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Q7.28(ii):
Write the equation of the reaction of hydrogen iodide with: methoxybenzene

Solution :

Step 1: Structural Analysis of Reactants

The reaction involves methoxybenzene (anisole) and hydrogen iodide ($HI$).

  • Methoxybenzene Structure: $C_6H_5-O-CH_3$. It consists of a methoxy group ($-OCH_3$) attached to a phenyl ring ($C_6H_5-$).
  • Cleavage Dynamics: In ethers with one alkyl group and one aryl group, the cleavage occurs specifically at the $O-CH_3$ bond, rather than the $O-C_6H_5$ bond.

Step 2: Mechanistic Pathway

The reaction proceeds via an $S_N2$ mechanism involving the following stages:

  1. Protonation: The lone pair on the ethereal oxygen atom attacks the electrophilic proton of $HI$, forming an oxonium ion:
    $C_6H_5-O-CH_3 + H^+ \rightarrow [C_6H_5-O^+(H)-CH_3]$ [Per the Brønsted-Lowry acid-base theory]
  2. Nucleophilic Attack: The iodide ion ($I^-$), being a strong nucleophile, attacks the less sterically hindered methyl carbon. The $C-O$ bond between the methyl group and oxygen breaks:
    $I^- + [C_6H_5-O^+(H)-CH_3] \rightarrow C_6H_5OH + CH_3I$

Step 3: Justification of Products

The $C_{aryl}-O$ bond is stronger and shorter due to partial double bond character arising from resonance, where the oxygen lone pair delocalizes into the benzene ring [Resonance Stabilization Effect]. Consequently, the nucleophile $I^-$ cannot attack the carbon atom of the phenyl ring. The primary products are phenol and methyl iodide.

Step 4: Chemical Equation

The balanced chemical equation is as follows:

$C_6H_5OCH_3 + HI \xrightarrow{\Delta} C_6H_5OH + CH_3I$

Visual Representation of Reaction Dynamics

Methoxybenzene (Anisole) C6H5 - O - CH3 + HI C6H5OH + CH3I (Phenol) (Methyl Iodide)

Final Solution: The reaction of methoxybenzene with hydrogen iodide yields phenol and methyl iodide: $C_6H_5OCH_3 + HI \rightarrow C_6H_5OH + CH_3I$


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