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

Monosaccharide acylation involves the reaction of monosaccharides at the oxygen position using acid chlorides or anhydrides, typically facilitated by pyridine as a base. This process forms polyester derivatives, commonly acetylated groups (OAC), through nucleophilic acyl substitution. The mechanism begins with pyridine deprotonating hydroxyl groups, generating nucleophiles that react with acyl groups, leading to a tetrahedral intermediate. The final product is a fully acetylated glucopyranose, showcasing the versatility of monosaccharides in organic synthesis.

General Reaction

Now let's talk about an O-site reaction on monosaccharides called monosaccharide acylation. Alright, guys, so monosaccharides have the ability to react at the O position or the oxygen position in several different ways. Specifically, we have a base-promoted reaction with acid chlorides or with anhydrides, you're able to form polyester derivatives of your glucose. Okay? So here what I've done is I've shown you OAc. Maybe you're a little bit not sure what that is. I'm going to explain in a second. It turns out that specifically for this reaction, even though lots of different bases would technically work, pyridine is a really good choice for reasons that I'm going to show you once I get into the mechanism. Pyridine is the most commonly used base by far, and it's probably the best choice. And then, specifically, even though any acyl group can be added, acetyl groups are the most common carbonyls that are added. Now acetyl groups are summarized or the condensed structure is OAc which I have drawn here. Just remember what an acetyl group is, guys, it's an O with a carbonyl with a methyl group. Okay. So, this, and then whatever that is attached to, this would be an OAc group. And OAc would just be O and then an acyl group after that. Okay, and notice that this is an ester, that's why we call it a polyester ester derivative. Okay.

So, guys, before we go through the mechanism, let's just go through the general reaction. So, my reaction has once again, I'm going to start off with beta-D-glucopyranose and I'm going to react it with a base. That base is going to turn these O's into good nucleophiles and then it can either react with an acid chloride or an anhydride to form these polyester derivatives. Now, guys, if you think about it, if you think what we've learned from the other parts of organic chemistry, it's not a far stretch to think about the mechanism that could do this because we know that it's possible for negative charges to come in and do nucleophilic acyl substitution with good leaving groups. So that's all that's going to happen here; this is going to be a NAS, nucleophilic acyl substitution mechanism. But don't worry, I'm going to show you that in a second. But I just want to give you a preview that the mechanism is quite easy to think about. All we're going to do is we're going to add either this group from the acid chloride or this group, which is the same thing from an anhydride. And we're going to add that to every single position. So our product, in this case, would be a fully acetylated glucopyranose. Okay. Also, I just want to make one point, this is not actually called a pyranose because it's not just an R group. It has other oxygens in it, so it's not called a pyranose, that's only when you specifically have an R group in that position, okay. So anyway, you guys know the general structure, let's look at the mechanism and see if it makes sense.

Mechanism

Alright guys, so here's the mechanism and what we see is that we're going to start off with our monosaccharide and we're going to react it with what I just drew out as pyridine. So this is what pyridine looks like and you might be saying but Johnny I mean I know that pyridine is a base, the way that pyridine works by the way is that it has a lone pair so that lone pair can come in and deprotonate any of these O's. In fact, it's going to do it 5 times. So what we're going to get is negatives everywhere. Okay? Now you might be saying, Johnny but is it specifically nucleophilic., Nucleophilic base. Why is that important? Guys, it's important because look at the type of molecule that we're going to use for the second step. In the second step, we're using either an acid chloride or an anhydride. Okay, what would happen if I had used O in the first step? Could O react with my anhydride? Hell yeah! It could hydrolyze that to a carboxylate. Remember back to the 3 rules of nucleophilic acyl substitution; the O could come in here and kick off the O and eventually, I would get a carboxylate. So I don't want to use bases that are going to react with my second step, which is why pyridine is a really good choice because pyridine, it's basic, but it would never actually like attack here. It's not a good enough nucleophile to do that. So Pyridine is a good choice. Now that I have these negative charges, how can they react with my acyl groups? Guys, this is just straight-up nucleophilic acyl substitution where the negative charge could come in. In this case, I'm doing it with an anhydride, and it could grab the bottom, kick electrons up to the top. So let's just draw this out for a second, Okay? So I'm going to draw pretty much the whole thing. So now what this is going to attach to is O and that O is going to be attached to let's see a carbon that has CH₃, that has O negative, that has O and then the rest of this thing here. Okay so this is the tetrahedral intermediate and all the other O groups are preserved so then I would still have O^-, O^-, O^-, O^-. Okay? So this is my tetrahedral intermediate and then what's going to happen is that this negative charge comes down and kicks out this carbon here or the good leaving group. So what I'm going to get at the end is a molecule that looks like this. So I still have all these O's in the same place. But then this one has no O attached to let's just do it this way attached to a double bond O and a CH₃ and then plus my leaving group which looks like this. Okay? And if you're confused about how I got that, just think that this O is this O here. This O is this guy right here. And this methyl group is this one here. So notice that this is what we also call an OAC group, and then this would just happen times 4. It would happen everywhere else so that you would

Here’s what students ask on this topic:

What is monosaccharide acylation and how does it work?

Monosaccharide acylation is a chemical reaction where monosaccharides react at the oxygen position with acid chlorides or anhydrides, typically using pyridine as a base. This process forms polyester derivatives, often acetylated groups (OAC), through nucleophilic acyl substitution. The mechanism starts with pyridine deprotonating the hydroxyl groups on the monosaccharide, creating nucleophiles. These nucleophiles then react with acyl groups, forming a tetrahedral intermediate. The final product is a fully acetylated glucopyranose, demonstrating the versatility of monosaccharides in organic synthesis.

Why is pyridine commonly used in monosaccharide acylation?

Pyridine is commonly used in monosaccharide acylation because it is an effective base that can deprotonate hydroxyl groups without reacting with the acylating agents (acid chlorides or anhydrides). Pyridine's lone pair of electrons makes it a good nucleophilic base, which is crucial for generating the nucleophiles needed for the nucleophilic acyl substitution mechanism. Additionally, pyridine does not hydrolyze the acylating agents, ensuring a smooth reaction process.

What are the common acyl groups added in monosaccharide acylation?

The most common acyl groups added in monosaccharide acylation are acetyl groups. These groups are represented by the condensed structure OAC, which stands for an oxygen atom bonded to a carbonyl group and a methyl group. The acetyl group is frequently used because it forms stable ester derivatives with the monosaccharide, resulting in fully acetylated glucopyranose.

What is the role of the tetrahedral intermediate in monosaccharide acylation?

The tetrahedral intermediate plays a crucial role in the mechanism of monosaccharide acylation. After the nucleophilic attack by the deprotonated hydroxyl groups on the acylating agent, a tetrahedral intermediate is formed. This intermediate then undergoes rearrangement, where the negative charge on the oxygen atom kicks out the leaving group, resulting in the formation of the final acetylated product. The tetrahedral intermediate is essential for the nucleophilic acyl substitution process.

How does nucleophilic acyl substitution occur in monosaccharide acylation?

Nucleophilic acyl substitution in monosaccharide acylation occurs when the deprotonated hydroxyl groups (nucleophiles) attack the carbonyl carbon of the acylating agent (acid chloride or anhydride). This attack forms a tetrahedral intermediate. The intermediate then rearranges, with the negative charge on the oxygen atom pushing out the leaving group, resulting in the formation of an ester bond. This process repeats for each hydroxyl group, leading to a fully acetylated monosaccharide.