Another form of dehydration uses POCl3 to convert alcohol to a good leaving group.
Concept: General features of dehydration with phosphoryl chloride.2m
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 a benzene ring with a nitrogen on it. 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 an alcohol and make a double bond. 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, Hofmann, stuff like that. And it turns out that for these reactions we're always going to favor the Zaitsev product. Zaitsev's rule still applies. So what that means is the most substituted product is the one that I favor.
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.
Concept: General features of dehydration with phosphoryl chloride.5m
Now I just want to show you guys the mechanism really quick so you'll know what's going on, OK? 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 IÕve got my phosphoric chloride that looks like this, now notice that it kind of looks like a carbon Uel except I have a phosphorus there instead of 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, OK? So if I had to give the phosphorus 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, OK? So what that means is my oxygen sees this and those lone pairs are nucleophilic so it attacks the phosphorus, OK? Now this phosphorous is attached to 3CLs 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, OK? What that's going to lead to a structure looks like this, it's going to be O with an H and then attached to this thing POCL2, OK? And obviously would get the CL leaving group just by itself, OK? Now are there any formal charges that I'm missing? Yes there's a positive charge rate here and in the second step Periodine or not the second step.... Sure second step Periodine does a deprotonation, OK? It turns out that the lone pair on Periodine is actually pretty basic, Ok? So itÕs a very very nucleophilic it's looking for something to give its electrons to so it can go ahead and deprotonate, OK? So in this next step what happens is that I basically get.... Opps Sorry I forgot this methyl group down there so what I basically get is OPOCL2 with the methyl and I wind up getting Periodine 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 POCL2 just made this oxygen an awesome leaving group so OH used to suck as a leaving group now it's an awesome leaving group, OK? So that means that this is my alpha carbon how many different beta carbons do I have? 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 and one doesn't do they both have at least one hydrogen? Yes, they do, this one has one hydrogen or two and this one has one so which one do I pick? Which one is going to be the one that I eliminate with? And turns out I'm going to eliminate with the one that's going to give Zaitsev products 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 Periodine, OK? I'm going to use another equivalent of Periodine and because it's of basic lone pair it's going to allow me to do this an elimination step and what I'm going to wind up getting as my organic product is a double bond like this plus I would get my OPOCL2 negative as a leaving group, OK? Oh sorry by the way I forgot to draw one arrow the leaving group hadn't left it this is E2 so if I make that that double bond I need to kick out the leaving group, I'm sorry about that obviously and I need to kick out the leaving group I would also get the CL negative from the beginning and then I'm also going to get 2 equivalent of protonated Periodine 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 get two of those because the first one happened in the first step with the deportation 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.
Nucleophilic Attack + Deprotonation
E2 β-Hydrogen Elimination