Animated Solution for Chemistry - Organic Chemistry: Main products formed during a reaction of 1-methoxy naphthalene with hydroiodic acid are

A Classic Ether Cleavage

Let's look at the reaction between 1-methoxy naphthalene and hydroiodic acid ($\text{HI}$). This is a classic example of the cleavage of an aryl alkyl ether by a strong acid. When you see an ether reacting with a strong hydrogen halide like $\text{HI}$ or $\text{HBr}$, you should immediately think about acid-base chemistry followed by a nucleophilic substitution.

The Acid-Base Dance (Protonation)

The first step is always protonation. The oxygen atom in the methoxy group has two lone pairs of electrons. It acts as a Lewis base and attacks the highly acidic hydrogen of $\text{HI}$. This is a very fast acid-base reaction.

This protonation gives us an intermediate oxonium ion. Notice the positive charge on the oxygen. This makes the attached carbon groups highly susceptible to nucleophilic attack because the oxygen desperately wants to pull electrons towards itself to neutralize that positive charge. We also have an iodide ion (${\text{I}}^{-}$) released into the medium.

The Crucial Choice (Nucleophilic Attack)

Now, the iodide ion is a strong nucleophile. It needs to attack a carbon to kick off the oxygen. But which carbon? It has two choices: the methyl carbon or the naphthalene ring carbon. Here is a major catch that you must remember.

The bond between the oxygen and the aromatic ring has partial double bond character due to resonance (the lone pairs of oxygen delocalize into the naphthalene ring). Plus, an $S_{N}2$ attack on an $s{p}^2$ hybridized aromatic carbon is sterically and electronically blocked. The electron-rich aromatic ring repels the incoming nucleophile.

So, the iodide ion takes the path of least resistance and attacks the less hindered, $s{p}^3$ hybridized methyl group via an $S_{N}2$ mechanism.

As the iodide ion attacks the methyl carbon, the carbon-oxygen bond breaks, and the electrons shift to the positively charged oxygen, neutralizing it. This beautiful cleavage leaves us with our final products: 1-naphthol and methyl iodide (${\text{CH}}_3\text{I}$).

What About Excess Reagent?

You might wonder, what if we used excess $\text{HI}$? Would the 1-naphthol react further to form an aryl iodide? Actually, no. Phenols and naphthols do not undergo further substitution with $\text{HI}$ because that aromatic carbon-oxygen bond is just too strong to break. So the reaction stops right here. This is a favorite concept for JEE, so make sure you never try to substitute an aromatic hydroxyl group with a halide using just $\text{HX}$!