(b) There is a different, thermally allowed cycloaddition of two ...

Question

(b) There is a different, thermally allowed cycloaddition of two butadiene molecules. Show this reaction, and explain why it is thermally allowed. (Hint: Consider the dimerization of cyclopentadiene.)

Thermal cycloaddition reaction of two butadiene molecules forming a dimer.

Accepted Answer

Hello everyone. Let's do this problem. It says the following two molecules can undergo a thermally allowed cyclo addition reaction show the reaction and explain why it is allowed. So the two molecules given are two molecules of two methyl beta 13 dying. So before we show the reaction, what do we need for a thermally allowed cyclo addition. What happens during a thermal cyclo addition? Well, we need at least three pi bonds in the reactants and then we are forming one pie bond in the products. So what is occurring is two pi bonds are being destroyed to form a cyclic compound. All right. And one of the most common examples of this is the deal's older reaction. So in a deal's alder reaction, the homo of molecule A must fill the limo of molecule B. All right. And how does that occur? We need symmetry allowed orientation of the orbitals in the homo, the highest occupied molecular orbital and the limo the lowest unoccupied molecular orbital. So they have to have symmetry allowed orientation, which means the phases have to overlap correctly. And we need to minimize the Homo luo gap. And if this all sounds a little bit confusing. Make sure you go back and watch Johnny's videos on thermal Cyclo edition. He does a fantastic job of going really in depth on all of this information. So let's use this information that we just reviewed to apply it to the reaction given to us. So I have two Dynes. One is going to be my 13 dying and one is my dienno file, right? That's what I need for this type of reaction. Since they are the same molecule, it doesn't matter which is which. But one is acting as the 13 dyne which has two pi bonds. And one is acting as the dienno file, which we only need one pi bond. One is going to act as an agent for forming that ring that we are trying to form. So let's draw the molecular orbitals for this structure so that we can show the homo luo overlap that is needed to perform this reaction. So we'll draw our two methyl beta 13 dying. And since there are four atoms that are participating in pie bonding, we'll draw four atomic orbitals, right? 12342 atoms in each double bond, and each of these atomic orbitals gets one electron and drawing that into our molecular orbitals. We know if there are four atomic orbitals, we have four molecular orbitals. So we have two bonding molecular orbitals which are lower energy and two anti bonding orbitals which are higher energy, the bonding orbitals are labeled si one si two, the anti bonding orbitals are si three star and si four star. So that star is the notation for an anti bonding orbital. So now if we use our principles to fill orbitals, we have four electrons and we start from si one, we fill 12 electrons, si two will have 34 electrons. So where is our homo our highest occupied molecular orbital? That'll be two, right? And the limo the lowest unoccupied molecular orbital is si three. So we want to show the overlap of these two molecular orbitals. Well, then we need to know what the molecular orbitals look like. So let's draw the molecular orbitals for these four energy levels. They each get four orbitals, these double lobed orbitals. And if you don't remember how to fill these, I'm going to show you. But if you need more explanation, Johnny has a video on the seven steps for how to fill molecular orbitals. So I'm going to use those steps. So in si one, our lowest energy level, we start by filling in all of the energy orbitals with the same phase. And according to Johnny's rule, we like to start with the first lobe in the top phase, last lobe also in the top phase. So as long as they're the same, they could all be the top or all be the bottom. But the first and last lobe should be the same. Now, as we move up the energy levels, we're going to keep the phase of that first lobe the same. So let's shade in the top lobe on all four energy levels. And then for the fourth lobe, the last lobe, we're going to alternate phases. So we started with the top phase filled. Now we'll shade in the bottom phase, then the top phase, then the bottom phase. OK. So now the next step is adding nodes and we add nodes to the N minus one amount. So the first energy level gets zero, second energy level gets one third, energy level gets two and fourth energy level gets three. So N minus one and we wanna add these nodes symmetrically, right. So in the first energy level, we have one node. So there will be two orbitals on each side of that center node, the third energy level, we have two nodes. So we'll have one orbital and then node, then two orbitals and then another node and then one orbital. And the last energy level will have nodes in between each orbital. OK? So now as we filled these orbitals with their phases from left to right, every time you hit a node you switch the phase. So if we're starting in energy level two, we have the first orbital is filled the top phase. So we continue shading that top phase, we hit the node, we switch and notice how it always lines up with the last orbital that we already filled. The third energy level, we start with the top phase, hit a node switch to the bottom phase, continue with the bottom phase hit a node switch to the top phase. And now in the fourth energy level, we're alternating every lobe. So top phase, bottom phase, top phase, bottom phase. OK. So we are only going to use P I two and SI three, right? Because that is our homo and luo and we're going to draw these in our structure and show how they overlap in order to allow this reaction to occur. So let's draw our structure, 1234 carbons with our methyl group. And I'll draw my four orbitals and let's make this one the homo. So our homo has two top phases filled, two bottom phases filled. So top lobe, top lobe, bottom lobe, bottom load. So now with our Lomo, we are only overlapping one pie bond with these two end orbitals. So we can orient our structure as follows with our P orbitals or pie orbitals rather. And this is our, so we have one phase up, two down and one up. So one up on the end two down. Oh I think I drew this incorrectly, right. My dine has this shape but it doesn't really matter. So we said up down, down, up and this is my limo. So do we have symmetry allowed orientation? Do we have bonding interactions? Yes. These phases line up, they are both the unshaded phase and these phases also line up, they are both shaded. So on carbons three and four, here they are both two bonding interactions. So this is thermally allowed because we have both bonding interactions if one of them lined up, but one of them did not, that would not be thermally allowed. OK. So this is the explanation for why this is a thermally allowed cyclo addition. And the problem also wants us to show the reaction that is possible. So since our structure, our dienno phyle has two pi bonds, we can actually have two different reactions, two different products. So let's draw our 13 dying and our dya file will line up one double bond in between the ends of our 13 dying. And I'll just draw the reactance of the second product while we are at it or the the reactance of the second reaction. So we're keeping our 13 dying the same and we can orient our dying a file towards the other double bond or the other direction. So this will give two different products. Let's draw those products. So in the first reaction, we are applying heat, right? And we have this double bond on our diana file is going to attack our 13 dying, shifting the double bonds, shifting the double bonds, we always have that three arrow mechanism. So now those carbons are going to form a ring where those three pi bonds are cyclohexane ring. We have our methyl group on the side. We formed a new pie bond here and we have this substituent group that will come off of our ring from our dienno file, which looks like this. Our second reaction also have heat. We have our three arrow mechanism, 123. So we're forming our cyclo heine again, the methyl and the pi bon is not changing. The only thing that is changing is the position of our substituent group since we oriented our diana file in a different way, so it will look like this. So these are the possible reactions for this thermally allowed cyclo edition. That's it for this problem. I hope this video is helpful and thank you for watching.

(b) There is a different, thermally allowed cycloaddition of two butadiene molecules. Show this reaction, and explain why it is thermally allowed. (Hint: Consider the dimerization of cyclopentadiene.)

Key Concepts

Cycloaddition Reactions

Thermal vs. Photochemical Reactions

Dimerization of Cyclopentadiene

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