Which of the following are correct? Correct any false statements
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Question

Which of the following are correct? Correct any false statements
c. A conjugated diene with an odd number of double bonds has a symmetric HOMO.

Accepted Answer

Hello everyone. Let's do this problem together. It says determine whether the following statement is true or false. If false, correct the statement, a conjugated poly with a symmetric homo highest occupied molecular orbital has an odd number of double bonds. And we are given four answer choices. A says true BC and D say false with different alternative statements. So let's draw an odd number, double bond conjugated dying and a conjugated dying with an even number of double bonds and draw their molecular orbitals and see which one is symmetric. So starting with an even number of double bonds. So a poly or an alkene with multiple double bonds that has an even number would be two double bonds. So we can draw a 13 conjugated dye. So a four carbon chain, 1234 with double bonds on carbons, one and three, this has four pi electrons. So we would have four atomic orbitals, one electron in each. And this translates into four molecular orbitals, right. We conserve the number of orbitals and these four orbitals would be labeled from bottom to top si one si two si three star and si four star. And we would have our four electrons filling molecular orbital si one and si two. OK. So where is our homo, our highest occupied molecular orbital is si two, our bonding orbital. So we want to look at the orientation of the molecular orbitals and see if they are symmetric or anti symmetric. So let's draw, we'll just draw the bonding orbitals. So we draw four orbitals. And remember Johnny's shading rules for shading molecular orbitals, The first orbital stays the same. So we'll draw the bottom one, the bottom phase shaded in and in si one, all the phases are the same. So they will all be shaded on the bottom inside two, the last orbital will have the opposite phase. So shaded on the top and there is one node in this molecular orbital. So we shade according to the orbital before and then flip once we hit a node. So the first one is shaded on the bottom already, we would shade the second one on the bottom, then we hit a node. So we flip and shade the top orbital and the last one is already shaded on the top. So what we're looking at is this symmetric does the left side mirror the right side of this blue line that I just drew. No. So this is going to be anti symmetric. So from this example, we would conclude that an even number of double bonds would be anti symmetric. OK. So we can eliminate choice C which says a conjugated pole with a symmetric homo has an even number of double bonds. OK. And we haven't looked at odd number of double bonds yet. We can also eliminate answer choice D because we don't usually talk about non conjugated polys with molecular orbital theory. So now let's draw a poly with an even or sorry, an odd number of double bonds. So let's say three double bonds. So we can draw 123456 carbon chain hexen. So double bonds on carbons, 13 and five similar process, we will have 246 atomic orbitals, one electron in each, which gives us six molecular orbitals, three bonding and three anti bonding. And we will only draw the orbitals for the bonding. Since we're looking at the homo labeled si one side two side three for the bonding, si four star si five star and si six star for the anti bonding and filling our electrons, we'll have two in si 12 in inside two and two in inside three, which makes si three our highest occupied molecular orbital or the homo for our try in. So let's draw our molecular orbitals. So this time we have six orbitals for each energy level. OK. So we keep the first orbital the same, we'll shade the bottom phase for all of these in each energy level. Then for the last orbital, we shade the same phase on the first energy level. And then opposite as we go up. So I one, the last orbital is shaded on the bottom side, two, the last orbital shaded on the top side three, last orbital shaded on the bottom. And then we apply our nodes. So we apply one node to si 22 nodes to side three. And we make those symmetrical. And now from left to right, we shade starting with the phase that we already have shaded. every time we hit a node we switch. So si one, all the phases are shaded on the bottom side. Two, the first three are on the bottom. Second three are shaded on the top in size three, the first two are shaded on the bottom. Then we hit a node second or sorry. 3rd and 4th are shaded on top. 5th and 6th are shaded on the bottom. So again, for symmetry, we are looking between orbitals three and four, I can draw a line. So now do the three orbitals on each side of this mirror. Are they mirror images of each other? Yes. So these are symmetric and that is for our poly with an odd number of double bonds, which is going to make our statement true, right? A conjugated poly with a symmetric homo has an odd number of double bonds. So that makes a the correct answer choice for this problem. That's it for this one. I hope this was helpful. And thanks for watching.

Which of the following are correct? Correct any false statements c. A conjugated diene with an odd number of double bonds has a symmetric HOMO.

Key Concepts

Conjugated Dienes

HOMO and LUMO

Symmetry in Molecular Orbitals

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