The increasing order of the rate of reaction | vedclass

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(D) The reaction with $HNO_3 / H_2SO_4$ is an electrophilic aromatic substitution (nitration). The rate of this reaction depends on the electron densi

Vedclass · Accepted Answer

$i < ii < iii$

Vedclass · Answer

(D) The reaction with $HNO_3 / H_2SO_4$ is an electrophilic aromatic substitution (nitration). The rate of this reaction depends on the electron density of the benzene ring. Electron-donating groups $(D)$ increase the electron density and thus increase the rate of reaction,while electron-withdrawing groups $(W)$ decrease the electron density and decrease the rate of reaction. In the given structures: $(i)$ has two electron-withdrawing groups $(W=2, D=2)$. (ii) has one electron-withdrawing group and three electron-donating groups $(W=1, D=3)$. (iii) has four electron-donating groups $(W=0, D=4)$. Since the rate of electrophilic substitution increases with an increase in the number of electron-donating groups,the order of reactivity is $(i) < (ii) < (iii)$.

The increasing order of the rate of reaction with $HNO_3 / H_2SO_4$ is:

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