Question

The major products formed by heating \( \text{C}_6\text{H}_5\text{CH}_2 - \text{O} - \text{C}_6\text{H}_5 \) with HI are

Hint:

For alkyl aryl ethers (like anisole or benzyl phenyl ether), the cleavage with HI always yields phenol and an alkyl iodide. The C-O bond attached to the benzene ring is too strong to break due to resonance.

Answer 1

Step 1: Understanding the Concept:
The reaction involves the cleavage of an ether by a strong acid (Hydrogen Iodide, HI). Ethers are cleaved by strong acids via nucleophilic substitution mechanisms ($S_N1$ or $S_N2$). The products depend on the stability of the possible carbocation intermediates and steric hindrance.

Step 2: Key Formula or Approach:
1. Protonate the ether oxygen to form an oxonium ion.
2. Determine which carbon-oxygen bond is more susceptible to cleavage by the nucleophile ($\text{I}^-$).

Step 3: Detailed Explanation:
The reactant is benzyl phenyl ether: $\text{C}_6\text{H}_5-\text{CH}_2-\text{O}-\text{C}_6\text{H}_5$

Step 1: Protonation by HI:
$\text{C}_6\text{H}_5-\text{CH}_2-\text{O}-\text{C}_6\text{H}_5 + \text{H}^+ \rightleftharpoons \text{C}_6\text{H}_5-\text{CH}_2-\text{OH}^+-\text{C}_6\text{H}_5$

Step 2: Cleavage. We must decide which $\text{C-O}$ bond breaks.
- Bond A: Between Oxygen and the Phenyl ring ($\text{O}-\text{C}_6\text{H}_5$). This bond is very strong and possesses partial double-bond character due to the resonance delocalization of oxygen's lone pair into the benzene ring. Furthermore, an $S_N2$ attack on an $sp^2$ hybridized aryl carbon is extremely difficult.
- Bond B: Between Oxygen and the Benzyl group ($\text{O}-\text{CH}_2\text{C}_6\text{H}_5$). This bond is weaker. Cleavage here can proceed rapidly via an $S_N1$ mechanism because it leads to the formation of a resonance-stabilized benzyl carbocation ($\text{C}_6\text{H}_5\text{CH}_2^+$). Even if following $S_N2$, the benzyl position is reactive.
Because Bond B is much weaker and its cleavage path is highly favorable, the iodide ion ($\text{I}^-$) attacks the benzylic carbon.
The bond breaks, yielding:
1. Benzyl iodide: $\text{C}_6\text{H}_5\text{CH}_2\text{I}$
2. Phenol: $\text{C}_6\text{H}_5\text{OH}$
Phenol does not react further with HI under normal conditions because its $\text{C-O}$ bond is too strong to be cleaved by halide ions.

Step 4: Final Answer:
The major products are Benzyl iodide and Phenol.