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

Options

A. `CH_(3)-underset(CH_(3))underset(|)CH-CH_(2)-O-CH_(3)`
B. `CH_(3)-CH_(2)-CH_(2)-CH_(2)-O-CH_(3)`
C. `CH_(3)-CH_(2)-underset(CH_(3))underset(|)CH-O-CH_(3)`
D. `CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-O-CH_(3)`

Explanation

To determine which ether will produce methyl alcohol (CH3OH) when treated with hot concentrated HI, we need to analyze the structure of the ethers provided in the options.

### Step-by-Step Solution:

1. **Understand the Reaction**: Ethers can undergo cleavage in the presence of strong acids like HI. The cleavage of ethers typically results in the formation of alcohols and alkyl halides.

2. **Identify Methyl Alcohol**: Methyl alcohol is also known as methanol (CH3OH). We need to find an ether that, upon cleavage, will yield methanol.

3. **Analyze the Ethers**: Without the specific ethers listed in the question, we can generally consider ethers of the form R-O-R', where R and R' are alkyl groups.

4. **Consider Possible Cleavage**: When HI reacts with an ether, it can break the ether bond (R-O-R') and form R-I and R'-OH. For the product to be methyl alcohol, one of the alkyl groups (R or R') must be a methyl group (CH3).

5. **Evaluate the Options**:
- If one of the ethers is methyl ether (CH3-O-R), treating it with HI will yield methyl alcohol (CH3OH) and R-I.
- If the ether is of the form R-O-CH3, it will also yield methyl alcohol upon cleavage.
- If neither of the alkyl groups is methyl, then methyl alcohol will not be produced.

6. **Conclusion**: The ether that contains a methyl group (either as CH3-O-R or R-O-CH3) will produce methyl alcohol when treated with hot concentrated HI.

7. **Final Answer**: Based on the analysis, the correct option that will produce methyl alcohol upon treatment with hot concentrated HI is the ether that contains a methyl group.