Alcohols are terrible leaving groups. So we’re going to learn an entire class of reagents whose sole job is to convert alcohol into better leaving groups.
Concept: Why do we need to convert Alcohol into a good leaving group?3m
As you guys might have figured out by now, alcohols are a pretty important functional group for organic chemistry. But there's one major limitation of alcohols and that's that they make terrible leaving groups. Remember that the definition of a good leaving group is something that once it leaves, it's stable.
Well, alcohol, after it leaves, it becomes OH-. OH- is the same as a hydroxide base, which is a very unstable molecule. It's a very strong base. That means that whenever we have an alcohol, we're a little bit stuck. We don't know exactly what to do with it because a lot of reactions in organic chemistry require leaving groups and alcohol isn't a good option.
But wait. There is a solution. It turns out that the major topic that we're going to discuss in this section is how to turn alcohol into a good leaving group. It turns out that there's two major options – like a fork in the road, we can take two major pathways and that they're both going to lead to awesome outcomes. They're both going to lead to alcohol being a much better leaving group. Let's go ahead and talk about the first one.
The first option that we have is to convert alcohol simply into an alkyl halide. Remember that alkyl halides have the molecular formula RX. The reason that they're such good leaving groups is because X negative, once it takes off is very stable. X could stand for iodine or bromine or chlorine. These are very electronegative atoms that don't mind having a negative charge. So alkyl halides are an awesome option.
But another option that we also have that we'll discuss in a little bit is sulfonate esters. Now you might have learned already that sulfonate esters make great leaving groups. I actually have talked about that before. But now what we're going to learn is how to actually turn an alcohol into a sulfonate ester so that it can become a great leaving group.
We’re aiming to turn alcohols into alkyl halides or sulfonate esters. Let’s explore the different ways to accomplish this.
The simplest way to theoretically convert alcohols into alkyl halides is just to react them with a strong halohydric acid (HX). There are some complications associated with this conversion however.
Can you predict what they are? (Three major complications given below).
Concept: Using HX acids via SN1 reaction.8m
So let's start off with this page. I just want to do the alkyl halides and later one we'll talk about sulfonate esters.
It turns out that the degree of your alcohol is going to determine which method you use to turn it into an alkyl halide. That has to do with the mechanism. The reagent might be the same actually, but the mechanism's going to be slightly different.
For a secondary or a tertiary alcohol, what's common to both of those is that they both make relatively good carbocations. If I were to kick the OH off, somehow make it stable and kick it off completely, I would get a carbocation that's pretty stable. It turns out that secondary and tertiary alcohols proceed with an SN1 mechanism. Remember that SN1 is a two-step mechanism that makes the carbocation first and then it's nucleophilically attacked.
Let's go ahead and kind of figure out what's going to happen. In the first step, what I'm going to do is I'm going to protonate my alcohol with my strong acid, HX. We're always going to use HX, a strong acid, to turn alcohol into a good leaving group. Let's just say that the X stands for Br. I have HBr.
In my first step what I'm going to get is alcohol grabbing a proton and kicking out the Br. What that's going to make is that's going to make a water where the alcohol used to be. Now keep in mind that that water is now way better. In terms of it's a much better leaving group.
What's going to happen in the next step – we just protonated. What happens in the next step is that just leaves completely on its own. Why? Because this is an SN1 reaction. Remember than in an SN1 reaction, you need to make a carbocation. What we're going to do now is we're just going to draw out our carbocation. The reason that we knew that it would make a carbocation is because if you'll notice, I started off with a tertiary alcohol. Remember that I told you tertiaries and secondaries are going to do SN1.
One question you guys might have is, “Johnny, the methyl group on the red structure used to be on the dash and now you just drew it on a stick. Did you do that on purpose? Did you mess up?” Actually, I did that on purpose. The reason is because remember, carbocations are trigonal planar. So you should draw everything on the same plane. It's actually not correct to keep it on the dash. You should just draw everything on the same plane.
Now you might have guessed it. What's going to happen to this positive charge? Well, keep in mind that it could rearrange. If it was unstable, it could have rearranged. Would this carbocation want to rearrange? It's already tertiary. It's happy. So let's go ahead and attack it with the Br. What I'm going to get at the end is an alkyl halide that looks like this.
Now one note, I purposefully didn't include stereochemistry here. The reason is because I don't know which side it's going to attack. It can attack the front or the back. Technically, you're going to get enantiomers from this attack. You're going to get a Br on the front and a Br in the back. But typically, that doesn't really matter too much because we're going to be usually making this leave anyway later.
