Question

Ethoxy benzene on reaction with hot and concentrated HI forms

• A. ethyl iodide and phenol
• B. ethyl iodide and iodobenzene
• C. ethyl alcohol and iodobenzene
• D. ethyl alcohol and phenol

Hint:

For any aromatic ether (Ar-O-R) reacting with HI, the aromatic Ar-O bond is an absolute "no-go" zone for cleavage because of its resonance stabilization. Phenol (Ar-OH) will always be one of your definitive products, while the alkyl fragment (R) converts into the alkyl iodide (R-I).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The objective is to identify the final organic products generated when ethoxybenzene (phenetole) reacts with a hot, concentrated solution of hydroiodic acid (HI).

Step 2: Key Formula or Approach:
When an alkyl aryl ether reacts with concentrated HI, protonation occurs at the ether oxygen atom to form an oxonium ion. The subsequent nucleophilic attack by the iodide ion ($\text{I}^-$) takes place via an $\text{S}_\text{N}2$ or $\text{S}_\text{N}1$ pathway. However, the bond between the sp$^2$-hybridized partial double-bonded phenyl carbon and the oxygen atom ($\text{C}_\text{aryl}\text{-O}$ bond) is exceptionally strong due to resonance and cannot be easily cleaved. Therefore, cleavage always happens exclusively at the alkyl-oxygen ($\text{C}_\text{alkyl}\text{-O}$) bond.

Step 3: Detailed Explanation:
1. Ethoxybenzene is $\text{C}_6\text{H}_5\text{-O-CH}_2\text{CH}_3$.
2. Upon treating with hot, concentrated HI, the ether oxygen is protonated: $$\text{C}_6\text{H}_5\text{-O-CH}_2\text{CH}_3 + \text{H}^+ \rightarrow \text{C}_6\text{H}_5\text{-\text{O} H}^+\text{-CH}_2\text{CH}_3$$ 3. The iodide nucleophile ($\text{I}^-$) attacks the less sterically hindered and weaker alkyl side ($\text{-CH}_2\text{CH}_3$) rather than the aromatic ring. This is because the phenyl-oxygen bond has partial double-bond character due to resonance delocalization of lone pairs into the benzene ring: $$\text{C}_6\text{H}_5\text{-\text{O} H}^+\text{-CH}_2\text{CH}_3 + \text{I}^- \rightarrow \text{C}_6\text{H}_5\text{OH} + \text{CH}_3\text{CH}_2\text{I}$$ Consequently, the structural products are phenol ($\text{C}_6\text{H}_5\text{OH}$) and ethyl iodide ($\text{CH}_3\text{CH}_2\text{I}$). Even with excess hot HI, phenol will not convert further to iodobenzene.

Step 4: Final Answer:
The reaction yields ethyl iodide and phenol, matching option (A).