[G O D I V A]

Which of the following are aromatic compounds? (Just number them 1,2,3,4 starting from the left)

Using Huckle's rule:

1 has 4π electrons and does not fit the (4n+2 rule) so eliminate that?
2 is not planar so eliminate that?
3 has 4π electrons and does not fit the (4n+2 rule) so eliminate that?
4 from the double bonds it has 8π electrons and does not fit the (4n+2 rule) so eliminate that?

So none of them would be aromatic?

Also would 4 be basic because of the nitrogens and being cyclic?

[Dan]

1 has 4π electrons and does not fit the (4n+2 rule) so eliminate that?

That depends on the hybridisation of the oxygen atom... I disagree with you.

2 is not planar so eliminate that?

Agreed

3 has 4π electrons and does not fit the (4n+2 rule) so eliminate that?

I disagree, what will the hybridisation be at the carbanion?

4 from the double bonds it has 8π electrons and does not fit the (4n+2 rule) so eliminate that?

I disagree, you're ignoring lone pairs again...

[G O D I V A]

1 has 4π electrons and does not fit the (4n+2 rule) so eliminate that?

That depends on the hybridisation of the oxygen atom... I disagree with you.

Oxygen would have 2 lone pairs so 8π which still does not fit the 4n+2 rule

3 has 4π electrons and does not fit the (4n+2 rule) so eliminate that?

I disagree, what will the hybridisation be at the carbanion?

sp2 hybridized?  I dont understand how hybridization helps me?


4 from the double bonds it has 8π electrons and does not fit the (4n+2 rule) so eliminate that?
[/quote]

I disagree, you're ignoring lone pairs again...


Ok so 14π bonds which follows the 4n+2 rule with n being 3?

[Markovnikov]

sp² hybridized atoms are planar, which means that one of their p-orbitals will be in the same plane and be able to hyperconjugate.

Maybe I'm counting the electrons in another way, but I don't get 14 electrons on #4.

[G O D I V A]

sp² hybridized atoms are planar, which means that one of their p-orbitals will be in the same plane and be able to hyperconjugate.

Maybe I'm counting the electrons in another way, but I don't get 14 electrons on #4.

So I get 1 to be aromatic, 2 is not, 3 is, and 4 i'm still a bit stuck.  I'm either getting 9π electrons so not aromatic or 10π electrons which is.

[Markovnikov]

How do you get an uneven numbers of electrons when you're calculating electron pairs?  

[G O D I V A]

How do you get an uneven numbers of electrons when you're calculating electron pairs?  

Huh? it's not uneven

4 electrons in each C=C bond so 4π bonds
4 electrons in each N=C bond so 4π bonds
1 lone pair on the nitrogen in the 5-ring so thats 1π or 2π?

[Dan]

I dont understand how hybridization helps me?

Hybridisation is key to the problem.

The system is aromatic if there are 4n+2 electron in a delocalised pi system. It seems to me that you are getting confused with the "n". The value of "n" can be any integer, it simply means that an aromatic system can have 2 (n=0), 6 (n=1), 10 (n=2), ..etc. etc. electrons in the delocalised pi system. I do'n understand what you are talking about when you say

4 electrons in each C=C bond so 4π bonds

To me, that makes no sense.

Pi symmetry is required, and you need to combine p atomic orbitals to make pi molecular orbitals.

In the case of furan - the oxygen does have two lone pairs, but only one of these is part of the pi system, the other is orthogonal to the pi system - meaning it does not overlap with it (see picture below).

So overall for furan with an sp² hybridised oxygen we have 2 pi bonds (2 electrons for each) and 1 lone pair (2 electrons) in the pi system. 2(2)+2 = 6 therefore aromatic.

The situation is similar for 3, and for 4 you can play exactly the same game - I count 4 pi bonds and 1 lone pair in the pi system (there are 2 lone pairs orthogonal to the pi system).

[Arctic-Nation]

Dan, I'm being a bit nitpicky here, but I've got to disagree with cyclopentadiene not being planar. It is planar, but it lacks the cyclic conjugation needed for aromaticity. Sorry.

[Dan]

No, you're absolutely right - the molecule is planar, my mistake. In my head I was thinking about the tetrahedral sp³ hybridisation of the CH₂ not being trigonal planar, I scanned the post too quickly. You're not nitpicking, that's an important correction.

[G O D I V A]

Alright. thanks a lot for all your help.  I understand it now