Week 25 of 52: Acetylide as a nucleophile

In Week 18, I covered the use of terminal alkynes as weak acids. At the end of that post, I casually remarked that the conjugate base of such an acid, an acetylide ion, can itself be used in reactions. Since that post went up, you've been waiting in agony for a post about a reaction using an acetylide ion. Your wait is finally over.

Perhaps I erred in the earlier post, actually. I mean, it is accurate that terminal alkynes are very weak as acids, but just leaving it at that seems sort of pointless. Oh look, we took some very strong base and we protonated it. Lots of things can do that. For Week 18, what I should have done was emphasize that the point of that reaction would be to prepare acetylides. We can then use the acetylides we just prepared in nucleophilic substitution reactions.

Overall, this reaction is a fairly straightforward example of the S_N2 reaction, which was the first post in this whole series. I did write about that reaction first because it was important. After mentioning nucleophilic substitution so much in these subsequent posts, I think it should start to become clear just why it's important.

The textbook I've been using for most of this takes the opportunity to use this reaction as a starting point for discussion of multistep synthesis, which we've only seen a little so far on this blog, and retrosynthetic analysis. The whole point of learning these reactions in the first place is to understand how to convert something into something else. It's transmutation. I hope to have time in the near future to create a reaction map for use here, charting all the functional groups I've covered and all the reactions with them that have been shown so far. If you've been paying attention, and I know you haven't, this should give you an idea of the big picture.

Week 12 of 52: Preparation of epoxides from halohydrins

I know I'm behind. Shut up. I'm also very busy. But I have a reaction for you. Learn it. Or else. Now, as you may have ascertained from the title already, this reaction is all about the preparation of epoxides. I introduced epoxides in the previous post, and of course you still remember them and love them. That's good. One way to make them is to use a halohydrin. "What's that?" you ask. Shut up and I'll tell you.

Halohydrins are themselves prepared from alkenes. But shut up. They look like this...
The first step of this reaction is a simple acid-base reaction in which a proton is stripped from that alcohol group. You've seen this before. It is not new. It is familiar. You are comfortable with it. What happens next is pretty cool: an intramolecular substitution reaction. The negatively charged oxygen forms a bond to the nearby carbon. The halogen is, of course, a leaving group. The end result is an epoxide: the oxygen is attached to both carbons, forming a strained ring.

Week 11 of 52: Breaking epoxides with nucleophiles

I think I missed a week and am behind on this. Whatever. I have no sympathy for you. Here is a weird reaction I arbitrarily chose. Learn it.

Epoxides contain a strained ring. It looks like this...
Look at those bond angles. Actually, don't. I mean, I just made that picture in a few seconds. It's not like it's accurate at all. But epoxide rings are strained. They're just waiting to pop open at any second if you give them reason to. Maybe. Actually, I made that up too. You probably shouldn't take this post too seriously. Just so you know.

One way for that ring to open is for a nucleophile to attack one of those α-carbons. Assuming that the reaction takes place in an aqueous environment, this leaves an alkoxide on the α-carbon that was not attacked by the nucleophile, which is protonated by the surrounding water.

The result of this reaction is that each of the carbons from the epoxide now has a different functional group attached to it. One has the nucleophile (whatever that was) and the other has an alcohol. This is probably useful for something. Ugh, sorry. That sounded lame. This isn't working. I need to change the way I do these posts. I need to make them better. This one sucks. I'm so sorry.

Week 9 of 52: Williamson ether synthesis

I'm sure that even you managed to deduce from the title that this post will cover a method of synthesizing ethers that is, for some reason, named after "Williamson." Good job. No, not really. I mean, no it wasn't really a "good job" that you figured this much out. Of course I mean that this is about the synthesis of ethers. "Williamson" turns out to be the person who invented this. Or discovered it. Whatever, I don't care which. You suspected all along that this was the case, but you couldn't be sure until I told you just now. Anyway, the Williamson in question was Alexander Williamson. He came up with this back in 1850, so you can safely assume that he is now dead.

Really, this is a simple S_N2 reaction, which is my way of saying that I won't be spending a great deal of time on this. But don't conclude that this is some minor, throwaway reaction I am lazily posting to keep up my weekly quota. You'd be wrong about that. Well, you'd be wrong about part of it anyway. This reaction is, to this day, the main way ethers are manufactured. Ethers are important for industrial stuff probably. I mean, I assume they are.

