aromaticity

[gomawum]

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View attachment aaa.JPG

why is this non-aromatic?
I thought they all have sp2 orbitals but huckle's law fail, so isn't it anti-aromatic?

because if it has 2 e- instead of a radical(1 e-), it's aromatic,
so I assumed that the radical e- is in the sam plan along with other 2 double bonds e-.

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i need to clarify this....

allantois has function of 'fetus respiration, excretion'
chorion has function of 'gas exchange b/t mom and fetus'
but later chorion become placenta and have function of 'respiration, exchange gas'
and allantois and yolk sac combine to form umbilcal cord to transfer 'nutrients, waste' from fetus

 

[raiden117]

So I'm not sure about this, but as far as I know, anti-aromatic compounds have only 4n pi electrons(instead of 4n+2 for aromatic compounds). In this case, there are a total of 5 pi electrons, which makes it non-aromatic despite the fact that it is planar and all atoms have p orbitals. Anyone agree?

 

[burgler09]

Actually it is NON-Aromatic, not anti aromatic. This is because it has 5n electrons. The radical counts as a single electron. If it had a lone pair of electrons that would make it sp3 hybridized.

 

[gomawum]

Actually it is NON-Aromatic, not anti aromatic. This is because it has 5n electrons. The radical counts as a single electron. If it had a lone pair of electrons that would make it sp3 hybridized.

I don't get it.
if it doesn't fit in huckle's rule (4n+2), it's supposed to be anti-aromatic as long as all the carbons on the cycle are conjugated sp2 and are all planar.

so you are saying that radical has sp3 character?
I thought it's sp2 character..

because one of the problem in topscore,
View attachment aaa.JPG

it said this was aromatic....

and i'm lost again..

 

[hoss19]

for aromaticity, go thru this checklist:

1. is it cyclic?
2. is it planar?
3. are all ring atoms sp2 hybridized?
*4. does the total number of pi electrons = 4n+2 where n=any interger?

if 1-4 are all yes, then molecule is aromatic.
if 1-3 are yes, but 4 is no then check to see if molecule has 4n pi electrons. If it does, then the moelcule is anti-aromatic. If it doesnt, then the molecule is just not aromatic.

gomawum: the first moelcule you posted (radical cyclopentadiene) is just not aromatic (or in burglars lingo, non-aromatic). Criteria 1-3 are all "yes" but it has 5 pi electrons. The second one that you posted (di-radical cyclobutene) is equivalent to cyclobutadiene because the radicals, if they were brought together, would yield a pi bond giving the molecule two double bonds for a total of 4 pi electrons. Thus, because criteria 1-3 are true and the molecule has 4(1) = 4 electrons, the moleule is anti-aromatic. Remember that RADICALS are sp2 hybridized (the p-orbital is esstentially empty b/c it only has 1 electron in it and therefore cannot be sp3). CARBOCATIONS are also sp2 hybridized.

hope this helps

 

[klutzy1987]

I don't get it.
if it doesn't fit in huckle's rule (4n+2), it's supposed to be anti-aromatic as long as all the carbons on the cycle are conjugated sp2 and are all planar.

so you are saying that radical has sp3 character?
I thought it's sp2 character..

because one of the problem in topscore,
View attachment 10057

it said this was aromatic....

and i'm lost again..

Oerhaps those electrons are on the outside and we dont count them. They are not in the same plane as the ring??? This may be why it is aromatic but im not 100% sure.

 

[bigstix808]

I don't get it.
if it doesn't fit in huckle's rule (4n+2), it's supposed to be anti-aromatic as long as all the carbons on the cycle are conjugated sp2 and are all planar.

1)not necessarily... anti-aromatic in only when there are 4, 8, 12, 16, etc, electrons (e=4n) in the pi orbital system when all other aromaticity rules hold true.
2) non-aromatic compounds are every other molecule that doesn't fit into eiter the aromatic, or anti-aromatic catagory

so you are saying that radical has sp3 character?
I thought it's sp2 character....

they are sp2 hybrids

because one of the problem in topscore,
View attachment 10057

it said this was aromatic....

and i'm lost again..

i'm not sure what topscore's reasoning is behind this question, but i went and asked an organic prof and she said 1) that molecule would never exist and 2) as hoss put it, those two radical electrons would reform a pi bond making this molecule anti-aromatic (4n=4; n=1).

Oerhaps those electrons are on the outside and we dont count them. They are not in the same plane as the ring??? This may be why it is aromatic but im not 100% sure.

also, if assuming that those electrons are on the "outside", making C3 and C4 sp3 hybrids, and thus "don't count", that would still not make this compund aromatic b/c then it is no longer conjugated as it would not be possible (2 adjacent sp3 hybrids breaks the pi orbital linkage and kills the conjugation)

 

[klutzy1987]

Forget what i said before, I am pretty sure that whereever yo go the problem from is wrong and i wouldnt give that source much credit. I a 99% sure that the molecule is not aromatic. I asked a very credible source and the source told me that the questoin is rediculous and the structure is not aromatic.

 

[Dencology]

oh, just to let you know that no one else picked this up, this question in Topscorere has been drawn wrong. if you read the explanation carefully, you will see that it said choice D is Cyclobutadiene which makes it with four pi electron and there should be a two double bonds. so the total number of double bonds are two and 4pi electron with falls under the category of antiaromatic.
choice C is cyclic, conjugated, sp2 but not planer so that is why it is not aromatic.
if it is cyclic, congugated, sp2 and if it has a 4n pi electron then it is not aromatic. if it does not then it is antiaromatic. please let me know what you think of this explanation.