Monosaccharides - Forming Cyclic Hemiacetals - Video Tutorials & Practice Problems

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.