Week 26 of 52: Hydrogenation of alkenes

This is an addition reaction because something is added across a double bond. It's also a reduction reaction because the alkene carbons are losing bonds to each other and gaining bonds to hydrogen. Oxidation/reduction is a very important concept in chemistry, but the terms are really quite terrible and deserve their own post or series of posts. However, when I took organic chemistry, oxidation and reduction were simplified for the sake of our reactions: carbon gaining bonds to more electronegative elements is oxidation and carbon gaining bonds to less electronegative elements is reduction. That simplification doesn't work for inorganic chemistry at all, but for our purposes, just note that this is a reduction.

Like other addition reactions, this one follows the pattern of RC=CR' → RZC=CZR', where "Z" is the thing being added across the double bond. Or to represent it pictorially, since you still don't get it for some reason, here's the reaction with hydrogen...
As you can see, it's not our most confusing reaction. This is a simple way to turn an alkene into an alkane, or to remove C=C bonds in general. The catalyst here is a metal, often palladium. As far as I know, the metal catalyst is mixed into charcoal to maximize the surface area for the reaction, but I would imagine that there are other methods used in some cases. If this is done with palladium, a typical abbreviation for the catalyst is "Pd-C" (standing for palladium on carbon).


Also, π-bonds and rings are known as degrees of unsaturation. Each π-bond or ring in a molecule counts as one degree of unsaturation. So replacing the double bond with bonds to hydrogen is a way of "saturating" the molecule. You've probably encountered this concept with saturated and unsaturated fats. But I explain no further. Good day to you.

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 24 of 52: Addition of water to alkynes

Last week's post show's how water adds across an alkene. An alkyne has the same kind of bond that an alkene does (alongside another bond, but more on that later or possibly never). So the reaction is similar for alkynes. However, the product of this reaction can potentially be something completely different (an alkene just becomes an alcohol).

My book points out that either a strong acid or a mercuric catalyst (I could explain what that means, but I don't feel like it) can facilitate a reaction in which water adds across one of the bonds, leaving a double bond and an alcohol on the same carbon. This is called an enol.
It's a hideous portmanteau of "alkene" and "alcohol." But nevermind that. What's important is that this sort of thing is unstable. So it does something super-cool. I am not kidding. This is tautomerization, and it's awesome.
The enol form of the molecule tautomerizes into the "keto" form (which could be either a ketone or an aldehyde). I won't spend any more time on tautomerization right now, despite how freaking cool it is.

Week 23 of 52: Hydration

I was about to write a post about the addition of water to alkynes, but I just realized that apparently I've yet to write one on addition of water to alkenes. This oversight on my part is unforgivable and you should berate me for it. Too late, as by the time you read this I will have already corrected my error in the form of a new post, this very post, in fact.

Hydration of an alkene is a specific case of an addition reaction. Unlike with hydrohalogenation, water does not itself provide a strong acid to attack the alkene. So we use sulfuric acid. Problem solved! The product is, of course, an alcohol.

The lone hydrogen tends to add to the less substituted carbon. The hydroxyl group adds to the other carbon. This is in accordance with something called Markovnikov's rule. But I have not explained this. How negligent of me.

Week 22 of 52: Halogenation of alkynes

I already covered addition of halogen to alkenes back in some previous week. I don't know. Look it up. I showed an alkene being turned into a vicinal dihalide. Alkynes are sort of like alkenes, but different. So this reaction is sort of like that one, but different. Everything is the same as everything else, only different. Amazing.

This reaction actually has two different products. That's because the halogen (either chlorine or bromine) gets added once, forming a trans dihalide. What that means is that there's still a double bond and that each carbon on the double bond gains a new bond to a halogen. Fine, here, it's like this...
Yes, the mechanism is complicated and you're freaked out and frustrated and you hate me right now, but shut up. I drew this piece of crap just for you, so be grateful. The first step here is addition, which forms that thing in the middle: a bridged halonium ion. This step isn't very fast, but the next step, a nucleophilic substitution, is.

But that's not all. This week is special for some reason, so you get two reactions instead of just one. No really. You see, the trans dichloride that is the product of this reaction will also react with chlorine. So given enough chlorine and time, the process repeats and we get a second addition reaction, just like before, but this time the initial substrate is a trans dichloride instead of an alkyne. The end product is a tetrachloride (or tetrabromide if this had been done with bromine).

So you do get two reactions, but really it's just up to two iterations of the same reaction. Still pretty cool, though.

Week 21 of 52: Oxidation of an alkylborane

I will properly introduce oxidation reactions at some point. Or perhaps not. I don't know. Anyway, this is one of them, although this is not a proper introduction to them. The product from the previous reaction, an alkylborane, is oxidized here, yielding an alcohol. Really, this kind of like a second addition reaction (although it isn't one). Like the hydroboration reaction, this one is seemingly simple, but has some caveats. However, this time the textbook mostly glosses over those caveats, so this post will be pretty brief.

The reagent used here is hydroxide in hydrogen peroxide. The bond to boron is replaced by a bond to hydroxide, yielding an alcohol. And that's it. Well, not really, but I'm leaving it at that, so there.

