(a) Sketch the pi molecular orbitals of hexa-1,3,5-triene.

Question

Accepted Answer

Hey everyone, let's do this problem and says draw the pi molecular orbital of Octa 1357 tetra. So when we're looking at pi molecular orbital's we're looking at pi bonds and we know pi bonds have P orbital's in them. Right? The overlap of p orbital's in addition to s orbital's. But here we're looking at specifically P orbital's. So first to know what the molecular orbital's are going to look like. We need to know the number of p orbital's we have and that's simply going to be the number of pi bonds times two. So for each double bond we will have two pi electrons or two p orbital's and then the number of p orbital's is going to equal the number of anti bonding. Or sorry, atomic orbital's and the number of molecular orbital's. Alright. So then when we're drawing our molecular orbital's, we know how many we need to draw. We're going to look at johnny's. Seven Rules if you don't know what I'm talking about, make sure you go watch johnny's Rules, seven rules on drawing molecular orbital's the first orbital in the molecular orbital's doesn't change the phases. So we're talking about the lobes on the orbital's of the first orbital in each energetic level. Okay, so those are not going to change the last orbital in the molecular orbital always changes the nodes Are going to equal N -1 where N is the number energy level that you're on. So we start at zero zero nodes for the lowest energy level. And you add one for each level. And remember that the nodes are symmetrical. Okay. And lastly orbital phases switch across nodes. Okay, so let's apply these two Octa Tetra in. So our structure looks like this eight carbons 12345678. With double bonds on carbons 135 and seven. So we see four double bonds which means we'll have four times two is eight P orbital's 12345678. So that is our atomic orbital's and each one gets one electron. And now drawing our molecular orbital's we'll have four bonding orbital's Pi one Pi Pi three Pi 4. And then for anti bonding orbital's high five star. Let me draw them all first pi six star Pi seven star and Pi eight star. Okay. And then our electrons will fill in from the bottom up using all of our electron configuration rules. So two electrons maximum. Then we move up to the next level. Next energetic level, Ending on Pi four. But we're looking at the molecular orbital's so let's draw those lobes. Okay, so now I have all my molecular orbital's drawn and labeled with their energetic level. So let's start filling in the phases. So the first rule that we said was the first orbital doesn't change. So let's draw them all shaded in on the top and then the last orbital is the opposite. It always changes. So it's going to start the same in the first energetic level. And then it's going to flip flop. So I mean it has to be the same phase as this one. But now we're looking at alternating these wands, so alternating top bottom, top bottom, top bottom. Okay now let's add our nodes. So we said pi one is going to be zero nodes then add one for each level and we add them symmetrically. So let's add where our nodes will go. So one note is going to be right down the middle, two nodes will have two, four and then another two, three. We're going to do for four. We're going to have one then too, Then a node and then another two and a node and then one Okay five. We're going to split right down the center and then put two on either side. So one and two And one and 2. Alright six. We have 1 1, 12, 111. There we go. 123456. And then seven is going to be one in between every pair. Every lobe I guess. Okay, so the highest molecular orbital is completely into bonding with a node at each overlap and it has the maximum energy So notice how the energy increases with the number of nodes increasing. So now let's fill in our phases so only change the phase when you're crossing a node. So there are no nodes in the first level. So they will all be at the top for pie too. We'll have the first four on the top and then there's a node. So the last four will be on the bottom. I always start with the first one and just go so top top node, bottom, bottom, bottom, bottom node top and notice how it's lining up with the first and the last nodes that we already filled in top top node, bottom, bottom node, top top node and bottom bottom top node, bottom bottom node, top top node, bottom bottom node, top, top bottom, top, top, bottom, bottom, top bottom, top node, bottom node, top node, bottom bottom node, top node, bottom node top and now we're alternating every lobe On the last one. Top bottom, top bottom. So that is it for our pi molecular orbital's for Octa 1357 tetra in. I'm going to add a note here that the red dashes represent the nodes and that makes be the correct answer for this problem. That's it for this one. I hope this video is helpful and thanks for watching

(a) Sketch the pi molecular orbitals of hexa-1,3,5-triene.

Key Concepts

Molecular Orbitals

Conjugation

Pi Bonding and Antibonding Orbitals

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