Monosaccharides - Cyclization

Concept: Concept

Video Transcript

Now it's time to dive deep on the most important reaction of monosaccharides and that is cyclization, So that as it turns out that in aqueous solutions monosaccharides are most stable in their Cyclic form and monosaccharides can form stable 5 and 6 membered rings and those have different names so a five carbon cyclic with sugar is going to be called a Furanose and a six carbon cyclic sugar is going to call a pyranose, there are no four carbon cyclic sugars or seven or greater because those aren't stable as rings anymore so really the only stable ones you have to worry about Furanose and Pyranose. So the nucleophilic addition of the penultimate alcohol to the electrophilic carbonyl leads to cyclization, now it doesn't always have to be the penultimate alcohol but usually it is so let me just kind of show you guys what we're talking about here this is DE glucose as you go should be well aware of this is the penultimate alcohol, this is the electrophilic carbon, remember that it has a positive partial positive so the nucleophilic addition meaning the O comes and attacks and kicks electrons up to the O that reaction right there is going to cyclize this straight chain glucose into one of two possible rings, OK? Now why would you get two possible rings? Well the carbonyl carbon if you think about the geometry of a double bond of an SP2 hybridized carbon it's trigonal planar, right? It's trigonal planar so that means that it can be attack either from the top or from the bottom, right? That O could either come around the top and attack or around the bottom and attack, well turns out that that is going to set up first stereoisomers that are possible when these monosaccharides cyclize two different C1 epimers are possible, now what are we talking about here? Let's just go through this really quickly, OK? The way that monosaccharides are numbered is always from the one carbon being the point of attack being the carbonyl because that's the top carbon, this is C2 3 4 5 and 6, OK? We're numbering carbons right now, OK? Well after they cyclize this is now going to be carbon 1 carton 2 carbon 3 carbon 4 carbon 5 and this is actually carbon 6 right here, the oxygen doesn't get a number because it's not a carbon, OK? So when I say that say that cyclization leads to 2 different C1 epimers what I'm saying is that you're going to have 2 different chirality at one that are possible and these two different epimers are actually specifically known as anomers so this is again this goes back to the very old tradition of sugar chemistry that it has unique names that are specific to sugars that only apply to carbohydrate chemistry so you need to know that the specific epimer at the C1 position, the two different positions are called anomers, OK? And what we have is the Alpha anomer that's possible and the beta in anomer that's possible, Ok? Now it turns out that the alpha anomer for specifically with DE glucose the alpha anomer is going to be the one that faces the OH down, OK? Meaning that the O must have attacked from the top because it attacked from the top and pushed the OH down, OK? The beta anomer faces the OH up which means that the carbonyl must have attacked from the bottom...I'm sorry the OH must have attacked the carbonyl from the bottom so that it would push this OH up, OK? Essentially guys this O here is the same as this O is the same as this O so telling us a little bit about the mechanism, it's saying that it attacked from the bottom pushing that OH up, OK? Now guys I don't want you to be overly concerned with being able to remember that, OK? I'm just trying to help it to make sense to you. Now there is a very easy way to remember what's the alpha anomer and what's the beta anomer and that's what I want to tell you guys now you can see there are some blank spaces, OK? So, this is the way it works the alpha anomer is the one that has the OH trans to the stereo descriptor C 5 carbon in this case it C 5 because that's the penultimate carbon of glucose and remember that we talked about that the penultimate carbon is the same as the stereo descriptor carbon it's the one that tells you whether it's a D or whether it's an L, remember that? So guys in this case notice that my carbon 5 is faced up, right? It has a group that's face up which means that my Alpha anomer should be the one that faces down because it's trans, OK?

