# Bond lengths in naphthalene

As I was reading Clayden: Organic Chemistry 2nd Edition, I came across an image of naphthalene showing its varying $\ce{C-C}$ bond lengths, however, I was unable to rationalize the bond lengths by drawing naphthalene's resonance forms. I have searched for a good explanation for the bond lengths shown but I have not found anything remotely convincing.

The 3 resonance forms of naphthalene are shown below however, these resonance forms are inconsistent with the bond lengths displayed by Clayden. From looking at these resonance forms I would not expect the $\ce{C-C}$ bond fusing the benzene rings to be the shortest since it has less double bond character compared to some of the other $\ce{C-C}$ bonds in the molecule.

## 1 Answer

You are assuming that all resonance structures contribute equally. The resonance structures on the right and left are equivalent but have one aromatic ring and one non-aromatic ring. This is in contrast to the center resonance structure, which has two aromatic rings and is more stable, thus giving the bonds between carbons 1-2; 3-4; 5-6; 7-8; and 9-10 more double bond character than the other bonds and making them shorter.

• Surely all the resonance structures are aromatic, since they all have a contiguous array of p orbitals containing 10 pi electrons, satisfying Hückel's "4n + 2" rule. Is it really correct to view two of the resonance structures as one aromatic ring connected to a non-aromatic ring? Aug 11 '18 at 20:10
• Also, this wouldn't explain why the C-C bond between carbons 9 and 10 is the shortest in length Aug 11 '18 at 20:51
• Due to symmetry constraints, only the one structure shown makes actually sense. The way resonance is used here is wrong. One should start from localised bonds, as they are shown, and then use resonance to justify that the single bonds are shorter than expected, and the double bonds slightly longer. Oct 6 '18 at 19:16
• @cяonus Naphthalene does not satisfy Hückel's rules; it is not a monocycle. Oct 6 '18 at 19:35