Q5 / 150 · MHT CET 2025 (20 Apr Shift 1) · Chemistry

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

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

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