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: Understanding the Concept:
The reaction of an alkane with a halogen in the presence of ultraviolet (UV) light proceeds via a free radical substitution mechanism. Halogenation, and specifically bromination, is highly selective regarding which hydrogen atom is replaced.
Step 2: Key Formula or Approach:
Approach: Analyze the types of hydrogen atoms in the alkane, determine the relative stabilities of the possible free radical intermediates, and use the high selectivity of bromination to predict the major product.
Step 3: Detailed Explanation:
Propane ($\text{CH}_3-\text{CH}_2-\text{CH}_3$) has two distinct types of hydrogen atoms: 1. Six primary ($1^\circ$) hydrogens on the terminal carbon atoms. 2. Two secondary ($2^\circ$) hydrogens on the middle carbon atom. During the propagation step of free radical substitution, a hydrogen atom is abstracted to form an alkyl radical. - Abstraction of a $1^\circ$ hydrogen forms a primary free radical ($1^\circ$ radical). - Abstraction of a $2^\circ$ hydrogen forms a secondary free radical ($2^\circ$ radical). The stability of free radicals follows the order: $3^\circ>2^\circ>1^\circ$. Therefore, the secondary isopropyl radical is significantly more stable than the primary n-propyl radical due to hyperconjugation. Bromination is highly selective and regioselective, heavily favoring the pathway that forms the most stable intermediate radical. Consequently, the major product is formed from the more stable secondary radical. \[ \text{CH}_3-\text{CH}_2-\text{CH}_3 + \text{Br}_2 \xrightarrow{\text{UV}} \text{CH}_3-\text{CH}(\text{Br})-\text{CH}_3 \, (\text{Major}) + \text{CH}_3-\text{CH}_2-\text{CH}_2\text{Br} \, (\text{Minor}) \] The predominant product is 2-bromopropane.
Step 4: Final Answer:
The predominant product formed is 2-Bromopropane.