Arrange the following in increasing order of rate of Nitration:

on the basis that the rate determining step is the formation of the carbonium ion, then, with benzene as the reference compound, the rate is faster when the substituent activates the ring (+/ and / or +R; o/p) and the rate is slower when the substituent deactivates the ring (-I and / or -R ; m). Halogen is an exception; it deactivates the ring (-I, + R ; - I > +R), but the orientation is of p. Hence, the order of increasing rate of nitration is :
$$\begin{gathered} \begin{array}{cl}\underset{(-\mathrm{I}, -\mathrm{R})}{{\mathrm{PhNO}}_2}<& \underset{(-\mathrm{I}, -\mathrm{R}); \\ -\mathrm{I} > +\mathrm{R})}{\mathrm{PhCl}}<\mathrm{PhH}={\mathrm{C}}_6{\mathrm{D}}_6<\underset{(+\mathrm{I})}{\mathrm{PhMe}}\end{array}\begin{array}{} \end{array} \end{gathered}$$
Since proton removal from the carbonium ion is fast, no kinetic isotope effect is observed when H is replaced by D. Hence, the rates of nitration of PhH and C₆D₆ are the same. Had a kinetic isotope effect been observed, this would mean that the second step had now become rate determining.