I am Stephen Bahl

Week 26 of 52: Hydrogenation of alkenes

2011-10-02T12:55:00.000-07:00

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

2011-10-01T17:38:00.000-07:00

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

2011-09-26T00:12:00.000-07:00

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

2011-09-23T18:21:00.000-07:00

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

2011-09-11T22:01:00.000-07:00

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

2011-09-03T18:16:00.000-07:00

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

2011-09-03T16:50:00.000-07:00

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...

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).
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.
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.
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

2011-09-02T17:30:00.000-07:00

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

2011-07-17T15:45:00.000-07:00

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

2011-07-09T16:23:00.000-07:00

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

2011-07-03T10:46:00.000-07:00

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

2011-06-28T20:53:00.000-07:00

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

2011-04-30T18:41:00.000-07:00

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

2011-04-25T09:12:00.000-07:00

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

2011-04-16T21:49:00.000-07:00

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

2011-03-24T21:15:00.000-07:00

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

2011-03-12T11:49:00.000-08:00

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

2011-03-04T20:16:00.000-08:00

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

2011-02-26T19:42:00.000-08:00

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

2011-02-22T18:14:00.000-08:00

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

2011-02-14T16:09:00.000-08:00

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

2011-02-03T19:45:00.000-08:00

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

2011-01-28T22:52:00.001-08:00

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

2011-01-21T23:35:00.000-08:00

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

2011-01-15T17:32:00.000-08:00

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

2011-01-03T21:50:00.000-08:00

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."

Reaction of the week

2010-12-27T22:35:00.000-08:00

I've been hesitant to commit to this, but now I suppose that I am, so go cry about it or something. The textbook I was using earlier is in Seattle right now anyway and I (at this point in time) am not. But my vision for this blog was never "Stephen Bahl condenses a textbook for you" or whatever. To be fair, I did post about some of my old labs and such. Alright, so it wasn't much, but surely it broke the monotony somewhat. And the textbook was really only ever a tool for me to introduce concepts from organic chemistry. That was what happened here. You're still confused? Try to keep up. I should explain. Fine. I will. What is wrong with you? Okay, new paragraph: go.

I am now stating it explicitly for you: this blog will change. It will continue to exist and it will continue to be written by me. It will continue to cover topics in chemistry. Those things will remain the same. They are not what is changing. Starting in January, the theme of this blog will be "reaction of the week." And I will keep it up for the duration of the year. This means that next year, I will blog about no less than fifty-two reactions. Fifty-two. 52. LII. Zweiundfünfzig. That's a lot. But it's happening. I've said I'm going to do it, so now it's too late to back down.

Not all of these will necessarily be reactions from organic chemistry. I won't mind throwing in the occasional inorganic reaction or whatever. But I have a stack of index cards in my desk right now with what appears to be upwards of seventy reactions from my organic chemistry class back in 2008. So yeah, it's not like I'm going to run out of reactions for this. 2011 shall be the year of the reaction. Or something. Oh yeah.

Polyketides

2010-12-18T23:21:00.000-08:00

I believe my previous post made some claim about marking calendars or some such thing that is now, in retrospect, quite ridiculous. Too bad. I thought it made more sense to extend my hiatus than to come back for another single post and vanish yet again. My vacation has been over for a while, but I've actually had a lot going on. Most importantly, I will be going to a new school in January (the University of Washington) and studying (among other things) inorganic chemistry. It's exciting. But because of this, I've considered shutting down operations here. I've been practically nonexistent on this blog recently, and if I'm too busy with school to ever post again, what's the point of trying to maintain this blog? Well, I'm recommitting myself to this endeavor. I know it seemed like I was doing that in my previous post. Fine. You've got me. I messed up. Just give me one more chance. Maybe. I have plans for this blog and I am almost sure that I know how I want to proceed, but I'll save it for the next post. If my plan works out, next year, despite me going back to school, will be this blog's busiest year so far by, well, a lot. There will be many posts. How many? At least fifty-two. No really, that's my plan.

But we'll save that for my next post, which will not be next year, but sometime soon. I'll try to have at least two more posts this year after the one you're reading now. So bear with me—until my next post, at least. I'm not quite ready for it. But it will be soon. Later this week, even. For now, I'd like to talk about a different subject entirely.

I did get to do some cool, science-related things on my vacation. You know, the one that ended in October that I was supposedly going to come back from and write a bunch of posts here right after that. Yeah, that one. The museums alone gave me a ton of material I could use here (but won't, for now anyway). There was also one, completely unexpected moment that is the inspiration for this post. After a bunch of crap we're not going to talk about right now, I arrived in the city of Bonn on, let's see, the October 3rd. The hostel I had booked was a weird one that didn't have its reception open until 5:00 PM. So I found it from the train station and walked down the street to get some lunch. I had a döner for lunch. I want a döner for lunch every day. They are so good. Almost too good. It's uncanny. Anyway, I was really thirsty, so I hunted down a store where I bought some ice cream and what I thought was orange juice but turned out to be more like orange soda. That sucked, because I hate carbonated beverages, but I was so thirsty that I drank it anyway. When I walked down to the hostel, there were some people talking. One of them was an American who mentioned doing biological research. I asked her about it. We ended up talking for a while, a lot of it about science.

It just so happened that a German biochemist who was there for some reason overheard our conversation. He talked a little about his own research on polyketides. I'd heard of them before, but that was about it. For the past two months, I've occasionally been amused at the realization that I "went all the way to Europe" to learn something. Of course, I knew I would learn things there and actively attempted to do so. But somehow, the realization that I learned some individual thing I could, hypothetically, have learned just as well here, but didn't learn it until I was there is amusing. I was recently made the comment that I had to go all the way to Austria to learn that grated horseradish is awesome. Anyway, I suppose that's how learning things always is. Even when we set out to learn things, we don't know what exactly it is we are going to learn.

From what I recall, he described work on a polyketide produced by bacteria or a series of homologous polyketides produced by bacteria, but the bacteria lived in completely different hosts: one in a beetle, one in a sponge, and one in some other thing I forget. Horizontal gene transfer for the win. Anyway, one thing he emphasized was that polyketides display tremendous variability. I'll spare you pictures because I was recently lambasting my organic chemistry textbook (the one I've been using a lot for this blog) for trying to scare students away with a picture of a big biological molecule. The important thing is that this variability, the incomprehensible number of forms molecules can potentially take is, well, the whole story of biochemistry. Structure determines properties. And those properties, if they're some of the right ones anyway, are what make life possible in the first place. Almost makes me want to switch to majoring in biochemistry...

Diastereomeric Alkenes

2010-09-09T00:12:00.000-07:00

Yes, it's a new post. I know. I hardly posted at all this year with school, then finished that in June, made two posts, and disappeared. No posts in July or August. It was not my intention to put this project on hiatus. I am sorry I failed you. I don't even know what kept me away for most of this time, but recently, there has been one thing: I am planning a trip. It's a pretty big one, actually. I'll be going to Germany. Since this is going to happen fairly soon and since my German is not so good, working on becoming as fluent in the language as I can be takes priority over writing new posts here. I guess this means I have to go on an actual, planned hiatus. Or, if you prefer, this means I'll have to extend the current hiatus, albeit interrupted by this solitary post. Whatever.

My primary reason for choosing German as a language to learn in the first place was its connection to chemistry. I haven't written about that here, so I'll explain. Germany as a region has long been a powerful contributor to the body of scientific knowledge. This was especially true in the nineteenth century, a formative period for many branches of chemistry. Because of this, in the twentieth century, in American colleges that required students to study a foreign language, chemistry majors traditionally chose German and were often encouraged to do so because it would allow them to access chemical literature that had only ever been published in German. The utility behind this is probably all but vanished these days, as any text that's of practical use to a chemist is probably available in English. But I enjoy history and the prospect of some day being able to read old chemistry texts in German has some sort of allure for me.

The German classes I took this year in school didn't do much in the way of making me comfortable with the language (the instructor mostly just played videocassettes from the 1990's), but they did make me really want to visit Germany. And so, here we are, with me abandoning you yet again. I'm really quite sorry about that.

Rest assured, this blog will be back with a vengeance. After the chapter I've been covering, there's material that I really like and I am excited about presenting it to you. Just seeing this material in the book makes me brainstorm different ways to cover it. There's some really cool chemistry to come once I finish this chapter, which I wish I'd already done over the summer instead of putting it off. So yeah, I should be back home and caught up on everything else by, let's say, October 29th or so. Mark your calendars for late October. Back with a vengeance. Really. More frequent updates. Better updates. Awesome chemistry. But not yet. You'll have to wait for my return. It will totally be worth the wait, though.

In honor of my trip, I'm going to skip ahead a bit to something with an obvious connection to organic chemistry's German roots. We'll return the to the material we were on shortly after I get back. And really, now might be as good a time as any to talk about this. By now, I'm sure you have a decent grasp of one type of stereoisomerism. But as you might have guessed, having a chiral center isn't the only way for stereoisomerism to occur. There are other ways that have nothing to do with a carbon atom bonded to four different groups. In fact, I initially wanted to write a post introducing all of the different ways for this to happen that I knew of, but I couldn't find a way to make it work. I did, once upon a time, say this, though...

Constitutional isomers often have dramatically different chemical properties. Their physical properties differ too. They might have different functional groups. In contrast, stereoisomers don't exhibit such bold differences. Two compounds that are stereoisomers of each other not only have the same atoms, but the atoms are connected in the same way. Their properties are almost identical. But the spatial positions of the atoms are different.

And a chiral center isn't the only way for that to happen. So, here's one of the other types of stereoisomerism, and it's a lot easier to demonstrate graphically in two dimensions than the type you already know about. There's just one thing I've probably never mentioned here that you need to keep in mind: unlike single bonds, there is no rotation along a double bond. Behold...
Can you spot the difference? Don't get too excited about it: everyone else notices it too. Same atoms. The atoms are connected in the same way, with an A and a B connected to a carbon that is double-bonded to another carbon also connected to an A and a B. But the spatial arrangement is different, no matter how we oriented these in three dimensions (try it if you want). We have a notation system (cis/trans) that makes this pretty easy, as seen in the drawing I just made. The "cis" version of this molecule has both A's on one side of the double bond and both B's on the other side of it. The "trans" version has an A and B on each side. They're diastereomers, which you recall from my last post means that they are stereoisomers, but not mirror images of each other. Easy, right?

But what if instead of just having two different kinds of groups, we have three or four? This notation system doesn't have a way to deal with those situations! For that, we need the E,Z system as my textbook calls it or Z-E Isomerie as the German Wikipedia calls it. And it's at this point that I realize I've gotten ahead of myself. In order to explain this, I need to use information that would come from a post I haven't written yet, probably the post I was supposed to have written if I didn't skip ahead to this section because I remembered it having something German in it. So despite my failure here, this still seems like a good post to end on before my trip.

If I had class, I'd rewrite this post and just make the whole thing be about cis/trans notation and save E/Z for later. Since I don't, I'll find some way to squeeze the German connection into this post. E and Z are really just more comprehensive versions of cis and trans, which themselves come from Latin instead of German: "cis" means "on the same side" or something like that and "trans" means "across" probably. In the E/Z system, each group is assigned a priority, but I haven't yet written about the rules for determining priority. They're the same ones that are used in the notation system for chiral centers, which is yet another Latin-based system. But since this simpler system already has the Latin words, for the comprehensive system, we use the German words. Crazy, I know. Eventually, I hope to show some examples of names of molecules with "E" or "Z" in them. "Z" stands for zusammen (together) and "E" stands for the word entgegen (against). I leave it to you to figure out which one corresponds to "cis" and which one corresponds to "trans."

Meso Compouds

2010-06-15T23:42:00.000-07:00

Remember way back when I said this?

A molecule that contains more than one stereogenic center might be chiral, but it might not. More on this later.

Well, it's later now. This post is on meso compounds. Please, no jokes about the name. Mostly because I am lazy, I will just start with the exact same example as this textbook: 2,3-dibromobutane.
And of course you spotted the tetrahedral stereogenic centers, right? We have a carbon attached to a methyl group, a hydrogen, a bromine, and another, identical carbon. So that's two chiral centers.

I don't think that I've mentioned it so far, but a molecule with two chiral centers can have, at most, four stereoisomers. I suppose that at this point I should introduce wedges and dashes to make three-dimensional interpretations of these stereoisomers, but I won't, so there. Actually, I should have done that a while ago. Fine, I'll get around to it at some point. Moving on...

Anyway, I'll probably just make a video to explain this, because it's even worse than trying to explain what a tetrahedral stereogenic center is using just words. But the short version is that we can arrange the bonds around both centers to have one version of the molecule, take its mirror image and have a pair of enantiomers, then take one of those forms and switch two bonds to have a third stereoisomer that is neither superimposable on either of the previous two molecules nor a mirror image of either of them. But this molecule is superimposable on its mirror image, so it has no enantiomers. It is achiral, even though it has two stereoisomers that are chiral. And that makes it a meso compound.

Also, the term for the relationship between stereoisomers that are not enantiomers is diastereomers.

Tetrahedral Stereogenic Centers in Cyclic Compounds

2010-06-13T14:12:00.000-07:00

A carbon atom that is part of a ring can potentially be a tetrahedral stereogenic center. It should be so obvious that I don't need to tell you this, but I'd better do it anyway: since two of the bonds on the potential center must attach in the ring, the two remaining bonds must be to two different substituents. Just to be safe, I shall illustrate this graphically.

So here is an example of a ring carbon that is not a tetrahedral stereogenic center...
And here is one that is...
That's pretty straightforward. Now, don't cry or anything, but that is not quite all there is to it. There is one more detail about this that sort of warrants a separate post. I think even you will find it rather easy, in principle. There is one further requirement in order for this hypothetical carbon atom to serve as a tetrahedral stereogenic center: there must be some difference between the two bonds in the structural sequence of the ring as we trace the path around it. No really, I worded it that way on purpose to dishearten you. It's actually not difficult.

We start with the central carbon atom and move along both ring bonds. Are the atoms that those two bonds attach to the same? And are the atoms that those atoms attach to the same? And so on. Eventually, both paths will converge (halfway across the ring). If both of those paths are identical, then the carbon in question is not a tetrahedral stereogenic center. However, if the paths are different, then the carbon is a tetrahedral stereogenic center.

And that's it! But just to be sure you don't forget about this, which you will anyway, let's demonstrate with an example. My textbook uses this example. Here's a compound that is achiral...
You've been practicing your nomenclature, right? So you already know that this is a skeletal structure for methylcyclohexane with one of the hydrogens drawn in for some reason. The reason is that the carbon we're focusing on is attached to that hydrogen, a methyl group (Me), and twice to the ring. But tracing both paths along the ring, we find that they are identical, arriving at a ring carbon attached to two hydrogens, a ring carbon attached to two hydrogens, and then meeting halfway along the bond between two ring carbons. So what we have is not a tetrahedral stereogenic center and this molecule is achiral.
If you only learned how to name compounds from this blog, you would not yet know that this is 3-methylcyclohexene. Don't worry about that. The important thing here is that, as before, we have a carbon attached to a methyl group, a hydrogen, and twice to a ring. However, this time, as we trace the paths of both ring bonds, going one way takes us to a ring carbon attached to two hydrogens and going the other way takes us to a ring carbon attached to one hydrogen and double-bonded to another ring carbon. The paths are not identical, so we have a tetrahedral stereogenic center, and this molecule is chiral.

