Glucose on prolonged heating with HI forms:

Question

Accepted Answer

Glucose on prolonged heating with hydroiodic acid (HI) undergoes a reduction process that ultimately leads to the formation of n-hexane. Here's a detailed explanation of why Option 3 is correct and why the other options are incorrect:

1. **Explanation for Option 3 (n-hexane):**
   - Glucose is a six-carbon sugar (hexose) with an aldehyde group. When glucose is heated with HI, it undergoes a series of reduction reactions. HI acts as a strong reducing agent, and during the reaction, the hydroxyl groups (-OH) in glucose are replaced by hydrogen atoms. This process effectively reduces the glucose molecule to a straight-chain alkane, which is n-hexane (C6H14). The conversion involves the removal of oxygen atoms and the addition of hydrogen atoms, resulting in the formation of n-hexane, a saturated hydrocarbon.

2. **Why Option 1 (Hexaiodo hexane) is incorrect:**
   - Hexaiodo hexane would imply that each carbon atom in the hexane chain is bonded to an iodine atom. However, the reaction of glucose with HI does not lead to the substitution of iodine atoms for hydrogen atoms in the final product. Instead, HI serves as a reducing agent, not as a source of iodine for substitution. Therefore, hexaiodo hexane is not formed in this reaction.

3. **Why Option 2 (Gluconic acid) is incorrect:**
   - Gluconic acid is formed by the oxidation of glucose, where the aldehyde group is converted to a carboxylic acid group. However, the reaction with HI is a reduction process, not an oxidation. Therefore, gluconic acid is not a product of this reaction. The conditions described (prolonged heating with HI) do not favor the formation of gluconic acid.

4. **Why Option 4 (Pentaiodo hexane) is incorrect:**
   - Similar to hexaiodo hexane, pentaiodo hexane would involve the substitution of iodine atoms into the carbon chain. However, the reaction with HI does not result in the incorporation of iodine atoms into the hexane structure. The role of HI in this reaction is to reduce the glucose molecule, not to iodinate it. Thus, pentaiodo hexane is not formed.

In conclusion, the correct answer is Option 3 (n-hexane) because the reaction of glucose with HI under prolonged heating conditions results in the reduction of glucose to n-hexane, a straight-chain alkane. The other options do not accurately describe the chemical transformation that occurs in this reaction.