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.
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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.
By definition, monosaccharides contain at least one carbonyl group and multiple alcohols. With that in mind, do you remember a reaction from the past that includes both of these groups? That's right, guys! It's the hemiacetal/acetal reaction. Let's see how this works with monosaccharides.
Monosaccharides - Forming Cyclic Hemiacetals
Video transcript
- Recall that the only stable hemiacetals are cyclic (5 and 6-membered rings)
Provide the mechanism for the cyclic hemiacetal formation of the following hydroxycarbonyl.
Problem Transcript
Do you want more practice?
More setsHere’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.
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