Question

The decreasing order of the rate of nitration of the following compounds is:
$I$. Benzene $II.C_6{D}_6$
$III$. Nitrobenzene $IV.$ Chlorobenzene

• A. $\left(I\right)>\left(II\right)>\left(III\right)>\left(IV\right)$
• B. $\left(I\right)>\left(II\right)>\left(IV\right)>\left(III\right)$
• C. $\left(I\right)=\left(II\right)>\left(IV\right)>\left(III\right)$
• D. $\left(I\right)=\left(II\right)>\left(III\right)>\left(IV\right)$

Answer 1

The correct option is C $\left(I\right)=\left(II\right)>\left(IV\right)>\left(III\right)$

Rate is faster when the substituent activates the ring ($+I$ or/and $+R$,o/p) and the rate is slower when the substituent deacivates the ring ($-I,-R;m$). Halogen deactivates the ring $(-I,+R,-I>+R)$ but the orientation is o/p
Rate of $C_6{H}_6=C_6{D}_6$, since no kinetic isotope effect is observed when $H$ is replaced by $D$.
Hence, the order is as given in (c)