Monosaccharides - Forming Cyclic Hemiacetals - Online Tutor, Practice Problems & Exam Prep

Monosaccharides, containing at least one carbonyl group and multiple alcohols, can undergo cyclization through a hemiacetal mechanism. This process involves the nucleophilic addition of an alcohol to a carbonyl, forming a stable cyclic hemiacetal, particularly in D-glucose, which results in a six-membered ring. The anomeric carbon plays a crucial role in determining the configuration of the resulting sugar. Understanding this cyclization is essential for grasping carbohydrate chemistry and the formation of more complex sugars like disaccharides and polysaccharides.

Monosaccharides - Forming Cyclic Hemiacetals

In this video, what I want to do is give you guys a little bit of context for what is going to be the most important reaction of monosaccharides, and that is cyclization. So, guys, by definition, monosaccharides always contain at least 1 carbonyl group, right? Right? Ketone or aldehyde usually, and multiple alcohols. Okay? Now remember, guys, we've already learned in the past how carbonyls and alcohols react with each other, especially in an acidic condition. So remember that in our carbonyls section of your textbook, we learned that the nucleophilic addition of 1 alcohol on a carbonyl produces a functional group called a hemiacetal. Remember that? And remember that then a second addition of alcohol to the hemiacetal produces a functional group called an acetal. And that's what I have drawn out here. Remember that the general structure was that your one OH comes in, and you get an R and OR group, so that would be this one right here. And then your second ROH can come in. It's not exactly that mechanism, but eventually it comes in and replaces the OH, and then you get this one over here. Okay, and that was the general structure of an acetal. Now remember, guys, that usually, hemiacetals are not stable. Remember that you usually never end with the hemiacetal; you go all the way to the full acetal. But there was only one exception to that. Remember that the only exception is a cyclic hemiacetal. And it is possible, if your OH comes from the same molecule as your carbonyl, that you can form a ring with your hemiacetal. And that one is actually stable. So let's draw out what that would look like. And the way that you could draw that out is by using the general structure. Remember that the general structure is that you have, OH, OR, and in this case, HR. Now why did I do that? Because this H comes from here, right? So it's the aldehyde H. This R is this one here. And then this R here is actually the R group on this O. Okay? So now all you have to do is plug in the R groups. So what are the R groups? Well, remember that the way any intermolecular reaction, you count out the distance from the nucleophile to the electrophile. So it would be 1, 1, 2, 3, 4, 5. So I know this would be a 5-membered ring. So then what I would do is I would erase this R, erase this R, and replace it with a 5-membered ring that looks like this. Okay? Where this is going to be atom 1, this is going to be atom 2, this is going to be atom 3, this is going to be atom 4, and this is going to be atom 5. Okay? So that is a very rough cyclic hemiacetal, but that is the correct answer to this nucleophilic addition. And this one is stable; it wouldn't add a second equivalent of alcohol because it's already stable like this okay? So it turns out that many monosaccharides can undergo this reversible intermolecular ring-forming hemiacetal mechanism, and this is what we call this whole part right here, the ring-forming hemiacetal mechanism is what we call cyclization. Okay? So here's an example of cyclization, D-glucose, which you guys should know very well by now what D-glucose is. It undergoes nucleophilic addition to form a cyclic 6-membered hemiacetal where basically, this O, it's always going to be the penultimate, not always but many times it's the penultimate OH, attacks the carbonyl and forms a ring. The size of that ring would be 1, 2, 3, 4, 5, 6. Now guys, I actually didn't number this according to nucleophile to electrophile; I numbered it based on the monosaccharide numbering that we know, the top one is 1, and then it goes down from there. But regardless, it's 6 right? There are 6 atoms in this ring, so then this is what it would look like as a cyclization, as a hemiacetal. What you would get is that this is now 1, 2, 3, 4, 5, and this is atom 6. Okay. It's not carbon 6, it's atom 6. And what we see is that what I labeled as the gray notice that it's gray down here in a gray box? This is that same atom here, so everything lines up. Okay? Now notice that what was in this gray circle before is now this guy down here. How did that happen? Because remember that a carbonyl after nucleophilic addition with alcohol, it turns into a hemiacetal. Where now I'm going to have, R group, H group, so those are your 2 groups that stayed from the beginning, and then you're going to have OH and then the OR group, which is that part right there. So basically you have the 4 groups from a hemiacetal. Now guys, by looking at this, you're not supposed to be able to draw this yet, don't worry. I'm just exposing you to the fact that cyclization happens through a hemiacetal mechanism. Is that fine? I'm going to explain how to get all those positions later, but for right now, just know that it happens through an acetal. Okay? Great. Let's move on.

