Question

An alkene \( X \) (\( C_5H_{10} \)) on reaction with \( HBr \) gave \( Y \) (\( C_5H_{11}Br \)). \( Y \) undergoes hydrolysis via an \( S_N1 \) mechanism. What is \( X \)?

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

- \( S_N1 \) reactions favor tertiary carbocations due to their stability.
- The order of carbocation stability: Tertiary \(>\) Secondary \(>\) Primary.
- Markovnikov’s rule applies: The Br adds to the most substituted carbon.

Answer 1

The Correct Option is D

Step 1: Understanding the Reaction 
- The given alkene X (\( C_5H_{10} \)) reacts with HBr to form Y (\( C_5H_{11}Br \)). 
- The formed Y undergoes hydrolysis via the \( S_N1 \) mechanism. 

Step 2: Identifying the \( S_N1 \) Mechanism 
- The \( S_N1 \) mechanism involves the formation of a stable carbocation intermediate. 
- Tertiary carbocations are more stable than secondary and primary carbocations. 

Step 3: Analyzing the Given Alkenes 
- Option (1): Forms a primary carbocation upon HBr addition (Least stable) 
- Option (2): Forms a secondary carbocation (Less stable) 
- Option (3): Forms a primary carbocation, unfavorable for \( S_N1 \) 
- Option (4): Forms a tertiary carbocation, which is the most stable 

Step 4: Conclusion 
- Since the most stable tertiary carbocation is formed in option (4), this is the correct alkene \( X \).