A community for students.
Here's the question you clicked on:
 0 viewing
Photon336
 one year ago
Explain the extradorinaiy stability of the allyl type radical in terms of (a) resonance and (b) pi orbital overlap delocalization.
Photon336
 one year ago
Explain the extradorinaiy stability of the allyl type radical in terms of (a) resonance and (b) pi orbital overlap delocalization.

This Question is Closed

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441762745720:dw 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.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0What I am having trouble understanding though is how to explain is why this is stable in terms of orbital diagrams

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0@abisar @abb0t @nincompoop

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0Like I know this is stable in terms of the resonance depiction and all but not in terms of molecule orbitals @Woodward ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0First 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 porbitals 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 porbitals 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Might 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.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0so SOMO is inbetween \[\Psi1, \psi2\]

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0I was wondering why there wasn't any nodal plane in psi 2

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0i have always had trouble with this concept but thanks for explaining it @woodward nicely done

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, 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 nonbonding 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.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.