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In electrophilic aromatic substitution nitration, what is the correct order of reaction rates for benzene rings with H, D, and T (hydrogen, deuterium, tritium) substituents?

1. k(C6H6) > k(C6D6) > k(C6T6)
2. k(C6H6) < k(C6D6) < k(C6T6)
3. k(C6H6) = k(C6D6) = k(C6T6)
4. k(C6H6) > k(C6D6) < k(C6T6)

Correct answer: k(C6H6) = k(C6D6) = k(C6T6)

Solution

In EAS nitration, the rate-determining step is attack of NO2⁺ on the aromatic ring to form the Wheland intermediate. Deprotonation (breaking C-H/D/T) is a fast step. Because the bond to H/D/T is not broken in the slow step, there is no kinetic isotope effect and all three rates are equal.

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