Provide the mechanism for the cyclic hemiacetal formation of the following hydroxycarbonyl.

Alright guys, so before we can practice drawing the cyclization mechanism of a sugar, why don't we practice drawing the hemiacetal mechanism on a much simpler molecule that just has the important pieces? It has the carbonyl, in this case, a ketone, and it has the OH on the end, but it doesn't have any of the OHs in the middle, which is just going to make this easier to draw. So it says provide the mechanism for the cyclic hemiacetal formation of the following hydroxyl carbonyl, meaning guys this is the same exact mechanism you learned in the acetal section of your carbonyls chapter, okay? So nothing I'm doing here is going to be new. This is a complete application of hemiacetals. So what's the first step of this mechanism guys? You guys remember? Remember that it's always protonation, right? So I'm going to go ahead and just start off with the molecule I've been given and I'm going to protonate it. Cool. Let's go ahead and redraw it underneath with the new proton. Now it has a positive and that OH is still there. What's the next step? Do you guys remember? We practice this so much, guys. It's going to be a resonance structure. So let's go ahead and draw that resonance structure. And remember that the way that the resonance structure forms is that your carbonyl can shift electrons up to the O. Meaning that now this OH is still here but I'm going to have a positive charge on the carbon instead of on the O. Of course, I'm going to put this in brackets to show that this is a resonance structure. And the whole reason that I like drawing that positive charge is because it gives me a very easy target for my nucleophile to show where it's going to nucleophilically add. Now usually I would take my first equivalent of alcohol from out here and I would attack and make the beginning of my hemiacetal. But remember guys, there's no alcohol here. This is not a regular hemiacetal, this is a cyclic hemiacetal where the only source of alcohol comes from the molecule itself. So that means that this OH actually would come in and that's what does the attacking. Okay, it's the O. Now if we want to figure out how big the new ring is going to be because this is intramolecular ring-forming, I have to let's number from nucleophile to electrophile, so it'd be 1, 2, 3, 4, 5, and 6. So I know I'm going to get a 6-membered ring. Okay? So this all happens reversibly, so I'm just going to put double-headed arrow, equilibrium arrows. And now let's draw what this new ring would look like. Now there are a lot of ways that you could draw this ring. I'm just going to go ahead and draw a 6-membered ring like this where the O came in attack from here and this is carbon 6. So like this is 6, 5, 4, 3, 2, 1. Now guys this is not normally the way that you number sugars or anything, this is not a sugar. I'm just trying to name my ring so it makes sense with what you're seeing here. Here. Does that make sense? Is that fine? How I kept 6 in the same position and then I just made a ring around it, okay? So now that we have those atoms numbered, we have to add our other groups to it, kind of plug and chug, right? So what other groups do we have? Well for sure 1 is going to have an H, so let's put a positive. For sure 6 is going to have a methyl and it's going to have an OH. Cool? Awesome. So are we done? No. We need to deprotonate. So let's go ahead and use our conjugate base of water to come in here and pull off that H and give us let's just put more, well I'll put them down here reversible arrows give us our final cyclic hemiacetal that now looks like this and like this. Cool? So see how this is a cyclic hemiacetal? Why? Because this carbon is attached to 2 R groups, RR. And it's attached to OH and it's attached to OR. So that's a hemiacetal and this is stable. Now guys, we're almost done but I just want to point something out. Is there a chiral center on this molecule? There wasn't on the original one but now there is. Isn't this chiral? Because it has 4 different groups on it. So I randomly drew the OH going up, but it could have also gone down, right? So let's go ahead and just draw that we would have another enantiomer but another stereoisomer which is that the same molecule could have occurred, but it's with the OH facing down and with the methyl facing up. Up. Cool? So do you guys see how you would get a cyclic acetal but with 2 stereoisomers. Oh my god, I can't spell. Okay. And this is very important because this is going to be very crucial once we actually start cyclizing sugars, those stereoisomers are going to be important. Cool guys, so hopefully before we talk about cyclizing actual sugars you guys are now more familiar, more refreshed on what the Hemiacetal mechanism is. Alright, cool. So let's move on to the next video.

