Hyperconjugation

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. In propene, the bond between the methyl carbon and its neighbour is a plain single bond. Yet it measures one hundred and fifty picometres, shorter than the carbon carbon single bond in ethane. So how can one single bond be shorter than another? Today's idea, hyperconjugation, explains it, and it comes up often in JEE, so let's begin.

0:27What is hyperconjugation?

So, look at the left of the figure, the ethyl cation; the C with a plus sign is the positive carbon. In hyperconjugation, the sigma electrons of a C, dash, H bond are delocalised, which means they spread out. Sigma is the Greek letter on screen, and every single bond, like this C H bond, is a sigma bond. With me? Now let me ask you this. Can any carbon hydrogen bond do this, or only special ones? Very good if you said only special ones. The C H bond must be on an alkyl group attached directly to an unsaturated system, like the C, double line, C on screen, a carbon carbon double bond. Or it must be next to an atom with an empty or half-filled p orbital, like the positive carbon in the figure. Clear? Here is what happens. The sigma bond overlaps sideways with the next-door p or pi orbital; pi is the other Greek letter. In the figure, the grey oval is the C H sigma bond, and the curved arrow shows its electrons spreading into the empty p orbital. Got it? It is a permanent effect, always present, and it is also called no-bond resonance. Let us see why, in propene. Okay?

1:35No-bond resonance in propene

Now the right of the figure, propene, also on screen: C H three, dash, C H, double line, C H two. The small numbers written low are subscripts. The arrow with two heads means resonance. In the second structure, one methyl hydrogen is shown as H plus, with a superscript plus up high. The double bond has moved, and the end carbon carries a negative charge, the small minus sign. Right? Now tell me, in that second structure, is the H plus bonded to any carbon at all? No, it is not, and that gives the name. There is no bond between that H plus and carbon, so we call it no-bond resonance. It is only a contributing structure, not the real molecule. With me? Only the hydrogens on the carbon next to the double bond can do this. We call them alpha hydrogens, and alpha is the Greek letter on screen. Propene's methyl group has three, so three such structures. Clear? Now, back to our puzzle. In the no-bond structure, what has happened to the bond between the methyl carbon and its neighbour? Yes, exactly! It has become a double bond. So in the real molecule this bond is partly double, and that is why it is shorter, one hundred and fifty picometres; p m means picometres. So our puzzle is solved. Got it?

2:53Count the alpha hydrogens

So, let us make this general. Alpha hydrogens sit on the carbon next to a carbon carbon double bond, a cation centre, C plus, or a radical centre, C with a dot up high; that dot is the unpaired electron. Okay? And the counting rule is simple: the number of hyperconjugation structures equals the number of alpha hydrogens. Right? Now let me ask you something. More alpha hydrogens means more structures, and more delocalisation. So is the species more stable, or less stable? Yes, exactly, it is more stable, absolutely! The double arrow on screen means implies. Clear? And this explains an important order. For carbocations and free radicals, tertiary is more stable than secondary, then primary, then methyl. The small circle up high means degree, and the greater than sign means more stable than. Methyl, C H three, has no carbon next to it, so no alpha hydrogens at all. With me?

3:53Alkenes and alkyl benzenes

Now let us look at alkenes. The more alkyl groups on the double bond, the more alpha hydrogens there usually are, and the more stable the alkene. And it gives out less heat when hydrogen is added, so its heat of hydrogenation is smaller. Okay? Here is the full order; capital R means any alkyl group. An alkene with four alkyl groups on the double bond is the most stable, then three, then two on the same carbon. That one is about as stable as trans, with the two groups on opposite sides; the wavy equals sign means about equal. Right? Then comes cis, with both groups on the same side. So be careful here: cis and trans both have two alkyl groups, but cis sits below trans. Then one alkyl group, and last of all ethene, with none. Clear? Now let us go to benzene. An alkyl group on the ring releases electrons by hyperconjugation and directs electrophiles to o and p, which means ortho and para. So tell me, which shows more hyperconjugation, methyl or tert-butyl? That is correct if you said methyl. Count the hydrogens on the carbon attached to the ring: three for methyl, two for ethyl, one for isopropyl, and zero for tert-butyl. The dash in front of each group is its bond to the ring, and the brackets group the methyl groups. And here is the trap: the plus I order from the inductive effect lesson is exactly the reverse. Never mix them up, okay?

5:24Example 1: ethyl, isopropyl, tert-butyl cation structures

Alright, example one. How many hyperconjugation structures do the ethyl, isopropyl and tert-butyl cations have? Pause the video and try it first. Done? Okay, let's check. In the ethyl cation, on the left of the figure, the positive carbon carries two hydrogens and its neighbour carries three. So, five structures? Do the two on the positive carbon count? No, they do not. We count one structure for each C H bond on the carbons next to the positive carbon, the C with the plus written above it. Only one methyl group is next to it, so three. Right? Isopropyl has two methyl groups on the positive carbon, and two times three gives six. Got it? Tert-butyl has three methyl groups there, and three times three gives nine. With me? So three, six and nine. More and more alpha hydrogens, and that is exactly the stability order: tertiary, then secondary, then primary. Clear?

6:23Example 2: stability order of the butene isomers

Now a slightly harder one. Arrange in order of increasing stability: but one ene, cis but two ene, trans but two ene, and two methyl propene. Here is a hint: count the alkyl groups on the double bond. Pause the video and try it. Done? Let's check. But one ene has an ethyl group on the double bond, with five hydrogens. So five alpha hydrogens, right? Careful, that is the trap: only the C H two right next to the double bond counts. So but one ene has one alkyl group, and just two alpha hydrogens. Okay? Now the other three. But two ene has a methyl group on each carbon of the double bond, and two methyl propene has two methyl groups on one carbon. So all three have two alkyl groups and six alpha hydrogens. Right? But cis and trans have the same count, so we use the order we learnt: cis is less stable than trans. With me? And from our order, two alkyl groups on the same carbon is about as stable as trans, so two methyl propene is about as stable as trans but two ene. So the answer is but one ene, less than cis but two ene, less than trans but two ene, about equal to two methyl propene; the less than sign means less stable than.

7:41Quick revision: alpha H count, cation and alkene stability

Clear? So, quick revision. Hyperconjugation means the C H sigma electrons spread into a next-door p or pi orbital. The number of structures equals the number of alpha hydrogens, and more alpha hydrogens means more stability. So for cations and radicals, tertiary beats secondary beats primary beats methyl, and more alkyl groups mean a more stable alkene. You have done really well today. Now try the practice problems on this page. All the best!