The substrate for this reaction is an alkyl halide. Yes, again. Why not? What's wrong with alkyl halides. I heard that you like them a lot. And you should. As you already realize, the halide acts as a leaving group here. But this time, the nucleophile is an alkoxide ion. Alkoxides are of the form R—O^-. They are typically prepared as salts. I could describe how, but it occurs to me that I can use that reaction to fill in another week, so you'll just have to wait. So cruel, I know.

Anyway, the alkoxide attacks, bumping the halide off and attaching to the α-carbon. So we get an ether. R—O—R'. Also, if the ether is unsymmetrical, we could potentially have either side be the alkyl halide or the alkoxide, but one of the two possible configurations is more efficient. If I revisit this topic in the future, you must remind me to explain that.

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 2 of 52: Unimolecular nucleophilic substitution

Because you read the previous post and totally didn't forget everything I said there, you already deduced that this nucleophilic substitution reaction (S_N1) occurs in two steps. Instead of the nucleophile attaching at the same time that the leaving group is removed, first the leaving group leaves, then the nucleophile attacks the "intermediate" and together they form the product of this reaction. In order for this to happen, the bond to the leaving group has to actually break on its own. If the intermediate would not be stable, this won't happen.

I started writing this post too late in the week for me to cover relative carbocation stability, so you'll just have to believe me when I tell you that it's—important. Yeah, that wasn't very convincing. Whatever. Shut up. Carbocations in which the carbon attached to the leaving group have more bonds to hydrogen atoms are less stable. If the carbon attached to the leaving group is attached to more carbons, the carbocation will be more stable. The greater the stability, the faster the S_N1 reaction.

Also note that unlike the backside attack of the S_N2 reaction, the leaving group in this case is already out of the way, so stereochemistry (if the carbon in question is a chiral center) is randomly split between both possible configurations. That means there will be an even mixture of both possible products, not that the each individual molecule will somehow be halfway between both possible products, obviously.

Week 1 of 52: Bimolecular nucleophilic substitution

I think this was the first reaction I learned in an organic chemistry class. Regardless of whether I'm right, it will be the first reaction in this series. Now, I know the name seems intimidating to you, because you're so pathetic. But I'll confess something: I didn't remember that name when I set out to write this post. At least I don't think so. It didn't really cross my mind. I'm used to just calling this reaction by the same name everyone else calls it. The more common name for this reaction is S_N2.

"Bimolecular" in this case refers to the fact that the reaction involves a collision of two molecules. Unlike some other reactions I'll be dazzling you with, this entire reaction happens in one step. One bond breaks at the same time as another bond is formed. Consequently, the thermodynamics of this follow the "second-order rate equation." I'll be gleefully ignoring that for now, so you can too, if you want. The important thing about it is that the rate at which this reaction occurs depends on the concentrations of both molecules involved (increase the amount of either in a system, and the rate of reaction speeds up).

"Nucleophilic" refers to the fact that one of the two reactants is, well, a nucleophile. Nucleophiles are attracted to positive charge. Remember: nucleii of atoms are positively charged. Here's the part where I could elaborate on the intricacies of nucleophilicity as a property, which molecules make good nucleophiles and which ones do not and why, but it turns out that I've been procrastinating on writing this post, so we're pretty much skipping that. Anyway, I will tell you that nucleophiles are often negatively charged particles, which should be obvious anyway.

"Substitution" means that the nucleophile replaces another group. The other group is aptly known as a leaving group. You know, because it leaves. To be a leaving group, an atom or group of atoms must be able to accept electron density. This leaves less electron density on the other side of the bond (which is to a carbon atom) and more exposed nucleus for the nucleophile to do its thing. The most popular leaving groups are halogen atoms, especially bromine and iodine (they're bigger, so the electron density is spread over a larger space). One thing that I didn't remember, but that my textbook deemed noteworthy is that "all good leaving groups are weak bases with strong conjugate acids having low pK_a values."

Another fun fact that can sometimes matter is that this reaction happens by "backside attack." As you may have noted, the nucleophile reacts with the carbon atom, not the leaving group and it wouldn't make sense for it to form a bond in the same spot where the bond to the leaving group is simultaneously breaking. This means that the stereochemistry of the carbon can be completely changed. It also allows for the pickup line: "Baby, if I were a reaction, I'd be S_N2, so I could attack your backside."