Question

When 2-methylbut-2-ene is treated with hydrogen chloride, the major product formed is ______.

• A. 3-chloro-2-methylbutane
• B. 2-chloro-2-methylbutane
• C. 2-chloro-3-methylbutane
• D. 2-chlorobutane

Hint:

Always analyze the stability of intermediate carbocations during alkene addition reactions. The stability order is $3^\circ > 2^\circ > 1^\circ$. The reaction will universally favor forming the most stable carbocation possible.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The addition of HCl to an unsymmetrical alkene follows Markovnikov's Rule.

Step 2: Formula Application:
The hydrogen atom attaches to the carbon with more hydrogen atoms, and the halogen attaches to the more substituted carbon to form a stable carbocation.

Step 3: Explanation:
In 2-methylbut-2-ene ($(CH_3)_2C=CHCH_3$), the two carbons of the double bond are C2 (substituted with two methyl groups) and C3 (substituted with one methyl group and one hydrogen). The $H^+$ attaches to C3, creating a stable $3^\circ$ carbocation at C2. The $Cl^-$ then attacks C2, yielding 2-chloro-2-methylbutane.

Step 4: Final Answer:
The major product is 2-chloro-2-methylbutane.