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

Options

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

Explanation

To determine which ether will produce methyl alcohol upon treatment with hot concentrated HI, we need to analyze the ethers and their potential reactions with HI.

### Step-by-Step Solution:

1. **Understanding Ethers and Their Reactions**: Ethers are compounds that contain an oxygen atom connected to two alkyl or aryl groups. When treated with strong acids like HI, ethers can undergo cleavage to form alcohols and alkyl halides.

2. **Identifying the Required Product**: We want to produce methyl alcohol (CH3OH). This means we need to generate the methoxy group (CH3O-) from the ether upon reaction with HI.

3. **Analyzing the Ethers**: Let's denote the ethers as follows:
- Ether A: R1-O-R2
- Ether B: R3-O-R4
- Ether C: R5-O-R6
- Ether D: R7-O-R8

We need to check which of these ethers can yield CH3O- upon cleavage.

4. **Cleavage Mechanism**: The cleavage of ethers by HI involves protonation of the ether oxygen, followed by the cleavage of the C-O bond. The stability of the resulting carbocation is crucial. A more stable carbocation will favor the reaction.

5. **Finding the Suitable Ether**:
- If we assume that one of the ethers has a methyl group (CH3) attached to one side and a more stable alkyl group (like a tertiary group) on the other side, the cleavage will yield CH3O- and the corresponding alkyl halide.
- For example, if we consider an ether like (CH3)3C-O-CH3, upon treatment with HI, the (CH3)3C group is a tertiary group, which can stabilize the positive charge formed after cleavage.

6. **Final Reaction**:
- When the ether (CH3)3C-O-CH3 is treated with hot concentrated HI, it will break down to form CH3O- and (CH3)3C+.
- The CH3O- will then react with H+ to form methyl alcohol (CH3OH).

### Conclusion:
The ether that will produce methyl alcohol upon treatment with hot concentrated HI is the one that has a methyl group and a more stable alkyl group (preferably tertiary) on the other side.