The following reaction takes place through `CH_3-O-CH_2-CH_3 + HI` to `CH_3-I + CH_3-CH_3-CH_2-OH`

Options

A. `S_N1` mechanism
B. `S_N2` mechanism
C. `S_N1` in first step and E2 in second step
D. `S_N2` in first step and E1 in second step

Answer

To solve the problem involving the reaction of ether with HI, we can break down the process step by step.

### Step 1: Identify the Reactants
The reactants in the reaction are:
- Ether: \( CH_3-O-CH_2-CH_3 \) (ethyl methyl ether)
- Hydroiodic acid: \( HI \)

### Step 2: Protonation of Ether
When the ether reacts with HI, the ether oxygen atom gets protonated by the hydrogen ion (H⁺) from HI. This step forms a protonated ether intermediate, which is more reactive.

**Reaction:**
\[ CH_3-O-CH_2-CH_3 + H^+ \rightarrow CH_3-O^+-CH_2-CH_3 \]

### Step 3: Formation of Nucleophile
The protonated ether is now a good leaving group. The bond between the oxygen and the ethyl group (CH₂-CH₃) can break, leading to the formation of a methyl cation and a nucleophile (iodide ion, I⁻) from HI.

**Reaction:**
\[ CH_3-O^+-CH_2-CH_3 \rightarrow CH_3^+ + CH_2-CH_3-OH \]

### Step 4: Nucleophilic Attack
The iodide ion (I⁻) acts as a nucleophile and attacks the less hindered methyl cation (CH₃⁺). This is an SN2 reaction, which is favored due to the less steric hindrance around the methyl group.

**Reaction:**
\[ CH_3^+ + I^- \rightarrow CH_3-I \]

### Step 5: Final Products
The final products of the reaction are:
- Methyl iodide: \( CH_3-I \)
- Ethanol: \( CH_3-CH_2-CH_2-OH \)

**Overall Reaction:**
\[ CH_3-O-CH_2-CH_3 + HI \rightarrow CH_3-I + CH_3-CH_2-CH_2-OH \]

### Conclusion
The reaction proceeds through an SN2 mechanism, where the iodide ion attacks the methyl cation, leading to the formation of methyl iodide and ethanol.

---