Chapter 24, Problem 24.62P

Interpretation Introduction

(a)

Interpretation:

The HOMO of one reactant and LUMO of the other reactant in the given reaction are to be drawn assuming one of the reactants is in its lowest excited state.

Concept introduction:

The bonding or antibonding character of an MO is determined by the number of constructive and destructive interactions. When the number of constructive (same phase) interactions exceed the number of destructive interactions, the MO is bonding. If the number of destructive interactions is larger, the MO is antibonding.

The highest energy bonding MO is the last one that contains electrons and is called the Highest Energy Molecular orbital (HOMO). The antibonding MO immediately above it in is the Lowest Energy Unoccupied Molecular Orbital, called LUMO.

For the molecule in its lowest excited state, one of the electrons from the ground state HOMO is promoted to the ground state LUMO. Therefore, the LUMO of the ground state, the first of the antibonding MOs now becomes the HOMO, and the LUMO is the second antibonding MO.

Answer to Problem 24.62P

The HOMO of one reactant and the LUMO of the other reactant for the three reactions are

Explanation of Solution

Reaction (i) is

Assuming the larger reactant to be the diene in its lowest excited state, i.e., the donor molecule in the Diels-Alder reaction, the HOMO of was determined by considering the different interactions between the six p orbitals of the carbons. The possible interactions are

With three $\text{π}$ bonds, the molecule will have three pairs in $\text{π}$ MOs. They will occupy the lowest three of the six $\text{π}$ MOs, i.e., all the bonding MOs. In the ground state, only the three bonding MOs are occupied. In the lowest excited state, one of the electrons from $\text{π3}$ MO is promoted to the $\text{π4}$ MO. So the $\text{π4}$ MO has now become the HOMO and $\text{π5}$ MO the LUMO.

The smaller of the reactants has only two carbons and one $\text{π}$ bond. Therefore, it will have two $\text{π}$ MOs, with only the bonding MO filled.

Thus, the HOMO and LUMO for this reaction are:

Reaction (ii) is

The reactants are the same. They both have four carbons, and therefore, p orbitals. The interaction of these will result in four $\text{π}$ MOs, two bonding and two antibonding. In the ground state, only the two bonding MOs are occupied. In the lowest excited state, one of the electrons from the ground state HOMO is promoted to the $\text{π3}$ MO. This makes it the HOMO of the excited state, and $\text{π4}$ the LUMO.

The MOs of the other molecule remain the same as in the ground state

Thus, the HOMO and LUMO in this case are

Reaction (iii) is

The first reactant has six carbons that contribute six p orbitals. The six $\text{π}$ MOs that result from their interactions are

Assuming this molecule is in the lowest excited state, one of the electrons from its ground state HOMO ($\text{π3}$) will be promoted to $\text{π4}$. Thus in the excited state, $\text{π4}$ is the HOMO and $\text{π5}$ the LUMO.

The second reactant has four carbons that contribute four p orbitals. The four $\text{π}$ MOs that result from their interactions are

Again, assuming that the larger molecule acts as the donor (diene) and the smaller as the acceptor (dienophile) molecule, the HOMO and LUMO that interact in the Diels-Alder reaction are

Conclusion

The $\text{π}$ MOs and the MO energy diagram is determined based on the overlap and phases of the contributing p orbitals

Interpretation Introduction

(b)

Interpretation:

The HOMO-LUMO interaction between the two reactants is to be drawn and if it will result in a thermally allowed or forbidden reaction is to be determined.

Concept introduction:

Diels-Alder reaction is a concerted reaction between a diene and a dienophile that produces a cyclic compound. It replaces two $\text{π}$ bonds with sigma bonds. The flow of all $\text{π}$ involved is in the same direction.

In terms of MO theory, the reaction requires that the overlapping orbitals of the two molecules be in phase so that constructive interference occurs.

The electron flow starts with the diene donating a pair of $\text{π}$ electrons to the dienophile. This means the HOMO of the diene overlaps with the LUMO of the dienophile. The interactions between the p orbitals of the carbon atoms result in the formation of an equal number of MOs, half of them bonding and half antibonding orbitals.

Answer to Problem 24.62P

The HOMO-LUMO interactions for the three reactions are:

Reactions (i) and (ii) are thermally allowed reactions. Reaction (iii) is a thermally forbidden reaction.

Explanation of Solution

Reaction (i) is

Based on the HOMO and LUMO from part (a), the HOMO-LUMO interaction for this reaction is

Both interactions are constructive. The net result is, therefore, a bonding interaction. Therefore, the reaction is thermally allowed.

Reaction (ii) is

Based on the HOMO and LUMO from part (a), the HOMO-LUMO interaction for this reaction is

Both interactions are constructive. The net result is, therefore, a bonding interaction. Therefore, the reaction is thermally allowed.

Reaction (iii) is

Based on the HOMO and LUMO from part (a), the HOMO-LUMO interaction for this reaction is

One of the interactions is constructive while the other is destructive, i.e., the net interaction is zero. Therefore, this reaction is thermally forbidden.

Conclusion

The HOMO-LUMO interaction between the reactants is determined based on the overlap and phases of the donor and acceptor MOs respectively and the feasibility of the reaction determined from the overall character of the interaction.