Question

When 2-Methylbut-2-ene is treated with hydrogen chloride, the major product obtained is

• A. 2-Chlorobutane
• B. 2-Chloro-2-methylbutane
• C. 3-Chloro-2-methylbutane
• D. 2-Chloro-3-methylbutane

Hint:

Markovnikov additions on alkenes always head towards the most stable intermediate. If a tertiary ($3^\circ$) position is directly available on the double bond, the halogen will end up there!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The reaction describes the addition of a hydrogen halide (HCl) to an unsymmetrical alkene (2-Methylbut-2-ene). We need to determine the major structural isomer obtained from this electrophilic addition reaction.

Step 2: Key Formula or Approach:
The addition of hydrohalic acids to unsymmetrical alkenes follows Markovnikov's rule. This rule dictates that the electrophile ($\text{H}^+$) adds to the double-bonded carbon containing the greater number of hydrogen atoms, leading to the formation of the more stable carbocation intermediate. The nucleophile ($\text{Cl}^-$) then attacks this stable carbocation.

Step 3: Detailed Explanation:
The starting material is 2-methylbut-2-ene: $\text{(CH}_3)_2\text{C=CH-CH}_3$. When HCl approaches the double bond, protonation can theoretically happen at two places: 1. If $\text{H}^+$ adds to C3, a positive charge develops at C2, generating a tertiary ($3^\circ$) carbocation: $\text{(CH}_3)_2\text{C}^+\text{-CH}_2\text{-CH}_3$. This carbocation is highly stable due to extensive inductive effect and hyperconjugation. 2. If $\text{H}^+$ adds to C2, a positive charge develops at C3, generating a secondary ($2^\circ$) carbocation: $\text{(CH}_3)_2\text{CH-CH}^+\text{-CH}_3$, which is less stable. \includegraphics[width=0.7\linewidth]{images/6sol.png}
Because the tertiary carbocation forms preferentially due to its superior thermodynamic stability, the chloride ion ($\text{Cl}^-$) selectively attacks C2. This yields 2-chloro-2-methylbutane as the exclusive major product.

Step 4: Final Answer:
The major product of the reaction is 2-chloro-2-methylbutane, which corresponds to option (B).