`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 given question, we need to analyze the reaction of the compound with excess HI. The compound is represented as:

**CH2OH - CH(OH) - CH2OH**

This compound has three hydroxyl (OH) groups. When it reacts with excess HI, the following steps occur:

### Step 1: Protonation of Hydroxyl Groups
The hydroxyl groups (OH) will donate their lone pairs to hydrogen ions (H+), leading to the formation of water (H2O) and leaving behind positively charged carbon centers.

**Reaction:**
- Each OH group reacts with H+ from HI, forming water and creating a carbocation at each carbon atom.

### Step 2: Formation of Iodide Ions
The iodide ions (I-) from HI will then attack the positively charged carbon atoms, replacing the hydroxyl groups.

**Reaction:**
- The carbocations formed will be attacked by I-, resulting in the formation of iodoalkanes.

### Step 3: Formation of the Double Bond
As the reaction proceeds, the structure can rearrange, leading to the formation of a double bond between carbon atoms. This is due to the elimination of I2 (iodine) as a byproduct.

### Step 4: Final Product Formation
The final product will be a compound where the hydroxyl groups have been replaced by iodine atoms, and there may be a double bond formed in the structure.

**Final Structure:**
The final product after all reactions will be:

**CH2I - CH - CH2I**

This can also rearrange to form:

**CH2=CH - CHI**

### Final Answer
Thus, the final product X is:

**CH2I - CH - CH2I** or **CH2=CH - CHI**