JEE Main PYQ | Reaction of 1-Methoxy Naphthalene with HI | Why 1-Naphthol & Methyl iodide is formed!

Описание: In this video, I provide a detailed mechanism and solution for Question 44 from the JEE Main 2021 exam (Paper I, Chemistry), held on 18th March 2021 (Shift 2).

🧪 The Question: "Main Products formed during a reaction of 1-methoxy naphthalene with hydroiodic acid are:"

(1) 1-Iodonaphthalene and CH3I

​(2) 1-Iodonaphthalene and CH3OH

(3) 1-Naphthol and CH3OH

(4) 1-Naphthol and CH3I

📝 Solution Breakdown: To solve this, we use the mechanism of Ether Cleavage by Halogen Acids (HI):

Protonation: The oxygen atom of the ether gets protonated by H + from HI, making it a good leaving group.

Bond Breaking Rule (Resonance vs Sterics):

The bond between the Oxygen and the Naphthalene ring has partial double bond character due to resonance (lone pair on O conjugates with the ring). This bond is very strong and hard to break.

The bond between Oxygen and the Methyl group (O−CH3) is a pure single bond and is weaker.

Nucleophilic Attack (SN2): The Iodide ion (I−) attacks the less hindered Methyl group.

Result: The bond breaks at the methyl side.

The Naphthalene part becomes 1-Naphthol (−OH stays with the ring).

The Methyl group becomes Methyl Iodide (CH3I).

✅ Correct Answer: Option (4) (1-Naphthol and CH3I).

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