3-Phenylpropene on reaction with $\mathrm{HBr}$ gives (as a major product)

3-Phenylpropene on reaction with $\mathrm{HBr}$ gives (as a major product)
  1. $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_3$
  2. $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_2 \mathrm{CH}_3$
  3. $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}$
  4. $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}(\mathrm{Br}) \mathrm{CH}=\mathrm{CH}_2$.

Solution

According to Markownikoff's rule, the negative part of the unsymmetrical reagent adds to less hydrogenated (more substituted) carbon atom of the double bond.

Asked in: NEET 2009 (Mains)

Practice more Halogen Derivatives questions on Aicharya