JEE Main 2026 — Apr 2, Shift 2 — Chemistry solution

JEE Main2026Apr 2, Shift 2Chemistry

The compound (X) on

(i) on heating in the presence of anhydrous AlCl $_{3}$ and HCl gas gives $2, 4$ -dimethyl pentane

(ii) aromatization gives toluene and

(iii) cyclisation gives methyl cyclohexane

The correct name of compound (X) is :

A
Hept-2-ene
B
Hept-1,3,5-triene
C
Heptane
D
Hept-2,4,6-triene

Solution

All three reactions are characteristic of $n$ -heptane (a straight-chain C $_{7}$ alkane):

(i) Isomerisation with anhydrous AlCl $_{3}$ /HCl converts $n$ -heptane to its branched isomer $2, 4$ -dimethylpentane.

(ii) Aromatisation (dehydrocyclisation) over Cr $_{2}$ O $_{3}$ /Al $_{2}$ O $_{3}$ at high temperature converts $n$ -heptane to toluene (loss of $4$ H $_{2}$ ).

(iii) Catalytic cyclisation produces methylcyclohexane (loss of $1$ H $_{2}$ ).

The triene options would not give these specific products under these conditions, and hept- $2$ -ene would not produce $2, 4$ -dimethylpentane on isomerisation.

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