Question

Among the following ethers, which one will produce methyl alcohol on treatment with hot concentrated HI?

• A.
• B.
• C.
• D.

Answer 1

The Correct Option is C

To determine which ether among the given options will produce methyl alcohol (methanol) upon treatment with hot concentrated HI, we need to understand the reaction mechanism involving ethers with HI.

When ethers are treated with hot concentrated HI, they undergo cleavage to produce alkyl iodides and alcohols. The cleavage occurs at the carbon-oxygen bond of the ether. The mechanism generally proceeds through the following steps:

1. The protonation of the ether oxygen.
2. The nucleophilic attack by iodide (I^-) resulting in the cleavage of the C-O bond.

The stability of the resulting carbocation often influences the site of cleavage, but primary alcohols usually form when possible if simple or symmetrical ethers are involved. In this context, methyl iodide and the corresponding alcohol are typically the products.

Among the given options, the correct answer can be understood through this reasoning:

Option Analysis:

• Option A: The structure needs to be analyzed to determine if it includes a methyl group attached to the ether linkage.
• Option B: Similar analysis approach as above.
• Option C: This is anisole (methoxybenzene), represented as C_6H_5OCH_3. Upon reaction with HI, it specifically cleaves to form methyl alcohol and phenol. The reaction is:

C_6H_5OCH_3 + HI \rightarrow C_6H_5OH + CH_3I

• Option D: Again, the structure would be assessed for a methyl group.

Thus, the ether that would produce methyl alcohol on treatment with hot concentrated HI is anisole (Option C).

Figure depicting the structure of anisole.

Conclusion: The correct answer is Option C, as anisole cleaves under the said conditions to provide methyl alcohol.