POCl3 Dehydration - Online Tutor, Practice Problems & Exam Prep

Phosphoryl chloride (POCl₃) and pyridine can effectively dehydrate alcohols without acid, preventing decomposition of sensitive molecules. This reaction favors the Zaitsev product, yielding the most substituted double bond. The mechanism involves nucleophilic attack by oxygen on POCl₃, leading to the formation of a stable leaving group. Subsequent deprotonation by pyridine and E2 elimination results in the desired alkene product, emphasizing the importance of regioselectivity in elimination reactions.

General Reaction of Dehydration with POCl3

So it turns out that there's another set of reagents that can also do a dehydration of alcohol. And these reagents aren't going to involve an acid at all. In fact, the reason that we would use this alternative reaction is because some molecules happen to be sensitive to acid and in the presence of acid, they could decompose. So instead of using that, for these what we're going to use is phosphoryl chloride. And what phosphoryl chloride looks like is basically POCl₃. That's what we're going to talk about right now.

So it turns out that POCl₃ in unison with pyridine, which what is that? That's basically just like a benzene ring with a nitrogen on it. Okay? And you don't have to necessarily memorize that right now. I just want you to recognize what it is. So POCl₃ and pyridine are going to do an elimination reaction on alcohol and make a double bond. Okay? So that's all you need to know right now in terms of the general reaction.

Now, any time that we're doing an elimination, that does bring up the whole possibility of Zaitsev, Hoffman, stuff like that. Okay? And it turns out that for these reactions, we're always going to favor the Zaitsev product. Okay? Okay. Zaitsev's rule still applies. So what that means is the most substituted product is the one that I favor. Okay? So in terms of the general reaction, as long as you just know that those two reagents with an alcohol give you a Zaitsev double bond, that's great.

Mechanism of Dehydration with POCl3

I just want to show you guys the mechanism really quick so you'll know what's going on. Okay? So basically, in my first step, what I have is a nucleophilic oxygen. Why is that nucleophilic? Because it's got 2 lone pairs. And then I've got my phosphoryl chloride that looks like this. Now notice that it kind of looks like a carbonyl except I have a phosphorus there instead of a carbon. So if I were to draw any dipoles, does that molecule have any? Yeah. There's actually a lot. There's a bunch of dipoles pulling away from the phosphorus. Okay? So if I had to give a charge, what do you think it would be? Would it be partially positive? Partially negative? What would make sense? This should have a very strong partial positive. It's going to be very electrophilic because all the electrons are basically getting sucked out of the phosphorus. Okay? So what that means is my oxygen sees this and those lone pairs are nucleophilic, so it attacks the phosphorus. Okay? Now this phosphorus is attached to 3 Cls and Cls are really good at leaving. After they leave and become a conjugate base, they're super stable. So we can just kick out any of the Cls that we want. Okay? What that's going to lead to is a structure that looks like this:

O H -P OCl 2

And obviously, we would get the Cl leaving group just by itself. Okay? Now, are there any formal charges that I'm missing? Yes. There's a positive charge right here. And in the second step, pyridine does a deprotonation. Okay? It turns out that the lone pair on pyridine is actually pretty basic. It's very nucleophilic. It's looking for something to give its electrons to, so it can go ahead and deprotonate. In the next step, what happens is that I basically get

O CH 3 -P OCl 2

and I wind up getting pyridine that's protonated and Cl negative. Does that make sense so far? Cool. So now, we're back here down to this bottom part where we're going to do an E2 Beta elimination. It turns out that the P OCl2 just made this oxygen an awesome leaving group. So O used to suck as a leaving group. Now it's an awesome leaving group. Okay? So that means that this is my alpha carbon. How many different beta carbons do I have if I'm going to do an elimination? Well, actually, this is beta and this is beta. Are they both exactly the same? No. They're not because one has a methyl, one doesn't. Do they both have at least 1 hydrogen? Yes, they do. This one has 2 and this one has 1. So which one do I pick? Which one's going to be the one that I eliminate with? And it turns out I'm going to eliminate with the one that's going to give me the Zaitsev product, so that would be green because green is going to give you the most substituted double bond. So in this last elimination step, what I do is I use another equivalent of pyridine. Okay? I'm going to use another equivalent of pyridine and because it's a basic lone pair, it's going to allow me to do this elimination step. And what I'm going to wind up getting is a double bond like this:

C = C

Plus I would get my O-P(OCl₂)^- as a leaving group. Okay? Oh, I'm sorry. By the way, I forgot to draw one arrow. The leaving group hadn't left yet. This is E2. So if I make that double bond, I need to kick out the leaving group. I'm sorry about that, obviously. I get 2 equivalents of protonated pyridines. I'm sorry. It's just going to have an H there. And since it has an H, that nitrogen is actually positive. And I'm going to use 2 of those because the first one happened in the first step with the deprotonation and the second one happened in the elimination step. So I would get all of these things, But guess what? What's the part I actually care about? Just the organic product.

Here’s what students ask on this topic:

What is the role of POCl₃ in the dehydration of alcohols?

POCl₃ (phosphoryl chloride) acts as a dehydrating agent in the elimination reaction of alcohols. It facilitates the formation of a good leaving group by converting the hydroxyl group (OH) into a better leaving group, which is then eliminated to form a double bond. This reaction is typically carried out in the presence of pyridine, which helps in deprotonation and further stabilizes the intermediate species.

Why is pyridine used in the POCl₃ dehydration reaction?

Pyridine is used in the POCl₃ dehydration reaction because it acts as a base to deprotonate the intermediate species formed during the reaction. This deprotonation is crucial for the elimination step, which leads to the formation of the double bond. Additionally, pyridine helps to neutralize the acidic by-products, making the reaction more efficient and preventing side reactions.

How does the POCl₃ dehydration mechanism favor the Zaitsev product?

The POCl₃ dehydration mechanism favors the Zaitsev product because the elimination step follows Zaitsev's rule, which states that the most substituted alkene is the preferred product. During the E2 elimination step, the base (pyridine) removes a proton from the β-carbon that leads to the formation of the most substituted double bond, resulting in the Zaitsev product.

What are the steps involved in the POCl₃ dehydration mechanism?

The POCl₃ dehydration mechanism involves several steps: (1) Nucleophilic attack by the alcohol's oxygen on the electrophilic phosphorus of POCl₃, forming an intermediate. (2) Deprotonation of the intermediate by pyridine, creating a good leaving group. (3) E2 elimination where pyridine removes a proton from the β-carbon, leading to the formation of a double bond and the Zaitsev product.

Can POCl₃ be used for dehydration of all types of alcohols?

POCl₃ can be used for the dehydration of primary, secondary, and tertiary alcohols. However, the reaction conditions and the efficiency may vary depending on the type of alcohol. Tertiary alcohols generally react faster due to the stability of the carbocation intermediate, while primary alcohols may require more stringent conditions. The reaction is particularly useful for alcohols sensitive to acidic conditions, as it avoids the use of strong acids.