Reaction Sequence of L-Glucose to Final Product Q

To determine the major product $\mathbf{Q}$ in the given reaction sequence, let us analyze each step systematically:

1. Reduction of L-Glucose: Heating L-glucose with red phosphorus and hydroiodic acid ($\text{HI}, \Delta$) results in the complete reduction of all the hydroxyl and carbonyl functional groups, yielding $n$-hexane: $\text{L-Glucose} \xrightarrow{\text{HI}, \Delta} \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_3 \quad (n\text{-hexane})$

2. Aromatization of $n$-hexane: When $n$-hexane is heated in the presence of chromium oxide ($\text{Cr}_2\text{O}_3$) catalyst at $775\text{ K}$ and $10\text{-}20\text{ atm}$ pressure, it undergoes catalytic reforming (cyclization and dehydrogenation) to form benzene ($\text{C}_6\text{H}_6$). $\text{CH}_3(\text{CH}_2)_4\text{CH}_3 \xrightarrow{\text{Cr}_2\text{O}_3, 775\text{ K}, 10\text{-}20\text{ atm}} \text{C}_6\text{H}_6 \quad (\text{Product } \mathbf{P})$ Thus, the intermediate compound $\mathbf{P}$ is benzene.

3. Photo-chlorination of Benzene: When benzene ($\mathbf{P}$) is treated with excess chlorine gas ($\text{Cl}_2$) in the presence of ultraviolet (UV) light, it undergoes a free-radical addition reaction across the double bonds (rather than electrophilic substitution): $\text{C}_6\text{H}_6 + 3\text{Cl}_2 \xrightarrow{\text{UV}} \text{C}_6\text{H}_6\text{Cl}_6 \quad (\text{Product } \mathbf{Q})$ This addition reaction destroys the aromaticity of the ring and yields 1,2,3,4,5,6-hexachlorocyclohexane (commonly known as benzene hexachloride or BHC).

Comparing the structure of $1,2,3,4,5,6$-hexachlorocyclohexane with the given options:

• Option (A) is hexachlorobenzene ($\text{C}_6\text{Cl}_6$), which would form via electrophilic substitution in the presence of a Lewis acid ($\text{AlCl}_3$).
• Option (D) represents $1,2,3,4,5,6$-hexachlorocyclohexane ($\text{C}_6\text{H}_6\text{Cl}_6$), a non-aromatic cyclohexane ring substituted with six chlorine atoms.

Therefore, the correct option is (D).