Video!

2010-03-25T01:07:00.000-07:00

Since you still don't get how tetrahedral stereogenic centers work, I made a video to help explain it. If that doesn't work, I don't know what to tell you.

Introduction to Determining Chirality

2010-03-07T19:08:00.000-08:00

Stereoisomerism can be a lot trickier to identify than constitutional isomerism. With this in mind, and to some extent because I am too busy to make really good posts right now but also want to keep this project moving, there will be a series of short posts on the subject, starting with this one.

In order for any of this to make sense, you need to understand what it means for a particular atom to be a tetrahedral stereogenic center. Don't panic. Just peruse the previous entry and make sure you grasp the concept I am describing. The pictures are probably best for this, but what we're basically dealing with are atoms attached four different groups. This is because when an atom (usually carbon) is attached to four different groups, it is not superimposable on its mirror image. And, if it helps any, this concept can be extended to macroscopic things in our everyday lives. The textbook contrasts gloves and socks. In a pair of socks, the two individuals are identical (usually). But in a pair of gloves, the right glove and the left glove are not interchangeable.

Chiral molecules are like gloves (or shoes, for that matter). Even though the properties of the isomers are virtually identical, they are, in principle different from each other and these differences can manifest in ways that are relevant to us. An obvious demonstration of this is in pharmaceuticals, where often only one of the isomers has the desired effect, but the drug is sold and administered as a mixture of both versions. I should do a post on the thalidomide incident. Not right now, though. But maybe later.

Anyway, this isomerism can show up in other types of situations and hopefully I'll soon get to some of them, but for now, we shall focus on chirality that arises from tetrahedral stereogenic centers. Here are some points to keep in mind about these types of chiral molecules.

1. A molecule for which the mirror image is superimposable is achiral. A molecule for which the mirror image is not superimposable is chiral.

2. A carbon that is bonded to four groups, none of which are identical to each other, is a stereogenic center (aka chiral center). This does not necessarily mean that the molecule itself is chiral as we shall see.

3. A molecule that contains exactly one stereogenic center is chiral. Always. No exceptions.

4. A molecule that contains more than one stereogenic center might be chiral, but it might not. More on this later.

Stereogenic Centers

2010-03-01T22:27:00.000-08:00

The fifth chapter in this textbook is all about stereochemistry. I considered skipping it, but decided against it. However, for now I am skipping a lot of the fourth chapter. It's not that I'm tired of alkanes, it's just that the remaining sections dealt with conformations and I'd rather get back to that stuff later.

While constitutional isomerism is interesting, most of the time we'll be discussing constitutional isomers in terms that they are completely different compounds. It's just important to keep in mind that they are made up of the same atoms in the same proportions. In case you've forgotten, the thing that makes compounds constitutional isomers is that the atoms are connected to each other in different ways for each molecule.

Constitutional isomers often have dramatically different chemical properties. Their physical properties differ too. They might have different functional groups. In contrast, stereoisomers don't exhibit such bold differences. Two compounds that are stereoisomers of each other not only have the same atoms, but the atoms are connected in the same way. Their properties are almost identical. But the spatial positions of the atoms are different.

There are multiple ways for molecules to exhibit stereoisomerism. But we won't go over all of them at once. Instead, we'll take this slowly. It seems only natural to start with the case of stereogenic centers, specifically tetrahedral chiral centers, but don't worry about those terms right now. The important thing to grasp is the concept.

I am a big fan of the written word. I strive to be as good as I can at communicating concepts verbally. However, this concept is just so much easier to convey using a picture. So here you go.
I made this in ChemSketch and it's supposed to be CHBrClF (a carbon attached to a hydrogen, a bromine, a chlorine, and a fluorine). It doesn't really matter what the things attached to the central carbon are, though, so long as they are all different things. They could be other atoms or even organic groups like methyl, ethyl, and so on. A carbon (or another atom) attached to four groups, none of them identical, is a stereogenic center. It is chiral because it is non-superimposable on its mirror image. Here's the mirror image.
I had to mess around with the program a bit to get this to work, but other than that, does it look exactly the same as the previous molecule? Yes? Look again. With the white ball (representing hydrogen, but whatever) on top, we can look down and starting from brown and going clockwise, we will necessarily have a different order for each of these. It's unavoidable. They're almost the same, but they're different in this one respect. A classic example is the difference between a right hand and a left hand. But for this type of stereoisomerism, all that we need is one central atom with four different groups attached to it. Usually, the central atom is carbon and one or more of the groups are part of an organic molecule (rather than just the single atoms used in my example).

Examples of Naming Cyclic Alkanes

2010-02-28T00:03:00.000-08:00

Once again, I steal some problems from my textbook and do them here.
It's a ring made of six carbons, so it's a cyclohexane. Only one of the positions in the ring has any groups attached to it, and both groups are methyl groups, so it's 1,1-dimethylcyclohexane.
Another cyclohexane, obviously. This one has two groups at two different ring positions. The positions are across from each other in a 1,4 relationship. But which group gets numbered "1" and which one gets numbered "4"? Well, one is a methyl group and the other is a butyl group (four carbons in a straight chain). Alphabetical order determines which one comes first, so this is 1-butyl-4-methylcyclohexane.
The chain is bigger than the ring this time. So this compound is, as far as naming goes, defined as a five-carbon chain with a group attached at the first carbon, and the group that is attached is a cyclopropane ring. Therefore, we have 1-cyclopropylpentane.
The ring is a cyclopentane. Three groups this time, and all of them methyl groups, which makes naming this easy. Almost so easy that you could do it by yourself. But how do we number these groups? It doesn't matter which way we count, the smallest number we can start with is 1. This compound is 1,2,3-trimethylpentane.
This one is trickier. We definitely have a cyclohexane ring, but what are those groups attached to it. Well, let's start with the smaller one. It's an isopropyl group. See that? Probably not. Well, I told you, so now you know. Isopropyl group. The other one has four carbons. You might remember that there are four such groups possible. And if you have really been paying attention, it's clear that this is a sec-butyl group. Alphabetical order again, but the only prefix that matters for that is "iso-." That means the sec-butyl group is first. So this compound is 1-sec-butyl-2-isopropylcyclohexane.

Well, that's way that I learned to name this. And it's even the name that my solutions manual gives. But ChemSketch generated a different name that I am guessing is the true systematic name using proper IUPAC rules. The only difference is that the groups can't be written as isomeric forms of their straight-chain versions. This makes the nomenclature a bit messier (but it also scales up nicely, while the shortcut I'm using doesn't.

And that, children, is how to name cycloalkanes. I don't actually know what topic I'll cover next. You'll just have to wait to find out.

Nomenclature of Cycloalkanes

2010-02-07T23:18:00.000-08:00

This will not cover all cycloalkanes. In fact, for now we're only dealing with compounds that have a single ring. But then I didn't really cover all acyclic alkanes either. But what you should have with this post is a basic idea of cycloalkane nomenclature.

Rings themselves are named by how many carbons they consist of. So, for example, this molecule...
...is cyclohexane. But you already know that, of course. I mean, you do, right? You'd better, seeing as I already told you that this is cyclohexane. Yes, it was back in October, but so what? I mean, you are supposed to read and remember everything I write here. You know, I'm getting the feeling that you're not being much of a team player here. Yeah, it sure seems like I'm the one doing all the work. Look, it's just cyclohexane. It's not complicated. It's a simple molecule with a simple name. Four syllables. That's not too many. Cyclohexane. Cyclohexane. Cyclohexane. And don't you forget it.

Speaking of earlier posts, in this one I showed cyclopropane. Unless you're as awful at geometry as you are at chemistry, you should be able to make the connection that if the triangle is cyclopropane and the hexagon is cyclohexane, a square is cyclobutane and a pentagon is cyclopentane. Yes, and a heptagon is cycloheptane and so on. All we're doing is using those chemical numeric prefixes I showed earlier and counting the numbers of carbon atoms making up the ring. You can count, right? You can at least do that much.

But watch out. Not everything in the molecule is necessarily part of the ring...
That is methylcyclohexane (in glorious 3-D). Seven carbons, but only six of them form a ring. The pesky seventh one is attached to the ring. And if you've already forgotten how skeletal structures work, the hydrogens attached to the carbons are not drawn in. All but one of the ring carbons has two hydrogens. One of them has only one hydrogen and is also bonded to that carbon outside the ring, which itself has three hydrogens. So it's a methyl group. Hence the name: methylcyclohexane.

Of course multiple groups could be attached to the ring. In that case, we use numbers. This compound...
...goes by the name 1-ethyl,3-methylcyclohexane. And if you think in terms of the rules you learned for acyclic alkanes, this makes sense. We have to number the positions on the ring somehow. So we're starting at first substituent alphabetically. Here, I'll even put the numbers in...
This system of numbering positions on rings will be used a lot in the future, so you should be comfortable with it. But it seems straightforward enough to me, so I'm not going to reiterate it further.

We might also end up with two groups attached at the same position on a ring. Not to worry...
That's 1,1-dimethylcyclopentane. The same general principles from naming acyclic alkanes still apply. This does run into limitations of course. I won't be covering those now. But I do think that I should to a follow-up post in which I name some examples from homework problems in the textbook. And just so that you can follow along, there is one more tiny little thing that you need to know about cycloalkanes. If a ring is attached to a hydrocarbon chain that is longer than the number of positions in the ring (like if a cyclopentane ring had an octane chain attached to it), the compound is named based on the chain (so that example I just made up would be 1-cyclopentyloctane).

Examples of Naming Acyclic Alkanes

2009-11-21T20:44:00.000-08:00

As promised, here are some right out of the textbook.

The first one is in condensed notation: CH₃CH₂CH(CH₃)CH₂CH₃

Since I am so good, I immediately recognize that the third carbon has a one-carbon branch. Other than that, this is a straight chain. But let's not get ahead of ourselves. We are doing this the right way. We start with the last part of the name. With no heteroatoms, this is a hydrocarbon. With no multiple bonds, it's an alkane. With no rings, it's an acyclic alkane. We know the name must end in "-ane." Next, what's the longest carbon chain? Five. So the parent name is pentane. Branches? Yes, at the third carbon (counting either way). And the branch is a methyl group. Therefore, the name of this compound is...

...3-methylpentane. And you know what else? I checked the answer in the study guide and I was right! Woo hoo, Stephen got something right. Anyway...

(CH₃)₃CCH₂CH(CH₂CH₃)₂

This one is harder. First we have three methyl groups attached to one carbon. That carbon links to another that links to another, which is attached to two ethyl groups. Which methyl group and which ethyl group is considered part of the longest carbon chain does not matter because the groups are identical (that is, the methyl groups are identical to each other and the ethyl groups are identical to each other). So adding those three carbons to the rest of the chain, we find that the longest carbon chain is six carbons long, so this is a hexane.

Which group gets priority? In this case, we go in alphabetical order. "E" comes before "M." So this should be...

...3-ethyl-5,5-dimethylhexane. Or not. Oops. I started at the wrong end. It's actually 4-ethyl-2,2-dimethylhexane. It's that instead of the one I thought it was because 2 is lower than 5. It doesn't matter that 3 is lower than 4 because the method that gives the lowest number period is the one that gets priority, not the one that gives the lowest sum or anything like that. I hope you learned your lesson. Moving on.

CH₃(CH₂)₃CH(CH₂CH₂CH₃)CH(CH₃)₂

A propyl group? No, that's part of the longest carbon chain. They're trying to trick us. Starting from the left we have a carbon and then a string of three more, so that's four in a row. Then there's another (five) with that propyl group branching off. If we count going up the propyl group we get three more (eight). If we treat the propyl group as a branch, we get another carbon with two methyl groups, one of which would be a branch, making the total length seven. Sneaky textbook. This is actually an octane.

If we start from the end of what's being labeled as a propyl group (but is actually part of the chain) we get a branch at the fourth carbon. Starting from the left makes it at the fifth, so we start from the end of the propyl group instead. The branch consists of three carbons and two of them are attached to the other, which is where the branch connects, so it's an isopropyl group, meaning the compound is...

...4-isopropyloctane. And I'm right. I rule.

Enough of these condensed structures!
I used MS Paint because it was a small one and it's kind of hard to make them look less awful on ChemSketch. Anyway, this one seems easy to me. Five carbons long means pentane. Two methyl branches at the second carbon and two at the fourth. Therefore...

...2,2,4,4-tetramethylpentane. And I am right again. Excellent.
It's seven carbons long, but there are a couple of different ways to arrive at that. The one that give the lowest number to a branch is the one that simply starts on the far left, for a methyl group at the second carbon. There's another one at the fifth carbon and an ethyl group at the third, so this is...

...3-ethyl-2,5-dimethylheptane. And I'm right yet again. Three in a row! Let's do one more.
I moved back to MS Paint again when I perhaps should not have. But ChemSketch was being annoying (it kept trying to put rings into this). Obviously this one is larger than the other ones so far, but the principle is the same. The longest carbon chain is ten. The fastest we can get to a branch with it is on the second carbon, again counting from the far left. From there we label the other branches and put them in the proper order. About that, the branch on the fifth carbon is a sec-butyl group. When alphabetizing the branch names, this is treated as a "B" and not as an "S." The same would be true for tert-butyl but not for isobutyl. Unnecessarily confusing, I know. But in this case it does slightly affect the name, which is...

...5-sec-butyl-3-ethyl-2,7-dimethyldecane. And that's pretty much all there is to it. The study guide I used to check my answers breaks the process into three steps.

Name the parent chain by finding the longest C chain.
Number the chain so that the first substituent gets the lower number. Then name and number all substituents, giving like substituents a prefix (di, tri, etc.).
Combine all parts, alphabetizing the substituents, ignoring all prefixes except iso.

It takes some getting used to, but this is the basis for how other compounds, even ones with multiple functional groups, are named.

Nomenclature of Acyclic Alkanes: Prefix

2009-11-21T02:05:00.000-08:00

I hope you have the other component of naming alkanes down, because I am never reviewing it again (just kidding). Now for the prefix. While the parent name identifies the longest carbon chain, the prefix tells us where on that chain branches occur and what the branches look like. Depending on how much branching (and what kind) is going on, the prefix may be anywhere from nonexistent (no branches, which we sometimes denote by using "n" as a prefix) to ridiculously long.

Firstly, the location of a branch is denoted using Arabic numerals. A branch at the second carbon in the longest carbon chain gets a "2" and a branch at the third carbon gets a "3" and so on. Some carbons in the longest carbon chain might have two branches. When that happens, its number gets used twice.

