[kelaklub]

The answer to the following question is given as choice C. I'm thinking this may be a typo. Or I don't know if my aromatic check rules are correct, because it seems that A, and D are also not aromatic by not having all the Carbons as being sp².

• Cyclic system.
• All the atoms in the ring must be sp² hybridized.
• There must be 4n + 2 Pi electrons in the ring (2, 6, 10, 14, 18, ...etc.).

[Dan]

The answer to the following question is given as choice C. I'm thinking this may be a typo. Or I don't know if my aromatic check rules are correct, because it seems that A, and D are also not aromatic by not having all the Carbons as being sp².

Your answer, C is correct, but it is not a typo.

In A the carbons are all sp2 hybridised, the minus charge resides in a purely p orbital, which is part of the pi system. For aromaticity you must have 4n+2 electrons IN THE PI SYSTEM. so for A we have 5+1=6 elecrons in the pi system => Aromatic

In D, again all the carbons are sp2, but the plus charge resides in an sp2 hybrid orbital and thus IS ORTHOGONAL TO THE PI SYSTEM and does not interact with it. As a side not, D is NOT a resonance stabilised carbocation for this reason! So, in the pi system we have the normal (for benzene) 6 electrons in the pi system => aromatic

The moral of the story - Don't forget orbital symmetry!

[kelaklub]

Oh man, I am so confused with this. In choice A, I assume the carbon with a negative sign has it because of two unpaired electrons and a bond to a hydrogen atom that is not visible, because....

Formal charge = Number of Valence electrons - number of bonds - number of non-bonding electrons.
                    = 4 - 3 - 2 = -1

I was taught that when determining hybridization, un-paired electron are counted as a "thing".

So from the two bonds to two other carbons, the bond to a hydrogen, and the unpaired electrons. Doesn't that make the carbon sp³ hybridized?

[Dan]

Oh man, I am so confused with this. In choice A, I assume the carbon with a negative sign has it because of two unpaired electrons and a bond to a hydrogen atom that is not visible

C has 4 electrons right? so C^- has 5 right? 3 of the 5 electrons are in the 3 bonds, that leaves 2 electrons. According to Pauli they are paired. So you have a c atom bonded to 3 other things, and a lone pair (it's isoelectronic with a secondary amine).

The hybridisation will give the most stable orbitals. In the case of this carbanion, it is more energetically favourable if the lone pair can be delocalised over the whole ring (resonance stabilised), dissapating the minus charge. It can only do this if the lone pair is in an orbital which can overlap with the pi bonds. In other words, the lone pair must have pi symmetry to do this => a purely P orbital.

So it is more favourable to hybridise to 3 sp2 orbitals and 1 p orbital than it is to hybridise to 4 sp3 orbitals because in the latter case, you don't get resonance stability. Molecule A is planar as a result.

[kelaklub]

Ok that makes alot of sense. I guess you shouldn't just really look at one atom, but rather at the entire structure. Since you were so helpful for choice A, can you also explain the sp² designation for the positively charged carbon in choice D. That obviously has only a double bond and a single bond connected to the carbon, so why is that not simply sp hybridized? I don't know how to state this correctly but is it because the electrons from one of the bonds in the double bond can move to plus (+) sign? Thanks.

[Yggdrasil]

The strong driving force of molecules to adopt aromatic electron configurations will cause molecules to adopt fairly unconventional electronic structures.  In the case of D, your reasoning is correct that a carbon with two ligands should be sp hybridized.  However, two factors disfavor the formation of an sp hydribized carbon:

1)  Geometric constraints.  Having a bond angle of 180 in a six membered ring is unfavored sterically.
2)  Aromaticity.  By allowing the carbon to be sp² hybridized and leaving one sp² orbital empty, the molecule becomes more stable due to aromaticity.

So, as you said before, you need to look at the entire structure.  In general, if an atom looks like it may be part of an aromatic system and it has lone pairs, a positive formal charge, or a negative formal charge, likely it will adopt an different hybridization than expected in order to allow its p orbitals to form an aromatic system.

Another strange example is benzyne which is benzene with a triple bond instead of one of the double bonds.  While triple bonds usualy involve two sp-hybridized carbons, benzyne contains a triple bond between two sp² carbons (primarily due to geometric considerations).  Strange, but true.

[kelaklub]

This has all been extremely helpful. Thanks guys.