The major products $P$ and $Q$ from the below reactions are : $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2 \stackrel{\mathrm{HBr}}{\longrightarrow} P$ $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2 \underset{\mathrm{H}_2 \mathrm{O}, \mathrm{H}_2 \mathrm{O}_2 / \mathrm{OH}^{-}}{\stackrel{\left(\mathrm{BH}_3\right)_2}{\longrightarrow}} Q$

• A

  
  
• Correct answer
  B

  $\begin{array}{ll}&&&&&&&P &&&&&&\quad Q\end{array}$
  
• C

  $\begin{array}{ll}P &Q \\ \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br} \quad \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}\end{array}$
• D

  $\begin{array}{ll}P & Q \\ \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br} \quad \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH}\end{array}$

Solution

In the reaction $(\mathrm{s})$, $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2 \stackrel{\mathrm{HBr}}{\longrightarrow}(P)$ $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2 \underset{\mathrm{H}_2 \mathrm{O}, \mathrm{H}_2 \mathrm{O}_2 / \mathrm{OH}^{-}}{\stackrel{\left(\mathrm{BH}_3\right)_2}{\longrightarrow}} Q$ The products are,
because,
Br add according to, Markownikoff's rule, while in
$\mathrm{H}$ and $\mathrm{OH}$ and according to anti-Markownikoff's rule due to formation of radicals.

Hence, the correct answer is option: 'B'.