Often, there are multiple possible places to start from. With alkanes, the correct starting carbon is the one which, when started from, yields the lowest possible number being named first. If we start counting on one end of a chain and the first number that comes up is for a branch at the fourth carbon, but counting from the other end of the chain would make our first branch be at the second carbon, then it is the end that would make the first branch be at the second carbon that is the correct starting point.

Also, numerals are separated from each other by commas and from the rest of the name by hyphens. That's not just for alkanes. That's a universal rule. Commit it to memory, slave.

Anyway, to specify how long a branch is, we use the wonderful numerical prefixes I introduced in my last post. You know, the ones that are mostly Greek, but not really. A branch that is only one carbon is a "methyl" group. Two carbons is an "ethyl" group, etc. A branch that is seven carbons long is a "heptyl" group (and since it's not part of the longest carbon chain, that means the longest carbon chain must be really long). This all works nicely for branches that are themselves straight. But what about branches that have branches of their own? That's the hard part. Kind of. In order for considerable branching to occur, the molecule itself has to be pretty big. I've never had to deal with such compounds myself. The textbook is covering substituents with up to four carbons and that's always been good enough for what I've had to do. There are not very many. Here we go...

Methyl group: R—CH₃
Ethyl group: R—CH₂CH₃
Propyl group: R—CH₂CH₂CH₃
Isopropyl group: R—CH(CH₃)₂
Butyl group: R—CH₂CH₂CH₃
sec-Butyl group: R—CH(CH₃)CH₂CH₃
Isobutyl group: R—CH₂CH(CH₃)₂
tert-Butyl group: R—C(CH₃)₃

If you find the condensed structures confusing for those four-carbon groups, here are some links to pictures (off-site) for the butyl variations...

Butyl, sec-butyl, isobutyl, and tert-butyl.

And that's all. Now you know how to name acyclic alkanes. Oh, one more thing. If two or more of the same type of branch exists in a molecule, those branches get named together and get a Greek numerical prefix just to confuse you even more. But really, that's it. Stay tuned for next time, where I'll do a follow-up post with some examples of naming alkanes using problems from the textbook. Oh wait, this isn't a radio. You can't tune anything. Whatever.

Nomenclature of Acyclic Alkanes: Parent Name

2009-11-14T00:14:00.000-08:00

I mentioned the IUPAC systematic nomenclature system before. I think I did, anyway. This project has been on hiatus for a while and I can't remember. But I'm back now! Really. I hope. Anyway, today we are going to learn how to name some alkanes. It's easy to do, and you need to know it to name other compounds. So learn it. I command you.

Let's start at the end. That's a good place to start, right? The last part of the name of any alkane is, get ready for this...

...it's "-ane." That should be quite easy to remember, even for you, because "alkane" itself ends in "-ane." If a compound is an alkane, its name ends in "-ane" and, conveniently enough, if a compound is not an alkane, its name will not end in "-ane." I know. Chemistry is so hard.

Next, we find the longest carbon chain. This is actually very easy, but teachers love trying to trick beginning students with odd drawings where they make part of the longest carbon chain look like a branch to people who are not paying attention. If this were a real chemistry class and I were the teacher (that would be bad), I would totally do this to you because I think it's hilarious. For now, I'll just give you the benefit of the doubt and assume that you are paying attention and can tell what the longest carbon chain in a molecule is.

Really? I shouldn't do that? Fine.
How long is the longest carbon chain? If you answered eight, congratulations, you did not fall for the dumbest trick in chemistry class. If you answered some other number, you were not paying attention or you cannot count or you're just a moron or something. I don't know. Shame on you anyway. You're bad (unless you got the right answer).

Once we know how long the longest chain is, we convert that into a numerical prefix, then attach it to our "-ane" suffix. Convert it into a numerical prefix? Yes, it's easy. No really. It is easy, just so long as you already know the Greek numerical prefixes—and use the Latin one for "nine" just to mess things up—and forget the first four prefixes and make up new special ones that are specific to chemistry. It was easy for me though! Here, I'll give you the first ten and we'll worry about going higher later.

1 = "meth"
2 = "eth"
3 = "prop"
4 = "but" (pronounced like the word "butte" just to confuse you even more)
5 = "pent"
6 = "hex"
7 = "hept"
8 = "oct"
9 = "non" (pronounced so that it rhymes with "tone" and not some other way)
10 = "dec"

Memorize them now. I command you. Done? Good. See, that wasn't so bad. Now, there's just one more tiny thing. Then we'll be all done and you'll know how to name acyclic alkanes. We have straight chains covered (unless they're longer than ten carbons long, but shut up). So a hydrocarbon that is a straight chain with five carbons would be "pentane" and one with nine would be "nonane" and so forth. Everything is fine, and then branches come and mess it all up. Not to worry: the IUPAC has an elaborate set of rules for us to denote where on a chain the branches lie and what the branches look like using prefixes and attaching them to the parent name (which simply describes the longest carbon chain. Well, it's elaborate enough that I'll save it for my next post, anyway. For now, just have the whole parent name part down.

Cycloalkanes

2009-10-18T18:39:00.000-07:00

I just found an error in my textbook. Seriously. The book even bolds its own error. The offending sentence reads...

Cycloalkanes have molecular formula C_nH_2n and contain carbon atoms arranged in a ring.

That is only true for cycloalkanes with just one ring. Cycloalkanes can have more than one ring, and each additional ring means two fewer hydrogens. And there are a lot of those. Try to keep up, textbook. Anyway, the examples the book then uses for cycloalkanes are all ones with just one ring. In fact, the examples in this section of the book have all of the carbons in the ring, but this is not necessary or even particularly significant.

The smallest cycloalkane ever is cyclopropane, with a molecular formula of C₃H₈ and a skeletal structure that looks like this...
Yes, it's a triangle. This really should not surprise you if you've been paying attention, which you haven't. I could show it in 3-D, but so far I haven't figured out a way to post my wonderful 3-D images here in this blog thing without them looking like crap (because I am pasting them into MS Paint and saving them there. If you want pretty pictures, go read a pretty pictures blog or something. I hear that they have those. Such things may be more suited to your intellect.

Cyclobutane's skeletal structure looks like a square. This should be easy to visualize. Same with cyclopentane and a pentagon. I have already shown the skeletal structure for cyclohexane here and here.

Also, the book claims that the largest known cycloalkane with a single ring has 288 carbon atoms. But this is in a problem asking for molecular formula (and obviously the molecular formula is C₂₈₈H₅₇₆) and I cannot tell if it is giving me authentic trivia or merely posing a hypothetical for the purposes of asking such a question and reinforcing the concept.

One last thing, which the book apparently omits in this section (although it will probably come up later) is ring strain. The carbon atoms are most stable at a certain bond angles. In the case of alkanes (and lots of other things, really), the ideal bond angle is 109.5° and all four groups attached to the carbon atom are equally far away from each other, forming a tetrahedron with the carbon atom in the center and each attached group in one of the corners. But when carbon atoms form rings, the bond angles become strained. This ring strain causes the molecule to be more reactive. Cyclopropane, with 60° angles between the carbons, has the most ring strain. After that, it becomes important to note that these rings are three-dimensional objects. They can be denoted with two-dimensional skeletal structures on paper, but are under no obligation to lie flat. So cyclobutane does not actually have 90° angles between its carbons, as "puckering" reduces strain and creates larger angles. Later in the chapter, this is explored for cyclohexane in particular, which has the most stable ring among cycloalkanes.

A Note on Complexity and Isomerism

2009-10-17T16:25:00.000-07:00

My textbook has a table with information that I did not include in my last post, but that may improve understanding of isomerism. In case it is not obvious, the number of isomers grows with the size of a molecule. In my last post, I showed the two isomers of butane. Larger alkanes have even more, because with more atoms, there are more ways to rearrange them. Small alkanes are easy to understand in this regard. A hydrocarbon with one carbon has no isomerism. The same is true for two or three carbons. When we get to four, as already demonstrated, there are two possibilities: a straight chain and one with a branch. Five carbons means three isomers. With seven carbons, we get nine isomers, which is still manageable, but then add a single carbon and there are eighteen isomers. The table ends with icosane (C₂₀H₄₂), which has 366,319 constitutional isomers.

And that is just acyclic alkanes. There are so many other things to consider, that the complexity is staggering. And that is why we have a systematic method of naming molecules. Anything else would get pretty impractical.

Constitutional Isomers Redux

2009-10-17T01:01:00.000-07:00

I suppose that my textbook introduces constitutional isomers in the alkanes chapter because alkanes are pretty straightforward and can ease one into the concept. Constitutional isomers can and do occur in other molecules. Isomerism is when two or more different compounds have the same molecular formulae. In other words, they have the same kinds of atoms and the same numbers of those atoms, but something makes them chemically distinct. Later on, we will explore stereoisomers, and it will be very exciting. But for now, we're looking at constitutional isomers, which differ in the way the atoms are connected to each other. Let's take a look at two molecules that are constitutional isomers of each other...

Name: n-butane (or just butane)
Molecular formula: C₄H₁₀Condensed structure: CH₃CH₂CH₂CH₃
In stunning 3-D:
Yes, I just figured out that I could render butane three-dimensionally with my nifty software. Anyway...

Name: isobutane (or 2-methylpropane)
Molecular formula: C₄H₁₀
Condensed structure: CH(CH₃)₃In glorious 3-D:
Both molecules have the same quantities of the same atoms. But the bonds are not identical here. A carbon bonded to two other carbons and two hydrogens is electromagnetically different from one bonded to three other carbons and one hydrogen. Also, the three-dimensional forms are quite different, and when the molecules interact with other bodies (including other molecules just like themselves) the results will be at least slightly different. Although very similar, these two compounds have different chemical and physical properties. They are more like each other than other compounds that have different atoms and other, more striking differences. Because of these facts, we use the term "constitutional isomers" to denote the relationship between these similar molecules.

But when it comes to properties, constitutional isomers are not always so similar to each other as those two. Some constitutional isomers contain different functional groups from each other and, if you remember the importance of functional groups like you should, this means they can have dramatically different chemical and physical properties...

Name: ethanol

Molecular formula: C₂H₆O

Condensed structure: CH₃CH₂OH

In brilliant 3-D:

It's an old friend: ethanol. I don't know how many times I've shown ethanol before, but you had better know that this is what it looks like. And if you managed to actually have some brain capacity, maybe you even remember that this compound is an alcohol, as it has a hydroxyl functional group. Easy, but here's a constitutional isomer of ethanol.

Name: dimethyl ether (or methoxymethane)

Molecular formula: C₂H₆O

Condensed structure: CH₃OCH₃

In spectacular 3-D:
Since the name has "ether" in it, you have deduced, unless you are a total idiot, that this is an ether (the name of the functional group is methoxy in this case). But the molecular formula is the same. The functional groups here are so unlike each other that reactions possible for one would be impossible for the other. Oh, and remember hydrogen bonding? Ethanol has it. Dimethyl ether cannot have hydrogen bonding because there is no hydrogen attached to the oxygen, so these two even have different intermolecular forces. In this way, two constitutional isomers can be quite dissimilar. What kind of atoms a molecule has and how many are very important, but the configuration of the bonds holding the atoms together in a molecule matters a lot too.

Edit: After posting this, I started going back to tag my posts. I noticed that way back in February, I wrote a post about constitutional isomers. I think this new post is better, but here is the old one. If you do not get the concept after reading this post, read the old one. If you still don't get it, tell me, I guess. It seems fairly simple to me and I think I did an adequate job of explaining it both times, but maybe I am wrong...

Cyclic and Acyclic Alkanes

2009-10-15T01:09:00.000-07:00

As I mentioned in my Functional Groups post, alkanes are hydrocarbon molecules with no π-bonds. They can be straight chains of carbons with attached hydrogens, or there can be branches or rings or both. All of the fourth chapter in my textbook is dedicated to alkanes. But the first part is just about getting acquainted with them. Alkanes are something of a baseline in organic chemistry. It's when functional groups are added that the chemical properties behind so much of our world come into play. Lacking functional groups, alkanes are not particularly reactive. They can react, though. And I know we'll come to that eventualy. There's a lot to learn from alkanes, though.

Firstly, let's distinguish between acyclic alkanes and cyclic alkanes. If it has a ring, it's cyclic. If it does not have a ring, it is acyclic. Simple, right? It better be. No, two rings is still cyclic. What counts as a ring? Oh, good question. A ring is pretty much what it sounds like. Three or more atoms bonded to each other with a loop that can be formed from the bonds between them. Carbon #1 is attached to Carbon #2 and Carbon #2 is attached to Carbon #3, which is itself attached to Carbon #1. Three atoms is the minimum, but larger rings are more common.

For an acyclic alkane, the number of hydrogens will always be two plus double the number of carbons. H = 2C+2. Actually, a little logic should demonstrate this point. No amount of branching chains changes the formula. But a single ring does. I shall illustrate with some examples. First, here is hexane...

Name: n-hexane
Molecular formula: C₆H₁₄
Skeletal structure:
Well, that's a nice, simple acyclic one. How about an acyclic alkane?

Name: Cyclohexane
Molecular formula: C₆H₁₂
Skeletal structure:
I Know I've shown this one at least once here, once upon a time. Hexagons should hopefully be pretty recognizable. And notice that it has two fewer hydrogens than the last one? That's because of the ring. What? You want to know how the ring makes it so that there are two fewer hydrogens in the molecule? Really? Look, just pretend we sever the bond between two carbons. Any two. Now those two carbons need a new bond to something else because, remember, carbon forms four bonds. So we stick a hydrogen onto each of them, and look at that, it's n-hexane, the same molecule I already showed you just before this one. Amazing. And that is why the ring makes it so that there are two fewer hydrogens than in an acyclic alkane. Simple.

Ribosome Rant

2009-10-09T00:16:00.000-07:00

I realize this is a departure from the content I normally post here, but I just started writing a rant on a different site and I think it really belongs here. The Nobel Prizes are being announced this week. The prize in chemistry went to Venkatraman Ramakrishnan, Thomas A. Steitz, and Ada Yonath for their work on the the structure of ribosomes. There's a sentiment that I've been seeing somewhat and it got me annoyed enough to actually write this. Here are some examples of the sentiment I am talking about...

I don’t care. For some reason, this year I’m not getting into Wednesday Madness nearly as much as I have in previous years. I’ll be happy if they give it to, uh, a chemist.

≡≡≡

Oh well. Here’s an idea. In lieu of giving out Nobel Prizes in Chemistry to achievements in chemistry (since they only seem to give it to actual chemists every other year anyway, it won’t be much of a stretch), let’s start handing them out to the authors with the best paper titles ever.

≡≡≡

As already announced biologists walked away with this year’s Nobel prize in chemistry once again, this time for work in determining the structure of Ribosomes.

≡≡≡

As chemists we would like to see the Nobel chemistry prize go to a chemist. Our Nobel hopefuls may be a measurable magnitude more chemically interesting, as measured by ChemFeeds, but there is more work for them to do until these topics become world renowned (which seems to be the dominant prerequisite these days).

