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Question

(B) The nitration of benzene involves the generation of the nitronium ion $(NO_2^{+})$ as the active electrophile. Step $1$: Sulfuric acid $(H_2SO_4)$

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$NO_2^{+}$

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(B) The nitration of benzene involves the generation of the nitronium ion $(NO_2^{+})$ as the active electrophile. Step $1$: Sulfuric acid $(H_2SO_4)$ acts as a proton donor and nitric acid $(HNO_3)$ acts as a base. The reaction is: $HNO_3 + H_2SO_4 \rightleftharpoons H_2NO_3^{+} + HSO_4^{-}$. Step $2$: The protonated nitric acid $(H_2NO_3^{+})$ loses a water molecule to form the nitronium ion: $H_2NO_3^{+} \rightarrow NO_2^{+} + H_2O$. This nitronium ion $(NO_2^{+})$ then attacks the benzene ring to form nitrobenzene.

The electrophile that participates in the nitration of benzene is:

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