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
What about a primary alcohol? Well if we have a primary alcohol then the only thing that really changes is the mechanism, primary alcohols are really good at having a backside and they're really bad at doing making carbocation so I means if I use HX once again I'm going to protonate in my first step, so what I'm going to wind up getting is OHH positive, OK? But if this left by itself and then obviously I would get the X leaving by itself so I'd get plus X negative but the problem is that this can't just leave and make a stable carbocation so instead we're going to do is we can do a straight up back side attack where the X hits the back side and kicks out the water, OK? Now the reason this is possible here but it wasn't possible with the other situation is because the primary alcohol has a much better backside, it has a good back side since it has a good back side it's easy for my X to just kick out the water all in one step so what we're going to get here is an Alkyl Halide once again but in this case my mechanism was different I used SN2 instead of SN1, alright? Same exact thing would apply if I used each HCL and the Lucas regent what I would wind up getting is 0H with then a zinc and 2 chlorines and a positive charge and what you would get in the second step is that my chlorine or not in the second step but well yea technically in the second step my chlorine would do a backside attack and kick out the entire complex so then in this case if I was reacting specifically with those reagents I would get a chlorine, alright? So really this isn't that bad I just went through the mechanism so you guys will understand it but really all you need to know is Hx it's really that easy you just take HX and you can convert an alcohol to an Alkyl halide, alright? Awesome so let's go ahead and move on to the next topic.
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
Since alcohols are terrible leaving groups we often want to convert them into Alkyl Halides and it turns out that there are two very common reagents that are used to turn alcohols into alkyl halides so let's go ahead and just look right into those so basically these reagents are SOCl2 and PVR3 and respectively they're going to turn alcohol into alkyl chlorides or alkyl bromides, alright? Now the only thing that's special about these is that these are always going to proceed through an SN2 mechanism, OK? So, what that means is that since its SN2 backside attack these are going to be limited to working on only primary and secondary alcohols, OK? So, if I have a tertiary alcohol I cannot use SOCl2 or PVR3 because I'm not going to get a backside attack remember the SN2 just means backside attack and remember that tertiaries suck at it because tertiaries don't have e a good backside so I needed to be either primary or secondary since this is a backside attack, what kind of stereo chemistry do think we could expect? We're always going to get inversion of configuration, OK? Remember that inversion of configuration and backside attack go hand in hand any time I say one I'm going to always be referring to the other so let's go ahead and analyze this mechanism, OK? I'm going to show you the general mechanism of SOCl2 but the same thing applies to PVR3, in fact I'm going to have you guys figure out the mechanism for PVR3 all on your own let's go and check out SOCL2 so overall the structure of SOCL2 or thionyl chloride is an S a double bond O and then two chlorines coming off the ends, OK? What's special about this molecule is that it actually has a very very electrophilic portion, OK? Think of this oxygen here as a nucleophile it has extra electrons that it wants to give away if it were to attack one of the atoms on those 4 on this thionyl chloride CL sulfur oxygen which would be the most susceptible towards nucleophilic attack? Or in other words which one has the biggest partial positive charge that would be attracted to the negative charge of the oxygen? And if you drop dipoles it becomes painfully clear that you have dipoles pulling in all directions away from the sulfur, OK? So, what that means that the sulfur is going to have a very significant partial positive charge so my arrow is going to start from my oxygen and attack the sulfur, OK? So, can I just leave it like that? Are we done with this mechanism or at least this part? No because if we make a bond we have to break a bond because sulfur cannot have five bonds, OK? it can either 4 or it can expand its octet to have six but it can't have five so the bond that we're going to break is the one the double bond leading up to the O so we're going to kick those electrons up to the O let's go ahead and draw the next step, so what I wind up getting is I wind up getting OH, OK? That H is still there from before but now that's attached to a sulfur an O negative, a chlorine and a chlorine, OK? So now in this next step......By the way we're forgetting a charge there should be a positive charge right there because oxygen doesn't like having three bonds, alright?
So now what's the next step? Well the next step is that this oxygen does not like having that negative formal charge and it actually has two really good leaving groups right there so what's going to happen is that this negative charge is going to go down and reform the double bond and kick out one of these chlorines, OK? So, what I'm going to get at this stage is I'm going to wind up getting a Leaving group that looks like this, O S double bond O CL with the H here and the positive charge, OK? So, it turns out that now this leaving group is awesome because that O has a direct positive charge which means that after it leaves it will be neutral which is great, on top of that the sulfur pulls electrons away so this just has an awesome there's an amazing leaving group so what do you think could kick it out? What do you think could attack the carbon and kick out this big leaving group? And it's going to be the Cl negative that just got kicked out, this CL that we kicked out is now this C L and what's going to wind up happening is that now we get the backside attack, now I'm going to get the Cl attacking that carbon and kicking out this entire thing as the leaving group so what I wind up getting is an alkyl halide in this case it's an alkyl chloride where my chlorine should be facing which direction? It needs to be on the dash the reason is because notice that my leaving group was on the wedge before so after I did the back-side attack I need to invert the stereo chemistry, I'm going to get that plus I'm going to get my leaving group which just looks like this, Cl on one side and OH on the other but guess what? Your chemistry professor does not really care about this one I'm just going to put a little X they don't care. about that what they really care about is your organic products or this guy over here especially that you know the stereo chemistry because it lets them know they really understand what's going, alright? So, I hope that mechanism wasn't too complicated I know it was a little weird how we kicked the electrons up to O and kicked out of chlorine but that's actually a really common pattern when it comes to double bonds to O and in Orgo 2 we're going to doing this mechanism all the time so it's actually really good that you're getting practice with it right away so that was the mechanism of Thionyl chloride, right? What I want you guys to do now is to use that knowledge to try to draw the mechanism for this entire reaction all on your own, I know that sounds really complicated but just so you guys know the PVR3 is going to react exactly the same way as the SOCL2 did in fact it's going to be a little bit easier, OK? And then NaN3 and it IÕm sorry the NaN3 is just a reaction that you should know from before that lets you integrate with this to try to figure out what's going on, OK? So even if you don't get the second part at least give the first step of this reaction your best shot and then I'll explain everything to you so go for it.
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
A few notes regarding the above video:
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.
Yes we know, there is an extra methyl in the original molecule in the video. However, we decided to remove it completely from the problem.
This allows for a backside attack to successfully take place in the last step (seen by the blue Nu-).
Here is a chart with the most important info you need to know for each conversion.
Concept: Comparing and contrasting the Alcohol Conversions.5m
So here's what I hope will be a hopeful cheat sheet for you guys to compare the different types of alcohols you could start with and the stereo chemistry that you'll get at the end, all of these reactions remember are going to convert alcohols to good leaving groups but they are going to proceed with different stereo chemistry so because this section has so much to do with stereo chemistry that's why I made this entire axis all about stereo chem, alright? So let's go ahead and start off with what if we want a good leaving group we want to convert alcohol to a good leaving group you want the end product to be racemic, OK? Well if that's the case then we could start with a secondary or tertiary alcohol and we could use Hx in an SN1 reaction, OK? Hx is the only reaction in SN1 that gives us 2 enantiomers that gives us a combination of products, OK? Now you might be wondering well Johnny why is it greyed out here? Why can't I also use that on a primary alcohol, OK? Well remember because primary alcohols don't form good Carbocation and in order to make a racemic mixture you need a good carbocation So I'm just going to write here can't make Carbocation and if you can't make the Carbocation that means you are going to do an SN2 mechanism not an SN1 so that's why it's greyed out, OK? Now one more thing I just want to add remember that what happens in the case of HCL? Can I also just use HCL by itself? No, remember we use or the Lucas reagent HCL over zinc and 2 chorines, the Lewis acid catalyst, OK? Cool so let's keep going what if IÕm starting with alcohol and I want to get inversion of configuration at the end, I don't really care what the leaving group is as long as it's inverted, OK? Well if it's primary alcohol then I can use Hx or SOCL2 and PVR3 so just you know SOCL2 and PVR3 are always good options for inversion, OK? Because of the fact that you get a backside attack every time, why did I only put the HX SN2 here? Why did I not put it in the secondary position.? Because if you guys remember HX will actually if you do it with a secondary it's actually going to wind up giving you an SN1 reaction so that's why we can't use it in the secondary position we can only use it in the primary position if you have a primary Alkyl Halide you will get an SN2, you will inversion but if that secondary remember you make your carbocation again and you would form a racemic mixture, OK? You might be wondering well Johnny why is this greyed out? Why can't I just do an inversion of configuration on a tertiary as well? Why can't I use SOCL2? Because remember you can't do a back side attack on a tertiary it's too crowded, OK? So I'm just going to put here bad backside since it's a bad backside I can never do an inversion of configuration on a tertiary, cool?
Alright so lastly what if I want retention of configuration? I don't really care what the leading group is as long as it's good and I don't want it to be alcohol obviously, well anytime you want retention of configuration you're going to use Sulfonyl chlorides to turn your alcohol into a Solphonate Ester, OK? This one's really easy because it doesn't matter what degree your alcohol is, it always works primary, secondary, tertiary it's always the same thing you're going to wind up getting a Solphonate Ester and it's that easy and then later on you can react it, OK? So hopefully that helps us to kind of organize the different ways that we can convert alcohol to good leaving groups, professors love to test on stereo chemistry here and that's why I really emphasized knowing the differences in stereo chemistry between these options, alright? So I hope that made sense to you guys let's go ahead and move on.
Provide the mechanism for the following reaction. Be sure to include all intermediates, formal charges and arrows depicting electron movement.
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.
Write a chemical equation for the reaction of 1-butanol with each of the following:
(c) Sodium bromide, sulfuric acid, heat
Predict the products of the following synthesis:
Predict the products of the following multi-step synthesis: