Question

OH CH3 8.

CH-CH3 HCI (1) CH3 CHCH3

CH2-CH3 (2) Cl [JEE (Main) 2021]

• A. (1)
• B. (2)
• C. (3)
• D. (4)

Answer 1

Answer: (A)

1

Answer 2

The reaction proceeds via an SN1 mechanism. The secondary alcohol is protonated, followed by the loss of water to form a secondary carbocation. This carbocation rearranges via a hydride shift to form a more stable tertiary carbocation. Finally, the chloride ion attacks the tertiary carbocation to yield the product. The initial reactant is 1-cyclohexyl-1-methylpropan-2-ol.

1. Protonation: The hydroxyl group is protonated by HCl.
2. Carbocation formation: Water leaves, forming a secondary carbocation at the carbon bearing the -CH(CH₃) group.
3. Rearrangement: A hydride shift from the adjacent carbon (bearing the -CH₃ group) to the carbocation center forms a tertiary carbocation at the carbon that was originally bonded to the hydroxyl group.
4. Nucleophilic attack: The chloride ion attacks the tertiary carbocation.

The final product is 2-chloro-2-(cyclohexyl)propane, which corresponds to option (1).