Ch. 12 - Alcohols, Ethers, Epoxides and ThiolsWorksheetSee all chapters
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Ch. 1 - A Review of General Chemistry
Ch. 2 - Molecular Representations
Ch. 3 - Acids and Bases
Ch. 4 - Alkanes and Cycloalkanes
Ch. 5 - Chirality
Ch. 6 - Thermodynamics and Kinetics
Ch. 7 - Substitution Reactions
Ch. 8 - Elimination Reactions
Ch. 9 - Alkenes and Alkynes
Ch. 10 - Addition Reactions
Ch. 11 - Radical Reactions
Ch. 12 - Alcohols, Ethers, Epoxides and Thiols
Ch. 13 - Alcohols and Carbonyl Compounds
Ch. 14 - Synthetic Techniques
Ch. 15 - Analytical Techniques: IR, NMR, Mass Spect
Ch. 16 - Conjugated Systems
Ch. 17 - Aromaticity
Ch. 18 - Reactions of Aromatics: EAS and Beyond
Ch. 19 - Aldehydes and Ketones: Nucleophilic Addition
Ch. 20 - Carboxylic Acid Derivatives: NAS
Ch. 21 - Enolate Chemistry: Reactions at the Alpha-Carbon
Ch. 22 - Condensation Chemistry
Ch. 23 - Amines
Ch. 24 - Carbohydrates
Ch. 25 - Phenols
Ch. 26 - Amino Acids, Peptides, and Proteins

Now that we’ve covered alkyl halides, sulfonyl chlorides can be used to convert alcohols into another great leaving group: sulfonate esters.

Concept #1: Learning the mechanism of Sulfonyl Chlorides.

Transcript

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-).