The nitration of toluene takes place at the ortho, para and meta positions, giving ortho-, para-, and meta- nitrotoluenes, respectively. Assume that the meta product is formed in negligible amount and that the ortho and para substitutions are irreversible first-order reactions. Let the initial toluene concentration be $[\mathrm{A}]_0$ and $k_1$ and $k_2$ be the ortho- and para- rate constants, respectively. Choose the correct statement(s).

• Correct answer
  A

  The rate of formation of the ortho substituted product is given by $k_1[\mathrm{~A}]_0 \mathrm{e}^{-\left(k_1+k_2\right) t}$, where $t$ is the time.
• Correct answer
  B

  If the rate constant for the ortho product formation is twice that for the para product formation, then $[\mathrm{o}-$ nitrotoluene $]=2[\mathrm{p}$-nitrotoluene $]$, at any time.
• Correct answer
  C

  Nitration at the ortho- and para- positions is favorable compared to that at the meta position as the ortho- and para- intermediates are more stabilised than the meta-intermediate.
• D

  If the methyl group of toluene is replaced by tertiary butyl group, more ortho product is expected.

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