Question

The major product formed in the following reaction is : 

• A. A
• B. B
• C. C
• D. D

Hint:

In the absence of peroxides, HBr addition to alkenes always follows Markovnikov's rule through the most stable carbocation intermediate.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The reaction involves the electrophilic addition of HBr to an alkene. This process follows Markovnikov's rule, where the electrophile ($H^+$) adds to the carbon with more hydrogens to form the most stable carbocation intermediate.
Step 2: Key Formula or Approach:
Markovnikov's Rule: Addition of a protic acid HX to an asymmetric alkene, the acid hydrogen (H) gets attached to the carbon with more hydrogen substituents, and the halide (X) group gets attached to the carbon with more alkyl substituents.
Step 3: Detailed Explanation:
The reactant is 2-methylbut-1-ene.
1. Electrophilic attack of $H^+$ on the double bond:
Attack at $C_1$ gives a $3^\circ$ carbocation: $CH_3-CH_2-C^+(CH_3)_2$.
Attack at $C_2$ gives a $1^\circ$ carbocation: $CH_3-CH_2-CH(CH_3)-CH_2^+$.
2. The $3^\circ$ carbocation is much more stable and forms preferentially.
3. Nucleophilic attack of $Br^-$ on the $3^\circ$ carbocation yields 2-bromo-2-methylbutane.
Step 4: Final Answer:
The major product is 2-bromo-2-methylbutane.