The addition of hydrogen bromide to 1,3-butadiene at two different temperatures give two products in different amounts as given below.
$\begin{array}{lll}<br>\text { 1,3-butadiene }+\mathrm{HBr} \\ \text { 1,3-butadiene }+\mathrm{HBr} & \xrightarrow{-80{ }^{\circ} \mathrm{C}} & \text { 1-bromo-2-butene }(\mathrm{p} \%)+\text { 3-bromo-1-butene }(\mathbf{q} \%)<br>\end{array}$

In these reactions, the first step is the addition of proton to one of the double bonds to form carbocation. At a lower temperature, the more stable carbocation is immediately attacked by bromide ion to form the product. At the higher temperature there is equilibrium among the products and the reactants, which decides the product. The correct statement(s) is/are

• Correct answer
  A

  $q \gt s$
• Correct answer
  B

  $\mathrm{r} \gt \mathrm{s}$
• Correct answer
  C

  1-Bromo-2-butene is more stable than 3 -bromo-1-butene
• D

  The activation energy required for the formation of 3-bromo-1-butene is higher than that for the formation of 1-bromo-2-butene

Hence, the correct answer is options: 'A', 'B', 'C'.