Question

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

• A.

  

• B.

  

• C.

  

• D.

  

Answer 1

The Correct Option is D

 The question asks which ether produces methyl alcohol when treated with hot concentrated HI. Let's analyze the options based on the mechanism of ether cleavage by HI:

1. An ether typically undergoes cleavage by hydrogen iodide (HI) to produce an alkyl iodide and an alcohol. The reaction usually follows the order: \(R-O-R' + HI \rightarrow R-I + R'-OH\). Here, alcohol is produced from the group that does not form the stable carbocation.
2. In the case of symmetric ethers like \(CH_3-O-CH_3\), the cleavage will produce methyl iodide and methyl alcohol. Similarly, for asymmetric ethers, the more substituted alkyl chain tends to form an iodide due to carbocation stability.
3. Considering the options provided, to correctly predict the ether producing methyl alcohol, we must identify the ether with a methoxy group \((-OCH_3)\). Based on the stability of intermediates, HI will cleave the methoxy group to yield methyl alcohol \((CH_3OH)\) and the corresponding alkyl iodide.
4. Examine the options:
   • First ether:would generate ethyl iodide and methanol upon cleavage.
   • Second ether:and so forth for other ethers until the correct one concluded.
   • Correct answer optionresults from methoxy substitution.
5. Therefore, the ether that produces methyl alcohol when treated with hot concentrated HI is:.