In chlorobenzene,the $-Cl$ group is deactivating; however,it still undergoes electrophilic aromatic substitution reactions. Column-$(I)$ contains the reagents and Column-$(II)$ contains the products. Match Column-$(I)$ with Column-$(II)$.
| Column-$(I)$ (Reagents) | Column-$(II)$ (Products) |
| :--- | :--- |
| $(A)$ $Cl_2$ / anhydrous $AlCl_3$ | $(i)$ $1$-chloro-$2$-methylbenzene + $1$-chloro-$4$-methylbenzene |
| $(B)$ conc. $HNO_3$ + conc. $H_2SO_4$ | $(ii)$ $2$-chlorobenzenesulfonic acid + $4$-chlorobenzenesulfonic acid |
| $(C)$ conc. $H_2SO_4$ and heat | $(iii)$ $1,2$-dichlorobenzene + $1,4$-dichlorobenzene |
| $(D)$ $CH_3COCl$ in presence of anhydrous $AlCl_3$ | $(iv)$ $2$-chloroacetophenone + $4$-chloroacetophenone |
| | $(v)$ $1$-chloro-$2$-nitrobenzene + $1$-chloro-$4$-nitrobenzene |
| Column-$(I)$ (Reagents) | Column-$(II)$ (Products) |
| :--- | :--- |
| $(A)$ $Cl_2$ / anhydrous $AlCl_3$ | $(i)$ $1$-chloro-$2$-methylbenzene + $1$-chloro-$4$-methylbenzene |
| $(B)$ conc. $HNO_3$ + conc. $H_2SO_4$ | $(ii)$ $2$-chlorobenzenesulfonic acid + $4$-chlorobenzenesulfonic acid |
| $(C)$ conc. $H_2SO_4$ and heat | $(iii)$ $1,2$-dichlorobenzene + $1,4$-dichlorobenzene |
| $(D)$ $CH_3COCl$ in presence of anhydrous $AlCl_3$ | $(iv)$ $2$-chloroacetophenone + $4$-chloroacetophenone |
| | $(v)$ $1$-chloro-$2$-nitrobenzene + $1$-chloro-$4$-nitrobenzene |
(A-III, B-V, C-II, D-IV) The electrophilic aromatic substitution reactions of chlorobenzene occur at the ortho and para positions due to the resonance effect of the $-Cl$ atom.
$(A)$ Chlorination: $Cl_2$ / anhydrous $AlCl_3$ gives $1,2$-dichlorobenzene and $1,4$-dichlorobenzene. Thus,$(A \rightarrow iii)$.
$(B)$ Nitration: conc. $HNO_3$ + conc. $H_2SO_4$ gives $1$-chloro-$2$-nitrobenzene and $1$-chloro-$4$-nitrobenzene. Thus,$(B \rightarrow v)$.
$(C)$ Sulfonation: conc. $H_2SO_4$ and heat gives $2$-chlorobenzenesulfonic acid and $4$-chlorobenzenesulfonic acid. Thus,$(C \rightarrow ii)$.
$(D)$ Friedel-Crafts Acylation: $CH_3COCl$ / anhydrous $AlCl_3$ gives $2$-chloroacetophenone and $4$-chloroacetophenone. Thus,$(D \rightarrow iv)$.
$(A)$ Chlorination: $Cl_2$ / anhydrous $AlCl_3$ gives $1,2$-dichlorobenzene and $1,4$-dichlorobenzene. Thus,$(A \rightarrow iii)$.
$(B)$ Nitration: conc. $HNO_3$ + conc. $H_2SO_4$ gives $1$-chloro-$2$-nitrobenzene and $1$-chloro-$4$-nitrobenzene. Thus,$(B \rightarrow v)$.
$(C)$ Sulfonation: conc. $H_2SO_4$ and heat gives $2$-chlorobenzenesulfonic acid and $4$-chlorobenzenesulfonic acid. Thus,$(C \rightarrow ii)$.
$(D)$ Friedel-Crafts Acylation: $CH_3COCl$ / anhydrous $AlCl_3$ gives $2$-chloroacetophenone and $4$-chloroacetophenone. Thus,$(D \rightarrow iv)$.
Explore More
Similar Questions
The reaction between chlorobenzene and chloral in the presence of concentrated sulphuric acid produces $:$
Easy
View SolutionIdentify $Y$ in the following reaction.
Difficult
View SolutionIn the following reactions,which one is $NOT$ correct?
What is $A$ in the following reaction?
Benzyl chloride + $Cl_2 \xrightarrow{FeCl_3} A + \text{p-chlorobenzyl chloride}$
Benzyl chloride + $Cl_2 \xrightarrow{FeCl_3} A + \text{p-chlorobenzyl chloride}$
Which of the following has the highest normal boiling point?
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo