`overset(OH)overset(|)(CH_(2))-overset(OH)overset(|)(CH)-overset(OH)overset(|)(CH_(2))+underset("excess")(Hi)toX`
What is X ?

Options

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

Answer

To solve the question, we need to analyze the reaction of the compound with excess HI. The compound has three hydroxyl (OH) groups attached to a carbon chain. Here’s a step-by-step breakdown of the reaction:

### Step 1: Identify the Compound
The compound can be represented as:
- CH₂(OH) - CH(OH) - CH₂(OH)

This compound has three hydroxyl groups attached to a three-carbon chain.

### Step 2: Reaction with HI
When we add excess HI (hydroiodic acid) to the compound, the hydroxyl groups will react with HI. The reaction involves the protonation of the hydroxyl groups, which makes them better leaving groups.

### Step 3: Protonation of Hydroxyl Groups
Each OH group donates its lone pair to H⁺ from HI, leading to the formation of water (H₂O) and leaving behind positively charged carbon atoms. This means that we will have three moles of H⁺ reacting with the three OH groups.

### Step 4: Formation of Carbocations
The protonation of the OH groups leads to the formation of carbocations (positively charged carbon atoms). The structure will now look like this:
- CH₂⁺ - CH⁺ - CH₂⁺

### Step 5: Nucleophilic Attack by Iodide Ion
The iodide ion (I⁻) from HI will now attack the carbocations. The I⁻ will preferentially attack the most stable carbocation, which is typically the one that is more substituted.

### Step 6: Formation of Alkyl Iodides
As a result of the nucleophilic attack, we will form alkyl iodides. The final products will be:
- CH₃ - CH(I) - CH₃ (from the central carbon)
- CH₃ - CH₂ - CH₃ (from the terminal carbons)

### Step 7: Final Product
The final product after the complete reaction with excess HI will be:
- Iodopropane (C₃H₇I) and possibly some di-iodinated products depending on the reaction conditions.

Thus, the final answer for X is:
- **X = C₃H₇I (Iodopropane)**