Question

Identify product ‘C’ in the following reaction:
\[ \text{Propylene dibromide} \xrightarrow[\text{Alcohol}]{\text{Zn},\,\Delta} A \xrightarrow{\text{HBr}} B \xrightarrow[\text{ether}]{\text{Na}} C \]

• A. 2-Bromobutane
• B. Isobutane
• C. 2,3-Dimethylbutane
• D. 1,2-Dibromobutane

Hint:

Wurtz reaction couples two identical alkyl halide molecules to form higher alkanes.

Answer 1

The Correct Option is C

Step 1: First reaction (dehalogenation).
Propylene dibromide on treatment with Zn/alcohol undergoes dehalogenation to form propene (A).
Step 2: Addition of HBr.
Propene reacts with HBr (Markovnikov addition) to form 2-bromopropane (B).
Step 3: Wurtz reaction.
2-Bromopropane reacts with sodium in dry ether to undergo coupling, forming 2,3-dimethylbutane.
Step 4: Conclusion.
Hence, the final product \(C\) is 2,3-dimethylbutane.