really long time and it it seems like every book or every professor has their own unique cocktail of reagents that they like to use for the Tollens' test. So sometimes you might see it drawn out like this, sometimes you might see it drawn out like this, they look very different from each other, but there are commonalities between them and what the commonalities are is that there's always going to be some elemental silver, there's always going to be some ammonia and there's always going to be some base. So what I want you guys to think about is I don't want you to memorize the exact order of the letters of the reagents because there are actually like 5 other ways that that can be drawn. You might see it simply as just like silver oxide ammonia, like NH₃. So I don't really want you to focus too much on the exact letters as much as the general picture that if you're reacting some kind of sugar with silver, ammonia, a combination of silver, ammonia and base that's going to be a Tollens' reagent. Okay? Now what does Tollens' do? It does the same thing as Bromine water but instead of just getting your altoic acid, you're also going to get a silver mirror on the outside of your test tube. So that it looks silver and you're going to know, oh there was a reducing sugar present. Okay? By the way, we're assuming like I said, this is a reducing sugar now. This is what we call a reducing sugar. Next, we have Benedict's and Fehling's test. Now Benedict's and Fehling's test actually do the same exact thing. They react with your reducing sugar, they turn it into an aldonic acid, they just have a different visual indication and the visual indication they have is that they turn into copper I oxide, Cu₂O and they form a brick red precipitate that settles to the bottom of the test tube. Okay. Now you might ask, well what's the difference between Benedict's and Fehling's? Guys, it doesn't matter. They're just slightly different complexes of copper II but as long as you see some kind of copper II it's going to be Benedict's or Fehling's. Okay? Now something that's interesting about this one is that it actually starts off blue and turns red at the end. So it's a kind of cool transformation that would you would know for sure that a reducing sugar was present, okay. Now slight disclaimer guys, not every professor, not every textbook, not every homework will require you to know all these 3. Some professors just care about Tollens', some professors just care about Benedict's, but I just decided to include them all in this lesson because it's like why not? They're all so easy, they're all so similar. It's like the more you know. Now you know that there are 3 different tests that all do very similar things, you should just know the difference in their visual indication. Okay. So one last thing I want to point out before we move on to kind of the bottom of the page, which is that one of the biggest misconceptions of reducing sugars is that students hear the term reducing sugar and they think that that means that the sugar is going to get reduced, right? But what are we saying? We're actually not saying that at all. What we're saying is reducing sugar is a sugar that can be oxidized. It's a sugar that has an aldose present like this one that can be oxidized. So it's important for you guys to know that that's a very easy trick that a lot of people mess up on, a lot of people don't understand it, but a reducing sugar is actually a sugar that can eventually be oxidized, okay? And then I guess one last disclaimer, which is that if you're paying really close attention, you might notice that ketones usually can't be oxidized very well. So you might be thinking Johnny, why would a ketose react in an oxidation reaction? Guys, because usually these reactions are in a base solvent or a base catalyst. So in base, we actually know that ketoses can tautomerize to aldoses. So what would wind up happening is that the ketose can react, it just has to tautomerize into the aldose first and then it can it can react through this reaction. So basically that's why we say that a reducing sugar is any sugar that's capable of forming an aldose or a ketose because if it's a ketose it can always just tautomerize back to the Aldose and then react. Okay. Guys, so we're done with this part. Now just to really briefly recap, Tollens', Benedict's and Fehling's test do the same thing as weak oxidation except that they provide visual cues and now what we're going to do in the next video is I'm going to go dive way deeper into the definition of reducing sugars because my definition right now is any sugar capable of forming a straight chain aldose or ketose, but there are actually a lot of different situations and a lot of tricky situations that you need to be aware of to know if it can form an aldose or a ketose. So in the next video, I'm gonna show you guys the definition of reducing sugars and how to predict if a sugar will be a reducing sugar or not. Let's move on to the next video.

