aromaticity

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inaccensa

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I recall in an earlier post, some one had mentioned that you don't count the lone pairs, if the cyclic structure satisfies Huckels rule. Is this the correct way to think of it? or is this just simplifying? For Pyridine, the lone pairs are not counted as a part of the conjugated system. It is aromatic,since the structure has 6pi electrons and nitrogen is sp2 hybridized. Kaplan says that you don't count them since they are perpendicular to the system. how do you determine that. For pyrrole, we do count the lone pairs. Thanks
 
Wait, so if there are two of these perpendicular ring structures side by side in a molecule, it's still aromatic?
 
But what does Kaplan mean by "the lone pairs in pyridine are perpendicular to the p orbitals."

the sp2 orbitals spread out in a flat triangle with the unhybridized p orbital that pi bonds sticks up vertically. (tee hee)
 
the sp2 orbitals spread out in a flat triangle with the unhybridized p orbital that pi bonds sticks up vertically. (tee hee)


duh, i understand that its perpendicular to the system, but what i meant was how do you know that. Won't you have to know the structure? Can I just say, since it the pyridine is conjugated and satisfies huckel's rule, I won't count the lone pairs?
 
duh, i understand that its perpendicular to the system, but what i meant was how do you know that. Won't you have to know the structure? Can I just say, since it the pyridine is conjugated and satisfies huckel's rule, I won't count the lone pairs?

it's kind of chicken-egg.. just go with what you just said.
 
duh, i understand that its perpendicular to the system, but what i meant was how do you know that. Won't you have to know the structure? Can I just say, since it the pyridine is conjugated and satisfies huckel's rule, I won't count the lone pairs?

It doesn't count the lone pairs because the Nitrogen is in an sp2 configuration and the lone pairs are forced into a perpendicular P orbital that cannot conjugate with the Pi electrons.
 
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