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Offline Cooper

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Determining Aromaticity
« on: February 07, 2013, 03:53:55 PM »
Hi, I am working on a problem and I am confused as to why my answer is not correct. I've posted a picture of the problem below, it is asking to determine which ions are aromatic.

I know i is not aromatic, because it only has 4 pi electrons, four is not a Huckel number. This is the same for iii.

ii has 6 pi electrons, which is a Huckel number (4n + 2; n = 1). But when I draw the polygon-and-circle test, it seems that four of its electrons lie in nonbonding orbitals. This means it is not aromatic.

iv has 6 pi electrons, which is a Huckel number and it passes the polygon-and-circle test. (I didn't draw that because it would be a pain on paint).

So I determined that only iv. is aromatic, but that's not an option!!  :-[ I am thinking the answer would be ii and iv because I know i and iii definitely aren't right, but I don't know why ii would be either.

Any suggestions?

Thanks!
~Cooper :)

Offline Cooper

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Re: Determining Aromaticity
« Reply #1 on: February 07, 2013, 05:44:05 PM »
Also, can someone explain to me why antiaromatic compounds are so unstable? I get that antiaromatic compounds are less stable than their aromatic counterparts because they don't have electron delocalization. But how come the cyclopentadienyl carbocation is even less stable than a cyclopentanyl carbocation? Isn't the carbocation spot on  a cyclopentadienyl cation an allylic spot? Wouldn't this mean that the compound is stabilized by resonance?

Thanks!
~Cooper :)

Offline Dan

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Re: Determining Aromaticity
« Reply #2 on: February 09, 2013, 12:53:47 PM »
But when I draw the polygon-and-circle test, it seems that four of its electrons lie in nonbonding orbitals. This means it is not aromatic.

I'm not sure that conclusion is valid. I thought to be aromatic the requirement is that you have a closed valence shell (i.e. all electrons paired - which you have.

Quote
how come the cyclopentadienyl carbocation is even less stable than a cyclopentanyl carbocation?

Do you have a source for that? I'm surprised by it as well.
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Offline Cooper

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Re: Determining Aromaticity
« Reply #3 on: February 09, 2013, 01:08:34 PM »
Thanks for the reply Dan. Hm, well then what's the point of the circle-and-polygon test? In my book it says if there are electrons in the nonbonding orbitals, it's not aromatic.

And the question with the cyclopentadienyl cation is from my homework. I'll post a pic below.

Thanks for your h.e.l.p!

~Cooper :)

Offline Dan

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Re: Determining Aromaticity
« Reply #4 on: February 09, 2013, 02:01:17 PM »
And the question with the cyclopentadienyl cation is from my homework. I'll post a pic below.

After a bit of poking in the literature, it certainly is true that solvolysis of 5-chlorocyclopentadiene is far slower than chloropentane.

I'm not sure I can come up with a good explanation. I guess that to avoid antiaromaticity there would be a distortion such that the cation does not overlap with the π-density of the alkenes - i.e. it is not resonance stabilised as in the case of 3-chlorocyclopent-1-ene. In comparison to the cyclopentyl cation, which is flanked by sp3 carbons, the cyclopentadienyl cation is flanked by sp2 carbons - these are inductively withdrawing compared to sp3 centres.

I'm not confident about this, it seems like a weak, hand-wavy argument to me. I suspect there is a good MO-based explanation - hopefully someone else can throw something in?
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Offline Cooper

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Re: Determining Aromaticity
« Reply #5 on: February 10, 2013, 11:05:30 AM »
I was talking to one of my classmates and he pointed out that the cyclopentadienyl cation would have two unpaired electrons in its orbitals. No wonder it's the least reactive in solvolysis; the cation is so unstable.

I think with the original question I have, I may have been misinformed. Can a compound still be aromatic if it has a full electron orbital in a non bonding orbital? It says to assume the compound is flat...would that affect it?
~Cooper :)

Offline fledarmus

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Re: Determining Aromaticity
« Reply #6 on: February 11, 2013, 09:12:20 PM »
I'm not sure exactly what you're asking when you say "Can a compound still be aromatic if it has a full electron orbital in a non bonding orbital?"

Are you asking if a lone pair of electrons in an orbital can participate in aromaticity? If so, the answer is yes. Look at all the aromatic nitrogen- and oxygen-containing ring systems.

And that is essentially what you have in your cyclobutene dianion - each of the two carbanions has a lone pair in a p-orbital which can overlap with the pi bond of the alkene, giving six electrons in a conjugated system. You would have the same electron populations if you replaced each carbanion with a neutral nitrogen atom, which might be easier to visualize.


Offline Cooper

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Re: Determining Aromaticity
« Reply #7 on: February 11, 2013, 11:35:55 PM »
I'm not sure exactly what you're asking when you say "Can a compound still be aromatic if it has a full electron orbital in a non bonding orbital?"

Are you asking if a lone pair of electrons in an orbital can participate in aromaticity? If so, the answer is yes. Look at all the aromatic nitrogen- and oxygen-containing ring systems.

And that is essentially what you have in your cyclobutene dianion - each of the two carbanions has a lone pair in a p-orbital which can overlap with the pi bond of the alkene, giving six electrons in a conjugated system. You would have the same electron populations if you replaced each carbanion with a neutral nitrogen atom, which might be easier to visualize.

Sorry, I meant "can a compound be aromatic if it has electrons in non bonding orbitals?"
~Cooper :)

Offline orgopete

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Re: Determining Aromaticity
« Reply #8 on: February 12, 2013, 06:04:26 AM »
Do you mean aromatic compounds like furan or pyrrole in which the additional electrons to total six are contained by the heteroatom?
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Offline Cooper

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Re: Determining Aromaticity
« Reply #9 on: February 12, 2013, 10:35:56 AM »
I mean like in my example above with the cyclobutadenyl dianion. It has electrons in non bonding orbitals, so if I was wondering if that would make it antiaromatic. (My textbook made it seem such.) I asked my professor today, though and he said no. :)
~Cooper :)

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