In the reaction `CH_(3)-overset(CH_(3))overset(|)CH-CH_(2)-O-CH_(2)-CH_(3)+HI overset("Heated")to`............
Which of the following compounds will be formed ?

Options

A. `CH_(3)-overset(CH_(3))overset(|)(CH)-CH_(2)OH+CH_(3)-CH_(2)-I`
B. `CH_(3)-overset(CH_(3))overset(|)(CH)-CH_(2)-I+CH_(3)CH_(2)OH`
C. `CH_(3)-underset(CH_(3))underset(|)(CH)-CH_(3)+CH_(3)CH_(2)OH`
D. `CH_(3)-underset(CH_(3))underset(|)(CH)-CH_2OH_CH+CH_(3)CH_(3)`

Answer

To solve the given reaction, we will analyze the interaction of the compound with HI upon heating.

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactant is `CH3-CH(CH3)-CH2-O-CH2-CH3` (an ether) and HI (hydroiodic acid).

2. **Protonation of the Ether**:
The ether oxygen has a lone pair of electrons which can accept a proton (H+) from HI. This results in the formation of an oxonium ion:
\[ CH3-CH(CH3)-CH2-OH^+ - CH2-CH3 \]
Here, the oxygen is now positively charged.

3. **Nucleophilic Attack**:
The iodide ion (I-) from HI acts as a nucleophile. In an SN2 reaction, the nucleophile will attack the least hindered carbon atom adjacent to the positively charged oxygen. In this case, the least hindered carbon is the one connected to the CH2 group.

4. **Formation of Products**:
The nucleophilic attack by I- on the CH2 group leads to the displacement of the OH group and the formation of two products:
- An alcohol: `CH3-CH(CH3)-CH2-OH` (propan-2-ol)
- An alkyl iodide: `CH3-CH2-I` (ethyl iodide)

5. **Final Products**:
The final products of the reaction are:
- Propan-2-ol (or isopropanol)
- Ethyl iodide

### Summary of Reaction:
\[ CH3-CH(CH3)-CH2-O-CH2-CH3 + HI \xrightarrow{\text{Heated}} CH3-CH(CH3)-CH2-OH + CH3-CH2-I \]

### Conclusion:
The products formed are propan-2-ol and ethyl iodide.