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How does Pyridine satisfy huckels rule?

I understand that benzene has 6 pi electrons because it has 3 double bonds.

When I look at pyridine I see 3 double bonds, plus a lone pair on the Nitrogen. Do you only consider the 3 double bonds pi electrons?

If so then why is pyrrole aromatic? It only has 2 double bonds in it structure.
 

RogueUnicorn

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double bonds are ALWAYS pi electrons. now, depending on how many pi electrons are present or not, the N can choose to donate a lone pair into the pi system. for example, in pyrrole, you see that there are two double bonds. as such, it's anti-aromatic. so the N atom donates its lone pairs by sp2 hybridizing, thereby bringing the number of pi electrons up to 6, and aromatizing the compound
 

loveoforganic

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Look at the orbitals. If the lone pair on nitrogen were to be in a p-orbital, you'd have sp hybridization and linear geometry. That doesn't work in a 6-membered ring. The lone pair on nitrogen in pyridine resides in an sp2 orbital.
 

minutemen11

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The general rule of thumb for these kinds of molecules is that the molecule will ALWAYS try to become aromatic if it can. So, pyrrol is aromatic because if it donated its lone pair, an aromatic compound results (there are now 6 pi electrons). Anyone have any exceptions to this rule??

I feel that this strategy is very useful when pressured during the MCAT
 

loveoforganic

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I've never heard of an exception to that, but I think I can think of one.

Azetes http://en.wikipedia.org/wiki/Azete

I think you could expan on that rule by saying that it will only donate it's lone pair to gain aromaticity if the heteroatom isn't already involved in a pi bond
 

RogueUnicorn

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I've never heard of an exception to that, but I think I can think of one.

Azetes http://en.wikipedia.org/wiki/Azete

I think you could expan on that rule by saying that it will only donate it's lone pair to gain aromaticity if the heteroatom isn't already involved in a pi bond
well that much is implied - if you have two p orbitals then the atom is sp hybridized, and an aromatic pi system necessarily means sp2 hybridization
 

McNCheese

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I've never heard of an exception to that, but I think I can think of one.

Azetes http://en.wikipedia.org/wiki/Azete

I think you could expan on that rule by saying that it will only donate it's lone pair to gain aromaticity if the heteroatom isn't already involved in a pi bond

I think this rule works because when the heteroatom participates in the pi bond, that pushes the electrons perpendicular to the delocalized pi electrons of the system.
 

inaccensa

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How does Pyridine satisfy huckels rule?

I understand that benzene has 6 pi electrons because it has 3 double bonds.

When I look at pyridine I see 3 double bonds, plus a lone pair on the Nitrogen. Do you only consider the 3 double bonds pi electrons?

If so then why is pyrrole aromatic? It only has 2 double bonds in it structure.

wouldn't pyridine be anti-aromatic? Since Nitrogen has a lone pair and is sp2 hybridized (this is an exception, any atom with one or more lone pair attached to a sp2 hybrized is also sp2 hybridized). Now nitrogen will place the lone pair in its unhybridized p orbital, since it is sp2 hybridized and has a vacant p unhybridized p orbital. Thus the total # of pi electrons boils down to 8 which gives us anti-aromatic compound.
 

loveoforganic

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Did you not read any of the replies to his question at all?
 

Omni

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You only count the lone pairs in order to satisfy huckel's number. If the molecule is all ready 2n+2 and has lone pairs, don't count them. If the molecule is just 2n and DOES have lone pairs, you count them in.
 

inaccensa

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You only count the lone pairs in order to satisfy huckel's number. If the molecule is all ready 2n+2 and has lone pairs, don't count them. If the molecule is just 2n and DOES have lone pairs, you count them in.

Thanks omni. That would make sense,since I see different answers!!!