Question:
The major products formed by heating $C_6H_5CH_2-O-C_6H_5$ with HI are
The major products formed by heating $C_6H_5CH_2-O-C_6H_5$ with HI are
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In aryl-alkyl ethers, HI cleaves the alkyl-O bond, not the aryl-O bond.
Updated On: Apr 24, 2026
- $C_6H_5CH_2I$ and $C_6H_5CH_2OH$
- $C_6H_5CH_3$ and $C_6H_5OH$
- $C_6H_5CH_2I$ and $C_6H_6$
- $C_6H_5CH_2OH$ and $C_6H_5I$
- $C_6H_5CH_2I$ and $C_6H_5OH$
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Verified By Collegedunia
The Correct Option is
Solution and Explanation
Concept:
Ethers react with HI to undergo cleavage of the C-O bond.
Step 1: Given ether.
\[ C_6H_5CH_2-O-C_6H_5 \] This is benzyl phenyl ether.
Step 2: Site of cleavage.
The bond breaks at alkyl side (benzyl carbon), because aryl-O bond is resistant to cleavage.
Step 3: Products formed.
Benzyl part gives benzyl iodide: \[ C_6H_5CH_2I \] Phenoxy part gives phenol after protonation: \[ C_6H_5OH \]
Step 4: Final answer.
\[ \boxed{C_6H_5CH_2I \text{ and } C_6H_5OH} \]
Hence, correct option is (E).
Ethers react with HI to undergo cleavage of the C-O bond.
Step 1: Given ether.
\[ C_6H_5CH_2-O-C_6H_5 \] This is benzyl phenyl ether.
Step 2: Site of cleavage.
The bond breaks at alkyl side (benzyl carbon), because aryl-O bond is resistant to cleavage.
Step 3: Products formed.
Benzyl part gives benzyl iodide: \[ C_6H_5CH_2I \] Phenoxy part gives phenol after protonation: \[ C_6H_5OH \]
Step 4: Final answer.
\[ \boxed{C_6H_5CH_2I \text{ and } C_6H_5OH} \]
Hence, correct option is (E).
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