Among the following statements on the nitration of aromatic compounds, the false one is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → The rate of nitration of benzene is greater than that of hexadeuterobenzene
Nitration is an electrophilic substitution in two steps. The nitronium ion \( \ce{NO2+} \) adds to the ring and forms the arenium ion; this slow step decides the rate. The ion then loses \( \ce{H+} \) (or \( \ce{D+} \)) in a fast step:
\( \ce{C6H6 + NO2+} \xrightarrow{\text{slow}} \ce{[C6H6NO2]+} \xrightarrow[\text{fast}]{-\ce{H+}} \ce{C6H5NO2} \)
The C–H (or C–D) bond breaks only in the fast step, so replacing H by D does not change the rate.
Option (1) The nitration of benzene is almost the same as that of hexadeuterobenzene — A true statement about nitration. The rate-determining step does not break a C–H or C–D bond, so the two rates are almost equal.
Option (2) The rate of nitration of toluene is greater than that of benzene — A true statement about nitration. The –\( \ce{CH3} \) group releases electrons (+I effect and hyperconjugation) and activates the ring, so toluene is nitrated faster.
Option (3) The rate of nitration of benzene is greater than that of hexadeuterobenzene — The false statement about nitration. There is no isotope effect here: benzene and \( \ce{C6D6} \) are nitrated at almost the same rate.
Option (4) Nitration is an electrophilic substitution reaction — A true statement about nitration. The attacking species \( \ce{NO2+} \) is an electrophile, and it replaces a hydrogen atom of the ring.
Note: The question is from beyond the NCERT textbook.