Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: asascl1992 on November 05, 2009, 01:38:26 PM
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Hello I am having trouble with this one homework question and was hoping you guys could be of assistance.
The questions asks "Assuming that all rings are planar, which molecule is antiaromatic?"
The possible answer choices are shown below:
http://img233.imageshack.us/img233/6743/20178302.jpg
From our class, I learned that for a molecule to be antiaromatic, the molecule must be planar, fully conjugated, and have a cyclic pi system with 4N pi electrons. Under those conditions, I thought B was the correct answer since it contains 4 pi electrons in a cyclic pi system. However, the answer is actually E and I have no idea why. I thought E was actually non-aromatic since it doesn't even have a fully conjugated pi system. Any assistance is greatly appreciated!
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Assuming that E is planar, then it can be antiaromatic. Since the boron has an incomplete octet and the neighboring nitrogen has a pair of electrons to share, you can draw a double bond between these atoms (though formal charges will exist on the boron and nitrogen, that won't matter). Now the structure will have 4 double bonds or 8-pi electrons, so it will be antiaromatic.
However, just as cyclooctatetraene would be antiaromatic, if planar, it is actually non-aromatic. Because compounds that are antiaromatic are higher in energy, if a structure can relieve unfavorable interactions by conformational changes, they probably do so.
Compound B is non-aromatic as the nitrogen is locked into an sp3 hybridization and thus cannot participate in any resonance structures.
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Thank you! I had two follow up questions to make sure I understand this correctly:
1) would a compound like cyclooctatetraene go from antiaromatic to nonaromatic by then taking pi bonds out of conjugations by bending them upwards or downwards to kind of create a boat-like shape?
2) Am I correct to assume if any element in a cyclic compound is sp3 hybridized, then the compound will be nonaromatic because that part of the compound can not participate in the pi system?
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yes to both
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Thank you! I had two follow up questions to make sure I understand this correctly:
1) would a compound like cyclooctatetraene go from antiaromatic to nonaromatic by then taking pi bonds out of conjugations by bending them upwards or downwards to kind of create a boat-like shape?
2) Am I correct to assume if any element in a cyclic compound is sp3 hybridized, then the compound will be nonaromatic because that part of the compound can not participate in the pi system?
1. Yes
2. No. Consider pyrrole, furan, and thiophene. These are all highly aromatic compounds. The heteroatom in each of these systems is "sp3 hybridized" using our conventional methods of determining hybridization. However because of delocalization of these lone pairs the heteroatom-carbon bonds have partial double bond character. This occurs any time a lone pair is delocalized with a pi bond. Just like cyclopentadienyl anion (conventionally containing an sp3 hybridized carbon) is aromatic and thus a highly stable anion.
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The heteroatom in each of these systems is "sp3 hybridized" using our conventional methods of determining hybridization.
I'm not sure this is constructive way of looking at the problem. I would say it is conventional to consider resonance stability when determining the hybridisation of an atom. The N in pyrrole is not sp3 hybridised since sp2 hybridisation confers additional aromatic stability. If the ring contains an element that cannot adopt the hybridisation required to form a planar pi system (for example, it is forced to be sp3 as in compound b in the original question), then it cannot be aromatic.
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The heteroatom in each of these systems is "sp3 hybridized" using our conventional methods of determining hybridization.
I'm not sure this is constructive way of looking at the problem. I would say it is conventional to consider resonance stability when determining the hybridisation of an atom. The N in pyrrole is not sp3 hybridised since sp2 hybridisation confers additional aromatic stability. If the ring contains an element that cannot adopt the hybridisation required to form a planar pi system (for example, it is forced to be sp3 as in compound b in the original question), then it cannot be aromatic.
You are correct. Most undergraduates are taught a specific way to figure out hybridization of atoms and using that approach (what I mean by conventional) it leads them to think that, for instance, any N with three bonds and a lp is sp3 hybridized.
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Re: sp3 hybridization
Everyone is correct on this. I did not use a correct term to explain why the quaternary nitrogen made Compound B non-aromatic. While sp3 in this case did render it non-aromatic, that term does not render nitrogen atoms non-aromatic in principle. I was attempting a short explanation that all atoms must be part of a resonance system, but that was not it.