In summation, the hydroboration-oxidation sequence takes us from an alkene to an alcohol, with the hydroxyl group bonded to the less substituted alkene carbon.

Week 20 of 52: Hydroboration

This one is an addition reaction and the product is the reactant for next week's reaction (which will actually not be in a different week at all, because I am catching up). I was going to do both reactions in a single post because these reactions don't take a long time to explain, they go together (one immediately follows the other), and the textbook does list them together before listing them separately. However, they are different reactions and I am going to err on the side of caution and split this across two posts. The first one (this one) is hydroboration of an alkene. The second post (the one right after this one) is oxidation of an alkylborane.

If you were really all that smart, you'd have gathered from the final sentence of the preceding paragraph that the product of this reaction is an alkylborane. This reaction is really quite simple...

Alkene + BH₃ → Alkylborane

Of course, that's only helpful if you know what an alkylborane is. Well, it's an alkane with a boron group of some sort attached to one of the carbons. Since this is an addition reaction and borane is adding across the double bond, one of the alkene carbons gets a hydrogen and the other alkene carbon gets a BH₂ group.

This is very simple, but the actual reaction is much more complicated, so perhaps it's best that I do this as its own post and note some of the ways this reaction is not as simple as it first appears...

1. BH₃ is a highly reactive gas and tends to form dimers (a molecule of borane will react with another molecule of borane to form diborane, B₂H₆).
2. Because borane and diborane are so reactive, they are impractical for this reaction. This problem is addressed by combining borane with a Lewis base to form a more stable complex. Tetrahydrofuran (pictured below) is apparently a favorite for this. I haven't yet talked about coordination complexes on this blog, so you have pretty much no idea what I'm talking about. So sorry.
3. The alkylborane formed through this reaction can still have the borane group react with other molecules of the original alkene (or any other alkenes that happen to be lying around). So the real product is a trialkylborane. The boron atom has three bonds, each one to a carbon that used to be an alkene carbon. The other three former alkene carbons have a hydrogen bound to them.
4. This reaction is regioselective. That's another topic that I suppose I've neglected on this blog. I guess I suck at this. In this case, what I mean is that the boron atom ends up on the less substituted carbon (if there is one). If the alkene is symmetrical, this doesn't matter. Otherwise, it has important implications on exactly what the product will look like.

Tetrahydrofuran:

Week 19 of 52: Halohydrin formation

I noted that epoxides could be formed from halohydrins, but still have not described how to form halohydrins in the first place. I aim to correct this oversight now. That's why this post is about halohydrin formation. If you'd been paying attention, which you haven't, you'd have seen this in the title.

The short version of this story is that exposing an alkene to a halogen and water yields a halohydrin. So the double bond in an alkene (C=C) becomes a single bond and one of the carbons gains a bond to a halogen and the other carbon gains a bond to a hydroxyl group. I just described the same thing twice and you still want a picture? Fine. I live to serve...
Making that picture just took valuable time that could have been spent playing Oblivion. I hope you're happy.

Week 18 of 52: Terminal alkyne as an acid

Last time I showed how to make an alkyne. So now of course you will want to make an alkyne into something else. Well, this reaction is only for terminal alkynes. If the triple bond is in the middle of a chain, like this R—C≡C—R, then it won't work. But when the triple bond is between the last two carbons of a chain (also the first two, because you can count either way), then it can act as an acid with the hydrogen at the end (it looks like this: R—C≡C—H) leaving and reacting with a base.

Terminal alkynes are weak acids. Very weak, actually. There's this big fancy chemistry explanation for why this is the case, but it might seem pretty intuitive to conclude that this reaction would require a very strong base, which is the case. And when I say strong here, hydroxide isn't strong enough. Amide is though, and of course there's the awesome hydride.

Once the terminal alkyne is deprotonated, it can act as a nucleophile known as an acetylide anion. This ion can then be used to react with an electrophile.

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 16 of 52: Halogenation

It's getting to me that I'm obviously rusty on this stuff. I don't like it. I see the phrase, "forming a vicinal dihalide" and I think to myself that I have no idea what a "vicinal dihalide" is. Have I ever even seen the word "vicinal" before? No matter, I just figured it out because of my magnificent intellect. A vicinal dihalide must be one in which the two halogens are bonded to adjacent carbons. A dihalide in which the halogens were bonded to carbons farther from each other would be some other sort of dihalide, presumably. I guess. As you can see, I'm not an expert. I'm just pretending to be one. Because pretending is fun.

This reaction is pretty simply though. Alkene + halogen yields vicinal dihalide. Wow, that is simple. Fine, here's a picture...
That's pretty good, if I do say so myself. Anyway, this reaction is normally only done with chlorine or bromine. Addition of iodine is often too slow to be practical and addition of fluorine is apparently explosive. Fun. Oh, and then there's this part about how dichlorides and dibromides formed this way are themselves used as reactants for the synthesis of alkynes. It looks like I have my next post all figured out...

