`CH_(3)-CH=CH_(2)overset((i)Hg(OA c)_(2)//H_(2)O)underset((ii)NaBH_(4))rarr X+Na rarr Y+CH_(3)Cl rarr Z+HI overset(0^(@)C)rarr A+B`
What are A and B?

Options

A. `CH_(3)-CH_(2)-CH_(2)-OH & CH_(3)O`
B. `CH_(3)-underset(CH_(3))underset("| ")"CH"-OH & CH_(3)I`
C. `CH_(3)-underset(CH_(3))underset("| ")"CH"-OH & CH_(3)OH`
D. `CH_(3)-CH_(2)CH_(2)I & CH_(3)OH`

Explanation

To solve the question step by step, we will analyze the reactions one by one and identify the products formed at each stage.

### Step 1: Oxymercuration of Propylene
The first reaction involves propylene (CH₃-CH=CH₂) reacting with mercuric acetate (Hg(OAc)₂) and water (H₂O). This is an oxymercuration reaction where the double bond is converted into an alcohol.

- **Reaction**:
\[ CH₃-CH=CH₂ \xrightarrow{Hg(OAc)₂/H₂O} CH₃-CHOH-CH₃ \]
- **Product X**: The product formed is isopropyl alcohol (CH₃-CHOH-CH₃), which is a secondary alcohol.

### Step 2: Reduction with Sodium Borohydride
Next, we reduce the alcohol using sodium borohydride (NaBH₄). This step replaces the mercury acetate group with a hydrogen atom.

- **Reaction**:
\[ CH₃-CHOH-CH₃ \xrightarrow{NaBH₄} CH₃-CHOH-CH₃ \]
- **Product X remains unchanged**: The product remains isopropyl alcohol (CH₃-CHOH-CH₃).

### Step 3: Reaction with Sodium
Now, we react the alcohol (X) with sodium (Na). Sodium reacts with the hydroxyl group to form a sodium alkoxide.

- **Reaction**:
\[ CH₃-CHOH-CH₃ + Na \rightarrow CH₃-CH(ONa)-CH₃ + \frac{1}{2} H₂ \]
- **Product Y**: The product formed is sodium isopropoxide (CH₃-CH(ONa)-CH₃).

### Step 4: Williamson Ether Synthesis
Next, we react the sodium alkoxide (Y) with methyl chloride (CH₃Cl) to form an ether through the Williamson synthesis mechanism.

- **Reaction**:
\[ CH₃-CH(ONa)-CH₃ + CH₃Cl \rightarrow CH₃-CH-CH₃-O-CH₃ + NaCl \]
- **Product Z**: The product formed is isopropyl methyl ether (CH₃-CH-CH₃-O-CH₃).

### Step 5: Reaction with Hydroiodic Acid
Finally, we react the ether (Z) with hydroiodic acid (HI) at 0°C. At low temperatures, the ether cleaves to form an alcohol and an alkyl halide.

- **Reaction**:
\[ CH₃-CH-CH₃-O-CH₃ + HI \rightarrow CH₃-CHOH-CH₃ + CH₃I \]
- **Products A and B**: The products formed are isopropyl alcohol (A: CH₃-CHOH-CH₃) and methyl iodide (B: CH₃I).

### Final Answer
Thus, the final products A and B are:
- **A**: Isopropyl alcohol (CH₃-CHOH-CH₃)
- **B**: Methyl iodide (CH₃I)