Reducing Sugars

So what is a reducing sugar? Alright guys, so I made it sound really complicated in my last video, it's not, there are just specific rules you need to be aware of. The first one is easy, a reducing sugar is any straight chain monosaccharide. So literally, it doesn't matter if it is an aldose or a ketose, if you see it in straight chain it is for sure a reducing sugar, okay? So that takes care of a lot of your examples. But it could also be any cyclic monosaccharide that's a ring disaccharide or sugar derivative that has at least one cyclic hemiacetal group present, okay. So hemiacetal, you guys should know that, but here's a reminder, a cyclic acetal would have OR groups on both sides. So you have OR OR. Remember that a cyclic hemiacetal would have an OR on one side, but then a free alcohol on the other and OH on the other side. This is basically present on a lot of different monosaccharide sugars. And if you have a hemiacetal present, that is a reducing sugar. Now you might be saying, well, why? What's the big difference? Why does it matter? Because guys, hemiacetal groups can be hydrolyzed to straight chain saccharides, whereas acetals cannot. Once you're in the acetal, this is actually what we call a glycoside. Remember glycosides guys, or glycoside? Remember that you put the carbon on there, and then it's pretty much locked in place unless you react it with a lot of acid and hydrolyze it off. But a cyclic hemiacetal is constantly hydrolyzing back to the straight chain, whereas an O-Glycoside is not. So that's why a cyclic hemiacetal will be a reducing sugar because we would expect that in a solution it's going to slowly hydrolyze back into the straight chain structure. Does that make sense? Whereas the other one as an O-Glycoside, it's locked into place unless we specifically want to hydrolyze it with acid. Okay? So guys let's go ahead and do the next question as an example.

Identify the following sugars

Identify the following sugars as either reducing sugars — you can just write RS — or non-reducing sugars — you can just write NS. Okay? Guys, if you haven't taken the time to already solve this, I would encourage you to pause the video and try to solve it. But if you already have, I'm going to go ahead and give you the answers now.

Okay. So let's look at the very first one. The very first one actually appears to be glucose. You don't have to necessarily memorize that, but just it looks like glucose to me. So is that a reducing sugar or not? What do you guys think? Well, I do see a hemiacetal. It's definitely not a straight chain, so that's important. It kind of fails the first test. It doesn't have a straight chain. But, does it have a free hemiacetal? Yes, it does. That hemiacetal would be right here. Okay. See how I have a carbon that has an O and an OH? Guys, it's always going to be the anomeric carbon by the way. The anomeric carbon has a hemiacetal that's present, so then this would be a reducing sugar, I'm going to put RS, okay.

By the way, you might be saying, “Johnny, aren't there lots of hemiacetals on this molecule?” Actually, no, there are no other hemiacetals. There are alcohols; even this is an alcohol but it's not close enough to the O here to be a hemiacetal. So all these other OHs are just alcohols. This is the only alcohol that's actually considered to be a hemiacetal. Cool. So let's move on to the next one.

This next one is tricky. So this next one is a disaccharide and disaccharides, you don't need to know a lot about them yet, but we do need to know if it's a reducing sugar or not. So is this disaccharide a reducing sugar? Does it have any free hemiacetal groups? Well, let's look at the left side of the sugar first. What I see is that I have this carbon, that's the anomeric carbon, and that anomeric carbon is attached to OR, OR and then OR, right? This is an O-glycoside. This is a sugar that has formed an acetal linkage. So is this a reducing sugar? What do you guys think? Do you see an acetal, right? So this would look more like this.

So very commonly a lot of students will say this is not a reducing sugar because it has an acetal present. But that's actually the wrong answer. This is a reducing sugar. And why is that? Because guys we have two functional groups. Remember I said you have to have at least one hemiacetal. This is not a hemiacetal, so that part can't react with the Tollens' reagent or whatever. But what do we have over here? Over here, we have a free hemiacetal. So since we have at least one free hemiacetal on the entire structure, this would still be a reducing sugar. So we would still write RS for this one. Tricky, right?

