Aromaticity

Transcript

0:00Introduction

Hello everyone! First, a small puzzle. Take two ring hydrocarbons that look almost alike: cyclopentadiene, a five-membered ring, and cycloheptatriene, a seven-membered ring. Each has one C H two group in its ring. Yet the p K a of cyclopentadiene is about sixteen, and that of cycloheptatriene is about thirty six, and a lower p K a means a stronger acid. That is twenty units apart, and every unit is a factor of ten! So why does the smaller ring give up its proton so much more easily? The answer is one idea, aromaticity. It comes up often in JEE, so let's begin.

0:41Hückel's rule: when is a ring aromatic?

So, when is a ring aromatic? There are four conditions, called Huckel's rule, the name in the heading. Number one: the species is cyclic, a ring, like every ring in the figure. Number two: planar, which means flat. Remember this word; it saves you from a trap soon. Okay? Number three: fully conjugated. Every atom of the ring must have a p orbital, so the p orbitals overlap all the way round, with no break. Right? The last condition counts electrons. Pi, the Greek letter on screen, stands for pi electrons, the electrons in the overlapping p orbitals. Benzene has three double bonds in its ring, and each holds two pi electrons. So how many pi electrons does benzene have? Count them yourself first, and then check the figure. It is six, yes, exactly! And number four says the ring must hold four n plus two pi electrons, the expression in brackets on screen. This is the famous four n plus two of Huckel's rule. Clear? Here n is zero, one, two, and the three dots mean and so on. So the magic numbers are two, six, ten, fourteen and eighteen. Benzene's six is four n plus two with n equal to one, so benzene is aromatic, as the figure shows. Got it?

2:01Anti-aromatic and non-aromatic

Now, what if a ring passes the first three tests but has four n pi electrons? Look at cyclobutadiene in the figure, the little square. Two double bonds give four pi electrons, which is four n with n equal to one. With me? Four, eight or twelve: such a ring is anti-aromatic. And that does not mean ordinary; it means specially unstable. So cyclobutadiene is anti-aromatic. Okay? Now cyclooctatetraene, the eight-membered ring in the figure. Four double bonds give eight pi electrons, four n with n equal to two. So it must be anti-aromatic, right? Before you say yes, find it in the figure. Which group is it in? No, no, careful, that is the trap! It sits with the non-aromatic rings. The figure says cyclooctatetraene is tub-shaped, so it is not planar. And a ring that is not planar, or has an s p three atom, with the small three written high, is non-aromatic. By folding into a tub, it avoids being anti-aromatic. Clear? So what does a non-aromatic ring do? It behaves like an ordinary alkene. Both rings from our puzzle are in this group, as the figure shows, because each has an s p three carbon in its ring. Got it? Now the stability order, with the greater than sign on screen: aromatic, greater than non-aromatic, greater than anti-aromatic. Anti-aromatic rings sit at the very bottom. Right?

3:32Shortcut: counting the pi electrons

Now the topper's shortcut for counting pi electrons. Each double bond in the ring gives two, so benzene gets three times two, which is six. Okay? Next, charged carbons. Each carbanion carbon, with a negative charge, has a lone pair, so it gives two. Each carbocation carbon, with a positive charge, has an empty p orbital, so it gives zero. With me? Now look at pyrrole in the figure: two double bonds, and a nitrogen carrying a hydrogen. Two double bonds give four, but the figure says six pi electrons. So where do the other two come from? From the lone pair on nitrogen, very good! So a heteroatom, meaning an atom other than carbon, gives two from its lone pair, but only when it has no double bond in the ring. Clear? So the N of pyrrole, the O of furan and the S of thiophene each give two. Furan has two double bonds giving four, plus two from oxygen, which makes six, so furan is aromatic, as the figure shows. Okay? Now pyridine, the six-membered ring with one nitrogen. Three double bonds give six, and the nitrogen has a lone pair, so add two more. Eight pi electrons, so pyridine is anti-aromatic, right? Now find pyridine in the figure. What does it say? Oh no, careful, that is the second trap! The figure says six, in the aromatic row. Why? In pyridine, the nitrogen is already part of a ring double bond, so its lone pair is not counted. Got it? That lone pair lies in the plane of the ring, outside the pi system. So pyridine has just six pi electrons, from its three double bonds, and it is aromatic, like benzene. Right?

5:18Example 1: classify cyclopropenyl, cyclopentadienyl ions

Now example one. Classify four ions: the cyclopropenyl cation and anion, a three-membered ring with one double bond, and the cyclopentadienyl cation and anion, a five-membered ring with two double bonds. Use the shortcut. Pause the video and try it first. Done? Okay, let's check. In the cyclopropenyl cation, the double bond gives two and the carbocation carbon gives zero. Two plus zero is two pi electrons, four n plus two with n equal to zero. So it is aromatic. Okay? Now the cyclopropenyl anion. Same ring, but the charged carbon now gives two. More electrons, so it is still aromatic, isn't it? Not at all, because two plus two is four, and four is four n, so the anion is anti-aromatic. Same ring; only the count changed the answer. With me? Now the five-membered ring. Two double bonds give four. In the cation, the charged carbon gives zero, so four plus zero is four pi electrons, and the cation is anti-aromatic. Clear? In the anion, four plus two makes six, with n equal to one, so it is aromatic, like benzene. So the answers are aromatic, anti-aromatic, anti-aromatic and aromatic. Keep that aromatic anion in mind; we need it next. Got it?

6:41Example 2: why cyclopentadiene is far more acidic

Now example two, our puzzle from the start. Cyclopentadiene, p K a about sixteen, is a far stronger acid than cycloheptatriene, p K a about thirty six. The small letter A, written low, stands for acid, and the wavy equals sign means approximately. Your job is to explain why the gap is so big. Here is a hint: look at the conjugate base. Pause the video and try it. Done? Okay, let's check. In both, the proton leaves from the C H two of the ring, with the small two written low. It leaves as H plus, the plus sign written high, and that carbon becomes a carbanion carbon. Right? Now use the shortcut, and count the pi electrons in each of the two anions. Take the five-membered ring first. Two double bonds give four, and the carbanion carbon gives two, so six. That is our aromatic anion from example one! Okay? Now the seven-membered ring. Three double bonds give six, and the carbanion carbon adds two, so eight. Eight is four n, so this anion is anti-aromatic, specially unstable. Got it? Now recall the rule from acids and bases: the more stable the conjugate base, the stronger the acid. Clear? So the conjugate base of cyclopentadiene is aromatic, while that of cycloheptatriene is anti-aromatic. And our puzzle is solved: the five-membered ring gives up its proton easily, because losing it makes an aromatic ring. See how much one count explains?

8:15Quick revision: 4n+2, anti-aromatic, non-aromatic, pi count

So, quick revision. Aromatic means cyclic, planar, fully conjugated, with four n plus two pi electrons. The same, but with four n, is anti-aromatic; not planar, or with an s p three atom, is non-aromatic. And to count, a double bond gives two, a carbanion carbon two, a carbocation carbon zero, and a lone pair two, only when that atom has no ring double bond. Now go and try the practice problems on this page. You will find them easy. All the best!