Week 17 of 52: Alkyne synthesis by two successive dehydrohalogenations

I've already covered the E2 mechanism by which an alkyl halide can be converted to an alkene. With a particularly strong base, a dihalide like the one shown in the previous post can undergo an E2 reaction twice, yielding an alkyne.
Like that! Or something. Note the use of sodium amide. I put it there because this reaction requires a very strong base. There's an explanation for this, but having just read over it, I find it beyond the scope of what I'm doing here (I definitely haven't introduced the concepts needed to understand it). So we'll just leave it at that. This reaction requires a very strong base.

Week 6 of 52: Dehydration using phosporus oxychloride and pyridine

This one is late. My apologies. But that means you'll get two reactions this week, and that's cool, right? Well, in this case, I'm going to have to make the post a brief one. I know, short and late. It's almost as though I'm not doing a very good job or something. But this is just going to be a specialized reaction. More important reactions should follow soon. And those posts will be longer. This one is really pretty much the same as last week. Well, now I guess it would be the week before last week, because I failed to update this blog last week. Whatever.

Last time, I showed how a strong acid can turn a bad leaving group (hydroxide) in to a good one (water), facilitating a specific type of elimination reaction in which water is eliminated by either an E2 or E1 mechanism, ultimately forming a double bond. Essentially, this can be used to turn an alcohol into an alkene. If the rest of a molecule won't react with the strong acid, it can even be used on a more complicated molecule forming a C=C bond at the area of interest, perhaps as one component of a series of reactions to form a desired product. But there's that caveat: this is only possible if the molecule will behave for us once we put it in a highly acidic environment. Many organic molecules will simply not do this. And that's where this reaction comes in. Phosphorus oxychloride and pyridine offer a way to dehydrate an alcohol by an E2 mechanism under basic conditions.

I will depict the mechanism for the first step of a reaction that converts cyclohexanol to cyclohexene. First, one of the lone electron pairs on cychohexanol's oxygen attacks the phosphorus in POCl₃, breaking chloride off to float away and never come back...

Next, pyridine reacts with the exposed proton from the alcohol: an acid base reaction...
Now, we have a good leaving group on that α-carbon for our E2 reaction. Note that there's already a base (pyridine) present. So it will remove a proton from a β-carbon...
And we have cyclohexene! Formed by the dehydration of cyclohexanol. Amazing, I know.

Week 5: Dehydration of an alcohol into an alkene

The title for this post is a bit unwieldy. I'm sure this won't be the last time that happens. I am using this title because there are multiple dehydration reactions, and there are even multiple dehydration reactions of alcohols. This post is only about a reaction in which an alcohol is dehydrated to form a π-bond between the α-carbon and the β-carbon (Greek letters are one of the most important tools in all of science and without them all sorts of bad stuff would happen somehow). I know, I know. You're confused. Again. That means it's time for a picture...
Like the reactions from the previous two weeks, this involves elimination. But those reactions involved the elimination of a halogen from the α-carbon and a hydrogen from the β-carbon. This one is called dehydration because water is removed from the alcohol. Water is lost. It's dehydration. Get it? Because that's what dehydration means. And you were probably already aware of this.

I know you probably aren't paying attention. But if you have been, you may be wondering how this happens. Surely the hydrogen on the β-carbon doesn't magically fuse to the hydroxyl group and form water because it wants to. So what's going on? How do we make an alcohol do this thing? The simple answer is acid. I know. It's awesome. Chemistry is awesome. Sulfuric acid works pretty well for this. You could use some other acid for some reason I suppose. It should be a strong acid though, because I don't traffic with weak acids.

And now an exercise for the reader. I'm serious. The acid protonates the oxygen in this reaction, forming water as a leaving group. You already know about leaving groups because they've been involved in all reactions I've done for this project (the reaction of the week one, not the whole blog) so far. The mechanism for the rest is either E2 or E1. Actually, I'll tell you that it's E2 for primary alcohols (the α-carbon is attached to only one other carbon) and that it's E1 for secondary and tertiary alcohols (the α-carbon is attached to two or three other carbons). So, from that, you should be able to figure out the rest on your own. Consider it a challenge. Well, maybe not. No, I'm not just being lazy here. I really think I've given you enough information in this and the posts on elimination reactions to see what's going on here. And it occurs to me that it may be better to try to leave some things intentionally only hinted at so that one can think about them, rather than omitting them entirely or simply spoon-feeding them to my readers (which is no one, but shut up). Well, go ahead then, deduce the rest of these reactions.

