Why is aniline aromatic?

[fluff]

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Why is aniline aromatic according to Huckel's rule? When I draw the structure out, I see six pi electrons around the benzene ring, and 2 pi electrons on the nitrogen (which is sp2 hybridized). This gives me a total of 8 pi electrons.

With 8 pi electrons, I find that aniline is antiaromatic. Where am I going wrong??

 

[sv3]

scratch that. Now I'm confused dammit. Perhaps the aromatic rules (cyclic, planar, huckel) only apply to atoms that are actually part of the ring...and not substituents?

 

[chemtopper]

You have to count no of pi electrons hybridization of ring atoms and not on the substitute .Huckel rule is applicable only delocalized pi electrons on the ring atoms .
NH2 is bonded to carbon of the benzene ring and here N is not atom of the ring.

One more point
N of the aniline is sp3 hybridized.

 

[loveoforganic]

I'm not sure I'd agree with that N being sp3 hybridized.. If that were the case then an amino substituent on a aromatic ring would be deactivating, while it is in fact incredibly strongly activating.

 

[rocketbooster]

the amine group on aniline is sp3. the lone pair of electrons counts as an electron dense region. hybridization is not only based on the # of bonds.

does the lone pair of electrons from the amine take part in the conjugated pi-system of the benzene ring? No, so they aren't counted. if the N had been included inside of the ring itself, then its lone pairs would have counted.

so, in other words, what chemtopper said. 😉

 

[sv3]

aromatic rules only apply to atoms that make up the ring ya?

I was confused as I knew NH2 can activate the ring and thus its e's can become delocalized, so wasn't quite sure.

 

[loveoforganic]

the amine group on aniline is sp3. the lone pair of electrons counts as an electron dense region. hybridization is not only based on the # of bonds.

does the lone pair of electrons from the amine take part in the conjugated pi-system of the benzene ring? No, so they aren't counted. if the N had been included inside of the ring itself, then its lone pairs would have counted.

so, in other words, what chemtopper said. 😉

How do you account for the activating effects of the amino substituent then? If the lone pair of the amine group of aniline was in an sp3 orbital, it would not be able to interact with the ring to increase electron density in the ortho and para positions. This can only happen if the lone pair is in a p orbital.

 

[G1SG2]

How do you account for the activating effects of the amino substituent then? If the lone pair of the amine group of aniline was in an sp3 orbital, it would not be able to interact with the ring to increase electron density in the ortho and para positions. This can only happen if the lone pair is in a p orbital.

It does interact with the ring in the arenium ion through resonance, but does not participate in the conjugated pi-system of the benzene ring for aromaticity.

 

[sv3]

when it donates its' electrons? Or is it always Sp2?

 

[loveoforganic]

Hm, I see what you're saying, and after a bunch of googling, I was starting to lean toward agreeing with what you were saying, but I decided to check the 13C/1H nmr spectra first.

http://riodb01.ibase.aist.go.jp/sdbs/cgi-bin/direct_frame_top.cgi
http://riodb01.ibase.aist.go.jp/sdbs/cgi-bin/direct_frame_top.cgi

The ortho/para carbons and the protons on those carbons are more shielded in both the 13C and 1H nmr spectra, which only makes sense to me if the nitrogen in aniline is sp2

 

[sv3]

NH2 on its own or connected to a non-aromatic group would be SP3. I think once it donates its electrons, then it becomes SP2. I don't see why this isn't the answer? (sort of akin to a carbocation changing its hybirdization). I think I'm just going to hope not to see this on the mcat....which means i will see it of course.

Besides, physics is manhandling me so I'm gunna get to that.....have fun here!

sv3

 

[loveoforganic]

It's not a 'once it donates its electrons' thing, if we're dealing with resonance structures.

 

[sv3]

for when an atom participates in resonance? Don't its non-bonded electrons move? What's that called technically? resonate?

But your right, donation is a term meant for lewis bases....thanks.

 

[loveoforganic]

Fundamental concept: Resonance structures are just a depiction we use to gain info about a compound. In actuallity, the compound that exhibits resonance is a hybrid of all of its resonance structures.

 

[fluff]

if you draw out the structure of aniline, the nitrogen is bonded to three atoms--one of the carbons in benzene and the two hydrogens. the nitrogen also has a lone pair. with three bonds and the lone pair, it would normally be sp3 hybridized.

however, the exception to this rule is that 1) if the nitrogen (or any other atom in question for that matter) has a lone pair and 2) that nitrogen is bonded to an sp2 hybridized atom, then the nitrogen is sp2 hybridized.

so i guess we are back to square one... why is aniline an aromatic amine?

 

[loveoforganic]

It's not an aromatic amine - it's an amine with an aromatic substituent. The phenyl substituent is aromatic, the molecule as a whole is nonaromatic. That's my opinion, at any rate. For example, I'd call pyrrole or pyridine an actual aromatic amine. For aromaticity you only consider the outside ring. If you consider pyrene, for instance, you count 16 pi electrons total within the ring system, but it's aromatic. This is because the double bond in the very center of the molecule is nonaromatic, and this can be proven if you attempt to hydrogenate pyrene - You will add one molar equiv of H2 very quickly.

