In the following reaction sequence, the major product Q is

The correct option is Option 4.

Let's break down the reaction sequence step-by-step to determine the identity of the major product Q:

Step 1: Conversion of L-Glucose to Benzene (P)
When L-glucose is heated with concentrated hydrogen iodide ($\text{HI}$), it undergoes complete reduction of all hydroxyl groups and the carbonyl group to yield $n$-hexane as the major hydrocarbon product:

$\text{L-Glucose}\stackrel{\text{HI,}\Delta }{\to }{\text{CH}}_3{\text{CH}}_2{\text{CH}}_2{\text{CH}}_2{\text{CH}}_2{\text{CH}}_3\text{(n-hexane)}$

When $n$-hexane is treated with chromium oxide (${\text{Cr}}_2{\text{O}}_3$) at 775 K under 10–20 atm pressure, it undergoes dehydrogenation and cyclization (aromatization) to form benzene:

${\text{C}}_6{\text{H}}_{14}\stackrel{{\text{Cr}}_2{\text{O}}_3\text{, 775 K, 10-20 atm}}{\to }{\text{C}}_6{\text{H}}_6\text{(Benzene, P)}+4{\text{H}}_2$

Step 2: Free Radical Addition to form Product Q
When benzene ($\text{P}$) is reacted with excess chlorine (${\text{Cl}}_2$) in the presence of ultraviolet (UV) light, it undergoes a free-radical addition reaction rather than substitution. The three conjugated double bonds in the benzene ring add three molecules of chlorine to produce 1,2,3,4,5,6-hexachlorocyclohexane (commonly known as benzene hexachloride, BHC, Lindane, or Gammaxene):

${\text{C}}_6{\text{H}}_6+3{\text{Cl}}_2\stackrel{\text{UV (excess)}}{\to }{\text{C}}_6{\text{H}}_6{\text{Cl}}_6\text{(Q)}$

The chemical structure of this addition product, 1,2,3,4,5,6-hexachlorocyclohexane, has chlorine atoms attached to all six carbons of a saturated cyclohexane ring, which corresponds precisely to the structure shown in Option 4.