How do you count pi electrons in aromatic compounds?

1 Answer
Jul 29, 2016

pi bonds are simply the second bond made in a double bond. Any pure double bond is one sigma/sigma and one pi/pi bond.

  • Since any one chemical bond (meaning only one line in bond line notation) contains at most two electrons, you can count two pi electrons per double bond, and ignore the sigma electrons.
  • If you see lone pairs, consider the molecular geometry, and only the pi electrons that are in the ring count towards aromaticity.

Here are some examples of rings that may or may not be aromatic:

Note that the only pi electrons I've counted are in the ring. The others are either outside of the ring or sigma electrons.

Counting from top to bottom, column-wise:

  1. Aromatic, because 4n + 2 = 6 pi electrons in the ring (with n = 1), planar, fully conjugated all around, and cyclic.

  2. Aromatic, because 4n + 2 = 6 pi electrons in the ring (with n = 1), planar, fully conjugated all around, and cyclic. The pi electrons in the double bond outside of the ring do not count towards the pi electrons one considers for aromaticity.

  3. Nonaromatic, because 4n + 2 ne 4 pi electrons, where n must be an integer. It's also not conjugated all around, so it's not antiaromatic. The pi electrons in the double bond outside of the ring do not count towards the pi electrons one considers for aromaticity.

  4. Aromatic, because 4n + 2 = 6 pi electrons in the ring (with n = 1), planar, fully conjugated all around, and cyclic. The lone pair is actually in a pure 2p orbital perpendicular to the ring. Don't be fooled, as the alkyl carbon has an implicit hydrogen.

  5. Aromatic, because 4n + 2 = 6 pi electrons in the ring (with n = 1), planar, fully conjugated all around, and cyclic. The lone pair is actually in a pure 2p orbital perpendicular to the ring, which means they count as pi electrons.

  6. Aromatic, because 4n + 2 = 6 pi electrons in the ring (with n = 1), planar, fully conjugated all around, and cyclic. Only one of the lone pairs is actually in a pure 2p orbital perpendicular to the ring, which means those count as pi electrons. The other lone pair is actually in a sigma (actually, sp^2) orbital, so it doesn't count. Thus furan is not antiaromatic.