Resonance

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. We draw benzene as a ring of six carbons, with three double and three single bonds. Now, a double bond is shorter than a single bond, so you would expect two different bond lengths, right? But measure them, and all six carbon carbon bonds come out exactly the same, one hundred and thirty nine picometres. So is our drawing wrong? Today one idea, resonance, clears this up. It comes up often in JEE, so let's begin.

0:32One Lewis structure is not enough

So, sometimes one Lewis structure cannot describe a molecule, and we write two or more structures, okay? But the atoms stay exactly where they are. Only the electrons move, the pi electrons and the lone pairs; pi is the Greek letter on screen. Look at the top left of the figure, the two Kekulé structures of benzene: the same six carbons, only the double bonds have shifted. With me? Now let me ask you this. Between the two structures there is an arrow with two heads. So does benzene keep flipping from one drawing to the other? No, it does not, and many students fall for this. There is only one real molecule, the resonance hybrid. The drawings are its contributing, or canonical, structures. In the figure, the hexagon with a circle inside, after the sign with three lines, is the hybrid. Clear? So write this line down. The double-headed arrow means resonance; it does not mean an equilibrium. Nothing is going back and forth, okay? And here is the beautiful part. The hybrid is more stable than any one of its contributing structures, even the best one. Right? That extra stability has a name. The energy difference between the hybrid and the most stable contributing structure is the resonance energy. We will calculate it in example two.

1:50When can resonance happen?

Got it? Now, resonance does not happen just anywhere. It needs conjugation, and all the atoms involved must lie in one plane. Okay? Conjugation means a pi bond next to another pi bond, or next to a lone pair. In methoxyethene, bottom right, the two dots on oxygen are a lone pair beside the double bond. That pair moves in, the pi electrons move out to the end carbon, and the second structure has a plus on oxygen and a minus on carbon. Right? Now look at the allyl cation, bottom left. What sits right next to its double bond? Yes, exactly, a positive charge, the small plus sign! That carbon has an empty p orbital. The pi electrons shift towards it, and the plus charge appears at the other end. And the last case is a pi bond next to an unpaired electron. Clear? Now the rules for the drawings. Every contributing structure has the same atoms in the same positions, the same total charge, and the same number of paired and unpaired electrons. Here is a trap: move even one atom, say a hydrogen, and it is not resonance at all; those are two different compounds. Okay?

3:00Bond length and bond order in the hybrid

Now, back to our puzzle. In the hybrid, a bond is neither a pure single bond nor a pure double bond. Its length lies in between the two. Right? When the structures are equivalent, we can put a number on it. Bond order is the total bonds between the two atoms in all the structures, over the number of structures, the fraction on screen. With me? So take one carbon carbon bond of benzene; C dash C on screen means that bond. In one Kekulé structure it is single, in the other it is double. So what is its bond order? One plus two, over two, gives one point five. Every bond in the ring is like this, so all six are identical, each one hundred and thirty nine picometres; p m means picometres. So the mystery is solved: the drawing was only one part of the picture. Clear?

3:52Which contributing structure counts more?

Now, the contributing structures do not count equally; the more stable ones contribute more. Rule one is more covalent bonds, and rule two is a complete octet on every atom of the second period, like carbon, nitrogen and oxygen. Okay? So look at methoxyethene again, bottom right. Both structures have the same number of bonds, and complete octets. So which one contributes more? The first one, the uncharged one, yes! That is rule three, less separation of charge. An uncharged structure beats a dipolar one, the one with a plus at one end and a minus at the other. Right? Rule four also goes against the second structure: the negative charge belongs on the more electronegative atom, the positive charge on the less. There it is the other way round, minus on carbon and plus on oxygen. Got it? Rule five is unlike charges close together and like charges far apart. Like charges on neighbouring atoms is the worst case of all. Clear? And finally, equivalent structures give the greatest stabilisation, like the two for benzene and the two for the ethanoate ion, at the top of the figure. The carbonate ion is another case, and it is in our first example. Okay?

5:01Example 1: carbonate, nitrate, ethanoate and C-O bond order

Alright, example one. How many equivalent resonance structures do carbonate, C O three with a two minus charge, nitrate, N O three with a minus charge, and ethanoate, C H three COO with a minus charge, have? The small numbers written low are subscripts, and the charges written high are superscripts. Also, what is the carbon oxygen bond order in carbonate? Pause the video and try it first. Done? Okay, let's check. In carbonate, carbon is bonded to three oxygens, and the double bond can be written to any one of them. So, three equivalent structures. Right? Nitrate is the same story, three oxygens, so three structures. For ethanoate, top right of the figure, the double bond goes to either oxygen, so two structures. With me? Now the bond order. Many students say one point five, just like benzene. Is that correct here? No, and this is the trap: benzene has two structures, but carbonate has three. In any one structure, carbon has four bonds to oxygen, one double and two single, shared over three carbon oxygen positions. So the bond order is four over three, not one point five. Clear?

6:17Example 2: resonance energy of benzene from hydrogenation

Now a slightly harder one. The heat of hydrogenation of cyclohexene is minus one hundred and twenty kilojoules per mole, and of benzene, minus two hundred and eight; k J mol to the power minus one means kilojoules per mole. Find the resonance energy of benzene. Here is a hint: cyclohexene has just one double bond. Pause the video and try it. Done? Let's check. Imagine benzene as its Kekulé structure, a ring with three ordinary double bonds; C, double line, C is a carbon carbon double bond. So three times minus one hundred and twenty gives minus three hundred and sixty kilojoules per mole. Okay? But real benzene, the hybrid, gives out only two hundred and eight. So is the resonance energy two hundred and eight, or three hundred and sixty? It is neither, so be careful; resonance energy is a difference. Three hundred and sixty minus two hundred and eight gives one hundred and fifty two. So benzene gives out one hundred and fifty two kilojoules per mole less. Right? So the resonance energy of benzene is about one hundred and fifty two kilojoules per mole; the wavy equals sign means approximately. The hybrid sits that much lower than its Kekulé structure, and that is why benzene is so stable. Got it?

7:35Quick revision: hybrid, ranking structures, resonance energy

So, quick revision. Same atoms, moving electrons, and one real hybrid. Rank the structures by bonds, octets and charges. And the hybrid has averaged bonds and extra stability, the resonance energy. Now go and try the practice problems on this page. All the best!