# Does the amine group participate in resonance in aniline? [duplicate]

When we draw resonance structures of aniline, 8 electrons participate in resonance, making it antiaromatic. But in our school textbook, it is said to be aromatic. Does the lone pair of $\ce{NH2}$ participate or not? If it doesn't, then it would make sense.

• Who said that 8 means antiaromatic? More importantly, when does this apply? – Ivan Neretin Sep 22 '17 at 5:05

The lone pair of nitrogen is also part of the extended π-system; the molecule itself is still aromatic. As I already explained in the question about fulvene, aromaticity is nothing you can determine based on simple rules, especially not counting electrons.
Structural indicators like bond lengths and angles give a good idea how well established resonance is. Additionally you can draw a qualitative molecular orbitals diagram and look if it is similar to benzene. Eventually the determination of aromaticity is an experimental one, for example with NMR spectroscopy.

From the crystal structure (M. Fukuyo, K. Hirotsu, and T. Higuchi, Acta Cryst. 1982, B38, 640-643. DOI: 10.1107/S056774088200363X) we know that the carbon-carbon bond lengths are all similar and between 136 and 141 pm. The NMR spectrum clearly shows the protons have aromatic shift, see chemicalbook.com.

The molecular orbitals of aniline (calculated at DF-BP86-D3BJ/def2-SVP; Gaussian 09 D.01) show that the nitrogen is part of the π-system. They are also very similar to benzene. Here are the orbitals of the π-system in decreasing energy. Virtual (unoccupied) orbitals are red/yellow, and occupied orbitals are blue/orange.

For aromaticity, you need several rules to be satisfied:

1) The compound must be cyclic;

2) It must be ring-conjugated (no cross-conjugation!);

3) The number of electrons on π-orbitals in ring conjugation must obey Huckel's rule and be equal to a number of (4n+2) electrons, where n is a whole non-negative number.

Al that means that we need to count only those electrons, which are in aromatic cycle, but we do not care for "exo" electrons.