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Chemistry Question 64 – JEE-MAIN 2026

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The compound (X) on (i) on heating in the presence of anhydrous AlCl3 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:

Consider the type of reactions described for compound (X) and the typical substrates that undergo these transformations.

Step 1: Analyze Cyclisation and Aromatization Reactions✦ Active

Condition (iii) states that cyclisation of compound (X) gives methyl cyclohexane. Methyl cyclohexane is a 7-carbon cyclic alkane. Condition (ii) states that aromatization of compound (X) gives toluene (methylbenzene), which is also a 7-carbon aromatic compound. Both these reactions are characteristic of n-heptane (C7H16) under catalytic reforming conditions (e.g., using Pt/Al2O3 catalyst), where n-heptane first cyclizes to methyl cyclohexane and then dehydrogenates to toluene.

Step 2: Analyze Isomerization Reaction○ Expand

Condition (i) states that heating compound (X) in the presence of anhydrous AlCl3 and HCl gas gives 2,4-dimethyl pentane. 2,4-dimethyl pentane is a branched 7-carbon alkane (C7H16). The reaction with AlCl3/HCl is a classic Friedel-Crafts type isomerization reaction for alkanes, where straight-chain alkanes rearrange to more branched isomers. n-Heptane is known to isomerize to various branched heptanes, including 2,4-dimethyl pentane, under these conditions.

Step 3: Identify Compound (X)○ Expand

All three reactions—cyclisation to methyl cyclohexane, aromatization to toluene, and isomerization to 2,4-dimethyl pentane—are consistent with compound (X) being n-heptane. While other 7-carbon hydrocarbons might undergo some of these reactions, n-heptane is the most direct and common substrate for all three transformations as described. Therefore, the correct name of compound (X) is Heptane.

💡 Teacher's Secret Hint

Consider the most direct and common reactions for each type of hydrocarbon when evaluating the options.

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