Question

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

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

Hint:

Remember the reactivity vs. selectivity principle: Chlorination is highly reactive but poorly selective (gives mixtures). Bromination is less reactive but highly selective (almost exclusively yields the most substituted product)!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks for the major product of the photochemical bromination of an alkane (propane).

Step 2: Key Formula or Approach:
Halogenation of alkanes in the presence of UV light proceeds via a free radical substitution mechanism.
Bromination is highly selective. The abstracting bromine radical will selectively remove the hydrogen atom that leaves behind the most stable carbon radical.
The stability of free radicals follows the order: Tertiary ($3^\circ$) $>$ Secondary ($2^\circ$) $>$ Primary ($1^\circ$).

Step 3: Detailed Explanation:
Propane ($\text{CH}_3\text{-CH}_2\text{-CH}_3$) has two types of hydrogen atoms:
1. Six primary ($1^\circ$) hydrogens on the terminal methyl groups.
2. Two secondary ($2^\circ$) hydrogens on the central methylene group.
Removal of a primary hydrogen forms a primary propyl radical.
Removal of a secondary hydrogen forms a secondary isopropyl radical, which is significantly more stable due to greater hyperconjugation.
Because bromination is highly selective for the more stable radical, the major product is formed by substitution at the central carbon.
The reaction is: $\text{CH}_3\text{-CH}_2\text{-CH}_3 + \text{Br}_2 \xrightarrow{UV} \text{CH}_3\text{-CH(Br)-CH}_3 + \text{HBr}$
The predominant product is 2-bromopropane.

Step 4: Final Answer:
The predominantly formed product is 2-bromopropane, which corresponds to option (A).