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NEET CHEMISTRYAlcohols, Phenols and EthersMedium

Question

The heating of phenylmethyl ether with HI produces:

A

Iodobenzene

B

Phenol

C

Benzene

D

Ethyl chloride

Step-by-Step Solution

The reaction of phenylmethyl ether (anisole) with hydrogen iodide (HI) involves the cleavage of the C-O bond.

  1. Bond Strength: The oxygen atom is bonded to a methyl group (alkyl) and a phenyl group (aryl). The bond between the oxygen and the benzene ring (Csp2OC_{sp^2}-O) has a partial double bond character due to resonance, making it much stronger and difficult to break compared to the bond between oxygen and the methyl group (Csp3OC_{sp^3}-O) .
  2. Mechanism: The reaction proceeds via the protonation of the ether oxygen. The iodide ion (II^-), being a good nucleophile, attacks the less sterically hindered methyl carbon via an SN2S_N2 mechanism.
  3. Products: This results in the cleavage of the OCH3O-CH_3 bond, forming methyl iodide (CH3ICH_3I) and phenol (C6H5OHC_6H_5OH). Iodobenzene is not formed because the strong aryl-oxygen bond remains intact.

Reaction: C6H5OCH3+HIΔC6H5OH+CH3IC_6H_5-O-CH_3 + HI \xrightarrow{\Delta} C_6H_5OH + CH_3I

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Alcohols, Phenols and Ethers. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYAlcohols, Phenols and Ethersheatingphenylmethylproduces

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