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o chem question
Started by ToothFairy423
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http://www.tamug.tamu.edu/mars/chem227/aromatics.htm
According to huckel's rule to be a aromatic a molecule must be conjugated and contain 4n + 2 pi electrons. Cyclooctane has 8 pi electrons and it is conjugated so it is antiaromatic (4n, n=2) pi electrons as opposed to aromatic 4n+2 pi electrons. Compounds that are non-conjugated are non-aromatic.
n must always be an integer.
According to huckel's rule to be a aromatic a molecule must be conjugated and contain 4n + 2 pi electrons. Cyclooctane has 8 pi electrons and it is conjugated so it is antiaromatic (4n, n=2) pi electrons as opposed to aromatic 4n+2 pi electrons. Compounds that are non-conjugated are non-aromatic.
n must always be an integer.
DrTacoElf said:
so it is safe to assume that topscore is wrong"???
ToothFairy423 said:
I'm pretty sure this question #98 on test 1 has been rewritten, that is that's what the scholarware site says http://www.scholarware.com/DATFix07.x.htm
But if it the question I'm thinking of, answer C is cyclooctatetraene, sure it has 8 pi electons, which, according to Huckle should make the molecule antiaromatic. But the molecule itself is not planar (it forms kind of a cube shape) and therefore the pi electrons do not interact. This molecule is stabe, and behaves as a normal alkane. My text classifies it as non-aromatic which is consistent with its chemical properties. It's not antiaromatic and why their website has this question updated.
So to answer your question, I believe the question has been updated.
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joy2dmd said:
wow thanks alot you have been so insightful
how do you know that it is a cube shape?" is this just something you memorized... i hope its stuff like this, that i dont know, and have no way of knowing, that doesnt mess up my score!
ToothFairy423 said:
No, I just looked up why answer C was not antiaromatic. It was an exception so I was curious. Topscore was great, like everyone says.. so when I missed a question, I looked it up. I think i was puzzled or asked scholarware. i think they sent me the solution to this one.
