Glucose when reduced with HI and red phosphorus gives

Options

A. n-pentane
B. n-butane
C. n-hexane
D. n-heptane

Answer

To solve the question of what glucose produces when reduced with HI (hydrogen iodide) and red phosphorus, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of Glucose**:
Glucose is a simple sugar with the molecular formula C₆H₁₂O₆. Its structure can be represented as:
\[ \text{CHO-CHOH-CHOH-CHOH-CH₂OH} \]
This shows that glucose has a chain of six carbon atoms.

2. **Understand the Reaction Conditions**:
When glucose is treated with HI and red phosphorus, these act as reducing agents. The red phosphorus helps to facilitate the reduction process, while HI provides hydrogen.

3. **Reduction Process**:
The reduction of glucose involves the conversion of the hydroxyl (-OH) groups and the aldehyde group (-CHO) into alkane structures. The reaction effectively removes the oxygen atoms and replaces them with hydrogen atoms.

4. **Formation of Hydrocarbons**:
Upon reduction, glucose will yield a hydrocarbon chain. The reduction of glucose leads to the formation of n-hexane (C₆H₁₄), which is a straight-chain alkane with six carbon atoms. The reaction can be summarized as:
\[ \text{C₆H₁₂O₆} \xrightarrow{\text{HI, Red P}} \text{C₆H₁₄ (n-hexane)} \]

5. **Other Possible Products**:
In addition to n-hexane, other products such as 2-iodohexane may also be formed due to the presence of iodine in HI. However, the primary hydrocarbon product from the complete reduction of glucose is n-hexane.

6. **Conclusion**:
Therefore, the main product formed when glucose is reduced with HI and red phosphorus is n-hexane.

### Final Answer:
Glucose when reduced with HI and red phosphorus gives n-hexane.

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