≡≡≡

And again the Nobel for Chemistry goes to "bio-chemists"....
Congratulations...but as a strictly synthetic organic chemist...I am a bit ticked off.
With all the biology and the nanoscience development in recent years, it'll be eons before an organic chemist wins the prize again.

Alright guys, the applicant to get into school to work on an undergraduate degree (I do already have my A.S. at least) has some news for you: biochemistry is chemistry. I find this reaction deplorable. Chemistry is all about atoms and the bonds between them, what things are made of and how they interact with each other. That is exactly what this prize was awarded for. Perhaps word has not yet reached the innermost confines of your biology-free ivory towers, but ribosomes are made out of atoms and ribosomes have bonds—lots of them. Ribosomes participate in chemical reactions. This really should go without saying.

I could be way off here, but I don't think I would see this in other branches of science. If an annual physics prize went to scientists who did work in astrophysics, would physicists complain that the astronomers are taking physics prizes? I think not, but maybe some of them would. Maybe some of them sequester themselves in ivory towers devoid of any science that is not their own particular specialization, just as apparently some chemists do. My impression is that many, if not most, physicists have a passion for the universe and its fascinating nature. They want to see the physics in everything. I want to see the chemistry in everything. And I'd like to think I'm in good company, but the reactions I've seen to this Nobel Prize have cast some doubts on that.

How arrogant must one be to think, "Only research in the area of chemistry that I focus on should win prizes"? Some might protest that this is an unfair characterization, but if one is willing to dismiss the entirety of biochemistry, I am more than willing to err on the side of assuming that one would go on to dismiss other purportedly unworthy subjects in a similar manner. This exclusive approach is the exact opposite of what I want to stand for. I want chemistry to be inclusive. If we excise some of it because it deals with biological molecules and can therefore be considered biology, we might as well excise the parts that deal with minerals and make that geology and so on until we have divided everything up and there are no more chemists, just former chemists working in other fields of science.

The ribosome people did not win because the biologists are taking over and they did not win because ribosomes are famous and other work was too obscure. They won because they did good chemistry that is of abundant benefit to humanity.

Strength of Intermolecular Forces

2009-10-04T02:00:00.000-07:00

This chapter in the textbook is quite long, but not all of it is well-suited to posts like these. A lot of this reviews concepts from general chemistry and has lots of pretty pictures and I don't want to spend too much time on things like melting point and solubility and soap. The soap thing is something I originally learned in high school and got to see repeated in two general chemistry classes in college and organic chemistry too. I might do a post on it, but for me, it's gotten kind of old. There is some really great material here. I especially like the explanations of biomolecules, but perhaps that's best reserved for later.

In short, I do want to write at least one more post on the odds and ends in the third chapter of my textbook. They will come soon if at all, because I am long overdue on starting the fourth chapter. Before I do either of those things, let's wrap up intermolecular forces.

The strength of intermolecular forces is, in ascending order...

London Dispersion: caused by fluctuations in charge density across the surfaces of molecules.
Dipole-Dipole: caused by permanent dipoles.
Hydrogen Bonding: Caused by extreme loss of electron density on hydrogen when bonded to oxygen or nitrogen (or fluorine, technically).

Note that ion-ion forces, which hold ions together in ionic compounds, could be compared to these forces, although ions are technically not considered molecules. Ion-ion forces are much stronger than any of these intermolecular forces.

Intermolecular forces have the following effects on physical properties...

Stronger intermolecular forces increase boiling point.
Stronger intermolecular forces increase melting point.

There's more, but I am skipping it because that's the important stuff and you'd forget the rest anyway. If something I omitted here becomes important later, I'll just blame the problem on you. It's either that or explain the thing when the issue comes up.

Hydrogen Bonding

2009-09-22T01:33:00.001-07:00

Hydrogen bonding is a special (and awesome) intermolecular force that can happen when hydrogen is attached to nitrogen, oxygen, or fluorine. All three of those are pretty small atoms and are also highly electronegative. They pull so much electron density away from the tiny hydrogen that the nucleus (a single proton) of the hydrogen is highly exposed and attracted to negatively charged things, letting it sort of stick to negatively charged bits of other molecules.

Fluorine is a halogen. I wrote about this, but it was so long ago that you probably forgot all about it. As a halogen, fluorine can only have a bond to one other thing. If that thing is hydrogen, then we have lots of hydrogen bonding fun, but there's only one compound for which this is possible, and that's HF (hydrogen fluoride), which isn't organic, so we won't be paying much attention to it right now.

Nitrogen and oxygen, unlike fluorine, can both be attached to hydrogen and have at least one bond to spare. So there are lots and lots of compounds that exhibit hydrogen bonding. Just look for hydrogen attached to nitrogen or oxygen. Hydrogen bonds are stronger than dipole-dipole forces and other intermolecular forces. In fact, they're partial covalent bonds, although still not nearly as strong as regular covalent bonds.

The classic picture for showing hydrogen bonds is with water, so I might as well just use one of those rather than trying to make my own picture. It's much easier and looks way better. Here's one...
Pretty much any property of water has something to do with hydrogen bonding. And it's important in lots of other compounds too. The structures of proteins and nucleic acids use plenty of hydrogen bonds. Hydrogen bonding is one of the reasons hydrogen is my favorite element. But just remember, it's stronger than other intermolecular forces.

Dipole-Dipole Forces

2009-09-08T01:20:00.000-07:00

Do you remember how electronegativity works? I posted about it, so you should. Here, just for you, I'll link back to that post. Electronegativity is necessary to understand polarity. And polarity is how these forces operate. An organic molecule with a functional group that contains an electronegative atom such as oxygen or nitrogen will likely have a permanent dipole. The heteroatom pulls negative charge toward itself (because it's electronegative, obviously). As described in the previous post, regions of the molecule can have differential charge. But a permanent dipole is much stronger than the fleeting changes responsible for London dispersion. Because of this, molecules that have dipole-dipole interaction experience stronger intermolecular forces than ones that have only London dispersion. Compounds with this property are said to be polar and ones that do not are non-polar. Consider these examples...

Acetone
Condensed structure: OC(CH₃)₂It has a permanent dipole that looks like this.
Those Greek letters represent partial charge. The electronegative oxygen pulls electron density toward itself, so it is the center of negative charge. The region opposite it, lying near the central carbon, is the most positively charged region of the molecule.

Carbon dioxide
Condensed structure: CO₂This time, when we draw the arrow through one oxygen, the other oxygen cancels it out. Despite having a highly electronegative element, carbon dioxide is non-polar.

Whether a solvent is polar or non-polar tells chemists a lot about its potential uses, and some reactions need one type of solvent or the other. Acetone is a well-known polar solvent, but water is the best known in this category. Many organic compounds such as n-hexane are commonly used as non-polar solvents. That's all for these at the moment. Just remember that in order for dipole-dipole forces to occur, the compound(s) must have permanent dipoles, and that dipole-dipole forces are usually much stronger than London dispersion. Of course, a molecule can have both. But the stronger forces are considered to override the weaker ones for all practical purposes that I've encountered.

London Dispersion Forces

2009-09-07T13:54:00.000-07:00

My next few posts will discuss intermolecular forces. The textbook put this information in the same chapter as functional groups, because functional groups play such a huge role in the type and strength of intermolecular forces a compound has. I don't know if this is the best approach, but I'm too lazy to figure out a different one and whatever, this works.

Hopefully the word "intermolecular" tips you off to the fact that these are forces that occur between molecules. If not, what is wrong with you. Intermolecular forces are much weaker than normal covalent or ionic bonds. If they weren't, they wouldn't be intermolecular because they'd be binding atoms as tightly as the bonds within the molecules and the molecules wouldn't really have distinct identities at all.

London dispersion forces are the weakest and most ubiquitous ones we'll look at. In this textbook and in the college chemistry classes I took, they're more often referred to as van der Waals forces. But from what I can tell, that term is actually more inclusive, so I'll be calling them "London dispersion forces" or "dispersion forces" or maybe "London forces" until I have some reason not to. Do note that it seems to be common to refer to these as "van der Waals" forces/interactions. They're also known as induced dipole–dipole forces.

London dispersion is so ubiquitous in organic chemistry in part because it's the basis for intermolecular attraction in hydrocarbons. The other forces don't come into play unless there are functional groups. It would be nice to have the picture in my textbook to convey what I'm talking about, because it really does seem better than what I'm finding on the web, but this ball & stick picture that uses hydrogen molecules should serve well enough...
Because of electromagnetic repulsion, positive and negative charge gather in regions of the molecule. The whole molecule is neutral in charge, but some parts of it are more positive and others are more negative. I like the methane example used in my textbook better, because it shows more randomness, but the principle is the same: the negative portion of one molecule is attracted to the positive portion of another molecule. And this interaction, magnified over massive numbers of molecules, gives the whole thing some cohesion. And when I say some, in the case of hydrogen at normal temperatures and pressures, it's not enough. That's why pure hydrogen is usually a gas. The same is true with methane. If it gets cold enough, as happens in that atmosphere of Titan, one of Saturn's moons, methane is liquid.

Not all hydrocarbons are gases at room temperature. The bigger they are, the higher the temperature has to be in order to excite the movements of the molecules enough to overcome London dispersion forces and liberate the molecules from each other. But it's not all because the molecules are simply bigger. They're also longer. And that means more surface area. My textbook points out that n-pentane has stronger dispersion forces than neopentane. Condensed structures should be enough to show why...

n-pentane: CH₃CH₂CH₂CH₂CH₃neopentane: C(CH₃)₄

The long, straight chain of n-pentane has more surface area than the bunched up neopentane. They have the exact same quantities of each atom, but this difference has significant effects on the physical properties of these molecules.

Finally, dispersion is also affected by the polarizability of atoms. In a large atom, the electrons are further from the nucleus and are more subject to disruption and fluctuation in electromagnetic forces than in a small atom, where the electrons are more tightly held.

Oh, and a fun trivia fact that everyone loves is that geckos hold onto surfaces using these forces (the bristles on their toes are small enough to do this and well distributed enough that enough surface area is there for the forces to be strong enough to hold the lizard's whole body even when walking upside-down). Seriously. Pretty cool, huh?

Visualizing Functional Groups

2009-09-07T01:03:00.000-07:00

I told you that I would post more functional groups and I meant it. But I also want to make the ones I've already introduced clear. And condensed structures can confuse people. I heard you're easily confused. So we'll spend a bit more time getting acquainted with these. I think I covered the hydrocarbons well enough, so we'll focus on functional groups that contain heteroatoms.

More about alcohols
The nature of the carbon that the hydroxyl group is attached to determines the type of alcohol here. First, there's a primary alcohol...
Note that the carbon attached to oxygen is attached to only one other carbon (in the R group). In a secondary alcohol, this carbon is attached to two other carbon atoms...
And as you might have guessed, in a tertiary alcohol, that carbon is bonded to three other carbons...
If you were thinking that a quaternary alcohol would be one in which that carbon is attached to four other carbons, you sure are dumb. Carbon is tetravalent. You remember that, don't you? It can't form five bonds. There is no such thing as a quaternary alcohol.

If you were wondering, whether an alcohol is primary, secondary, or tertiary has important implications for its chemical properties, hence the distinction. This is also the case with amines, as I already alluded to, but in that case, it's how many carbons the nitrogen is attached to that determine which type of amine the molecule is. So there's some cool new information for you. But now for some clarity on material I already covered in my last post.

Visualizing aldehydes & ketones
Here's the Lewis structure of an aldehyde...

And here, for contrast, is a ketone...
Notice the big difference: with an aldehyde, the oxygen is at the end of a chain and with a ketone, the oxygen is attached to a carbon that is somewhere in the middle of a chain. These structures both have a "carbonyl" group and their chemical properties are often similar, but they can be different in important ways and this distinction is certainly worth remembering.

Carboxylic acids and friends
Carboxylic acids get several other classes of compounds grouped with them as "derivatives of carboxylic acids" quite literally because carboxylic acids can be used to make these other compounds. I won't cover all of them because there's a whole chapter on this stuff and it's way later in my textbook. But because you're slow, I worry about your ability to even deduce the general appearance of these groups from a condensed structure. So here's a carboxylic acid...
Like the aldehydes and ketones, there's a carbon double-bonded to an oxygen and single-bonded to an R-group. But the fourth bond isn't to hydrogen or another carbon. It's to oxygen, which itself is attached to hydrogen. Remember acidity? You know, that thing the last chapter was all about and such. And maybe you even remember that in my "Aspirin" post I said, of the carboxylic acid, "This arrangement of atoms makes it easy for a certain reaction to occur. That reaction is a Brønsted-Lowry acid-base reaction." Really, it's not familiar. Whatever. That proton can totally come off.

Since I like functional groups so much, here are some more in condensed structure...

Acyl halide
R—COX (like a carboxylic acid, but with the second oxygen replaced by a halogen)

Imine (imino group)
R=N—R' (these come in multiple varieties and I haven't really studied them yet)

Peroxide (peroxy group)
R—O—O—R' (the oxygens are actually attached to one another)

Nitrile (cyano group)
R—C≡N

Enough. We will now cover new functional groups as they come up.

Functional Groups

2009-09-05T00:10:00.000-07:00

Functional groups are structures within molecules that contribute to the properties of that molecule. In his excellent book, The Same and Not the Same, Roald Hoffmann cites the concept of the functional group as something in chemistry that is not reducible to the physical laws that affect it. Functional groups as concepts seem to be uniquely chemical. I, at least, can't think of anything that's more than superficially analogous. And for organic chemistry, they're hugely important.

C—C bonds and C—H bonds are extremely common in organic molecules. If you remember my post showing how skeletal structures work, you should recall that these bonds are only noted with points and intersections for the former and are left to inference with the latter. Such structures are "skeletal" because they really do show the hydrocarbon skeleton of a molecule. All those C—C and C—H bonds are usually pretty stable. They can contribute to the chemistry of a molecule, but not nearly so prominently as functional groups.

I won't attempt to list every functional group here, because there are lots and lots of them and your puny brain would probably die or something. But I will list some of them. First, I want to introduce a new notation. Actually, I'm not sure if I already introduced it, but I'm too lazy to go back and check and you probably forgot about it anyway. The letter "R" is often used to denote the rest of a molecule apart from a functional group, especially if what remains is a plain old hydrocarbon. This can be convenient for situations when it's only the functional group we care about and drawing the rest of the molecule would be impractical or wouldn't even make sense. If we do this more than once though, and the groups being condensed are not identical, using "R" to denote both would be inappropriate, but "R" for one and "R'" (R prime) for another is fine.

So for an example, I've decided to use 2-butoxyethanol because I've used it to clean graffiti off the walls in the restroom at work.
Of course, while this is the stuff I was using, other molecules that change one or more atoms are easily possible. What if we decided only to focus on part of the molecule? We might do this...
What is "R"? Is it still a chain of four carbons with nine hydrogens? Maybe. Or maybe it's still a straight chain, but one carbon longer than that now. Maybe there's a ring. Maybe it has multiple branches. Maybe there are dragons. No one knows! It's a mystery. Exciting, I'm sure. And that's how "R" works. It's a placeholder. It saves space. "Here be dragons" might work too, but "R" is the standard. I have no idea how it became that way, actually.

Aliphatic hydrocarbons
This term comes from the Greek aleiphas, meaning "fat." Actually, many of the properties fats have come from the long hydrocarbon chains they possess. If an aliphatic hydrocarbon has no π-bonds (that is, no double or triple bonds), it is an alkane. Branches and rings might be present and do affect the properties of alkanes, but they're still called alkanes, although a compound with a ring might be said to be a cycloalkane.

But if π-bonds are present, no matter how few there are or how big the rest of the molecule is, it's not an alkane anymore. A double bond means that the molecule is an alkene. So using what we learned earlier, a molecule that contains this functional group: R₂C=CR₂ (different groups are all "R" here because noting them all with superscripts would be ridiculously clunky) is an alkene. The double bond counts as a functional group. Specifically it is the alkenyl functional group.

Similarly, a triple bond is a functional group. Something with R—C≡C—R' functional group is an alkyne (no matter how many double bonds or single bonds it has). And this is an alkynyl functional group.

Aromatic hydrocarbons
The only hydrocarbons I know of that are not classed as aliphatic are ones containing aromatic rings. You might be anticipating this from the trend with what I've said regarding alkenes and alkynes, but the presence of even a single aromatic ring in an otherwise aliphatic molecule means that the compound is considered aromatic and not aliphatic. Aromaticity is a tricky concept though, and we're not covering it just yet. So for now, the only aromatic ring we'll concern ourselves with is the benzene ring.
It is not a cycloalkene, even thought it might look like one. I mentioned in an earlier post that this type of ring is resonance stabilized. The electrons are evenly distributed around the whole ring. How this happens and whether a particular ring is aromatic or not are subjects for later posts. But this ring, the benzene ring, is aromatic. If it's treated as a functional group, it's the phenyl group. To abbreviate when I'm using condensed structures rather than skeletal structures, I'll probably use "Ph" for "phenyl." So something with this functional group would be R—Ph. Other people sometimes just abbreviate a benzene ring attached to something else as C₆H₅ but I'll try not to do that so as to avoid confusion.

Another well-known aromatic hydrocarbon with its own common name is toluene. It's just like benzene, but where benzene has six hydrogens, one attached to each carbon, toluene replaced on hydrogen with a methyl group, a carbon bonded to three hydrogens. So it's like R—Ph with "R" being "CH₃" except this is a special case and gets its own name. If toluene acts as a functional group itself, with one of the hydrogens on the carbon outside the ring being replaced by a bond to the rest of molecule, and I'll draw a picture just to be sure you're following me...
The functional group is actually not phenyl. It's a benzyl group. If this confuses you, good. I like confusing you. Try to remember that this is a benzyl group and that a benzene ring attached to something is just a pheny group. I know both "benzene" and "benzyl" start with "benz-" and you're really tempted, but don't. Just don't.

Now for some more functional groups...

Alkyl halide (halo group)
R—X (where X is a halogen). It could be an alkyl fluoride, alkyl chloride, alkyl bromide, or alkyl iodide depending on which halogen is attached.

Alcohol (hydroxyl group)
R—OH

Ether (alkoxy group)
R—O—RAmine (amino group)R—NH₂ (primary) or R₂NH (secondary) or R₃N (tertiary)

Thiol (mercapto group)
R—SH

Sulfide (alkylthio group)
R—S—R'

Aldehyde (carbonyl group)
R—CHO (for those of you who can't deal with condensed structures very well, the carbon is double-bonded to the oxygen, bonded to the hydrogen, and bonded to the R group, so it could also be R—CH=O)

Ketone (another carbonyl group)
R—CO—R' (or R—C=O—R'). The difference between an aldehyde and a ketone is that in an aldehyde, the carbobyl group is on the end of a chain, but in a ketone, it's in the middle of a chain.

Carboxylic acid (carboxyl group)
R—COOH (like an aldehyde with the hydrogen on the carbonyl carbon replaced by an hydroxyl group, alternatively it's like an alcohol with an oxygen double-bonded to the terminal carbon)

Ester (ester group)
R—COOR' (like a carboxylic acid, but with the hydrogen on the oxygen replaced by a hydrocarbon group).

Amide (carboxamide group)
R—CONH₂ (primary) or R—CONHR (secondary) or R—CONR₂ (tertiary). Not to be confused with amines, which don't have that oxygen double bonded to the carbon that nitrogen is attached to.

What's that? You want more? Fine. Next time, I'll post some more functional groups for you.

The Next Chapter is About Organic Molecules and Functional Groups

2009-08-29T02:04:00.000-07:00

I am so excited for this. With the huge hiatus between the "Acid Strength" post and the "Aspirin" post, I hadn't bothered to look at what the next chapter of my text book had in store for this project. Would I skip the chapter? Go over some of it and move on as quickly as possible? Functional groups are one of my favorite things in the whole world and I can hardly wait to write about them. All chemistry is interesting, but functional groups are what truly fascinate me and I hope to study them when I go back to school. But even before then, I'll be doing a little reading up on them while writing posts here. In the meantime, have a little patience for me. I really am back and giving this project my attention, but I want this next post to be my best one so far. I love functional groups.

Aspirin

2009-08-21T16:22:00.000-07:00

It's been a while and I need to get a feel for this. I'm also not done with the chapter on acids and bases. So I thought it would be cool to show you aspirin. My book does it, although there's some other stuff before it that I sort of went over, kind of, more or less, already. This will also be an opportunity for me to try to use a program other than MS Paint to render something. We'll see how it works...

Well, that seems to work. I had to make it on this ChemSketch program and then copy the thing into MS Paint in order to upload it here, but whatever. Much easier than drawing everything from scratch.

Now let's make sense of this. There are a few functional groups here, and we'll learn all about functional groups some day. Won't that be exciting? The one we're interested in now is the carboxylic acid group. One of the carbons is attached to an oxygen by a double bond and a second oxygen by a single bond, with that second oxygen itself being attached to hydrogen. This arrangement of atoms makes it easy for a certain reaction to occur. That reaction is a Brønsted-Lowry acid-base reaction. Here's a mechanism...
Okay, those arrows look terrible. I'll have to work on that next time. Anyway, that's the same as the mechanism for the Brønsted-Lowry acid-base reaction I covered in my last post. This reaction can happen in the human body, assuming aspirin gets into the body in the first place. Aspirin is not found in nature. I've sometimes heard it said that aspirin was found in willow bark, but that's incorrect. Willow bark contains salicylic acid. Aspirin is acetylsalicylic acid. The stuff in willow bark would look just like aspirin, but the group attached to that oxygen on top would be replaced with a hydrogen. Aspirin became the analgesic of choice, as opposed to salicylic acid or another derivative of it, because it lacked the irritating side effects of those drugs.

So the reaction can and does occur in the human body. When this happens, the conjugate base, or acetylsalicylate, is formed. The conjugate base is ionic (with a negative charge on the oxygen that gave up hydrogen) and cannot cross cell membranes. Fortunately, this isn't a huge problem because, like other Brønsted-Lowry acid-base reactions, the reaction that turns aspirin into its conjugate base is reversible. My textbook notes that it's the acid that is present in the stomach, and the base that is present in the intestines. This should be pretty intuitive. When conditions are highly acidic, aspirin, which is only a weak acid, even when it does protonate something, will get protonated right away by the acid around it. In basic conditions, there's not much acid around to do the reverse reaction and what is there might be even weaker than aspirin as an acid anyway.

So yeah, aspirin.

Acid Strength

2009-05-17T13:19:00.000-07:00

Acid strength refers to the potential of the acid to lose a proton. Acids that give up a proton easily are stronger. Acids that don't readily give up a proton are weaker. Here, I'll show you a reaction mechanism for an acid dissolving in water. I know you don't know what those are, but if you paid attention the first time I showed you, then you would. It's not my fault you weren't paying attention. Grow up. So yeah, mechanism...
For a little review, I'll note that the first reactant (A—H) is the acid. Water, the second reactant, is the base. And the products are the conjugate base of the reactant acid and hydronium, the conjugate acid of water, respectively. This reaction is reversible, so the hydronium could protonate the conjugate base of the original acid and leave us with the reactants again. But the equilibrium favors whichever side of the equation has the weaker acid. This might seem intuitive, actually. The acid that more readily gives up a proton (the stronger acid) will do so more and therefore will show up less. But unless we know which acid is weaker, that is, unless we have a way to measure acid strength, the knowledge doesn't really help us. Fortunately, quantifying the acid strength is possible. But this goes a bit beyond the scope of the textbook I'm using, which assumes the student remembers certain things from general chemistry. Besides, it involves math. So I'll just tell you that in organic chemistry, the figure that is typically used is pK_a.That's the opposite of the common logarithm of the acid dissociation constant. The acid dissociation constant is determined by multiplying the equilibrium concentrations of the products and dividing this by the equilibrium concentration of the acid. Easy, right?

So I totally don't remember how those equilibrium concentrations are determined (with instruments, I guess). I just look up the pK_a of the acid in question using a table of pKa values. That way, all I need to know is that the lower the pK_a, the stronger the acid. My textbook states that typical pK_a values for organic acids range from 5 to 50. I'm not sure where those numbers came from, but whatever. Be aware that some organic acids are not typical. Also, in lab, I dealt with inorganic acids all the time. Acids with negative pK_avalues are considered "strong" acids. Actually, Wikipedia says that strong acids are those with pK_avalues less than -2. Whatever. What it means for an acid to be "strong" is that essentially all of it will lose its protons to water. In other words, the equilibrium completely favors the products and none of the original acid is present in any concentration. So really, the strongest acid that exists in water is hydronium. Anything stronger just protonates the water to form hydronium.

Acid-Base Definitions

2009-05-13T02:12:00.000-07:00

Do you know what acids and bases are? Seriously? I know I used to think I did and I totally didn't. I mean, I knew some examples of acids and bases, but the chemistry behind them was completely unknown to me even after I took chemistry in high school and even when I started taking it in college. But finally, in one class I was introduced to three acid-base definitions. I knew there were more, but I had no idea how many. This post will introduce some of them. Really, I've only ever used two of them and those two, which happen to coincide with each other a lot, are the only two that will be important here, but I think the historical definitions are interesting and this is my blog or whatever and so I get to make a post including them.

Lavoisier definition
In case you didn't know who Lavoisier was, he was one of the founders of chemistry and you are not worthy. Through a series of experiments, he determined that oxygen combining with other elements could lead to "acidic" (the word comes from the Greek word for "sharp") properties. He extrapolated from this that oxygen was the element that contributed the acidity. This is where oxygen got its name, which means "acid-generating." There was a small problem with this: Lavoisier was wrong. Considering that he basically invented chemistry, I think he's allowed to be wrong every once in a while.

Liebig definition
You know who Liebig was too, right? Because he was another great chemist. Anyway, this was sort of the first real definition. A Liebig acid is a molecule that contains at least one hydrogen atom that can be replaced by a metal. This was only a definition for acids. Back then, bases were loosely defined as the opposites of acids. Known acids were generally liquids, so a base was whatever compound would react with an acid to neutralize it into a solid salt. Such reactions are known as acid-base reactions and we'll deal with them other posts pretty soon.

Arrhenius definition
This was one of the three definitions I originally learned and it was the main definition everyone used for a long time. But that was in the past. The distant past. Like before I was born, even. Probably before you were born too. Arrhenius acids are molecules that, in water, lose hydrogen atoms, generating hydrogen ions in the solution. Well, we now know that they're actually hydronium ions. Oh, hydronium is important. You should know what it looks like. Here, I'll paint one for you.
If you weren't stupid, you'd remember what a water molecule looks like. Man, I'm not posting water again here just for you. Go back and find the post where I did show it or something. Anyway, this is like water, but with another hydrogen. Oxygen normally only forms two bonds. I guess I never talked about formal charges or whatever, but the oxygen is positively charged now. Really. We'll talk about it later if you want. Hydronium is properly written as H₃O⁺. But you should be aware that a common shorthand is just to just write H⁺.

So those are Arrhenius acids, but there are also Arrhenius bases. They are molecules that, in water, generate hydroxide ions. You want a hydroxide ion? Here you go.
Note that when it has oxygen has one too many bonds, it is positively charged, but when it has one too few bonds, it is negatively charged. Unlike H⁺, hydroxide ions actually can and do exist in water. The shorthand for hydroxide is ⁻OH. That's how I would notate it if I were writing stuff down by hand, but superscripts and subscripts, although necessary for chemical notation, can be annoying to do on Blogger, so I'll probably stick to just typing "hydroxide" most of the time.

Now might be a good time to mention that in aqueous systems (systems with water as the solvent—I'm not going to elaborate on this for now), hydronium and hydroxide act as a sort of currency of acidity and bacisity. When a reaction takes place, the actual atoms from the acid/base aren't the ones that are participating. I'll illustrate this with a classic acid-base reaction...

HCl + NaOH → NaCl + H₂O.

So, hydrochloric acid and sodium hydroxide yield sodium chloride and water. Sodium chloride is the salt in this case. In other reactions, other salts would be formed. A salt and water are the products of Arrhenius acid-base reactions. But keep in mind that water is the solvent in which this whole thing is taking place. HCl is a gas and NaOH is a solid. When we do this reaction in a lab, we're likely to just mix samples of water that have the compounds dissolved in them. The bond between the chlorine and the hydrogen breaks and the hydrogen reacts with a water molecule to form hydronium while the chlorine atom becomes a chloride ion and sits there dissolved in the water. The bond between sodium and oxygen likewise breaks and we're left with hydroxide and a sodium ion, which also sits there dissolved in the water.

The hydronium ion produced by dissolving the acid in water can and will react with another water molecule. But the product of that reaction leaves the original hydronium ion turned into an ordinary water molecule and the water molecule attacked by hydronium as the new hydronium ion. This process occurs repeatedly, but without really changing anything because the number of hydronium ions remains constant. The same is true with the base. If hydroxide reacts with a water molecule, the hydroxide gains a proton and is now a water molecule while the water molecule it reacted with lost a proton and is now a hydroxide ion. The charges are rapidly transferred from one water molecule to the next, but they remain there. This tranfer would continue for a long time, but when we mix the two solutions, in addition to reacting with water, the hydroniums and hydroxides can react with each other. Whenever this reaction takes place, the two ions neutralize each other and we're left with only water (H₃O⁺+⁻OH → 2H₂O). This reaction also produces heat, so if you do it at home for some reason, keep that in mind so that you don't die or whatever. If you were wondering, the sodium and chloride ions stay dissolved. You have seen what happens when you add sodium chloride to water, right? Seriously, you'd better have. If not, go do it right now.

So that's the Arrhenius definition. There are several other definitions, some of them relevant to certain fields, but two definitions are by far the most popular and important, so I'll introduce them now.

