Q5 / 150 · MHT CET 2025 (20 Apr Shift 1) · Chemistry
The reaction of propane with bromine in presence of UV light predominantly forms
- A1-Bromopropane
- Correct answerB2-Bromopropane
- C1,2-dibromopropane
- D1,3-dibromopropane
Solution
The reaction between propane and bromine under UV light is a free radical substitution mechanism. The key selectivity arises during hydrogen abstraction, where bromine radicals target propane's six primary and two secondary hydrogens.
The secondary radical, $\mathrm{CH_3 CH\cdot CH_3 }$, is more stable than the primary radical due to hyperconjugation and inductive effects. Bromination is highly selective, favoring formation of the more stable radical with lower activation energy.
Thus, 2-bromopropane forms predominantly.
$\mathrm{CH_3 CH_2 CH_3 + Br\cdot \rightarrow CH_3 CH\cdot CH_3 + HBr}$
$\mathrm{CH_3 CH\cdot CH_3 + Br_2 \rightarrow CH_3 CHBrCH_3 + Br\cdot}$
The major product is 2-bromopropane.
Hence, the correct answer is option: 'B'.