Glycerol on being heated with an excess of `HI` produces

Options

A. allyl iodide
B. propene
C. glycerol triiodide
D. 2-iodopropane

Answer

To solve the question regarding the reaction of glycerol with excess HI, we will follow these steps:

### Step 1: Identify the structure of glycerol
Glycerol, also known as glycerin, has the chemical formula C3H8O3. Its structure can be represented as:
\[ \text{CH}_2\text{OH} - \text{CH}(\text{OH}) - \text{CH}_2\text{OH} \]

### Step 2: Reaction with HI
When glycerol is heated with an excess of hydroiodic acid (HI), it undergoes a substitution reaction where the hydroxyl (–OH) groups are replaced by iodine (I). Since there are three hydroxyl groups in glycerol, we will need three moles of HI to fully convert glycerol into its iodinated form.

### Step 3: Formation of triiodopropane
The reaction can be represented as:
\[ \text{C}_3\text{H}_8\text{O}_3 + 3 \text{HI} \rightarrow \text{C}_3\text{H}_5\text{I}_3 + 3 \text{H}_2\text{O} \]
This results in the formation of 1,2,3-triiodopropane (C3H5I3).

### Step 4: Decomposition of triiodopropane
1,2,3-Triiodopropane is unstable and can further decompose. One of the iodine atoms can be eliminated, leading to the formation of allyl iodide (C3H5I):
\[ \text{C}_3\text{H}_5\text{I}_3 \rightarrow \text{C}_3\text{H}_5\text{I} + \text{I}_2 \]

### Step 5: Further reaction with HI
The allyl iodide can react again with excess HI, leading to the formation of 2-iodopropane (C3H7I):
\[ \text{C}_3\text{H}_5\text{I} + \text{HI} \rightarrow \text{C}_3\text{H}_7\text{I} \]

### Step 6: Final product
The final product after all reactions with excess HI is 2-iodopropane (C3H7I).

### Conclusion
Thus, the product formed when glycerol is heated with an excess of HI is **2-iodopropane**.

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