Monosaccharides have the ability to react at the –O position in several different ways. The simplest of these is called exhaustive alkylation.
Concept: General Reaction4m
In this video we're going to discuss an O position reaction of monosaccharides called alkylation. So, guys monosaccharides have the ability to react at the O position or the oxygen position in several different ways, and the simplest of these is simply called exhaustive alkylation, and what exhaustive alkylation forms is it forms for ether groups or for ethers and an acetal, okay? So, we form four ethers in an acetal, let me just show you right now, this is what the product looks like, you have these four OR groups and then this OR it looks like it's an ether but guys is actually attached to a carbon, that's attached to another OR, right? So, whenever you have two OR's is attached to the same position you don't call that a diether you call it an acetal, okay? Now, the reagents for this reaction are very straightforward and they actually just resemble Williamson ether synthesis, okay? Remember, the general idea behind Williamson ether synthesis is you take an alcohol then it's some kind of base you deprotonated. So, I'm just going to put B minus, you protonate it and then you react it with something like an alkyl halide where it can do a backside attack and you get an OR group, does that kind of, it's off the screen, but does that ring a bell a little bit? Williamson ether synthesis, you're just turning the OH into a nucleophile and then attacking an alkyl halide and guys, that's actually what we're doing here that's one of the sets of reagents we can use.
So, there are several ways that your textbook shows that alkylation is possible but I'm going to show you guys the three most common ways, which are just RX in base, that's literally what I just to showed you right now, any type of base in the presence of alkyl halide will perform a Williamson ether synthesis or in the presence of at least a primary alkyl halide, if it's secondary or tertiary it won't work because SN2 won't be powerful enough, also another leaving group that's possible in base is something similar to a sulfonate ester. So, remember that sulfonate esters were also good leaving groups, right? So, if you have something attached to sulfonate ester or sulfates group, that would work. So, this another good leaving group that can be attacked and then lastly would be a leaving group in silver oxide, which has a slightly different mechanism that I'm going to go over specifically, okay? But regardless, what's the general mechanism? the general mechanism is that you take once again work with our beta-d-glucose pyranose and we expose it to some kind of catalyst it's usually base but it could be silver oxide. So, that's why I just put catalyst and what that's going to do is it's going to turn those OH groups into good nucleophiles, okay? And then it's going to attack, the R, kick off the x and what we're going to wind up getting is a fully alkylated, fully alkylated because every single O has been alkylated beta D-glucopyranoseside. Now, why is this important? Well, guys, we're going to discuss this later but anytime you have an R group coming off of the anomeric carbon this turns from being called a pyranose to a pyranoside or a ringoside, whatever ring it is, furanoside, regardless oside tells us that we have some kind of R group coming off of the anomeric position okay, sorry guys, ready for the mechanism? cool. Let's do it in the next video.
Concept: Base-Promoted Mechanism5m
So, let's start off with the easier mechanism of the three, which is the base promoted mechanism and guys I am being strategic about my use of the word promoted here, I'm not saying base catalyzed and that's because you may not necessarily get all of your OH negative back, the net ionic equation should be the same, meaning that for every minus I have on one side I'm going to have minuses on this side as well but it's just that instead of being an OH minus maybe it will be like the conjugate base of this thing. So, that's why it's promoted because you're not necessarily cattle up, you're not necessarily generating the base at the end, okay? Cool. So, deprotonation this part is easy, we can use pretty much any base, it doesn't have to be OH minus I'm just using that as our standard base but there are so many bases that can be used for this, it could be NaH could be LDA, it could be NH2 minus, whatever, just a strong source of anions, okay? A strong negatively charged base and what I can do is it can fully deprotonate the entire molecule. So, what I'm going to wind up getting is a bunch of negative charges, okay? Now, these might not all happen all at once maybe they happen one at a time and react one at a time but it doesn't really matter, the most important thing is that this happens five times, okay? I'm going to put here times five, cool? Now, that I have these negative charges they're able to do a backside attack, that's the whole point, they're able to do an SN2 mechanism, okay? Now, specifically the alkylating agents that I used here is actually on the similar to a softening ester, it's a sulfate, group and the reason that this one is important is it's a little bit harder, it's a little bit harder to visualize than alkyl halide and that's why I'm using it because I want to show you guys.
