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JEE Main levelSingle correct optionDifficulty 2 of 3hc210

In the nitration of benzene with the mixed acid, the slowest (rate-determining) step is:

AFormation of $NO X_{2} X^{+}$ from $HNO X_{3}$ and $H X_{2} SO X_{4}$
BAttack of $NO X_{2} X^{+}$ on the ring to give the sigma complex (arenium ion)
CLoss of a proton from the sigma complex
DProtonation of benzene by $H X_{2} SO X_{4}$

Answer: B

The attack of the electrophile destroys aromaticity and forms the high-energy arenium ion, so it has the highest barrier. Loss of the proton restores the aromatic ring quickly.

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