# Rate of nitration of C6H6, C6D6 and C6T6

Why the reaction rate (nitration) of $\ce{C6H6}$, $\ce{C6D6}$ and $\ce{C6T6}$ are equal? Can anyone explain this using simple words?

• I suspect the rates seem to be the same within an error of measurement. – andselisk Nov 8 '17 at 5:33
• If you measured the rates carefully, you would probably see an inverse kinetic isotope effect (that is, the reaction will very slightly speed up as one goes from $\ce{C6H6}$ to $\ce{C6T6}$) because nitration involves an $sp^2$ to $sp^3$ change in the rate determining step. – levineds Nov 8 '17 at 7:03
• In simple words, H, D, and T are isotopes*, and from chemical point of view they are just the same. Only if you measure them with really, really great precision, you may notice some differences. – Ivan Neretin Nov 8 '17 at 8:04

The rate-determining step is the attack of the benzene aromatic ring onto the electrophile, and the loss of aromaticity is what causes the instability of the transition state, resulting in a higher activation energy and thus a slower rate of reaction. The subsequent deprotonation is after the rate-determining step and thus will not affect the rate of reaction, despite C-T > C-D > C-H bond strength.

In nitration of benzene or $\ce{C6D6}$, the rate determining or slowest step is the formation of the nitronium ion $\ce{NO2+}$ and not the formation of sigma complex as in the other electrophilic aromatic substitutions, therefore the rate is unaffected by the leaving of a hydrogen or deuterium atom.