JEE Advanced 2024 — Paper 2 — CHEM-I solution

JEE Advanced2024Paper 2CHEM-I

In the following reaction sequence, the major product Q is

Solution

Step (i): Reduction of L-glucose ( $C_{6} H_{12} O_{6}$ ) with HI under heat removes all hydroxyl groups and produces $n$ -hexane, $C_{6} H_{14}$ .

Step (ii): $n$ -hexane on aromatisation over $Cr_{2} O_{3}$ at $775$ K and $10 - 20$ atm gives benzene by cyclisation and dehydrogenation. So intermediate $P$ is benzene.

Final step: Benzene treated with excess $Cl_{2}$ in the presence of UV light undergoes free-radical addition (not electrophilic substitution) across all three double bonds, giving $1, 2, 3, 4, 5, 6$ -hexachlorocyclohexane (commonly known as BHC or benzene hexachloride).

Hence the major product $Q$ is the fully saturated hexachlorocyclohexane shown in option (D).

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