Q13 / 75 · JEE Main 2026 (08 April Shift 2) · Chemistry
The major product of which of the following reaction is not obtained by rearrangement reaction?
- A
- Correct answerB
- C
- D
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:
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:
Hence, the correct answer is option: 'B'.