Why does the reaction between the diazonium ion and 2-naphthol occur primarily on the 1 and not the 3 position

Question

Can someone also draw out the initially-formed carbonium ion for attack at each position placing the + charge on the C atom at 2 to look at conjugation in the distribution of the remaining C-C double bonds? This is given as a hint for me to answer the above question. Thanks!!

Answer 1

The short answer is that the carbocation intermediate in the attack at C-1 has better resonance structures than the intermediate in the attack at C-3.

Attack at C-1: If you place carbocation on C-2, the other ring is aromatic, and there is one double bond conjugated to the aromatic ring.
Attack at C-3: If you place carbocation on C-2, the other ring is not aromatic (it only has two double bonds). In fact, it is not possible to move the double bonds around and get an aromatic structure in this intermediate.

Answer 2

First off, according to the Gilbert N. Lewis theory of bonding, the most likely structure of naphthalene has C=C between C-1 and C-2, and between C-3 and C-4. If a diazonium ion attacks at C-1, then the electrons of the C-1,2 double bond shift out to nitrogen. Electrons from oxygen of -OH shift in to form C=OH+. That takes care of "C atom at 2." But if the diazonium ion attacks at C-3, then the electrons shift from C-4 to C-2, and this is unlikely.