Question

The reaction of propane with bromine in presence of UV light predominantly forms

• A. 1-Bromopropane
• B. 2-Bromopropane
• C. 1,2-dibromopropane
• D. 1,3-dibromopropane

Hint:

Radical stability order: $3^\circ > 2^\circ > 1^\circ$. Bromine is more selective than Chlorine, so it almost always picks the more stable position.

Answer 1

The Correct Option is B

Step 1: Concept
Free radical halogenation of alkanes involves the substitution of hydrogen by a halogen. The stability of the intermediate free radical determines the major product.

Step 2: Meaning
Predominantly refers to the major product formed in a regioselective reaction.

Step 3: Analysis
Propane ($\text{CH}_3\text{CH}_2\text{CH}_3$) has primary and secondary hydrogens. Bromination is highly selective. The secondary free radical ($\text{CH}_3\dot{\text{C}}\text{H}\text{CH}_3$) is more stable than the primary free radical ($\text{CH}_3\text{CH}_2\dot{\text{C}}\text{H}_2$). Therefore, bromine prefers to substitute the hydrogen on the second carbon.

Step 4: Conclusion
2-bromopropane is the major product due to the higher stability of the $2^\circ$ radical. Final Answer: (B)