Week 8 of 52: Cleaving ethers with hydrohalic acids

Well, first we had reactions of alkyl halides, then a reaction of an alcohol to an alkyl halide. Alkyl halides are so much fun that you definitely want to learn more reactions involving them. I know I would, if I didn't already know all of them (that last phrase may not actually be true).

As you may have guessed, cleaving an ether means breaking one or both of the bonds to the oxygen, which also breaks the chain at that point. Don't think of ethers as particularly unstable, because most of the time they are not. But in the right environment, that oxygen can be the weak link in a chain (and when cleavage occurs, that's where it happens). Hydrobromic and hydroiodic acid are one way to provide that environment, protonating the oxygen in an acid-base reaction. Did I mention that this reaction involves a nucleophilic substitution mechanism? That should be a big hint.

Still don't get it? Well, I haven't talked about ethers much, so you probably just aren't used to them. But remember how we can turn a bad leaving group into a good one? Of course you do. Well, that's what happens here. Twice. The oxygen is protonated, and a bromide or iodide reacts with one of the α-carbons by nucleophilic substitution. I'll emphasize that, yet again, this is S_N1 in the case or secondary or tertiary α-carbons and S_N2 in the case of primary or methyl α-carbons. It's important and I don't think I've been emphasizing it enough so far, but now it's in bold, so you are not allowed to ever forget it.

Conveniently enough, this leaves us with one alkyl halide (with the carbon chain on the side that underwent nucleophilic substitution) and one alcohol (the other carbon chain keeps the oxygen, which is now bonded to hydrogen). Also conveniently, the alcohol undergoes nucleophilic substitution by the reaction we covered last week.

And that's it. In conclusion, we go from R—C—O—C—R' to R—C—X and R'—C—X (with water as a byproduct). Keep in mind that this reaction works because the acid provides protonation of the oxygen, which creates a leaving group, and also because the acid provides a halide to act as a nucleophile.

Ugh, and we're still a week behind.

Week 7 of 52: Conversion of alcohols to alkyl halides by hydrohalic acids

Yes, I'm still a week behind schedule. I know. Someday I'll even catch up. But not yet. Last time, I introduced a specific variation on elimination. So now let's try one for substitution. Are you excited? I know I am. This reaction is pretty fantastic, but it might not be what you're used to. Instead of using the properties of nucleophilic substitution to replace a halogen with something else, we're going to replace something else with a halogen. It's backwards!

As you almost certainly do not recall, halogens make good leaving groups and hydroxide makes a good nucleophile. So this really does seem backwards. How can we have a nucleophilic substitution reaction in which something that is ordinarily a good nucleophile is the leaving group and something that is ordinarily a good leaving group is the nucleophile. The answer, of course, is that we cheat. Come on, isn't that obvious? What might not be obvious is just how we are going to go about cheating. No, I'm just kidding. That's obvious too. No?

Fine. Remember how we can dehydrate alcohols? I mean, the last two reactions have been about that. We turn that bad leaving group into a good one. So here's your first hint: we'll turn that hydroxyl group into a good leaving group by protonating it with a strong acid. You get it now, right? No. Here's another hint: the title of this post mentions hydrohalic acids. That's right, hydrochloric, hydrobromic, hydroiodic. HCl, HBr, HI.

Do I really still have to spell it out for you? The acid protonates the oxygen, creating a good leaving group, then the halide attacks the molecule as a nucleophile. This occurs by an S_N2 mechanism for primary alcohols and an S_N1 mechanism for secondary and tertiary alcohols, of course. Isn't it great?

Also, bromide and iodide are strong enough nucleophiles for this, but chloride can require a catalyst for the reaction to progress. But no more about that for now.

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.