JEE Main Chemistry Question (2026) — Solution

Step-by-step solution

In option (A), the Friedel-Crafts alkylation of benzene with 1-chloropropane involves the formation of a primary carbocation, CH_3CH_2CH_2^+. This undergoes a 1,2-hydride shift to form a more stable secondary carbocation, CH_3CH^+CH_3, which then attacks the benzene ring to yield cumene. Thus, rearrangement occurs. In option (B), the acid-catalyzed dehydration of tert-butyl alcohol proceeds via the formation of a tert-butyl carbocation, (CH_3)_3C^+. Since it is already a highly stable tertiary carbocation, it does not undergo any rearrangement and directly loses a proton to form 2-methylpropene. In option (C), the isomerization of n-hexane in the presence of anhydrous AlCl_3 and HCl proceeds via the formation of carbocations that undergo skeletal rearrangements to form more stable branched isomers like 2-methylpentane. In option (D), the acid-catalyzed dehydration of neopentyl alcohol initially forms a primary neopentyl carbocation, (CH_3)_3CCH_2^+. This undergoes a 1,2-methyl shift to form a more stable tertiary carbocation, (CH_3)_2C^+CH_2CH_3, which then loses a proton to yield 2-methyl-2-butene. Thus, rearrangement occurs. Therefore, the reaction in option (B) is the only one where the major product is not obtained by a rearrangement reaction. Answer: