Chemistry 3B Lecture 7 - Chemsitry 3B Lecture 7 Thursday...

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Chemsitry 3B Lecture 7 Thursday February 12 th , 2009 Aromatic molecules have to be cyclic, completely conjugated= p orbital one very atom in the cycle, and planar. All of those molecules are cyclic and completely conjugated, and they even look planar. But lets define planarity: if your ring is 3,4,5,6, or 7 membered and is completely conjugated, you can assume it is planar. These three criteria suggest we are missing something that was added by Huckel. Huckels Rules= in order for a compound to be aromatic, it must also have 4n-2(pi) electrons where n= an integer (as well as the other qualifications). 4n+2 represents the number of pi electrons. The pi MO of benzene: we will go with the energy diagram of benzene. Benzene has 2 sets of degenerate bonding MOs, and 2 sets of degenerate antibonding MOs. We have 6 p orbitals in the cycle of benzene and we have 6 new MO’s. The degenerate levels are due to the benzene being cyclic and completely conjugated (it makes them different than other alkeles). Pi energy diagram for benzene: there are 6 electrons, and all 6 pi electrons are in bonding orbitals which is a good thing. There is an easy way to come up with what the pi energy diagram level looks like for cyclic, completely conjugated, and planar molecules. These are called frost diagram, or the inscribed polygon method. You draw a circle, and inscribe your polygon into it. In this case it is a hexagon for benzene. One of the vertices always has to be at the bottom of the circle. The midpoint of the circle constitutes an area where everything above it is pi* and everything below it is pi. Levels at the midpoint will always be non bonding.
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This note was uploaded on 02/04/2010 for the course CHEM 3BL taught by Professor Chunmei during the Spring '08 term at University of California, Berkeley.

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Chemistry 3B Lecture 7 - Chemsitry 3B Lecture 7 Thursday...

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