Glucose on prolonged heating with HI, forms

Options

A. n-Pentane
B. n-Hexane
C. lodopentane
D. lodohexane

Answer

To solve the question regarding the reaction of glucose with HI (hydriodic acid) upon prolonged heating, we can break it down into a step-by-step process:

### Step 1: Understand the Structure of Glucose
Glucose is a six-carbon aldose sugar with the molecular formula C6H12O6. Its structure can be represented as:
\[ \text{CHO-CHOH-CHOH-CHOH-CHOH-CH2OH} \]

### Step 2: Identify the Reaction with HI
When glucose is heated with HI, it undergoes a process called reductive cleavage. The HI acts as a reducing agent, breaking the glycosidic bonds and converting the glucose into simpler hydrocarbons.

### Step 3: Prolonged Heating Effect
Prolonged heating with HI leads to the complete reduction of glucose. The reaction conditions favor the formation of straight-chain alkanes. The carbon atoms from glucose are rearranged and reduced to form n-hexane.

### Step 4: Determine the Product
The reaction of glucose with HI and red phosphorus results in the formation of n-hexane, which has the molecular formula C6H14. The structure of n-hexane can be represented as:
\[ \text{CH3-CH2-CH2-CH2-CH2-CH3} \]

### Conclusion
Thus, the product formed from the reaction of glucose on prolonged heating with HI is n-hexane (option B).

### Final Answer
The correct answer is **B: n-hexane**.

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