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.
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 cyclisation. So, guys by definition monosaccharides always contain at least one carbonyl group, 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 it's but in an acidic condition. So, remember that in the carbonyl section of your textbook, we learned that the nucleophilic addition of one alcohol on a carbonyl produces a functional group called a hemiacetal. Remember that? and remember that then a second equivalent 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 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 to 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 for, 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, let's try to draw what would that look like and the way that you could draw that out is by using the general structure. Remember, the general structure is that you have OH, let me put it on the other side, is that you have OH OR and then 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 weight any intermolecular reaction you count out the distance from the nucleophile to the electrophile so it would be one 1, 2, 3, 4, 5. So, I know this would be a five membered ring. So, then what I would do is I would erase this R erase this R and replace it with a five membered ring that looks like this, okay? Where, this is going to be carbon or this is going to be atom one, this is going to be atom two, this is going to be atom three, this is going to be atom four and this is going to be atom five, okay? So, that is a very ugly cyclic hemiacetal but that is the answer, that is the correct answer to this to this nucleophilic addition, and this one is stable, it wouldn't actually 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-glusoce is, it undergoes nucleophilic addition to form a cyclic six-member hemiacetal where basically this O it's always going to be the penultimate, not always but man times it is 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 already know the top one is one and then it goes down from there but regardless it's six right, there's six atoms in this ring so 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? Not carbon 6 is 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, we're now I'm going to have R group, H group. So, those are your two groups that stayed from the beginning and then you're going to have OH and then the OR group, which is that guy right there. So, basically you have the four groups form 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 a hemiacetal, okay? Great, let's move on.
Problem: Provide the mechanism for the cyclic hemiacetal formation of the following hydroxycarbonyl.6m
4-Hydroxy- and 5-hydroxyaldehydes exist primarily as cyclic hemiacetals. Draw the structure of the cyclic hemiacetal formed by each of the following:
a. 4-hydroxybutanal b. 4-hydroxypentanal c. 5-hydroxypentanal d. 4-hydroxyheptanal
Propose a curved arrow reaction mechanism for the following acetal formation reaction.