 

[chemtopper]

I'm not sure I'd agree with that N being sp3 hybridized.. If that were the case then an amino substituent on a aromatic ring would be deactivating, while it is in fact incredibly strongly activating.

NH2 on aromatic ring is in Sp3 when it has its electrons intact on it.When it undergoes resonance with pi electron system of the ring at that time it acquires sp2 hybridization.
Group is a strong activator because it can give its lone pair by resonance to the ring and at that time N acquires a positive charge on it and forms a double bond with ring carbon.
We show aniline structure with lone pairs, so I am right that N is sp3 hybridized.We don't show aniline with positive charge on N and N=C of the ring.Do we???
So I am right that N of aniline is sp3 hybridized .

 

[loveoforganic]

We also don't typically draw ketones as enols, nor do we draw the lesser resonance structure as the typical image, but hey, it happens. Could you explain the NMR spectra? Also, the only way the nitrogen can donate it's lone pair into the ring is if that lone pair is in a p orbital.

 

[Adamska]

^we do draw ketones as enols when the enol form is preferred according to reaction conditions.

 

[loveoforganic]

Even under acidic/basic conditions, I believe the tautomerization just becomes faster - equilibrium isn't shifted. When we draw a ketone in its enol form, it's typically because that's the reactive species, not necessarily the favored species. That's beside the point though.

 

[Adamska]

If the enol tautomer give an aromatic structure, for instance, then it's preferred.

 

[rocketbooster]

haha, this is a good question btw!!

i Just took my MCAT and there was a question on aromaticity pertaining to where the N was positioned on the ring! honestly, I think I missed it but....hahaaha yea...know this!

 

[loveoforganic]

If the enol tautomer give an aromatic structure, for instance, then it's preferred.

Good example^^ hadn't thought of that.

 

[kia ora]

You have to count no of pi electrons hybridization of ring atoms and not on the substitute .Huckel rule is applicable only delocalized pi electrons on the ring atoms .
NH2 is bonded to carbon of the benzene ring and here N is not atom of the ring.

One more point
N of the aniline is sp3 hybridized.

yep - you only count the pi electrons in the ring (6). 6 = 4n + 2, so n=1 and thus aniline is aromatic. rocketbooster, I know the recent mcat question to which you refer in your post. without getting into specifics, in general I will say: if the N is in the ring, then its lone pair will count towards the pi electrons used for Huckel's rule ONLY if that lone pair is needed to make the molecule aromatic. (ie, if n=0, 1, 2... without the lone pair, then don't count the lone pair; if the lone pair is needed to satisfy Huckel, count it). essentially, if a molecule can be aromatic it will be. and if there is an sp2 carbon in the ring, then the molecule is not aromatic

I 😍 o-chem!!!😍

 

[sv3]

Hi Kia Ora,

Below, did you mean SP3? Nice explanation btw

sv3

yep - you only count the pi electrons in the ring (6). 6 = 4n + 2, so n=1 and thus aniline is aromatic. rocketbooster, I know the recent mcat question to which you refer in your post. without getting into specifics, in general I will say: if the N is in the ring, then its lone pair will count towards the pi electrons used for Huckel's rule ONLY if that lone pair is needed to make the molecule aromatic. (ie, if n=0, 1, 2... without the lone pair, then don't count the lone pair; if the lone pair is needed to satisfy Huckel, count it). essentially, if a molecule can be aromatic it will be. and if there is an sp2 carbon in the ring, then the molecule is not aromatic

I 😍 o-chem!!!😍

 

[loveoforganic]

Yes, sp3. And to clarify a little bit what kia said - if a heterocycle is double bonded to the heteroatom, the lone pair cannot take part in the resonance, as it doesn't lie in the correct orbital. You can only consider using a lone pair to gain aromaticity if the heteroatom isn't double bonded within the ring system.

I feel like that didn't come out very clearly - if that's the case I can try and draw a picture or something.

 

[kia ora]

oops, I did mean sp3 - typing FAIL! thanks!

and loveoforganic, your explanation seems crystal clear - thanks for clarifying what I meant to say...yep, if the N is in the ring and is singly bonded to each adjacent ring carbon, then you count its lone pair as pi electrons if they are needed to achieve aromaticity. if it's double bonded to one of the ring Cs, you can't, as you very clearly explained

 

[VanBrown]

just as a bit of friendly advice... you should know ALL of the conditions for determining aromaticity.

1_ planar molecule
2_ fully conjugated ring
3_ obeys Huckel's rule 4n+2, know how free electrons work.

 

[VanBrown]

oh and anilline is aromatic because the lone pair on the N group is outside of the ring structure.. therefore they do not play a role in Huckel's Rule.