Topscore aromaticity question

[jheidenr]

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Can anyone explain this to me? I thought that an anti-aromatic compound is a cyclic, conjugated polyene that posseses 4n electrons. Topscore says the answer is D, I though it should be C (Kaplan even uses answer C in their book as an example of an anti-aromatic compound....am I missing something?):

 

[weiy01]

Can anyone explain this to me? I thought that an anti-aromatic compound is a cyclic, conjugated polyene that posseses 4n electrons. Topscore says the answer is D, I though it should be C (Kaplan even uses answer C in their book as an example of an anti-aromatic compound....am I missing something?):

Yeah, I thought this was a bad question.

Answer "A" is not conjugated therefore it is antiaromatic.
"B" is definitely aromatic.
Both "C" and "D" do not satisfy Hukel's therfore it is antiaromatic.
"E" is not cyclic therefore it is antiaromatic.

To me, everything is antiaromatic but "B."

😡

 

[mg777]

to be antiaromatic the compound must meet all the requirements of an aromatic, EXCEPT the number pi electrons is equal to 4n INSTEAD OF 4n +2. This still makes this a bad question, compound C is hypothetical and does not exist but if it did it would be antiaromatic all day long. It has 8 pi electrons, which is equal to 4(2) or 4n. Compound D also fits this criteria since it has 4n pi electrons or 4(3)=12, however I beleive this compound is aromatic. Regardless, i m 99% sure you wont have a question asking you which is ANTIAROMATIC on the DAT.

 

[sinned]

you know topscore can make mistakes as well...humans made that test afterall, C is definately antiaromatic

 

[ems5184]

Is there a difference between anti-aromatic and something just not being aromatic? If so, what is/are the difference(s)?

 

[ScottW3]

I thought the structure in C isn't planar. I vaguely remember reading that it would form a sort of boat configuration. This may be the reasoning behind Topscore's answer.

 

[jheidenr]

Is there a difference between anti-aromatic and something just not being aromatic? If so, what is/are the difference(s)?

Yes, I think mg777 put it pretty clearly:

to be antiaromatic the compound must meet all the requirements of an aromatic, EXCEPT the number pi electrons is equal to 4n INSTEAD OF 4n +2.

An non-aromatic compund either isn't planar, isn't cyclical, or doesn't have the right amount of pi electrons

 

[ScholarWare]

Kaplan does not cover aromaticity completely. It is just a high level review. Here is the solution. Make sure you study from your organic chemistry text book also. Good luck.

T1, OC, Q98 [Solution]
Aromaticity. In order for a molecule to be aromatic or anti-aromatic it must be cyclic, and closed-conjugated (every other double bond). This rules out answers A and E. B is aromatic. Using Huckel?s rule, aromatic compounds have 4n + 2 pi electrons (where n is an integer) and anti-aromatic compounds have 4n. Yet, C does have 8 pi. However, the molecule itself is not planar and therefore the pi electrons do not interact. C is a stable molecule and behaves as a normal alkane. Most textbooks classify it as non-aromatic which is consistent. Therefore the best answer is D, cyclobutadiene which is anti-aromatic.

 

[SloppyJoe]

is there a way to get the solution for the topscore test?

 

[jheidenr]

No, you can find the correct answer on the test details, but your on your own to figure out why you got somethign wrong. I guess that has its pros and cons. Most notably, you will learn more by researching the topic and figuring out where you went wrong...but it can be very frustrating when you get stuck.

 

[mg777]

Yet, C does have 8 pi. However, the molecule itself is not planar and therefore the pi electrons do not interact. C is a stable molecule and behaves as a normal alkane. Most textbooks classify it as non-aromatic which is consistent. Therefore the best answer is D, cyclobutadiene which is anti-aromatic.

The molecule in C, cyclooctatetraene, is drawn in a hypothetical planar fashion making it anti-aromatic along with the other antiaromatic (d). This problem should be corrected on future versions of Topscore Pro to eliminate confusion for Scholware customers.