htkim6680

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Apr 5, 2008
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Need help with huckles rule. Been a while since I took ochem.

I understand the 4n+2 deal, but how do you know if an electron is conjugated or not? Is there a simple rule?
 

Dannelid

7+ Year Member
Jun 9, 2009
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Basically, if you see two double bonds separated by a single bond or a cation/anion on the carbon adjacent to a double bond, it will be conjugated.
 
Jul 11, 2009
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Basically, if you see two double bonds separated by a single bond or a cation/anion on the carbon adjacent to a double bond, it will be conjugated.
Then if the cation/anion is conjugated does that mean that you count it as a pi orbital in Huckles rule? I always get confucsed with aromaticity whenever they start throwing N/O/+/- into these rings.
 

Dannelid

7+ Year Member
Jun 9, 2009
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well, any conjugated system uses p orbitals, but I think you're thinking of pi electrons, which is what huckel's rule talks about.

Because a cation has lost electrons, the + charge wouldn't really take part in the 4n+2 rule.

An anion, however, would take part, because it has an electron pair in a p orbital.



Take pyrazole, for example. If the proton on N-1 were removed, you'd have an anion, and there would be an electron pair in a p orbital perpendicular to the plane of the molecule, which would count as a pi electron, bringing the total to 6 pi electrons, making the anion aromatic.

note: you wouldn't have to remove the H from the N for pyrazole to be aromatic because whether or not its an anion, it has an electron pair (not shown in the structure given) in a p orbital. So this is a case where both the pyrazole and its anion are aromatic
 
May 22, 2009
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From what I understand, you gotta look at whether the e-pair is perpendicular to the plane of the ring or in the plane. For example, if the N or O have a double bond attached to them, they're in SP2 hybridization and the e-pair is gonna be in the plane of the ring and won't conjugate with the other pi electrons, so it doesnt count (The right N in the image above), but if they have single bonds, and are thus sp3 hybridized (like the left N in image above), the e-pair is perpendicular to the plane of the ring and so can take part in conjugation and stabilization.

Hope that makes sense ....