Now let's look at the other ring the other may notice that my carbon five once again is going up but the beta anomer is the one that faces the anomeric oxygen Cis to the stereo descriptor meaning that should face the same direction so it should also face up, OK? So this is the way you can remember it that the stereo descriptor decides which one is alpha and which one is beta, trans to the stereo descriptor is alpha Cis to the stereo descriptor is beta, now that you have that in mind let's look at this reaction actually kind of take this all in what we notice is that in aqueous solution very little of the monosaccharides actually looks like the Fisher projection that we've been drawing, the straight chain really represents a very tiny fraction of what the sugar actually looks like, the monosaccharides it is either going to look like the cyclized Alpha anomer or the cyclized beta anomer, now specifically for glucose the beta anomer predominates it's about 64 percent of the beta anomer and about 36 percent of the alpha anomer part of this explanation actually lies in equatorial preference, do you guys remember the principle of equatorial preference and cyclohexane and how larger groups like to be equatorial? See how my OH is equatorial? That is going part of the reason that this percentage is higher than that one because it likes to exist in the equatorial form but guys as much as I wish that was always true there are exceptions to this and there are going to be some sugars that actually prefer the alpha anomer for other reasons so just based on this topic you are not going to be expected to know which one is the major anomer, OK? All you need to know is that there are two different anomers, you need to know how to name them and how to draw them, OK? That's really all I'm concerned about for right now, now let's also just take in the nomenclature because the nomenclature has gotten significantly more complicated so remember guys that what you guys know up to this point is that a DE glucose is a glucose with the penultimate alcohol facing towards the right but once it cyclizes is it makes the six-membered ring, right? So that means that DE glucose now turns into DE glucose Pyranose, why is that? Because this stands for the six-membered ring, Right? Six membered rings, cool. Now where is the beta coming from? The beta comes from the anomer so this is how we name which anomer it is so the beta anomer once again is the one that Cis to the stereo descriptor and the alpha anomer over here is the one that's trans so one their you guys are going to notice is that the nomenclature is going to get more complicated as we keep changing the shapes of these of these monosaccharides, cool. So that is one thing that you might really be confused about and that you shouldnÕt know how to do at this point is to determine where the other OHs are going because one thing that we've already discussed is that the anomeric OH can go either up or down because it's the C1 epimers, right? But what about this OH on 2? how do I know the draw down and not up? How about this OH on 3 how I know to draw up and not down? Where are is this coming from? Are you just supposed to memorize it? Guys we have a system it's easy you're going to be fine let's go ahead and learn it. The system it's just a really simple word that I used to describe it down right, OK? So downright is just a random word but it perfectly describes what we see on the cyclization of monosaccharides so all you can think about is that If the OH is on the right of your monosaccharides in the Fisher projection then it's going to point down in the ring, OK? So it's two facing towards the right it's going to point down and that means that if it's facing left on the ring it's going to face up in every...Sorry left on the straight chain it's going to face up in the ring so how does this play out? Let's erase a few numbers let's erase a few...I'm just going to try to simplify this so this is my OH that's important now let's number our other carbons so we have a carbon 1 2 3 4 5 and 6 once again and now we can go carbon by carbon so 2 is to the right which means that it should be facing down and indeed it is, 3 is to the left which means it should be facing up and it is, 4 is to the right and that means that it should be facing down and that's it, that's as far as your rules get you to, your last one the penultimate one the stereo descriptor does not follow that rule so how do I know whether I'm going to face it up or down? Well the penultimate one has a special rule which is that I have it right here your stereo descriptor if it's a D sugar it's going to face up on the ring so if it's a D sugar it's going to face up and what did I tell you guys about the D sugars? They're the most common almost everything is going to be a D sugar so in this case was this a D sugar? Yes it was it was a D Sugar which means that my penultimate one here should face up if it had been L glucose then it would face down that's where the rule comes from for your penultimate or stereo descriptor carbon once again as I don't want to confuse you I put here C5 that's only for this example because it happens to be a Hexose but if it had been a smaller monosaccharides then it would not be C5 it would be whichever one is the penultimate the last chiral carbon the stereo descriptor, OK? So guys just one more thing that I want to give you a preview of before we move on to the practice problem which is that the process of turning a straight chain into a ring is called cyclization but then there's an opposite process that we're going to discuss more later call mutorotation, what is mutorotation? mutorotation is the ability for one anomer to turn into the other anomer by going back into the straight chain so mutorotation would be a smaller process where beta turns into alpha and then where Alpha turns into beta by going through the straight chain and shuffling the OH up and now, OK? Cool guys let's move on to a practice problem.

Problem: Draw the β-anomer predicted through the cyclization of D-mannose