Now you might also be wondering, “Johnny, what was the point of this?” Well, I'll tell you the point, the point is that alcohol wasn't going to do anything. It was just going to sit there forever, so by reacting it with HX, what we were able to do is turn it into an alkyl halide, which is much more convertible. Alkyl halides are much more functional. They can do a lot more things. I just made my alcohol more functional by turning it into an alkyl halide.
Now, there is one more note that I have to say which is that the X in HX can equal two things. It can either equal iodine or it can equal bromine. What about chlorine and fluorine? Well, just so you know, fluorine is too weak to react. I'm just going to put F doesn't happen. It can't be F. HF is not a strong enough acid to make this happen.
But how about HCl? Well, HCl, the Cl is still a pretty strong nucleophile, but it's not strong enough to make this reaction happen in full yield, in a high yield. So what we usually do is if we really want to have a Cl instead of the Br, we're going to couple that with a Lewis acid catalyst that is zinc and two chlorines together. What that's going to do is it's actually going to make the leaving group stronger.
Let me just show you really quick how that works. By the way, this together, having HCl and the zinc complex together is going to be called the Lucas reagents. Some professors don't care that you know that. Some books don't really teach it. But, in general, it is kind of widely known as the Lucas reagent. Really it's going to do the same exact thing as what we just did, except it just has one extra step which is this.
Basically what we do is we take our alcohol. I'm going to draw it a little bit smaller. And what's going to happen is that the Lewis acid, remember that a Lewis acid is a proton acceptor. I'm sorry. A Lewis acid is an electron pair acceptor. It has an empty orbital. What's going to happen is that the Lewis acid is actually going to couple to the oxygen. The oxygen is going to donate electrons to the zinc. What that's going to make is something that looks like this, where I now have O, H, zinc with two chlorines and that O is going to have a positive charge.
What's great about that is that now that is a much better leaving group than just water by itself. Remember that over here I also had OH2+. But this one's even better. What that means is that the Cl when it comes in, it's going to have an easier time attacking there. What's basically going to happen is that it's going to have an easier time leaving by itself. So once it leaves, what I'm going to wind up getting is a carbocation that is then easier for my Cl to attack.
That's the first situation I want to tell you guys about.
This is the predominant mechanism for strong halohydric acids with 2° and 3° alcohols.
Concept: Using HX acids via SN2 reaction.3m
This is the predominant mechanism for strong halohydric acids with 1° alcohols.
Complications: Strong HX acids conversions come with three major complications.
These complications render this method mostly useless. Sorry not sorry.
Don’t you wish there a method to convert alcohols into alkyl halides without all these complications? Ah, looks like we’ve got just the reagents for you!
Concept: Learning the mechanism of SOCl2.8m
These reagents proceed through SN2 mechanisms, so they will only work on 1° and 2° alcohols. That said, we expect inversion of configuration from this conversion.
Example: Predict the mechanism of PBr3, and draw the final product.7m
Now that we’ve covered alkyl halides, sulfonyl chlorides can be used to convert alcohols into another great leaving group, sulfonate esters.
Concept: Learning the mechanism of Sulfonyl Chlorides.8m
Alcohols are really bad leaving groups. So we're often in the position where we want to convert the alcohol to a better leaving group and one of the options that we can use is sulfonate esters. It turns out that sulfonate esters are really the ultimate leaving groups of organic chemistry because they're so extremely stable after they leave. They do an amazing job balancing out that negative charge and delocalizing it.
So how do we turn an alcohol into a sulfonate ester? All we do is we just use the chloride of that sulfonate ester. So we're going to use a sulfonyl chloride to convert the alcohol into a sulfonate ester. I'll show you guys how to do that in just a second.
First of all, what is a sulfonate ester? If you guys remember, the general structure was just that you had basically an S with two double bond O's. I'm just going to draw a line here. Then you had an R group. A sulfonate ester would have basically an O on this side, an O where the Cl is, then an S, the two double bond O's and then your R group. Now the identity of the R group is just going to change the name of the sulfonate ester. In general, all of these molecules can be categorized as sulfonate esters, but they do get individual names.
For example, the easiest situation would just be what if it's just a methyl group? Well, then that's called a mesyl or a mesylate sulfonate ester or mesylate. How about if it's a benzene ring with a methyl group? Well, then that's called a tosyl. Basically, if the R is a benzene and a CH3, then that's a tosylate. Then, finally, if it's a CF3, so I just replaced three H's with F's, that's called a triflate. Really easy. I don't even expect you guys to memorize all three of these, just to be aware that these are all sulfonate esters and they all react almost identically.
