Aromaticity question

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Aromaticity question

Chemistry
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Last one got cut out choice D second question:
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For the second question which one of those isn't aromatic
Are you familiar with hückles rule?
Sorry @Photon336 i was at work all day! So for the second one (the ones with the crown compounds) if you look at the Huckel's rule which is 4n+2 the one that doesn't follow is B which follows the 4n rule so it's non aromatic.
As for the first I'm stuck. A follows the 4n rule but I looked it up and it said that it was aromatic so I don't really know for that one. B I'm pretty sure that one is aromatic as well but it could be that one. C I don't think it's that one cause it has the overlapping p orbitals because I'm assuming at the top of the point is a pair of lone electrons. And D follows the 4n+2 rule unless I managed to count wrong which I definitely could have.
hey I am totally new to this . But can I try one? For te 1st one I think it is the answer A is it correct?
@Rushwr im pretty sure A is the correct answer as it doesn't follow Huckel's rule of 4n+2 it goes by the 4n rule meaning it's non aromatic
Yeah that's what I was thinking too. I still haven't learned these!
http://openstudy.com/users/taramgrant0543664#/updates/55a9d8e5e4b038ffab0bb4f7 I made a small tutorial about aromaticity if you want to check it out
Thank you so much @taramgrant0543664
No problem!!!
We can rule D out for sure right away because it follows 4n+2 rule, it's cyclic planar, and fully conjugated system.
I think C would be aromatic because 4n+2 n =1, and those electrons participate in the pi system. also that's cyclic conjugated and planar.
|dw:1437845946077:dw|
I feel like it's between A and B
Ya I was stuck between A and B but I know that doesn't follow the 4n+2 rule
A doesn't follow the 4n+2 rule
@taramgrant0543664 @Rushwr I just looked it up answer is A
So then B follows the 4n+2 rule so I need to go over examples of aromatic compounds with charges on them since they are a little tricky
Awesome I hope I have the guessing ability ! anyways thank you @Photon336 cuz I learned alot here @taramgrant0543664 Thank you for teaching more than I knew !
"Hukels rule cyclic conjugated polyene may be aromatic if there are 4n+2 pi electrons. this is true for compounds b,c,d but compound A contains only a 4n pi system and is therefore not aromatic"
I guess if something has 4n pi electrons it's considered not aromatic even if it is cyclic, conjugated, that's what threw me off
but in your tutorial you said something about ANTI aromatic, that was a compound with 4n pi electrons right?
It's tricky with the charges that's what got me questioning and whatever site told me that A was aromatic needs to fix their site
So aromatic compounds An aromatic molecule is: Flat, Cyclic, Completely conjugated and the number of π electrons must obey Huckel’s rule of 4n+2. An anti-aromatic molecule is much like the aromatic compound in that it is also Flat, Cyclic, and Completely conjugated but the difference is that the number of π electrons follows the formula of 4n instead of the 4n+2. A non-aromatic molecule is: NOT flat, NOT cyclic, and NOT completely conjugated. Some examples include the typical alkane and/or alkene such as butane or hexane.
A is anti aromatic not nonaromatic
The book never made reference to that .... i think that's the issue it said not aromatic it's kind of ambiguous
for the second one I posted,the answer is B. because it has 4n pi electrons not 4n+2
So there is a chart like thing that you could follow my computer won't let me load it but if you search aromatic nonaromatic anti aromatic it should pop up and it goes like if it follows this rule follow this arrow go to this box if it doesn't go to this box
for choice D |dw:1437846783821:dw| Just another question. For something to have lone pairs but participates in the pi system does it have to be sp^2 hybridized atom? I think there was a formula for the hybridization wasn't sure what it was: that sulfur atom is number of atoms bonded + Lp 2 atoms + 2Lp -1 = sp^3
There are 6 pi electrons. One of the lone pairs on the sulfur is in the same plane as the two double bonds.
ah ok I see
ever hear of this? found this formula for getting the hybridization of an atom 1/2 ( V + M - C + A ) V = Number of valence electrons M = Monoatomic atoms bonded to central atom C = Cationic charge A = Anionic charge
Never heard of that before
Seems like it should be YMCA lol!!!!
I just found it right now lol. so for the sulphur atom 6 valence electrons 2 oxygens bonded to it No charges positive or negative so it be comes (6+0+0+2)(1/2) = 4 and 4 corresponds to sp3 hybridization. 2=sp 3=sp2 4=sp3 5=sp3d 6=sp3d2 7=sp3d3
ahahaha
you'll never forget it with that mnemonic
So when it says charges how would that work? In the previous example if sulfur had a charge it would count but what happens if oxygen had a charge?
Haha I know!!
well if oxygen had a +1 charge you would add +1 if something had a -1 charge you would subtract one from the total.
Ok I think I understand it I'll have to look it up I've never heard of that before but if it works then it can make things easier!
thing is I didn't either, just found it one a site, but it's like usually they ask questions such as what's the hybridization of this or that atom and it's kind of unclear
Ya for sure

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