So, what can happen is that even though you might not know which side to react with, you've got a carbon with three hydrogen's that have a good back side, right? And then instead of having an x usually you would see this CH3-X, right? We're used to thinking that x is a leaving group but what about all this? isn't this whole thing a really good leaving group? it is, because once you post a negative charge on it it's going to be resonance stabilized. So, that's exactly what happened, you basically just do a backside attack on one of the methyl groups and kick the electrons out to the O. So, it's just an amazing, amazing leaving group, and by the way, you didn't have to pick that side, you could have also picked this side and would have been the same thing. So, what that's going to do is it's going to put a CH3 right here, and then it's going to give you plus OSO3CH negative, right? That's off the screen again, I'm sorry, So what I was trying to show you is that, plus OSO3CH3,3 and that would be like my conjugate base now, but it doesn't have to stop there, it would happen over and over again, so it would be SN2 times 5, so it keeps happening and I would get CH3 here, CH3 here, CH3 here and CH3 here. So, we get a fully alkylated glucopyranoside because once again it's called a pyranoside because I have an R group attached to the anomeric position, cool? By the way, I could have just replaced, I used the kind of weird leaving group, because I wanted you guys to get practice, but I could have just switched it with R-X and it would have been the same thing, it just would have been instead of that weird leaving group it would have been you grab the R and kick out the x, but the net result would have been absolutely the same, okay? Cool, I hope that made sense. Now, let's move on to the silver oxide mechanism.
Concept: Silver Oxide Catalyzed Mechanism4m
So, now let's try out the silver oxide mechanism. So, guys unlike the base catalyzed mechanism or promoted mechanism where you're making the nucleophile stronger by giving it a negative charge, it's actually kind of opposite with a silver oxide mechanism, we're not going to touch the alcohol, the alcohol is going to stay neutral, we're just going to make the leaving group such a great leaving group that is going to end up getting attacked, okay? So, how does this happen? Well, the way that silver oxide looks is it's an oxygen, let's just draw right here, it's an oxygen, let's attach to two silver atoms, okay? And there's a very, there's a pretty strong dipole whereas a o is really negative, okay? So, the o is going to have partial negative and then these are going to have partial positives, okay? This is the O is more electronegative than the silver that makes sense. Remember, oxygen is way more on this side of the periodic table and silver is like in the middle section. So, it's not very electronegative, cool? Well, it turns out that what can happen is that it's going to make a partial bond to the x and it's going to basically make the x more negative because it's going to attach to it and some of that negative energy, some of that negative character is going to be donated to the x, and what that means is it's going to make this R more positive. So, if I were to maybe draw it up in a line where it makes more sense it would look like this, r with a partial bond to x, X with a partial bond to O and then o with the two Ag's and the way that the specific partial charges work is that this is partial negative, which is making this partial negative because it's attaching to it, which is making the r more positive than usual and by being super, super positive it's going to basically make the Oh more conducive to attacking, that which is going to have more of a reason to attack because that R is more positive than usual and then it's just going to kick off the leaving group. So, it's going to happen at the end is that we're going to get an R here, whatever R it was for the alkyl halide, we'll get an r, and then always to do is just multiply that times 4, okay? So, you would end up doing this four times and getting R, R, R, R, okay? Now, guys I actually doubt you're going to be responsible for this mechanism, most likely just you'll to recognize the reagents, which are silver oxide and a silver oxide and an alkyl halide or a good leaving group but I just wanted to throw this in there just in case so that you guys would understand a little bit better how is silver oxide catalyzes the reaction the alkylation reaction, cool? Awesome guys, so we're with this problem, let's go ahead and once the next video.
Draw the expected product of the reaction of the following sugars with excess methyl iodide and silver oxide.
Propose a mechanism for methylation of any one of the hydroxyl groups of methyl α-D-glucopyranoside, using NaOH and dimethyl sulfate.
The mechanism of glycoside formation is the same as the second part of the mechanism for acetal formation. Propose a mechanism for the formation of methyl β-D-glucopyranoside.
Provide the missing reagents, intermediates, or products in the boxes provided.
Please predict the product of the following reaction.