Animated Solution for Chemistry - Organic Chemistry: The number of resonance structures for N is :

Deprotonation of 2-Naphthol

Let's dive into this fascinating problem. We are given 2-naphthol reacting with sodium hydroxide ($\text{NaOH}$). Sodium hydroxide is a strong base, and it will abstract the acidic proton from the hydroxyl group, leaving us with the 2-naphthoxide ion.

Our goal is to find the total number of resonance structures for this ion. This requires us to systematically track the delocalization of the negative charge across the entire bicyclic naphthalene system.

Delocalization in the Right Ring

We start with our initial structure, the 2-naphthoxide ion. Notice the negative charge localized on the oxygen atom. This specific arrangement of double bonds in the naphthalene ring is one of its Kekule structures. Let's call this Structure 1.

Now, the real magic begins. The lone pair on the oxygen atom drops down to form a carbon-oxygen double bond ($C=O$). To maintain the octet rule, the adjacent carbon-carbon double bond ($C_1=C_2$) breaks, and the $\pi$ electrons are pushed onto $C_1$, creating a carbanion. This is our Structure 2.

Branching Paths

The negative charge doesn't just sit there; it continues its journey. The lone pair on $C_1$ moves to form a double bond with the bridgehead carbon, $C_{8a}$. This forces the central double bond ($C_{4a}=C_{8a}$) to break, pushing the electrons onto the other bridgehead carbon, $C_{4a}$. We now have our Structure 3.

From this bridgehead carbon, the path branches. The charge can either return to the right ring or cross over into the left ring.

The Second Kekule Structure

Let's take the first path, moving back into the right ring. The negative charge forms a double bond between $C_4$ and $C_{4a}$, which pushes the $\pi$ electrons of the adjacent bond onto $C_3$. This gives us Structure 4.

Continuing on this path, the negative charge on $C_3$ moves to form a double bond between $C_2$ and $C_3$. This pushes the electrons from the carbon-oxygen double bond back onto the oxygen atom. Look at what we have! It's another Kekule structure, our Structure 6.

Invading the Left Ring

Now, let's rewind to Structure 3 and take the second path. Instead of going right, the negative charge at the bridgehead carbon ($C_{4a}$) delocalizes into the left ring. It forms a double bond with $C_5$, pushing the $\pi$ electrons onto $C_6$. This is Structure 5.

The charge continues its tour around the left ring. The lone pair on $C_6$ moves to form a double bond with $C_7$, which in turn breaks the next double bond, pushing the electrons onto $C_8$. We have arrived at Structure 7.

Almost there! The negative charge on $C_8$ now moves to form a double bond with the top bridgehead carbon, $C_{8a}$. This action pushes the $\pi$ electrons back into the right ring, landing squarely on $C_1$. This gives us Structure 8.

For the grand finale, the negative charge on $C_1$ forms a double bond with $C_2$, which pushes the electrons back onto the oxygen atom. We have generated the third and final Kekule structure of the naphthoxide ion. This is Structure 9.

So, let's count them all up. We found 3 distinct Kekule structures where the charge is on the oxygen, and 6 structures where the charge is delocalized onto various carbon atoms ($C_1$, $C_{4a}$, $C_3$, $C_6$, $C_8$, and another $C_1$ variation).

That makes a total of 9 resonance structures. A beautiful display of extended electron delocalization!