Here’s what students ask on this topic:

What is the process of cyclization in monosaccharides?

Cyclization in monosaccharides involves the formation of a cyclic hemiacetal. Monosaccharides contain at least one carbonyl group (either an aldehyde or ketone) and multiple hydroxyl groups. During cyclization, a hydroxyl group within the same molecule acts as a nucleophile and attacks the carbonyl carbon, forming a ring structure. This process is particularly important in D-glucose, where the hydroxyl group on the fifth carbon attacks the carbonyl carbon on the first carbon, resulting in a six-membered ring. This ring structure is stable and is a key feature in the chemistry of carbohydrates.

Why are cyclic hemiacetals more stable than linear forms in monosaccharides?

Cyclic hemiacetals are more stable than their linear counterparts in monosaccharides due to the formation of a ring structure, which reduces the molecule's overall energy. The intramolecular reaction that forms the ring minimizes steric strain and allows for more favorable interactions between atoms. In the case of D-glucose, the formation of a six-membered ring (pyranose form) is particularly stable due to the optimal bond angles and reduced torsional strain. This stability is crucial for the biological functions of sugars and their ability to form more complex carbohydrates.

How does the nucleophilic addition of alcohol to a carbonyl group lead to hemiacetal formation?

The nucleophilic addition of an alcohol to a carbonyl group involves the alcohol's hydroxyl group attacking the electrophilic carbonyl carbon. This reaction forms a tetrahedral intermediate, which then stabilizes to form a hemiacetal. In the context of monosaccharides, this process occurs intramolecularly, where a hydroxyl group within the same molecule attacks the carbonyl carbon, resulting in a cyclic hemiacetal. This mechanism is essential for the cyclization of sugars, such as the formation of the six-membered ring in D-glucose.

What role does the anomeric carbon play in the cyclization of monosaccharides?

The anomeric carbon is the carbonyl carbon that becomes a new stereocenter during the cyclization of monosaccharides. In D-glucose, for example, the anomeric carbon is the first carbon. When the hydroxyl group on the fifth carbon attacks this carbon, it forms a new chiral center, resulting in two possible configurations: α and β anomers. The configuration of the anomeric carbon determines the properties and reactivity of the resulting cyclic sugar, making it a crucial aspect of carbohydrate chemistry.

What is the significance of the penultimate hydroxyl group in the cyclization of monosaccharides?

The penultimate hydroxyl group, which is the hydroxyl group on the second-to-last carbon in a monosaccharide, plays a critical role in the cyclization process. In D-glucose, this is the hydroxyl group on the fifth carbon. During cyclization, this hydroxyl group acts as the nucleophile that attacks the carbonyl carbon, leading to the formation of a cyclic hemiacetal. The position and reactivity of the penultimate hydroxyl group are essential for determining the size and stability of the resulting ring structure, such as the six-membered ring in D-glucose.