a. For conjugated systems with two, three, four, five, six, and s...

Question

a. For conjugated systems with two, three, four, five, six, and seven conjugated p-bonds, construct quick MOs (just draw the lobes at the ends of the conjugated system as they are drawn on pages 1220 and 1221) to show whether the HOMO is symmetric or antisymmetric.

b. Using these drawings, convince yourself that the Woodward–Hoffmann rules in Table 28.1 are valid.

Accepted Answer

Hey, everyone, let's do this problem and says, draw the end lobes of molecular orbital's of conjugated pi bonds in the following to show whether the homo is symmetric or anti symmetric one for a poly in with even number of pi bonds and two for a polyethylene with dye or sorry, odd numbers of pi bonds. And part B asks, do these mos follow the Woodward Hoffmann rules given below. And we'll get into that once we get to that part of the problem. So first, let's focus on part a drawing the end lobes of molecular orbital's of conjugated pi bonds in these structures. So I'm going to use examples of these in order to get to where we're trying to go. So what does a poly poly in with even numbers of pi bonds look like? Well, the simplest example would be this dying, right? And those P orbital's would look something like this, right? So drawing the energy levels of the molecular orbital's, we would have 1234, right? Because we have 1234 orbital's. So that would be pi one pi two and then the second half our anti bonding orbital pi three star pi four star and those would look have electrons two electrons in each one. So that makes pi to our homo, right, our highest occupied molecular orbital, which is what we're talking about, right? We want to look at how the end lobes are in the homo energy level. So what do the lobes look like? Well, we'll start with pie one and we won't draw the anti bonding ones here because we're not talking about those. So we know in pie one, the lowest energy level, all of these lobes are in the same phase, right? So the same color is on the same side. But as we increase energy levels, we get one node for each energy level that we increase. So pi two is going to have one node and we always draw the nodes symmetrical. So to fill in the phases, we start with the same for the first lobe. So shaded on top and keep going until you have a node and then flip. So then the last two will be in the opposite phase. So since this is our home. Oh right, this is what we're going to apply to our structure. So we'll do the first two starting from the left shaded on top, Last two on the right shaded on the bottom. So notice we're looking at the end lobes of the molecular orbital's so that would be carbon one and carbon four. So notice how these are anti symmetric right, one of them has the shading up and one has the shading down. So they are in the opposite phases which makes these anti symmetric. So we can conclude that a polyethylene with even number of pi bonds, we'll have the anti symmetric and lobes. Alright. So let's do the same thing but for a polygon with odd numbers of pi bonds. So let's just add one to make it three right, 123, 1, 23. So that is one, 234, 5, 6 lobes. So we have pie, one pie too Pi three for bonding Pi four star, Oops Pie five star. And I keep doing that pi six star for anti bonding. But again, we're looking at the highest occupied molecular orbital which will be 123456 electrons that will be pi three. So drawing our lobes 123456, we start shading from the top and will do all the same phases. Now, the second level We add one node which will be down the center to make it symmetrical. So we'll start from the top shade and then we hit our nodes. So we switched to the bottom 123. And now Pi three is the one that we're looking at for our homo And we need to have two nodes. So after every two orbital's, we have a node. So starting on top 12, then we flip go to the bottom 12, then we flip, go to the top 12. So this is what we're going to use for this problem, right? Our homo, so drawing that on our structure, we start from the top one to flip, bottom 12, flip top one to flip. And again, looking at our end lobes, if we flatten out this structure, right, to make it a straight line instead of curved like this, it would look something like this, right. So the phases are both on the top, the same phases are both on the top, the blue color. So we can conclude that for a polyethylene with odd number of pi electrons pi bonds rather not pi electrons pi bonds, we will have symmetric globes. Okay. So that is it for part A, let's go on to part B which asks do these mos follow the Woodward Hoffmann rules given below. So let's read the Woodward Hoffmann rules. They say for an even number of pi bonds, an electro cyclic reaction occurs thermally with a con rotary ring closure and photo chemically with dis rotary ring closure and vice versa for an odd number of pi bonds, electro cyclic reaction thermally with a dis rotary ring closure and photo chemically with a con rotary ring closure. So let's look at these types of ring closures. So when we have a thermal electro cyclic situation, thermal electro cyclic, our molecular orbital's will look like the ones that we just drew. So for even number of pi bonds, they will be the anti symmetric up down. So they will rotate like. So which is con rotary. And our odd number of pi bonds which are symmetric are going to rotate dis rotary. So they're going to sort of fall towards each other. But the situation is different in photochemical electro sit click. And why is that? Well, we know these two situations are different because in photochemical reactions, we have an excited state molecular orbital, Right? The Homo jumps up one energetic level to where the loom A was excited state. I like your orbital's. So that's why it is the opposite like the rules state. So when we have an even number of pi bonds, those orbital's are going to look like the odd number in the thermal situation. And the odd number A pi bonds oops odd, I forgot to write number odd number pi bonds will be anti symmetric. So the rule states that for an even number of pi bonds in a photochemical situation, it will have dis rotary ring closure which we see that is true dis rotary. And for odd number pi bonds, it will be con rotary. So to answer the question for part B, the Woodward Hoffmann rules are valid. What happened? The rules are valid for both types of conjugated systems. So these highlighted in green are the answers for this problem which makes a the correct answer choice. That's it for this one. I hope this video is helpful and thanks for watching.

Key Concepts

Conjugated Systems

Molecular Orbitals (MOs)

Woodward–Hoffmann Rules

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