`CH_(3)OC_(2)H_(5)` and `(CH_(3))_(3)COCH_(3)` are treated with hydroiodic acid. The fragments after reaction obtained are

Options

A. `CH_(3)I+HOC_(2)H_(5), (CH_(3))_(3)C-I+HOCH_(3)`
B. `CH_(3)OH+C_(2)H_(5)I, (CH_(3))_(3)Cl+HOCH_(3)`
C. `CH_(3)OH+C_(2)H_(5), (CH_(3))_(3)C-OH+CH_(3)I`
D. `CH_(3)I+HOC_(2)H_(5), CH_(3)I+(CH_(3))_(3)-C-OH`.

Explanation

To solve the problem, we need to analyze the reactions of the two ethers with hydroiodic acid (HI) and determine the fragments produced after the reaction.

### Step-by-Step Solution:

1. **Identify the Ethers**:
- The first ether is **ethyl methyl ether**: \( CH_3OC_2H_5 \).
- The second ether is **tert-butyl methyl ether**: \( (CH_3)_3COCH_3 \).

2. **Reaction of Ethyl Methyl Ether with HI**:
- Ethyl methyl ether reacts with hydroiodic acid (HI).
- The reaction proceeds via an **SN2 mechanism** because the ether has a primary (ethyl) and a methyl group.
- The iodide ion (I⁻) will attack the less hindered methyl group (due to steric hindrance from the ethyl group).
- The products formed are:
- Methyl iodide (\( CH_3I \))
- Ethanol (\( C_2H_5OH \))

3. **Reaction of Tert-Butyl Methyl Ether with HI**:
- Tert-butyl methyl ether reacts with hydroiodic acid (HI).
- This reaction proceeds via an **SN1 mechanism** because the ether contains a tertiary alkyl group (tert-butyl).
- The ether bond breaks, forming a stable tertiary carbocation.
- The iodide ion (I⁻) will then attack the carbocation.
- The products formed are:
- Tertiary butyl iodide (\( (CH_3)_3CI \))
- Methanol (\( CH_3OH \))

4. **Summarizing the Fragments**:
- From the first ether \( CH_3OC_2H_5 \), we get:
- \( CH_3I \) (Methyl iodide)
- \( C_2H_5OH \) (Ethanol)
- From the second ether \( (CH_3)_3COCH_3 \), we get:
- \( (CH_3)_3CI \) (Tertiary butyl iodide)
- \( CH_3OH \) (Methanol)

5. **Final Answer**:
- The fragments obtained after the reaction of \( CH_3OC_2H_5 \) and \( (CH_3)_3COCH_3 \) with hydroiodic acid are:
- \( CH_3I \), \( C_2H_5OH \), \( (CH_3)_3CI \), and \( CH_3OH \).