Week 15 of 52: Hydrohalogenation

I was getting caught up. And then I stopped. I blame school. And myself. Mostly school. but I am not giving up. I missed May and June, but I will get caught up by September. And you will read it. We are making this happen. A lot. I'm not sure quite how, though. Aside from being busy with school, I'm also finding this project harder now. I've lost track of which reactions I've written about. I've forgotten a lot of reactions. This is not good. I haven't been taken chemistry and I haven't been focused on it. Enough whining.

Hydrohalogenation is a good word. I like it. Before I went on this stupid, two-month hiatus, I wrote about electrophilic addition. Hydrohalogenation is a specific case of electrophilic addition. This textbook says, "Hydrohalogenation is the addition of hydrogen halides to alkenes to form alkyl halides." And of course you remember that alkyl halides themselves can be used in substitution reactions. And there's even elimination! You could do an addition on an alkene to make an alkyl halide and an elimination on that alkyl halide to make it back into an alkene! It would be useless, but I think it would be fun.
That's an image I made. It depicts the reaction. Obviously.

Week 14 of 52: Electrophilic addition

The catching up continues furiously. Or maybe just aggressively. With a scowl-like expression at the very least. I don't feel like doing this right now, but I am forcing myself to, alright? I could force myself to do my actual schoolwork, but I'll do that later. Yes, I'm procrastinating on my schoolwork by writing a summary of a reaction. It's not that weird. There are weirder people. Plus, this hardly even counts because I'm padding it with nonsense like, well, pretty much this whole paragraph. So there's that...

An electrophilic addition reaction involves the breaking of a π-bond and the formation of two σ-bonds. For now, let's keep it simple and consider alkenes. These reactions also work on other molecules, like alkynes (hydrocarbons with at least one triple bond), but we'll move on to them later (or never).

And electrophile is sort of the opposite of a nucleophile. And you already know about nucleophiles because I already explained them. Remember?

Nucleophiles are attracted to positive charge. Remember: nucleii of atoms are positively charged.

Well, electrophiles are attracted to negative charge. And, as we all know, electrons are negatively charged. Alkanes consist of C—H σ-bonds and C—C σ-bonds. But in alkenes, there is at least one C=C bond (a π-bond). The double bond is stronger than the single C—C bonds are by themselves, but the π-bond portion of that double bond is significantly weaker and sort of more spread out. The electron density is more exposed to attack. And like nucleophiles, electrophiles attack.

I won't provide a list of common electrophiles right now. Maybe some other time (probably not). However, here's the general form of an electrophilic addition...
And there would be an electrophile in there somewhere, which would probably take up two of those new bonds that formed. You'll hopefully become more comfortable with this over the next month or so. I plan to post a few specific versions of addition reactions on alkenes, so perhaps May will be the month of addition reactions. Well, I'm actually still behind, so that doesn't really work. But shut up.

Week 13 of 52: Epoxidation of ethene

I am still behind, but I have resolved to catch up. This project will not die until I want it to. And I don't want it to. Not yet, anyway. I am so dedicated that I am starting a new post while sitting in my classroom. Class starts in fifteen minutes or so as I am typing this sentence, so I won't finish it yet. Shut up. Obviously I don't have an organic chemistry textbook in front of me, which has been my traditional method of searching for and selecting reactions to post here. And because of that, this time, we get a reaction not from the textbook at all. Pretty cool, huh?

This week's reaction, or the reaction for whichever week I'm on now, is another epoxide synthesis. Rather than working on an entire functional group (like the halohydrins from last week or last post or whatever), this one is specific to a single molecule: ethene. Being limited thusly is detrimental to a reaction's usefulness, assuming we want to construct a toolbox of reactions. However, this is still an important reaction because oxirane, the epoxide produced from it, is used extensively in industry. The reaction goes something like this...

7H₂C=CH₂ + 6O₂ → 6C₂H₄O + 2CO₂ + 2H₂O

In case it wasn't clear, the product (other than water and carbon dioxide) is oxirane, the smallest and simplest of the epoxides.

This reaction is mediated by a silver catalyst. Have I explained catalysts before? No? Too bad. Anyway, even though this doesn't work for larger alkenes, it's still important because oxirane is an important precursor in the manufacture of a lot of other things, most prominent among them being ethane-1,2-diol (ethylene glycol).

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 10 of 52: Preparation of alkoxide salts from alcohols

Last week, I showed the Williamson ether synthesis, a nucleophilic substitution reaction in which an alkoxide ion attacks an alkyl halide. I noted that the preparation of the alkoxide itself was another reaction and that I would save this reaction for later. Well, it's later now. I think. Anyway, if you actually bothered to read the title, which really isn't all that impressive of an accomplishment, you would knot that I am indeed using alkoxide preparation as this week's reaction.

This is an acid-base reaction. I hope you remember how those work, because I shan't be reviewing it. Go back and find those posts yourself. Or don't. Whichever. Even if you are a bit familiar with acid-base reactions, you might find this one slightly peculiar. You might notice that an alcohol is not normally a good acid. That's why this time, we're using a super-strong base. Sodium hydride is one of my favorite bases ever and it's a fairly standard one for this. Probably. I think. So we'll use that as our example. Behold, a reaction:

CH₃CH₂O—H + NaH → CH₃CH₂ONa + H₂

See how awesome sodium hydride is?

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 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?

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."