Monosaccharides - Alkylation - Online Tutor, Practice Problems & Exam Prep

Exhaustive alkylation of monosaccharides involves the formation of four ether groups and an acetal through mechanisms like Williamson ether synthesis. The process utilizes a strong base to deprotonate alcohols, enabling nucleophilic attacks on alkyl halides or sulfonate esters. The silver oxide mechanism enhances the leaving group, facilitating the reaction without altering the alcohol. Ultimately, these reactions yield fully alkylated glucopyranosides, crucial for understanding carbohydrate chemistry and the significance of the anomeric carbon in glycoside formation.

General Reaction

In this video, we're going to discuss an O-position reaction of monosaccharides called alkylation. 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. Exhaustive alkylation forms four ether groups or four ethers and an acetal. We form four ethers and 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 it's actually attached to a carbon that's attached to another OR. So whenever you have two ORs attached to the same position, you don't call that a diether, you call it an acetal. The reagents for this reaction are very straightforward and they actually just resemble Williamson ether synthesis. Remember the general idea behind Williamson ether synthesis is you take an alcohol, then in some kind of base you protonate it, so I'm just going to put B minus, you deprotonate 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 ring a bell a little bit? Williamson ether synthesis. You're just turning the OH into a nucleophile and then attacking an alkyl halide. That's one of the sets of reagents we can use. 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. 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. Another leaving group that's possible in base is something similar to a sulfonate ester. Sulfonate esters were also good leaving groups. If you have something attached to sulfonate ester, or sulfate group, that would work because it's 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. The general mechanism is that you take once again let's work with our beta-D-glucopyranose and we expose it to some kind of catalyst. It's usually base, but it could be silver oxide, so that's why I put just catalyst. And what that's going to do is it's going to turn those OH groups into good nucleophiles, 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 beta-D-glucopyranoside. Why is this important? 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, phoranoside. Regardless, "oside" tells us that we have some kind of R-group coming off the anomeric position. Are you guys ready for the mechanism? Cool. Let's do it in the next video.

Base-Promoted Mechanism

So let's start off with the easier mechanism of the 3, 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- 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-, maybe it will be like the conjugate base of this thing. So that's why it's promoted because you're not necessarily catalytic; 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-. 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^-, whatever. Just a strong source of anions, okay? A strong negatively charged base. And wha

Silver Oxide Catalyzed Mechanism

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 the 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 it's going to end up getting attacked. Okay? So how does this happen? The way that silver oxide looks is it's an oxygen, let's just draw it right here, it's an oxygen that's attached to 2 silver atoms. Okay? And there's a pretty strong dipole where the O is really negative, okay? So the O is going to have a partial negative charge, and then these are going to have partial positives. Okay? Because 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 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 out 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 2 Ags. 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 positive, it's going to basically make the O more conducive to attacking. That O 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 what'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 all we have to do is multiply that times 4. Okay? So you would end up doing this 4 times and getting RRRR. Okay?

Now guys, I actually doubt you're going to be responsible for this mechanism. Most likely, you just need to be able to recognize the reagents, which are silver oxide and an alkyl halide or a good leaving group. But I just wanted to throw this in there just in case, so you guys would understand a little better how the silver oxide catalyzes the reaction, the alkylation reaction. Cool? Awesome guys, so we're done with this problem. Let's go ahead and move on to the next video.

Here’s what students ask on this topic:

What is exhaustive alkylation in monosaccharides?

Exhaustive alkylation in monosaccharides involves the formation of four ether groups and an acetal. This process typically uses a strong base to deprotonate the alcohol groups in the monosaccharide, turning them into nucleophiles. These nucleophiles then perform nucleophilic substitution reactions (SN2) with alkyl halides or sulfonate esters, resulting in the formation of ether linkages. The reaction is crucial for fully alkylating the monosaccharide, particularly at the anomeric carbon, which transforms the molecule into a glucopyranoside. This process is essential for understanding carbohydrate chemistry and the role of the anomeric carbon in glycoside formation.

How does the Williamson ether synthesis apply to monosaccharide alkylation?

The Williamson ether synthesis is a key mechanism in monosaccharide alkylation. In this process, a strong base deprotonates the alcohol groups in the monosaccharide, converting them into alkoxide ions, which are strong nucleophiles. These alkoxide ions then perform nucleophilic substitution (SN2) on alkyl halides, resulting in the formation of ether linkages. This method is particularly effective for primary alkyl halides. The overall reaction leads to the formation of fully alkylated monosaccharides, with ether groups replacing the original hydroxyl groups.

What role does silver oxide play in the alkylation of monosaccharides?

Silver oxide plays a unique role in the alkylation of monosaccharides by enhancing the leaving group rather than altering the alcohol. In this mechanism, silver oxide makes the leaving group more electrophilic, facilitating its departure and making the carbon more susceptible to nucleophilic attack by the alcohol. This results in the formation of ether linkages without the need to deprotonate the alcohol. The reaction typically involves silver oxide and an alkyl halide or another good leaving group, leading to the formation of fully alkylated monosaccharides.

Why is the anomeric carbon important in monosaccharide alkylation?

The anomeric carbon is crucial in monosaccharide alkylation because it determines the formation of glycosides. When an R group attaches to the anomeric carbon, the monosaccharide transforms from a pyranose to a pyranoside. This change is significant because it affects the molecule's chemical properties and reactivity. The anomeric carbon's involvement in forming glycosidic bonds is essential for understanding carbohydrate chemistry, particularly in the synthesis of complex carbohydrates and glycoconjugates.

What are the common reagents used in the alkylation of monosaccharides?

Common reagents used in the alkylation of monosaccharides include strong bases like NaH, LDA, or NH₂^- to deprotonate the alcohol groups, turning them into nucleophiles. Alkyl halides (R-X) are typically used as the alkylating agents in a Williamson ether synthesis. Additionally, sulfonate esters or sulfates can serve as good leaving groups. Silver oxide is another reagent that can be used to enhance the leaving group, facilitating the alkylation without altering the alcohol. These reagents collectively enable the formation of fully alkylated monosaccharides.