Using the rules described in Section 21.3, draw the molecular orb...

Question

Using the rules described in Section 21.3, draw the molecular orbital picture of hexa-1,3,5-triene. Label the HOMO and LUMO.

Accepted Answer

Hello everyone. Let's do this problem together. It says provide the molecular orbital diagram and label the homo and luo for Octa 246 trine. We are given the structure of this compound. We see it is an eight carbon chain with three double bonds on carbons, 24 and six. So let's review the steps to providing the molecular orbital diagram. First, we want to draw the atomic orbitals and how do we know the number of atomic orbitals to draw it equals the number of pi electrons in the structure. And then once we draw the atomic orbitals, we fill them with electrons and they will have one electron each because they are all in the same energy level, right? OK. And then we want to draw the molecular orbitals and the number of molecular orbitals equals the number of atomic orbitals. So now when we draw the orbitals, we draw six molecular orbitals for each energy level. And when we are filling in the phases of the molecular orbitals, we follow these steps. So we don't change the first orbital phase, but we will alternate the last orbitals phase, then we will apply nodes to each energy level. And the number of nodes for that energy level will be N minus one where N is the number of that energy level. And we apply those nodes symmetrically. And once we provide those nodes, we will use those as our guide for filling in the remaining phases of those molecular orbitals. And lastly, we will fill in our electrons in the molecular orbitals following off Bar's principle, which means that we fill in those electrons from lowest energy level to highest energy level. And then we will also note or identify the homo and the limo as mentioned in the problem, recall that Homo is the highest occupied molecular orbital and Lomo is the lowest unoccupied molecular orbital. All right. So let's follow all these steps for this specific problem. So drawing the atomic orbitals, the number of pi electrons, how many pi electrons are in this structure? We have three double bonds, each double bond will have two pi electrons. So we have 246 pi electrons, which means we will have six atomic orbitals, all the same energy level. And we said we have one electron in each atomic orbital. Now we draw the molecular orbitals and we draw these vertically and from bottom to top lowest energy level to highest energy energy level, we label these side one side, two si three size four, size five and size six. But for the top half of the molecular orbitals, these are the anti bonding molecular orbitals. So we also notate those with a star in the bottom three are the bonding molecular orbitals. OK. So now we want to draw our actual molecular orbitals. So we will draw six molecular orbitals and we will draw it like it is on our structure. OK. Let me copy this five more times. OK. So now let's fill in the phases on our molecular orbitals. So we said that we don't change the first orbital phase. So let's shade in the top lobe for the first orbital. And for step B, we said alternate the last orbital phase, we keep the first energy level the same. So we will also fill in the top lobe. But now we alternate. So bottom, first side two, top, first side three, bottom first si four, top first sci five and bottom four si six. Now to fill in the rest of the orbitals, we want to apply the nodes. So si one will have zero nodes, right, one minus one is zero, side 22 minus one is one, we'll have one node and so on. So side three will have two nodes, side four will have three nodes, side five will have four nodes size six will have five nodes and let's apply those symmetrically. So side two will have one right down the center. Si three will have two nodes that so three sections with two molecular orbitals. Then we have one orbital A node, two orbitals. A node two orbitals, a node si five, we have four nodes. So one orbital, a node, one orbital, a node, two orbitals, a node, one orbital, a node, one orbital and si six has five nodes. So one in between each orbital. OK. So now we fill in the rest of our phases from left to right. And every time we hit a node, we switch the phase. OK. So si one has no node. So we will fill in the top lobe for all of the orbitals side two, we start on the top for the first three. Then we hit a node. So we switched to the bottom phase size three. We have top top nude bottom, bottom nude top top side four, we have top phase a node, bottom phase, bottom phase node, top phase, top phase I five, top phase node, bottom phase node, top phase, top phase node, bottom phase node, top phase size six. We're alternating every single one. So top bottom, top, bottom, top bottom. OK. So lastly, we just need to add our electrons to the molecular orbitals that we drew before. So each orbital can get two electrons. So si one will have two, si two will have two, si three will have two and that uses up all six of our electrons. So where is the highest occupied molecular orbital? That will be SI three, right? And our lowest unoccupied molecular orbital will be side four. So that is it for this problem. I hope this video was helpful and thanks for watching.

Using the rules described in Section 21.3, draw the molecular orbital picture of hexa-1,3,5-triene. Label the HOMO and LUMO.

Key Concepts

Molecular Orbitals

HOMO and LUMO

Conjugation

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