Explain the extradorinaiy stability of the allyl type radical in terms of (a) resonance and (b) pi orbital overlap delocalization.
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i get the first part, that the tertiary radical is quite stable especially when it's adjacent to a double bond, this gives two resonance structures.
What I am having trouble understanding though is how to explain is why this is stable in terms of orbital diagrams
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Like I know this is stable in terms of the resonance depiction and all but not in terms of molecule orbitals @Woodward ?
First off there's a slight error to fix with your resonance structure, and for an added bonus I'll also compare it with the anion and cation allylic positions: |dw:1441769296098:dw| Notice I'm using single barbed arrows for the radical, and that there are some subtle differences between them.
Now for the orbital picture, if we look at the p-orbitals of three atoms side by side like this: https://upload.wikimedia.org/wikipedia/commons/6/6f/AllylMO.png
Then you can see that all three of these p-orbitals can create one continuous \(\pi\)-orbital to bond together, since the lobes are all the same color, which means the sign of the wave functions at those atoms are the same. This means they can constructively interfere when added together (LCAO) so that when you square the wave function you get a higher probability of finding electrons there.
I don't know if that exactly answers your question or not though
Might be fun to play around with this, click the buttons and stuff \(\Psi_1\), SOMO, and \(\Psi_3\) are the orbitals, SOMO means Singly Occupied Orbital, it's sorta like an inbetween of HOMO and LUMO.
so SOMO is in-between \[\Psi1, \psi2\]
I was wondering why there wasn't any nodal plane in psi 2
i have always had trouble with this concept but thanks for explaining it @woodward nicely done
Yeah, it's apparently because the wavefunction is a superposition of both wave functions, so they cancel each other out due to being opposite phase there, and that destructive interference is the non-bonding orbital. Funny I guess, I can never be to sure that what I'm saying is correct when it comes to these QM things though.