HCl with an alkene X reacts in accordance with Markovnikov’s rule to give 1-Chloro-1-methylcyclohexane. The st

Question

HCl with an alkene X reacts in accordance with Markovnikov’s rule to give 1-Chloro-1-methylcyclohexane. The structure of alkene (X) is:

Accepted Answer

1. **Analyze the Reaction:** The reaction involves the electrophilic addition of HCl to an alkene (X) to form a tertiary alkyl halide: 1-Chloro-1-methylcyclohexane.
2. **Markovnikov's Rule:** The rule states that the negative part of the addendum (Cl⁻) attaches to the carbon atom possessing fewer hydrogen atoms, while the hydrogen (H⁺) attaches to the carbon with more hydrogen atoms. Mechanistically, the reaction proceeds via the most stable carbocation intermediate [NCERT 11th, Ch 13, Sec 13.3.4; NCERT 12th, Ch 10, Sec 10.4.2].
3. **Product Analysis:** The product has a chlorine atom and a methyl group on the same carbon (C1) of the cyclohexane ring. This implies the intermediate was a stable tertiary carbocation at C1.
4. **Identify Precursors (Alkene X):**
   - **Case 1 (1-Methylcyclohexene):** Protonation of the double bond (between C1 and C2) occurs at C2 to generate the stable tertiary carbocation at C1. Attack by Cl⁻ yields the product.
   - **Case 2 (Methylenecyclohexane):** Protonation of the exocyclic double bond (between ring C and external CH₂) occurs at the CH₂ group to generate the same stable tertiary carbocation at the ring carbon. Attack by Cl⁻ yields the same product.
5. **Conclusion:** Both 1-Methylcyclohexene and Methylenecyclohexane can serve as Alkene X.

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