[hz862000]

Sorry for posting twice, I just read the rules.  I was just wondering if [16] annulene would be aromatic if it has a 2+ charge.  I think it's a trick question or something because it has a circled 2+ in the middle of the ring, but it also shows the alternating double bonds around the ring (8 in all).  I know that when they circle the charge it usually just means that there's a charge alternating around the ring, but the double bonds aren't drawn in too in the other examples. I'm just having trouble interpreting this and my book doesn't have any examples set up this way.  It does have a [16] annulene with a circled 2+ saying it's aromatic, but the double bonds aren't drawn in.

[Yggdrasil]

Think about whether the [16] annulene ion would satisfy all of the criteria for aromaticity.  Remember that aromatic compounds must:
(1) contain a ring of sp² hybridized atoms that form a conjugated pi system
(2) have a planar structure with all of the sp² hybridized atoms in the ring lying in the same plane
(3) 4n+2 electrons in the conjugated pi system

if the [16] annulene structure fulfills these criteria, then it is aromatic.

[hz862000]

I'm confused on a couple parts:  If the double bonds are fixed as they're drawn, then that would mean that two of the C are sp hybridized, and therefore not sp2.  However, if the way it's drawn just means that the double bonds are in alternating positions, then it would be aromatic, because the ones that don't have the double bonds nearby would have a positive charge and be sp2 hybridized.  I don't know which way it's intended. 

[Yggdrasil]

Don't focus too much on how molecules are drawn.  When molecules have systems of conjugated pi bonds, our conventional means of drawing bonds never accurately reflects the true position of bonding electrons in a molecule.

As a side note, why would some of the carbons be sp hybridized in the [16]annulene cation?  Do you mean sp³?

[hz862000]

Yeah, I guess I'm thinking about it way too much.  It would be aromatic.  I did mean sp though because I was thinking that there were 8 alternating double bonds and a 2+ charge which would make 2 of the C have only 2 bonds, since they would have 3 to begin with.  I was taking the drawing too literally.  Thanks for all of your help.  It's really great that you guys volunteer your time to help people on here.

[Yggdrasil]

Remember that all the carbons in [16]annulene have hydrogens attached.

Here's how I would explain the sittuation with [16]annulene.  In [16]annulene, every carbon is bonded to two carbons and a hydrogen.  Each carbon also participates in a double bond with a neighboring carbon.  Together, these say that each carbon is sp² hybridized and has one electron in the unhybridized p-orbital.

So far, [16]annulene fulfils the first two criteria for aromatic molecules.  It has a ring of sp² hybridized carbons that are coplanar.  But, since each carbon contributes one electron to the conjugated pi system, the conjugated pi system has 16 electrons, which does not fulfil the 4n+2 pi electron criteria (14 or 18 would, however).

But, what happens if we take away two electrons?  The two carbons would lose the electron in their p-orbitals, generating two carbocations.  As usual, the carbocations are sp² hybridized and have an empty p-orbital.  So, in your [16]annulene cation, you still have a coplanar ring of sp² hybridized atoms.  Furthermore, the conjugated pi system now has 14 electrons which fulfills the third criterion for aromaticity.