The increasing order of the rate of reaction | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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:

Similar Questions

Consider the following reactions:
$Ethylbenzene \xrightarrow[(ii) H_3O^{+}]{(i) KMnO_4-KOH/\Delta} Y$
$Propylbenzene \xrightarrow[(ii) H_3O^{+}]{(i) KMnO_4-KOH/\Delta} Z$
What are the structures of $Y$ and $Z$?

Which of the following compounds undergoes the Friedel-Crafts reaction?

Which of the following is anti-aromatic?

The product $(Z)$ of the following reaction is

Identify the molecular formula of pyridine from the following.

Vedclass Test Series

Exam Paper Generator

Online Exam Module