Brønsted-Lowry definition
This is the definition that seems to be used the most in my textbook and introductory chemistry courses. It is more inclusive than the Arrhenius definition. A Brønsted-Lowry acid is a proton donor. A Brønsted-Lowry base is a proton acceptor. One big difference between this and the older Arrhenius definition is that water is not necessarily present as a solvent. It can be and often is, but it's not necessary. The most important difference to keep in mind might be that not every Brønsted-Lowry base will lose hydroxide. Every Brønsted-Lowry acid must have a hydrogen atom that can be lost in order to give off a proton, but the bases need not look anything like Arrhenius bases. What a Brønsted-Lowry base does need is the ability to form a bond to a proton. That means it needs free valence electrons that can be recruited to form the bond. Lone pairs of electrons work best, but electron pairs in π-bonds also have basic potential.

My next few posts will be dealing with Brønsted-Lowry acids and reactions with them, so this is the definition to pay the most attention to. Hopefully, you'll get a feel for what chemicals act as good acids and bases and how acid-base reactions work. For now, definitely remember that an acid is a proton donor and a base is a proton acceptor.

You might have deduced that once an acid has donated a proton, what's left is capable, under the right circumstances, of accepting a proton and recreating the original acid. Likewise, a base that accepts a proton now has a proton that could be donated to something else, recreating the original base. These are conjugate acid/bases. That is, whenever an acid-base reaction occurs, the acid becomes its conjugate base and the base becomes its conjugate acid. The reaction could potentially reverse. Here's one reaction...

H—A (acid) + :B (base) → :A⁻ (conjugate base) + H—B⁺ (conjugate acid)
Now we'll reverse the reaction...

:A⁻ (conjugate base) + H—B⁺ (conjugate acid) → H—A (acid) + :B (base)

The system can bounce back and forth between these two. So where does that leave us? The rule I learned in general chemistry and used all the time in my organic chemistry classes was that the equilibrium favors the side with the weaker acid. Unhelpful if you don't know which acid is weaker, but we have ways of figuring that out.

Under the Brønsted-Lowry definition, acidity and basicity are relative to what something is reacting with. Water, for example, is an acid when it's reacting with hydroxide (it donates a proton) and a base when reacting with hydronium (it accepts a proton). Also, you might have deduced that the conjugate acid of water is hydronium and the conjugate base of water is hydroxide. If you deduced that, good job. You get a gold star.

Lewis definition
This definition was formulated by Gilbert N. Lewis. You do know who he was, right? A Lewis acid is an electron pair acceptor. A Lewis base is an electron pair donor. This definition is more inclusive than the Brønsted-Lowry definition. A base still needs free valence electrons to donate, but the acid in the reaction no longer needs to provide hydrogen. So while all Lewis bases are also potential Brønsted-Lowry bases, not all Lewis acids are Brønsted-Lowry acids.

I'll say more about Lewis acids later. This material is in the last section of the acids and bases chapter in my textbook and it's time to move on to some specific Brønsted-Lowry acid-base reactions.

Acids and Bases

2009-05-07T01:17:00.000-07:00

I finished Chapter 1 of my textbook in March and I should have started Chapter 2 some time ago. I don't understand why Chapter 2 is the one on acids and bases. It looks like it's sandwiched between Chapter 1 and Chapter 3 because this material needed to go somewhere early in the book and there was nowhere else for it. But maybe I'm wrong. I'll cover this material, just because I'm not sure.

My textbook is only covering two definitions of acids and bases. But there are others. Looking at Wikipedia, there are some I've never even heard of. I think understanding these would be useful, and it's something that's new to me. This means a bit of a delay in this blog getting actual content because I'll have to do some research online.

Mixed Melting Point

2009-05-06T01:54:00.001-07:00

I am back from hiatus and I resolve to post here more often. If you follow my Livejournal (you know, the only Livejournal that belongs to Stephen Bahl) you already know that I have been slacking off on internet-related things lately because I've been reading actual books printed on paper. I know, it's weird. So that's the main reason I haven't updated here in well over a month.

I notice that my last entry ends with this ...

For reasons I'll explain in my next post, if my product had not been the same compound as the salicylic acid from benzene, even if the melting points were nearly identical, the tube with the mixture would melt at a lower temperature and over a broad range, as opposed to melting all at once. Since my product was pure, all three samples melted sharply at 160°C.

What? Why did I say I'd explain that in my next post? I think I knew that I would go on hiatus with this project and put that there as a way of punishing myself or something. Well, I suppose I could explain this. Using pictures would help, but I'm going to try to do it without pictures because I like words so much and also because I'm lazy. Here we go...

The melting of a crystal involves the excitation of the particles forming the crystalline structure. In the case of organic compounds, the particles are organic molecules and the crystal is a specifically shaped arrangement of these molecules based on the intermolecular forces at work in the molecules. Now, the reason I say that I am punishing myself by explaining this now is that I haven't yet done a post on intermolecular forces yet. It's not a particularly difficult topic, but rather than going into the details, I'll just simplify and state that intermolecular forces cause molecules to interface with one another. This has a direct effect on the phases of matter (solid, liquid, gas, and the others). A crystal is solid because the intermolecular forces are strong enough to make the molecules stick together in the crystalline pattern. Adding heat excites these molecules and the intermolecular forces are no longer strong enough to hold them together, so the molecules slide against each other and bump around making a big mess or something.

The change in phase from solid crystal to liquid mess is, as you hopefully already know, referred to by the scientific term "melting." The precise temperature at which a pure crystal of a given compound melts is referred to as the compound's melting point. Different compounds, because they have different intermolecular forces based on their structures, have different melting points. So let's say we have Compound A with a melting point of 160°C (that's 320°F for those of you who still don't know the Celsius scale).

We have a sample that we think is Compound A. We can melt it and see that it melts at 160°C just like Compound A should. But now let's say that this sample is actually Compound B. Just by luck, Compound B, although made of completely different molecules than Compound A, happens to also have a melting point around 160°C. If we do a proper melting point analysis, we won't be fooled. Here's why.

We take some of Compound A and some of Compound B, mix them together, stir the powder up a bit and put some of our A/B mixture into a melting point sample tube (a thin tube made of glass, like I desribed in the last post), then use the melting point apparatus. As noted, this is a mixture. We don't have a pure, uniform crystal of one molecule, but pockets of two different molecules jammed up against each other. They interfere with each other's intermolecular forces and rather than a sharp melting point at 160°C, we'll probably see regions of softening and melting appear over a broad range of temperatures well below 160°C (because some areas will have a more even mixture meaning little crystalline structure while other regions might be isolated pockets of nearly pure A or B).

This technique is called a mixed melting point. Makes sense, right?

Preparation of Salicylic Acid from Methyl Salicylate

2009-03-15T17:32:00.000-07:00

I just realized that, although I've skipped a little, I'm done posting material from the first chapter of the textbook. At this rate, it will take me the rest of my damn life to finish the whole book. As much as I'm anxious to move on to the next chapter, right now I'm going to deal with something more important. It's something that I wanted to write about before even starting with material from my textbook, but it occurred to me that you'd need knowledge of notation to follow this. Well, if you actually read the posts about notation of structure, you should now have that knowledge (unless I'm a bad teacher or you're just stupid or something). What I'll be writing about this time is an experiment I did in my first quarter of organic chemistry. I was going to say that it was the very first experiment in the class, but I checked my lab notebook and I was wrong. This was the fourth experiment in the class.

The lab instruction manual has these silly "scenario" sections for each lab. Sometimes they were excessively silly and other times they were actually rather interesting. I'd like to quote this on in its entirety.

The new-age pharmaceutical company Natural Nostrums manufactures drugs from "natural" starting materials. For example, the company manufactures a painkilling drug it advertises as "organic aspirin" starting with methyl salicylate, which occurs naturally in wintergreen oil. Most commercially marketed aspirin is manufactured starting with benzene, a product of petroleum refining. An intermediate in both of these syntheses is salicylic acid.

Natural Nostrums claims that its aspirin, which is supposedly more natural than aspirin made from benzene, has fewer side effects than ordinary aspirin. Critics have accused the company of false and misleading advertising, asserting that salicylic acid made from methyl salicylate is no different than salicylic acid made from benzene, and that the resulting aspirin is no better than any other aspirin.

Les Payne, Director of Operations for the Association for Safe Pharmaceuticals (ASP), is investigating the company. He just shipped your supervisor a sample of salicylic acid manufactured from benzene and a bottle of methyl salicylate that one of his agents obtained from the chemical stockroom at Natural Nostrums. Your assignment is to prepare salicylic acid from methyl salicylate and find out whether or not it differs from salicylic acid made from benzene.

The main purpose of this experiment, in my class, was to introduce laboratory synthesis of one molecule from another (it was our first true synthesis lab—the previous three dealt with identification, extraction, and both purification/identification respectively). But the "scenario" for the experiment touches on what I consider one of the most fundamental concepts in chemistry: the properties of a chemical are caused by the the atoms comprising the chemical and the bonds between them. There's the law of definite proportions, but that's just part of it. Chemistry assumes, because we've tested it countless times, that provided all of the atoms are there in the same proportions and are of the same isotopes, the bonds are in the same places and arranged in the same way, two molecules are not just indiscernible from each other, but identical. Molecules don't "remember" where the atoms making them up used to be. Carbon 12 is carbon 12. Water is water. Salicylic acid is salicylic acid and if I mixed up pure salicylic acid synthesized from benzene with pure salicylic acid synthesized from methyl salicylate, no one could separate the one from the other or tell me which source a randomly chosen molecule came from.

I don't think I've ever seen anyone else articulate just that, even though every chemist knows it to be true. In all my chemistry classes, the closest thing I've seen to anyone pointing out this fundamental fact that really forms the basis for all chemistry is that quote from my lab instruction manual. Maybe people think it's so simple and obvious that it's not worth mentioning. I strongly disagree. Scams like the fictional one described by that scenario really do exist, and people only fall into these traps because they know nothing about chemistry, meanwhile the part of chemistry that makes it clear why these scams are wrong, the same part that's fundamental to all work done in chemistry, isn't considered worth noting.

Then there's homeopathy. Don't get me started on that. Enough ranting though. On with the science! You do want to know how we made salicylic acid from methyl salicylate, right? Of course you do. Well, to begin, here's methyl salicylate.

You might want to just not worry about the ring for now. It's important, but it's not changing into anything else in any reactions I'll be covering for this post. If you are curious, know that the ring has resonance stabilization: the electrons that make up the double bonds in that image are actually evenly distributed across the whole ring. It's called an aromatic ring. That's not because it has a strong aroma. It's for historical reasons. But in this particular case, the molecule actually does have a strong aroma. Methyl salicylate is the chemical that creates the "wintergreen" smell. That's probably a big reason why my professor chose it. When you're doing labs with chemicals that smell bad all the time, ever once in a while it's nice to do a lab with something that smells nice. Anyway, that's methyl salicylate. Here's salicylic acid.

So both of them have that aromatic ring and the hydroxyl group (OH) attached to one carbon with a second group attached to an adjacent carbon. But with methyl salicylate, the group is —COOCH₃ whereas with salicylic acid, the group is COOH. In other words, the difference is that the oxygen the carbon is singly bonded to is attached to a hydrogen or a methyl group, depending on which compound this is. In order to go from one to the other, we need some sort of chemical reaction to change that hydrogen into a methyl group. Chemistry! Oh yeah.

What we do is put melthyl salicylate in a vial, then add an aqueous solution of sodium hydroxide. I wrote a bit about ions a while back, but if you don't remember, ions tend to dissociate from each other in water. The cation (positively charged) is sodium. The anion (negatively charged) is hydroxide, or ⁻OH. At any point, the hydroxide could rip a proton (the nucleus of a hydrogen atom) from a water molecule (this would be an acid-base reaction with water as the acid and hydroxide as the base) and become a water molecule itself, but then the water molecule that it reacted with would be hydroxide, so the total amount of hydroxide ions stays the same.

This mixture is then heated under reflux. What that means in this case is that the mixture is boiling, but there is a glass tube called an air-cooled condenser at the top of the vial. As the vapors from the mixture rise up the tube, the air around them, being cooler than them, absorbs their heat (thermodynamics for the win), and causes them to become liquid again (condensation), at which point they run back down the glass tube into the reaction vial. And all the heat adds kinetic energy to the molecules, which speeds up the rate of reaction. I will now show you the reaction using a mechanism.

Reaction mechanisms show how bonds are formed and broken in reactions using arrows that always start at a pair of electrons and show what those electrons go to. In this case, the electrons from hydroxide form a covalent bond to carbon, breaking the π-bond in the C=O bond, so that the carbon is attached to four things: the ring, the methoxy group (–OCH₃), the hydroxyl group (–OH), and the now negatively charged oxygen that used to be double-bonded, as shown below.But the product of this reaction is an unstable reaction intermediate. It probably exists for only a fraction of a second before the electrons on that oxygen atom form a new bond to the carbon. For reasons that won't be explained right now, the bond that carbon must lose is the one to the methoxy group, so we get this reaction.
Why, look at that. It's almost salicylic acid. Almost. There's just one pesky detail remaining. Being in such basic conditions (lots of hydroxide floating around), the hydroxyl group attached to the ring lost a proton. As long as the conditions are basic, we have no way of putting that proton back on. There's a simple solution to this: douse the whole thing with sulfuric acid. Yes, I'm serious. It's what we did. Well, it wasn't really "dousing." We added more than enough sulfuric acid to neutralize the sodium hydroxide and the disodium salicylate (the name of that last product). At that point, we get salicylic acid. Conveniently for us, salicylic acid is a solid that precipitates out of the solution, so we just filter it out (we use a funnel with filter paper over a vacuum system to pull the liquid through and leave the solid crystals on the filter paper). Finally, we pour cold water onto the filter to wash away any remaining traces of the original solution. Then we just let our crystals dry and we have pure salicylic acid.

We can analyze the purity of our compound by melting point analysis. There are different ways to do this. The method I used was to take three very thin glass tubes (open on one end and rounded at the other) and pressed the open ends of the tubes against crushed powder from crystals, tapping the tubes to shake the samples to the bottoms of them. I put the tubes in a melting point apparatus, basically a heated chamber with a display indicating the temperature and a magnifying glass letting me see the tiny samples in the tubes with ease (I still had to wear my glasses, but shut up). One tube contained my product. One tube (the control) contained salicylic acid from benzene. And one tube contained a mixture of my product and salicylic acid from benzene. For reasons I'll explain in my next post, if my product had not been the same compound as the salicylic acid from benzene, even if the melting points were nearly identical, the tube with the mixture would melt at a lower temperature and over a broad range, as opposed to melting all at once. Since my product was pure, all three samples melted sharply at 160°C.

Electronegativity

2009-03-04T02:14:00.000-08:00

That last post was pretty long, huh? I'll bet it was too long for you. Well, you're in luck. This post is going to be a short one because I'm tired and electronegativity seems pretty easy to explain. It should be easy to explain, even to someone like you. That's why I'm skipping some stuff in this book that looks like it's about molecular orbital theory. Do you think you can handle molecular orbital theory? Yeah, maybe you can. I don't know. But for now, we're not going to worry about it.

