The rate of o-nitration of the following comp | vedclass

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(A) The rate-determining step in electrophilic aromatic substitution (nitration) is the formation of the sigma complex (arenium ion). In the case of $

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$I > II > III$

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(A) The rate-determining step in electrophilic aromatic substitution (nitration) is the formation of the sigma complex (arenium ion). In the case of $o-$nitration of toluene and its deuterated derivatives,the $C-H$ (or $C-D$) bond at the ortho position must be broken or involved in the transition state. Due to the primary kinetic isotope effect,the $C-H$ bond is cleaved more easily than the $C-D$ bond because the $C-H$ bond has a higher zero-point energy and a lower activation energy for the rate-determining step. Therefore,the presence of deuterium at the ortho position slows down the rate of nitration. Compound $(I)$ has two $H$ atoms at the ortho positions,$(II)$ has one $H$ and one $D$ at the ortho positions,and $(III)$ has two $D$ atoms at the ortho positions. Thus,the rate of $o-$nitration follows the order: $(I) > (II) > (III)$.

The rate of $o-$nitration of the following compounds,$(I)$ toluene,$(II)$ $2-D-$toluene,and $(III)$ $2,6-D_2-$toluene,is in the following order:

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