This is why I wanted to go more in-depth with you guys because reducing sugars are very commonly tested and you need to know what the tricks are to know whether one is reducing or not. So this once again is a reducing sugar not because of the first acetal linkage but because of the second hemiacetal on this side.

Now I'm going to take myself out of the screen for the last one. So guys, this is a straight chain monosaccharide. So is it a reducing sugar or not? By the way, I want to point out this is a ketose. Is that a big deal? Can you reduce, can you oxidize a ketose? Totally. Remember I said it's any straight chain monosaccharide, it doesn't matter whether it is a ketose or an aldose because I assume that in base, in the solution that's provided, it's going to tautomerize back to the aldose and then it's going to react with the oxidizing reaction. So I would also say that this is a reducing sugar. So far, all three of our answer choices here were reducing sugars, but they were all reducing sugars for slightly different reasons and there were some tricks associated with them. Cool?

Alright guys, so I hope this made sense. Let's move on to the next video.

Identify the following sugar as reducing sugars (RS) or non-reducing sugars (NS)

• A. Reducing sugars
• B. Non-reducing sugars

Identify the following sugar as reducing sugar (RS) or non-reducing sugar (NS)

• A. Reducing sugar
• B. Non-reducing sugar

Identify the following sugar as reducing sugar (RS) or non-reducing sugar (NS)

• A. Reducing sugar
• B. Non-reducing sugar

Identify the following sugar as reducing sugar (RS) or non-reducing sugar (NS)

• A. Reducing sugar
• B. Non-reducing sugar

Here’s what students ask on this topic:

What is a reducing sugar?

A reducing sugar is any sugar that can act as a reducing agent due to its free aldehyde or ketone group. This includes all monosaccharides in their straight-chain form and any cyclic monosaccharide with at least one cyclic hemiacetal group. These sugars can be oxidized to form aldonic acids, which is a key characteristic used in tests like Tollens, Benedict's, and Fehling's. These tests provide visual cues, such as a silver mirror or a brick-red precipitate, indicating the presence of a reducing sugar.

How do Tollens, Benedict's, and Fehling's tests identify reducing sugars?

Tollens, Benedict's, and Fehling's tests identify reducing sugars by oxidizing them and providing visual cues. The Tollens test uses a combination of silver, ammonia, and base to oxidize the sugar, resulting in a silver mirror on the test tube. Benedict's and Fehling's tests use copper(II) ions to oxidize the sugar, forming a brick-red precipitate of copper(I) oxide (Cu₂O). These visual changes confirm the presence of reducing sugars, which can be oxidized to form aldonic acids.

What is the difference between a hemiacetal and an acetal in the context of reducing sugars?

A hemiacetal has an OR group and a free OH group on the same carbon, while an acetal has two OR groups. In the context of reducing sugars, hemiacetals can revert to their straight-chain form, allowing them to participate in oxidation reactions. Acetals, also known as glycosides, are more stable and do not easily revert to the straight-chain form, making them non-reducing sugars. This distinction is crucial for identifying reducing sugars, as only those with hemiacetal groups can be oxidized in tests like Tollens, Benedict's, and Fehling's.

Why are ketoses considered reducing sugars even though they are not easily oxidized?

Ketoses are considered reducing sugars because they can tautomerize to aldoses in basic conditions. This tautomerization allows the ketose to convert into an aldose, which can then be oxidized. In tests like Tollens, Benedict's, and Fehling's, the basic environment facilitates this conversion, enabling the ketose to participate in the oxidation reaction and produce the visual cues associated with reducing sugars.

What visual cues indicate the presence of reducing sugars in Benedict's test?

In Benedict's test, the presence of reducing sugars is indicated by a color change from blue to brick-red. The test starts with a blue solution containing copper(II) ions. When a reducing sugar is present, it reduces the copper(II) ions to copper(I) oxide (Cu₂O), which forms a brick-red precipitate. This color change confirms the presence of a reducing sugar, as it indicates the sugar has been oxidized.