Electronegativity is a property atoms have. Atoms that are highly electronegative have a high affinity for electrons. Atoms that are not very electronegative have a low affinity for electrons. Is that simple enough for you? Here. I'll personify the atoms for you, just like every high school science teacher on the planet has ever done. Atoms that are highly electronegative want electrons. They will take them. Atoms that are not very electronegative are more likely to lose their electrons to those mean, electronegative bullies. Now I feel all dirty.

Anyway, electronegativity is inversely proportional to atomic radius. So smaller atoms tend to be more electronegative than bigger ones. There are various explanations for this, but I took general chemistry a long time ago. I usually look at it in terms of "shielding" although technically, there's more to it. For an example, let's compare oxygen to sulfur. They're in the same group and have similar chemical properties, after all. Now, oxygen has eight protons, but sulfur has sixteen. And if you were good and read the second post I made here, you know that protons reside in the nucleus of the atom. So all those protons in sulfur are presumably twice as attractive to any electrons as the only half as many protons in oxygen. If that were the whole story, we'd expect sulfur to be more attractive to electrons. But sulfur also has twice as many electrons sitting in its orbitals. And they take up more space. Hence the mention of atomic radius. The sulfur atoms doesn't just have more subatomic particles. It is spatially bigger. And that large, negatively charged space has a repelling effect on other electrons. It's not strong enough to override the attractive effect the nucleus has on electrons. It doesn't shield the nucleus that much. But the effect is enough to make sulfur's electronegativity lower than oxygen's.

Of course, not all elements are in the same group as oxygen. And that matters because the number of valence electrons is also very important when it comes to electronegativity. One very simple way to look at electronegativity is that it increases as one moves to the right on the periodic table (ignoring the very last column for reasons that are obvious if you take general chemistry) and decreases as you move down the periodic table. The most electronegative element of all is fluorine. Second is oxygen, followed closely by chlorine. Fourth is nitrogen followed closely by bromine, with iodine being sixth. Whenever one of these elements forms a covalent bond with some other element I didn't just name, the electronegative one will have more of the electron density from the covalent bond.

By the way, carbon is more electronegative than hydrogen, but only by a little bit. It's not usually enough to concern us. But when oxygen or nitrogen (or any of the halogens) is in a molecule, electronegativity becomes important.

Notation of Molecular Structure

2009-03-01T02:00:00.000-08:00

I'm skipping some stuff in this textbook that's about molecule shape, bond angle and other important stuff. I'll get back to it later. Remind me to do that. Right now, I get to go over drawing organic molecules. Awesome. No really, this will be good to do because once this is out of the way, I can assume that you understand how these structures work and move on to whatever I want to. So it's very important that you understand this material. It will let you understand the notation I'll be using from now on. Fortunately, it's easier than you might think: I'm only introducing two structures here.

Condensed Structures:

Condensed structures are what they sound like. They take molecules like the ones I've been showing in previous posts and convey all of the structural information in a short line of text. They're not really practical in certain cases, but for most small molecules, they're easy to write and understand. I'll be using them when I can because I won't need MS Paint or anything like that. Let's start with some rules...

All of the atoms are drawn in, but single bonds usually aren't.
Atoms are drawn in next to atoms they are bonded to.
Parentheses enclose groups of atoms that are all bonded to the same atom.
Pairs of electrons are omitted.
Read the structure from left to right and remember that every carbon must be tetravalent.

Don't get it? Don't worry. Here are some examples.

First example: n-butane
Molecular formula: C₄H₁₀Lewis structure:

Condensed structure: CH₃CH₂CH₂CH₃Notice the molecular formula, while telling us what atoms are in the molecule, doesn't tell us where they are. The Lewis structure does that. So does the condensed structure. We don't actually get the bonds drawn for us, though. Starting from the left, we have a carbon and three hydrogens. The assumption is that the hydrogens are bonded to the nearest atom, and since each hydrogen can only bond to one thing, that means that the first carbon is bonded to all three of them, leaving one bond open. Next there's another carbon, which must then be attached to that first one, meaning that the first carbon is tetravalent and the second one must still have three more bonds. Two hydrogens are written in after it, so they're both attached to that second carbon, leaving it with one more bond, and so on. With practice, it becomes quite easy to convert between regular Lewis structures and condensed structures, but you probably won't practice, so I'm not sure how easily this will come.

One note on this condensed structure is that it could be further condensed. When the same sequence is repeated, it's acceptable to enclose it in parentheses and use a subscript to note the number of repetitions. So an alternate condensed structure for this molecule would be CH₃(CH₂)₂CH_3.I don't want to be confusing here, but it's the first example the textbook used. Oh well, it does provide the lesson that there isn't necessarily just one condensed structure for a molecule.

Second example: isobutane
Molecular formula: C₄H₁₀Lewis structure:
The hydrogen atoms are crowding into each other here, which is why I had to cut off two of the bond lengths. In reality all of the hydrogen-carbon bonds would be identical except for the one on the hydrogen attached to the middle carbon. But that's a topic for a more advanced post. I should have just made the carbon-carbon bonds dashes and the hydrogen-carbon bonds hyphens in order to avoid the crowding, but I already made this stupid picture and I'm not recreating it now, so you're stuck with it.

Condensed structure: CH(CH₃)₃

The molecular formula is the same as the previous example. But the condensed structure is completely different. It shows that the first carbon has a hydrogen attached to it, which takes up only one bond, so it has three left. Then there are three CH₃groups (that's what the parentheses are for) all attached to the same atom, which must be that first carbon. So in our two examples so far, we've used parentheses in two different ways. In the first example, they enclosed a chain of the same repeated group and in the second example, they enclosed identical groups all attached to the same single atom.

Third example: 2-butene Molecular formula: C₄H₈
Lewis structure:

Condensed structure: CH₃CH=CHCH₃
We still depict multiple bonds, which is why I was overjoyed when I realized that I could use the "≡" symbol, already having "=" to represent double bonds. Everything else in this example should be familiar from the two previous examples. The hydrogens are listed directly after the carbons they're bonded to and each carbon is listed after the carbon it's bonded to.

Fourth example: methyl acetate
Molecular formula: C₃H₆O₂Lewis structure:
Don't worry about the name here. Nomenclature of molecules will come later. Now, if you stayed awake for the post on Lewis structures, you should understand the structure of this molecule.

Condensed structure: CH₃CO₂CH₃

Are you getting the hang of this yet, or will you be a failure forever? We start on the left with a carbon that has three hydrogens attached, then attach the next carbon to it, which apparently has two oxygens attached to it. But how do we know if they're attached by single or double bonds? Well, like I already said, each carbon must be tetravalent. The second carbon is already attached to the first carbon, so that's one. It can't have two double-bonded oxygens attached to it, because that would be five bonds to a carbon atom. It could have a single bond to each oxygen and a bond to the next carbon, but that would leave both oxygen atoms with only one bond and oxygen forms two bonds (uh, unless it's a free radical or an ion or something, but this isn't, so shut up). The only possiblity left is that the second carbon is double-bonded to one oxygen and single bonded to the other, leaving that oxygen with one more bond left, which goes to the next carbon. And finally that last carbon is attached to three hydrogens. Easy.

That's how the textbook does it, but it's not how my professor did it. He would draw the condensed structure like this: CH₃COOCH_3.This is fine, but I had to be careful, because I kept thinking that this version of a condensed structure was depicting a peroxide, in which the oxgygen atoms actually are attached to each other. So don't make that same mistake. And remember to make sure that the atoms have the right amounts of bonds. If that structure really were depicting a peroxide, it would mean that the second carbon is only forming three bonds.

Skeletal Structures:

I like skeletal structures. They're a lot easier to draw than condensed structures, but they're pretty much impossible to type, as far as I know, so I'll have to use MS Paint or something. Many compounds have rings. Using skeletal structures makes depicting rings easy. They're also good for large, branching compounds and the like. They can also be used for smaller compounds down to ones with only two carbon atoms. Again, we'll go over some rules...

Any time there is a corner where two line segments meet or a point where a line segment just ends, that represents a carbon atom.
Hydrogens attached to carbons are not depicted. You're supposed to be able to figure out how many hydrogens are attached to each carbon all by yourself. You can do that right? I mean, what are you, a child?
Atoms that are not carbon or hydrogen (also known as heteroatoms) are drawn in and any hydrogens attached to them are drawn in too.
Triple bonds mess with the first rule a little bit because they force the atoms involved to form a straight line, so there won't be any corners, but know that if no heteroatoms are shown there, the things forming the triple bond must be carbons.
Skeletal structures are awesome.

Since you've already mastered condensed structures, I'm not going to waste my precious time drawing the Lewis structure for this first example. Figure it out yourself.

First example: n-hexane
Molecular formula: C₆H₁₄
Condensed structure: CH₃CH₂CH₂CH₂CH₂CH₃Skeletal structure:
Yeah, I was too lazy to try to make it on MS Paint, so I spent more time than it would take me to just draw that damn thing looking around the web for an image I could use and trying to get it the right size. I still managed to make it pretty small and puny, but you get the idea. In the future, I'll either get this figured out or just go all sloppy and draw my skeletal structures in MS Paint.

Both ends are carbons and each corner is a carbon atom. If you can count, you'll realize that's six carbon atoms all in a straight chain. And if you're not stupid, by now you realize that the carbons in the middle of the chain still need two bonds each, so they'll all be bonded to a pair of hydrogens, while the carbons on the ends of the chain will need three hydrogens. And look, that perfectly matches the condensed structure. And it has the same numbers of both types of atoms as the molecular formula. It's almost as though this is science or something.

Second example: cyclohexane
Molecular formula: C₆H₁₂
Condensed structure: Fool, you cannot draw a condensed structure for cyclohexane (it has a ring).
Fine then, Lewis structure:
Yeah, I know. It looks like crap. My awful drawing skills are preserved for all the world to see. Moving on...

That sure is ugly, so let's see the skeletal structure:

Yeah, that's right. It's a hexagon. I'm pretty sure even you are smart enough to know that a hexagon has six corners. Each one represents a carbon. And since each carbon is bonded to two other carbons, that means there must be two hydrogens on each carbon.

That's enough about skeletal structures for now. They can get a lot more complicated, but I'll try to ease you into it, rather than saying, "Look, it's capsaicin."

Constitutional Isomers

2009-02-23T23:27:00.000-08:00

Earlier today I said that I was going to do a post on isomers tomorrow. Well, I'm so anxious to talk about isomers that I am starting it early. Only half an hour left until tomorrow anyway as I'm typing this sentence, so maybe I won't finish it until it's tomorrow. We'll see.

Constitutional isomers are compounds that have the same atoms, but arranged in a different structure. This is completely different from resonance structures, because both isomers are real molecules and can have very different properties. So don't get the two mixed up. Remember, resonance structures have double-headed arrows between them. Isomers don't. Because I'm lazy, I'll just use the first example I find in my book...
Both molecules have the same atoms. They each have three carbons, one oxygen, and six hydrogens. But in molecule A, the double bond is between two of the carbons and the oxygen is bonded to a hydrogen. In molecule B, the double bond is between the oxygen and the carbon it's attached to, while the other carbons each have three hydrogens.

Remember how it's bonding that really makes molecules what they are? These compounds have different bonding structures, so I would expect them to have very different chemical and physical properties.

Addendum: People should know what common or important molecules look like. I just realized, as I was about to close my textbook, that "molecule B" is, in fact, acetone, a chemical you might be familiar with. Acetone is an important solvent. I used it in the laboratory all the time in my chemistry classes. It's also in paints and stuff. And it's used to make acrylic glass. As for the other molecule (A), its name is 2-propenol and I'm pretty sure it's just an enol form of acetone, so basically it is highly unstable and will typically turn into acetone by itself. But that's a topic for another day.

Stephen Bahl Has Triple Bonds

2009-02-23T22:15:00.000-08:00

So I totally realize that I can use a symbol for a triple bond. Yep. It makes things easier. Like here's ethyne, also known as acetylene...

H—C≡C—H

I chose that molecule because I didn't even need MS Paint to do it. Awesome. I'm so excited to have that symbol at my disposal...

≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡

Awesome. The symbol is actually available because of its meaning in logic and mathematics, not because of anything to do with chemistry. But I'll take what I can get.

Resonance Stabilization

2009-02-23T02:28:00.000-08:00

Once again, it's been a while. I had so much fun with the Lewis structures post that I wanted to make a new post the very next day, but other things kept getting in the way. Well, I'm not putting it off anymore. My introduces resonance after introducing Lewis structures, so I suppose I'll follow its lead...

Many molecules, both organic and inorganic, are resonance stabilized. Such molecules cannot be represented by a single Lewis structure. For that, we can use resonance structures—if we want to. I don't really like resonance structures myself. But you know, whatever works. Before I go any further with resonance structures, I must explain why they're used at all. They exist because of the delocalization of electrons. Delocalized electrons aren't sitting on any one atom and they're not locked into any one bond. That's why we use the word "delocalized." It's like they're spread out over multiple atoms as one big forcefield of negative awesomeness. I don't know. Let's just move on to the example.

The example my book uses is this anion. Don't worry about its name because I don't know its name either. It's the conjugate base of formamide and it's an anion, but beyond that, I have no idea. Here's are the resonance structures...
Yes, the circled "-" sign is a negative charge. That was obvious, right? That double-headed arrow indicates that these are resonance structures. The important thing to realize is that these structures aren't two molecules. They're representing the same molecule in two different ways. The electrical charge is in two different places, but all of the atoms are in the same places. The first structure makes the oxygen the center of negative charge. The second structure makes the nitrogen the center of negative charge. In actuality, the negative charge is distributed between those two (and in this case, it will be slightly more centered on the oxygen because oxygen is more electronegative than nitrogen).

Resonance structures are not real. My professor compared it to describing a rhinoceros, to someone who had never seen one, as a cross between a unicorn and a dragon. Yeah, that doesn't make any sense. He realized that after he said it. But he was trying to think of a more everyday example of using two fictitious things to describe one real thing. Resonance structures aren't always in pairs, though. Many molecules have three resonance structures.

So yeah, not real. Resonance structures. The bond between the carbon and the oxygen isn't actually a double bond. And the bond between the carbon and the nitrogen isn't a double bond either. It's more like the bond between carbon and oxygen is a little bit more than a 1.5 bond and the bond between the carbon and the nitrogen is a little bit less than a 1.5 bond. I don't know the actual numbers. It's possible to take a measurement, but I don't have the equipment or the expertise, so shut up. We'll just pretend that it's exactly 1.5 on both, even though I know that it isn't. Actually, why didn't this stupid book just use an example where that was the case? I mean, there are cases like that. This molecule isn't one of them. Anyway, it's more like those bonds are each 1.5 bonds. But Lewis structures don't have a way of representing fractional bonds, so we need a way to represent what's going on here. Some day, I'll fix the notation in chemistry. Until then, we're stuck with resonance structures.

This post was probably kind of confusing. I know I get resonance structures. But maybe you don't. Remind me to help make this post more clear. Right now I need sleep. I'm totally going to write about isomers tomorrow.

