# Basic sites on Indene ion

Answer key says that A is the most basic site but I don’t understand the concept behind that.

To assess the basic site, I’d normally look at the molecule as is and assess the current stability of the atom in question. The least stable would be my strongest base.

Looking at this question both the atoms properties are the same (C with lone pair) so resonance is the next thing I would look at. Both of these rings have resonance which moves (delocalizes) the negative charge around. So I would think since the volume of resonance is greater in the bigger ring, A is more stable thus making B the stronger base.

What am I misunderstanding or what concept am I not considering?

I believe the concept assessed here is the idea of aromaticity. You must first be able to recognise that the lone pair held by each carbon atom with the negative formal charge is in a different type of orbital. In the carbon atom of the 6-membered ring, it is in an $$\ce {sp^2}$$ hybrid orbital that is in a plane perpendicular to the $$\pi$$ electron cloud of the indene molecule. This lone pair is not delocalised into the indene $$\pi$$ system. On the other hand, in the carbon atom of the 5-membered ring, it is in an unhybridised p orbital that is parallel to the other p orbitals of the indene molecule. This lone pair is, in fact, delocalised into the indene $$\pi$$ system. As such, it is much less available for donation to an acid.
Protonation at sites C and D would also be unfavourable due to destruction of aromatic character. Between sites A and B, I have explained above that the lone pair at A would be much more available for donation due to lack of delocalisation in to the $$\pi$$ system. Hence, the preferred site of protonation would be site A.