Week 4 of 52: Unimolecular elimination

Apparently January will have been the month of reactions on alkyl halides by bases and nucleophiles. I didn't initially set out to do it that way, but in retrospect it does make sense. I have no idea what reaction I'll start February with. Oh well, it's not time for that yet. It's time for unimolecular elimination.

If you've been paying close attention (which you haven't), you'll probably have deduced that this has the "unimolecular" feature of the S_N1 reaction and the "elimination" feature of the E1 reaction. So you practically know what will happen just from the name, you clever reader, you. But just in case you aren't that bright, which let's face it, is pretty likely, I'll offer a brief explanation...

• This is a two-step reaction.
• In the first step, the bond between the α-carbon and the leaving group breaks. The leaving group leaves.
• The departure of the leaving group results in the formation of a carbocation.
• In the second step, a base removes a proton from a β-carbon.
• As in the E2 reaction, the electron pair from the broken C-H bond forms a π-bond between the α-carbon and the β-carbon.

Still confused? Let's try some pictures. One note: with the S_N2 and E2 reactions, I just assumed that we were dealing with primary alkyl halides. With the unimolecular reactions, a carbocation is formed, so these reactions will be fastest with tertiary alkyl halides. So I'll draw some more groups, just to show that. So, here's the first step...
That's the same first step for both the substitution reaction and the elimination reaction. It's the second steps that are completely different. Here's S_N1...
And you already know exactly what the E2 reaction will look like now, but I'm showing you anyway, just on the off-chance that you really are that incompetent...

Week 3 of 52: Bimolecular elimination

The bimolecular elimination reaction is typically just referred to as the "E2" reaction. It has some things in common with the S_N2 reaction featured in the first week. Both are bimolecular: the reaction is driven by a collision of two molecules. Both involve the concept of a leaving group: an atom or group of atoms that can accept electron density. A classic example is an alkyl halide. In both the E2 reaction and the S_N2, something attacks the alkyl halide, and the halogen is broken away as an anion. I was being lazy for the first two weeks of this and didn't use any illustrations, so let's throw in a picture for this...
I say that's totally a step up from my previous use of pictures in this blog. Anyway, in the S_N2 reaction, which I covered in the first week, not this week, the "something" would be a nucleophile and it would attack the carbon that the halogen is bonded to, the α-carbon. The bond between the α-carbon and the halogen would break and a new bond would be formed between the nucleophile and the α-carbon. Here's a reaction mechanism. Oh, I'm leaving out the hydrogens bonded to carbon in these pictures because I want to. But realize that the α-carbon has two hydrogens attached to it that are just sitting there, not doing anything...
This week's reaction has some important differences. Firstly, the "something" is a base. Basicity and nucleophilicity are similar concepts and the same thing can behave in both ways. A nucleophile attacks a relatively exposed area of positive charge, the nucleus of the α-carbon. A base participates in a traditional acid/base reaction, reacting with a proton. No bond is formed between the base and the alkyl halide. Instead, a bond is broken. Here, I'll show you, but this time, I need to draw some hydrogens...
All of the electron density in the C-H bond at that β-carbon (it's called a β-carbon because it's adjacent to an α-carbon) is dumped onto the carbon. This simultaneously forms a double bond between the α-carbon and the β-carbon and drives the leaving group (the halogen) away. More importantly, I think making this image has placed me officially beyond all redemption.

Emphasizing this yet again, rather than providing new information about this reaction, because I know you can only handle so many new things at once: in nucleophilic substitution, the nucleophile replaces the leaving group, hence the name. In elimination, a π-bond (double bond) is formed between two carbons while the leaving group and a proton are removed, hence the name. Easy, right?