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.
Monosaccharides - Alkylation - Online Tutor, Practice Problems & Exam Prep
Monosaccharides have the ability to react at the –O position in several different ways. The simplest of these is called exhaustive alkylation.
General Reaction
Video transcript
Base-Promoted Mechanism
Video transcript
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
Video transcript
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.
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