Identify the reaction

Glucose is a six-carbon polyhydroxy aldehyde. On prolonged heating with HI (hydroiodic acid), especially in the presence of red phosphorus in classical reductions, HI acts as a strong reducing agent.

What HI does to glucose

Glucose contains:

• one aldehyde group,
• five hydroxyl groups.

On prolonged treatment with HI:

• all the oxygen-containing groups are removed,
• the carbon chain remains intact,
• the molecule gets converted into the corresponding straight-chain hydrocarbon.

Carbon skeleton of glucose

Glucose has 6 carbon atoms in an unbranched chain.

Therefore, complete reduction/deoxygenation gives the corresponding 6-carbon alkane: $\text{C}_6\text{H}_{14}$ which is $n$-hexane.

Check options

• A: 6-iodohexanal — not the final product on prolonged heating with HI.
• B: $n$-Hexane — correct; complete reduction of glucose gives straight-chain hexane.
• C: 1-Hexene — alkene is not the final product here.
• D: Hexanoic acid — oxidation product, not reduction product.

Conclusion

Hence, glucose on prolonged heating with HI gives: $\boxed{n\text{-Hexane}}$