That's what a sulfonate ester is. What's a sulfonyl chloride? Sulfonyl chloride just means remove the O and put a chlorine right there. So how is this going to work? Basically, just so you guys know, this general structure stands for any of these three. It could be my tosyl chloride, my mesyl chloride or my trifyl chloride.
In this case, let's just start off with the first step of the mechanism. What's going to happen? Well, my oxygen once again, on the alcohol is going to be considered my nucleophile because it's got extra electrons – not extra, but it's just got electrons that are freely able to donate. If I were to attack one of the atoms on this sulfonyl chloride, which of them is the most likely to be attacked by a negative charge? In other words, which one is the one with the partial positive? Which one is the one that's the most electrophilic?
If you draw dipoles, we're quick to find that the partial-positive has to once again rest on the central atom or the sulfur. What that means is that my arrow is going to do this. Now sulfur can either have four bonds or six bonds. In this case, it had six. But it can't have seven. So if I make a bond, I do have to break a bond.
The way that's going to break – well, I'm just going to move these electrons up to the O. What that's going to do is it's going to make a structure that looks like this, where my O is still on the front. It's still attached to an H, but now that's going to be attached to S, O-, double bond O, Cl. Everyone following that? All the formal charges in? No, I do have to draw a positive charge right there. Cool.
So can you guys predict what the next step might be? Do I have any good leaving groups present? I do. I have a Cl. So what I could do is I could just reform the double bond and kick out the Cl. What that's going to give me now is the same stereochemistry. I'm still in the front. Still have an H. But now what I have is a molecule that looks like this.
And I'm missing the R group. I never drew it. Wow. Just a second. There's an R group here. You might be wondering, “Johnny, what happened to the R group before?” It was there. I just never drew it. Please draw that into your notes. You should have single bond O, double bond O, Cl, R and then the other O. That should add up to six. If you want, you can count it up. It does add up to six. And this also adds up to six. I have that plus I have the Cl leaving group. Perfect. Then I have a positive charge here.
So now the interesting part is that this molecule right here is actually pretty stable. This is my sulfonate ester. That's really it. I'm not going to keep going to another step. I'm just going to keep it like this. This is my sulfonate ester. Later on, I can react with it.
Now the sulfonate ester should be the neutral compound. So what's going to happen is that my Cl or whatever – it's usually a Cl, whatever my anion is, is going to take away that H. Eventually, what I should get at the end is a neutral sulfonate ester that looks like this: O, S, double bond O, double bond O, R.
The reason this is so stable is because basically there are no formal charges anymore, so I can keep this sulfonate ester around as long as I want, just like an alkyl halide. I can keep an alkyl halide around. And then when I finally want to react with it, I could introduce a nucleophile, let's just use general Nu- and then I could do a backside attack. But it's cool that we can just keep the sulfonate ester around. Now we just converted alcohol to a really good leaving group, which is the sulfonate ester.
Up here there was a blank that I skipped earlier. It's actually really important. Basically, because this reaction all takes place on the O and notice that the O never gets attacked during the entire mechanism for sulfonate ester. This reaction is going to proceed with retention of configuration because I'm never doing a backside attack during the course of the reaction. I'm always just ending with the sulfonate ester and then I'll figure out what to do with it later. Later on, could I do a backside attack? Sure, but that's a different reaction.
During this reaction, to get from here down to here, from the OH to the sulfonate ester, notice that the O never moved. The O is still in the same exact place it was before, so we call that retention of configuration. Once again, this topic, this section is all about stereochemistry, so you need to remember that sulfonate esters proceed with retention of configuration, whereas thionyl chloride and phosphorous bromide are going to proceed with backside attack or inversion of configuration.
So that's it for this. Now what I want to do is I just want to quickly show you guys a summary of everything that we just learned.
This conversion proceeds without the use of SN2, meaning that we expect retention of configuration.
Here is a chart with the most important info you need to know for each conversion.
Concept: Comparing and contrasting the Alcohol Conversions.5m
Provide a reasonable mechanism for the transformation below. Hint: This occurs in three steps, all of which are Lewis Acid/Base reactions. Fill in the missing intermediates A and B. Be sure to provide the arrows to show electron flow.
Predict the reagent for the following reaction.
Predict the products of the following synthesis: