Cycloadditions Pt3 Flashcards
In a cycloaddition, how do the orbitals interact?
- The HOMO of the diene interacts with the
- LUMO of the dienophile
How to electron donating group affect the energy of orbitals?
They increase the energy of orbitals
How do electron withdrawing groups affect the energy of orbitals?
Lowers the energy of the molecular orbitals
What is the overall effect of these substituents
They lower/decrease the HOMO/LUMO gap
We can apply these same ideas of “mixing” the orbitals of ethene and butadiene, with those of allyl cation and pentadienyl anion
Draw the molecular orbital diagrams for the allyl cation
We can apply these same ideas of “mixing” the orbitals of ethene and butadiene, with those of allyl cation and pentadienyl anion
Draw the molecular orbital diagrams for the pentadienyl anion
We can start off by mixing the LUMO of the allyl cation and ethene to find the intermediate
What does this look like?
- Where we have two orbitals combining it creates one larger one
- An orbital in ethane goes in node on the allyl cation
- Carbon 1 has the largest orbital coefficient
We can then mix the HOMO of the pentadienyl anion and butadiene to find the intermediate
What does this look like?
- Carbon 1 produces a large orbital due to mixing two medium sized orbitals
- Carbon 3 produces a slightly smaller orbital than 1 due to mixing one small and one medium orbital
- The largest lobe is on carbon 1
When we have mixed the orbtials, what is next?
We want to match up the largest orbital coefficients, i.e. the largest lobes of the two respective π system - this determines regioselectivity
Why does the orbital coefficients determine regioselectivity?
- Matching up the sizes of the orbtial coefficint that determines the chemical outcome
How do the orbital overlaps affect geochemical outcome (i.e. exo/endo)
- For the endo product, you can have a secondary orbital overlap between carbons 3 on butadiene and 3 on ethene (small bonding interaction)
- For the exo product, there is no secondary orbital interaction
When considering alternative substitution patterns, the best way to start is considering the suitable resonance forms. Do this for the following diene
- It is best to push from the electron-donating group first
- Note: negative charge build-up on carbon 1. There is no possibility to get negative charge on carbon atom 4
Once you have drawn resonance structures, you must now choose a suitable all-carbon approximation to best reflect the electron distribution:
The effect of the electron donating group (OMe) on the diene’s orbital coefficient sizes can be approximated by splitting the simplified all-carbon analogue into butadiene and the propenyl anion
How do the carbon align?
Draw the molecular orbital diagrams for the butadiene