For each of the unsaturated systems below, indicate whether or not it is aromatic, and briefly explain how it meets or fails to meet the requirements for an aromatic system. Note: in each case, be sure you put in the appropriate number of non-bonding electrons on each atom to give the indicated formal charge.
 
 

Remember the requirements for a system to be aromatic. It must:

• be cyclic;

• planar;

• each atom in the ring must have a p orbital that can overlap with those on the adjacent atoms;

• the # of p electrons in the system must = 4n + 2, where n = any integer (0, 1, 2, etc.); This is known as Huckel’s Rule (the p electrons are those in the p bonds and in the p orbitals that are part of the p system; if electrons on an atom are in an orbital that can’t overlap with the orbitals of the p system, those electrons don’t count).

1.

This molecule is aromatic. The oxygen is sp² hybridized; one pair of electrons on the oxygen is in a p orbital that is part of the p system. This means each atom in the ring has a p orbital, and there are a total of 6 p electrons. (The other pair of electrons on the oxygen is in an sp² hybrid orbital, which is perpendicular to the p orbital, and thus cannot interact with the p system.
 
 

2.

This molecule is also aromatic. The positively-charged carbon is sp² hybridized, and has an empty p orbital that is part of the p system. There is a total of only 2 electrons in the p system (the pair from the C/C p bond), which fits Huckel’s rule where n = 0.
 
 

3.

This molecule is also aromatic. The negatively charged carbon is sp2 hybridized, with the electron pair in a p orbital which is part of the p system. This gives a total of 6 p electrons, satisfying Huckel’s rule.
 
 

4.

This molecule is not aromatic. If the nitrogen were sp² hybridized, its p orbital could overlap with those of the carbons, but there would be a total of 4 electrons in the p system (2 from the nitrogen and 2 from the C/C p bond). This would not fit Huckel’s rule. Instead, the nitrogen is sp3 hybridized as it would be in a regular amine.
 
 

5.

This molecule is not aromatic. While it is cyclic, could be planar, and each atom in the ring has a p orbital, there are a total of 8 p electrons. This doesn’t fit Huckel’s rule.
 
 

6.

This molecule is aromatic. The positively charged carbon is sp² hybridized, and has an empty p orbital that is part of the p system. The 3 C/C p bonds provide 6 electrons for the p system, satisfying Huckel’s rule.
 
 

7.

This molecule is also aromatic. The nitrogen is sp² hybridized, contributing a p orbital (the one it uses to make the p bond to carbon) to the p system. The non-bonding pair of electrons on the nitrogen is in an sp² orbital which is perpendicular to the plane of the p system. Thus each atom in the ring contributes one p orbital to the p system, which contains a total of 6 p electrons.
 
 

8.

This molecule is not aromatic. One atom in the ring is sp³ hybridized (it makes 4 sigma bonds), so there is no way it can contribute a p orbital to the p system.
 
 

9.

This molecule is also not aromatic. While the carbon with the positive charge is sp² hybridized and has a p orbital to contribute to the p system, there are only 4 p electrons (2 from each of the 2 C/C p bonds), which doesn’t fit Huckel’s rule.
 
 

10.

11.

12.