Glucose on prolonged heating with HI gives:

We start with glucose, whose open-chain (acyclic) form may be written as $$\mathrm{HO{-}CH_2{-}(CHOH)_4{-}CHO}$$. In this structure every carbon except the aldehydic $$\mathrm{CHO}$$ carries a hydroxyl ($$\mathrm{OH}$$) group.

When glucose is treated with an excess of hydroiodic acid (HI) and heated for a long time, the reaction medium is strongly reducing because HI is continually regenerated from $$\mathrm{I_2}$$ by the phosphorous that is usually present. The overall effect of this “HI & red P” system is to:

1. Convert every hydroxyl group ($$\mathrm{OH}$$) into an iodo group ($$\mathrm{I}$$) by the substitution reaction $$\mathrm{R{-}OH \;+\; HI \;\longrightarrow\; R{-}I \;+\; H_2O}$$

2. Reduce every iodo group ($$\mathrm{C{-}I}$$) to a hydrogen atom by the reduction step $$\mathrm{R{-}I \;+\; HI \;\xrightarrow[\;]{\text{red P,\;heat}}\; R{-}H \;+\; I_2}$$ (the phosphorous reconverts $$\mathrm{I_2}$$ back to HI, so HI acts as the ultimate reducing agent).

Let us now apply these two changes carbon by carbon in glucose.

• First, each of the five secondary alcohol groups $$\mathrm{-CHOH-}$$ becomes $$\mathrm{-CHI-}$$ and then $$\mathrm{-CH_2-}$$ after full reduction.

• Second, the primary alcohol end $$\mathrm{HO{-}CH_2-}$$ is similarly converted to $$\mathrm{CH_3-}$$.

• Finally, the aldehyde carbon $$\mathrm{-CHO}$$ is reduced to a $$\mathrm{-CH_3}$$ group because the carbon-iodine intermediate that forms is again fully hydrogenated under the prolonged reducing conditions.

Therefore every carbon in the six-carbon chain of glucose ends up bearing only hydrogen atoms, giving the completely saturated straight-chain hydrocarbon $$\mathrm{CH_3{-}CH_2{-}CH_2{-}CH_2{-}CH_2{-}CH_3}$$, which is n-hexane.

No carbon-carbon bonds are broken or newly formed; all hetero-atoms (oxygen and iodine) are simply removed as water and iodine, leaving the intact six-carbon skeleton fully hydrogenated.

Thus on prolonged heating with HI, glucose is converted into n-hexane.

Hence, the correct answer is Option B.