Treatment of 2-bromo-2-methylbutane with sodium ethoxide in ethanol at $70^{\circ} \mathrm{C}$ gives a mixture of two products $\mathrm{X}_1$ and $\mathrm{X}_2$ through one step processes. The energy profile diagrams of the two processes are shown below.

• Correct answer
  A

  Product $\mathrm{X}_1$ is 2-methyl-1-butene.
• B

  Product $X_2$ is formed faster than $X_1$.
• Correct answer
  C

  If the same reaction is carried out using potassium tertbutoxide in tertbutanol at $75^{\circ} \mathrm{C}$, the major product will be 2-methyl-1-butene.
• D

  In the reaction using sodium ethoxide in ethanol large amount of 2-ethoxy-2-methylbutane is formed.

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