During nitration of benzene, the attacking electrophile is

• A

  $\mathrm{NO}_3^{-}$
• B

  $\mathrm{NO}_2^{-}$
• Correct answer
  C

  $\mathrm{NO}_2^{+}$
• D

  $\mathrm{HNO}_3$

Solution

During nitration of benzene the attacking electrophile is $\mathrm{NO}_2^{+}$. It is formed by the reaction between $\mathrm{HNO}_3$ and $\mathrm{H}_2 \mathrm{SO}_4$ :

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