Lewis Structures

2009-02-08T22:39:00.000-08:00

I am not pleased with the pace at which I've been updating this blog so far. I am still getting the hang of it and I think with more practice I'll do better, but right now it's frustrating. Checking out my textbook again, I'm not sure what to do about the whole "homework" thing. The first chapter has 84 problems. I could do them all. I do have the solutions manual if I get stuck and I generally remember this stuff. That's not the issue for me. I don't want to do these problems. They look too easy. At least from what I've skimmed through, this is really basic stuff. I don't want to plod through it before moving on to the problems I really need to work on. I apparently thought that I would work on problems and write here about the concepts I'm studying. But I'm just not motivated to do 84 problems that will mostly be really easy anyway. Maybe I'll make some sort of split where I work on the problems later in the book and write about the simpler stuff. I don't know if that could work.

Enough about me. The first chapter of this book sure seems to think that Lewis structures are important. So we'll talk about Lewis structures. This would all be so much easier if I could write stuff by hand. The QWERTY keyboard does not, to my knowledge, really do Lewis structures. I know I could easily go over Lewis structures in some sort of classroom setting, but here on the web, I don't have even have a whiteboard. Or do I?
Well, it's a toy I'll need to practice with (practice, practice, practice--it's all about practice).

So with Lewis structures, my book cites three general rules

1. Draw only the valence electrons.
2. Give every second-row element an octet of electrons, if possible.
3. Give each hydrogen two electrons.

Also, a line represents a covalent bond. As aslways, each covalent bond is made up of two electrons. Here's a Lewis structure for methane.

I could make all the bonds the same length, as they are in the real molecule, but I'm really bad at drawing and crap, so you'd better get used to it now. Also note that this is not what the molecule actually looks like. This is just the Lewis structure. The hydrogens repel one another and so the configuration they'll be in is the one where each hydrogen is as far away as possible from each other hydrogen. Last time I checked, the world is not two-dimensional, so the actual molecule ends up with a tetrahedral shape. Each hydrogen is in one corner of the tetrahedron and carbon is in the center. I'll save you some time and tell you that the angle between any two of those bonds is 109.5°. But on paper, it's a whole lot easier to just ignore the whole third dimension thing and draw the Lewis structure. Next, let's try the Lewis structure for water.

The two non-bonding pairs of electrons are represented as dots. They're in pairs because that's how electrons roll. They use the buddy system. Actually, it's because each atomic orbital holds two electrons, but that's not what we're worried about right now. The molecule has a bent shape because non-bonding pairs take up more room than bonding pairs due to greater repulsion force and compress bonding electrons closer together. You did know that, right? Please say you knew that. Anyway, the point of Lewis structures isn't really to represent bond angles. But it is possible to have the atoms line up in a straight line. Here's carbon dioxide.
See? The carbon has no non-bonding electrons, so they can't compress anything and the whole molecule has a linear shape. Both oxygen atoms have non-bonding pairs, but think about it. What would they compress? Yeah, the linear shape keeps those electrons as far from each other as they can possibly get. And that's Lewis structures. How about one more? I'll make it one of your favorite organic compounds.

Vitalism and the Origins of Organic Chemistry

2009-01-31T23:01:00.000-08:00

Time flies. I can't believe how long it's been since my last post here. I am slacking a little, but part of the reason that posts here have been infrequent is that I've been spending more time with friends, which is something I had a goal of doing. So I'm not too upset. And I am here, right now, updating this blog. So here we go...

In order to appreciate the distinction between organic chemistry and the rest of chemistry, it take some appreciation for the history of the science. Organic chemistry is frequently defined as being the chemistry of carbon or the chemistry of compounds containing carbon.

My textbook puts it rather simply:

Organic chemistry is the chemistry of compounds that contain the element carbon.

Wikipedia is a bit more specific:

Organic chemistry is a discipline within chemistry which involves the scientific study of the structure, properties, composition, reactions, and preparation (by synthesis or by other means) of chemical compounds that contain carbon.

What neither of these simple definitions tell us is that not all compounds containing carbon are considered organic. My textbook doesn't seem to mention it, but looking up "organic compounds" on Wikipedia reveals this:

For historical reasons discussed below, a few types of compounds such as carbonates, simple oxides of carbon and cyanides, as well as the allotropes of carbon, are considered inorganic.

This is because of the now defunct concept of vitalism. People believed that organic matter and inorganic matter were fundamentally different (as an aside, some attribute this to Aristotle, but I haven't looked into it). Carbonates, oxides of carbon, and cyanides (and carbides, another class of inorganic carbon-containing compounds) are all found naturally outside living systems (in minerals, for example). Other carbon-containing compounds were only known to be associated with life. Then this whole view of things got wrecked in 1828 by Friedrich Wöhler doing this reaction:
Pretty cool, huh? Wöhler combined ~~ammonia with a solution of cyanic acid~~ ammonium chloride (dissolved in water) with silver cyanate and got, as a product, urea, which was previously only known to be produced by the kidneys of animals (mammals and some other animals produce urea as a waste product). Although this didn't immediately strike the deathblow for vitalism (which has, in some form or another, survived to this day, although thankfully not among chemists), it laid the foundation for organic chemistry as a field (these compounds were now something that could potentially be synthesized in laboratories and there wasn't necessarily any essential "life force" that was generating them).

The distinction between organic chemistry and inorganic chemistry is important today because organic compounds have some specific properties of their own that can be studied in detail and because organic chemistry is so important in biological systems. Being biological systems ourselves, we have an interest in organic chemistry as it relates to our health. Pharmacy is one branch of applied organic chemistry.

Addendum:

I forgot to mention this, but before taking organic chemistry, I was under the impression, from general chemistry, that organic compounds had both carbon and hydrogen. My organic chemistry professor pointed out early on that technically, not all organic compounds have hydrogen. For example, carbon tetrachloride (CCl₄) is considered an organic compound, but has no hydrogen. However, most organic compounds do contain hydrogen.

Here's some of what you'll need to know in order to have an idea of what I'm talking about...

2009-01-19T20:11:00.001-08:00

I can't relate all of the fundamentals of chemistry in a single post. A lot of what I'd like to get across would best be done with drawings, images, graphs, and such. I'm too used to using words, but I'll try to provide some images, starting with this very post. Well, let's introduce the periodic table.

I hope you've seen it before. And if you haven't, what is wrong with you? There a lot to say about this table. It's one of the most important developments in science. It's fascinating and I recommend learning about it, but if you find it intimidating now, you can take solace in the fact that I'll only be using a small portion of it for this blog. Here, I'll show you.

See? It's the same table, but I erased everything we don't really care about. Some of the elements I erased from the table will actually be important in the future, but not for a while. We'll cross that bridge when we come to it. I also shaded in the elements we care about most. And out of the ones I didn't shade in, each one is pretty similar to the ones it shares a column with. That means we're really looking at about eleven different things, rather than over a hundred.

Of course, there's the issue of what these elements actually mean and why they're placed on the table the way they are. Each element is a substance that has its own unique atom. Atoms have three building blocks: protons, neutrons, and electrons. Neutrons are the heaviest. They have some important properties, but for our purposes they don't do much besides sit there in the middle of the atom, a region called the nucleus. Right beside them are the protons. Every atom has a specific number of protons, which is its atomic number. The number of protons is what makes an atom have the chemical properties that it does. And the table organizes them according to this number. You can see that hydrogen (atomic symbol H) is the very first one. Hydrogen atoms each have one proton. Helium (atomic symbol He), which I erased in the second table, is second and has two protons. Lithium (atomic symbol Li) is third and has three protons. This continues for the whole table. Every proton has a positive charge, often represented as +1.

Atoms also have electrons. But electrons don't sit in the nucleus. They buzz around in an area surrounding the nucleus. Electrons are much smaller than protons, but each one has a negative charge that cancels out the positive charge from a proton, often represented as -1. This means, of course, that an atom with the same number of electrons as it has protons will have no charge (it's neutral). Electrons interact in three-dimensional geometric structures called "orbitals" that I happen to find very annoying to draw. I'll cover those at some point in the future. What's important is that these orbitals take up a lot of space. Even though the nucleus is much heavier than all of the electrons combined and would take up more space than all of them, the negatively charged area of the orbitals engulfs the nucleus. I've heard the size difference likened to a golf ball sitting in the middle of a stadium, with the golf ball being the nucleus (with almost all of the mass) and the walls of the stadium being the outermost electrons.

The electrons on the highest or outside orbital are valence electrons. They're the ones that form bonds to other atoms. Chemistry deals with the countless ways in which bonds interact, where they can form, what it takes to break them, etc. Electrons in this "valence shell" typically exist in pairs (there's a reason for this, but it's another thing that I'm skipping for now). With this in mind, I'll go over the interactions we expect to see in bonds with our important atoms.

The alkali metals:

The alkali metals we're worried about are lithium (Li), sodium (Na), and potassium (K). And really, they're all pretty similar to each other, which simplifies things for us. Alkali metals are very reactive. They'll typically form ions. Ions are like atoms, but with either fewer or more electrons than the number of protons in the nucleus. This happens because the atoms either lose or gain valence electrons. An atom that has lost valence electrons is called a cation and has a positive charge (because it has more protons than electrons). An atom that has gained valence electrons is called an anion and has a negative charge (it has more electrons than protons). Alkali metals all have one valence electron. When they react, they become +1 charged cations. The most famous example of an ionic compound is sodium chloride, common table salt. The sodium cations and the chloride anions (anions of chlorine) are attracted to each other because they're oppositely charged (you did know that opposites attract, I hope). We won't be dealing with metals much, because this is organic chemistry. But it is important to understand ions.

The alkaline-earth metals:

The alkaline-earth metals we're worried about are beryllium (Be), magnesium (Mg), and calcium (Ca). When I was taking organic chemistry, I dealt with them even less than the alkali metals (and I didn't deal with berylium at all). But like the alkali metals and some other metals, these can come up when we get into organometallic chemistry (which won't be for a while). The only important thing to keep in mind about these metals is that they have two valence electrons and will react to form cations by losing both. I used sodium chloride as an example above. The equivalent to that here would be magnesium chloride, which is less well-known, but it is used to make tofu and is sometimes used as a dessicant (it absorbs water from the air). Because it needs to lose both of its valence electrons to become a stable cation, magnesium chloride crystals actually have pairs of chloride ions attached to the magnesium ions so the ratio of anions to cations is 2:1, whereas in the case of sodium chloride, the ratio was 1:1. Enough about metals, though. We'll be dealing almost entirely with non-metals.

Boron:

Boron is of some importance. When I reach material that deals with boron, I'll give more details on boron. For now, just note that boron has three valence electrons (it has two other electrons in a lower orbital, and only valence electrons are used in bonding). So boron will form three bonds. When these bonds are to other non-metals, they're generally covalent bonds. Covalent bonds are links between atoms generated by sharing of valence electrons. Boron forms three.

Hydrogen:

Hydrogen is a very simple case. Hydrogen atoms can bond to only one atom each. That is, each hydrogen atom can form only one bond. If two hydrogen atoms are bound to each other, it's a molecule of hydrogen gas. But most of the bonds we'll be seeing between hydrogen and other elements will be with hydrogen and carbon, nitrogen, or oxygen. Again, the rule for hydrogen is simple: hydrogen can only form one bond. Hydrogen can also form ions, but we'll deal with those later.

Carbon:

Here's where things get more complicated. Carbon has four valence electrons and forms four bonds. That means any given carbon atom could easily be bonded to four other atoms. But that's using only sigma bonding. There's also pi bonding. Rather than worry about that for now, just remember that double and triple bonds are also possible. So a carbon atom can be bonded to two, three, or four other atoms. Two bonds would mean that is forming either a triple bond and a single bond or two double bonds. Three bonds would mean that the carbon is forming a double bond and two single bonds. Four bonds would mean that it has single bonds to four different atoms.

A double or triple bond to hydrogen is impossible. What I said earlier about hydrogen only forming one bond also excludes double and triple bonds (those require electrons too). But carbon is capable of forming bonds to other carbon atoms, single, double or triple. Long chains or rings are possible. It's this versatility that makes organic chemistry unique.

Nitrogen:

Nitrogen has five valence electrons. One pair will always stay with the atom, so it can form up to three bonds. A nitrogen atom with a triple bond to another nitrogen atom forms a molecule of nitrogen gas, and you should know that this gas makes up most of the air you breathe. Triple bonds between nitrogen and carbon are also possible, as are double and single bonds.

Oxygen:

Oxygen has six valence electrons. Two pairs will always stay with the atom, so it can form two bonds, either two single bonds or one double bond. No triple bonds are possible with oxygen. Oxygen double-bonded to oxygen forms a molecule of oxygen gas, which makes up most of the rest of the air you breathe (the part that isn't nitrogen).

The halogens:

The halogens (the only ones that matter, anyway) are fluorine (F), chlorine (Cl), bromine (Br), and iodine (I). The differences will eventually come up, but as far as the fundamentals go, they're all interchangeable and when using atomic symbols to represent atoms, "X" in organic chemistry means one of these elements. They have seven valence electrons. Three pairs will always stay with the atom. One electron can participate in covalent bonding.

Silicon:

We won't be using silicon much. For now, all you need to remember is that, like carbon, silicon is tetravalent. Again, this means that it forms up to four bonds.

Phosphorus, and sulfur:

These two get a bit tricky. Phosphorus behaves a lot like nitrogen and sulfur behaves a lot like oxygen, but they can do other things too. These elements can form more than four bonds. Even though phosphorus will usually form three like nitrogen and sulfur will usually form two like oxygen, both can, in certain situations, form several bonds. So while you should expect not to see a carbon with five bonds to it, this can totally happen with these two elements.

That's all for now. I should probably cover some more, but I want to post this now.

Welcome

2009-01-12T23:44:00.000-08:00

This is, obviously, my first post on this blog. And what a marvelous start we have: two instances of self-reference in the first sentence. One week ago, I expressed my intention of starting a blog like this one.

I still don't know exactly what this blog is going to look like, but my intention is for it to be didactic. This will require me to know things, but knowing things is exactly what I've always wanted to do. We'll see if it works. I am very serious about this.

For now, I'll be writing about organic chemistry. I want to focus on making this accessible to people. If you know basic chemistry, you should be able to follow this blog easily. If you don't know any chemistry, at all, well, you should really learn some, because it's good stuff. If something is unclear, please ask me about it and I'll try to expound on the topic. Like I said, I want this to be accessible to people. I'll also try to write some "review" posts that cover basics. I don't really have any such topics planned yet.

For those of you who don't know much chemistry or have forgotten most of what you did know, later this week, I'll make a post that will cover some fundamentals. That is all for now. I hope this blog provides some value, either for you or for myself, or maybe both of us.

Edit: I was out of the house more than I anticipated this week. I was actually planning on making two more posts here, but that number turned out to be zero. The one about fundamentals will be arriving sometime next week, though. I could have tried to do it tonight, but I'd rather make it late than inadequate.