Question

List - I	List - II
a)$Ar{\stackrel{+}{N}}_2{X}^{-}\underset{HBr}{\overset{CuBr}{\to }}$	1)ArF
b)$Ar{\stackrel{+}{N}}_2{X}^{-}\underset{\Delta }{\overset{HB{F}_4}{\to }}$	2)ArH
c)$Ar{\stackrel{+}{N}}_2{X}^{-}\stackrel{C_2{H}_{5}OH}{\to }$	3)ArOH
d)$Ar{\stackrel{+}{N}}_2{X}^{-}\stackrel{H_{2}O}{\to }$	4)ArBr

• A. $a-4,b-1,c-2,d-3$
• B. $a-1,b-2,c-3,d-4$
• C. $a-2,b-3,c-4,d-1$
• D. $a-3,b-4,c-1,d-2$

Answer 1

The correct option is A $a-4,b-1,c-2,d-3$

$Ar{\stackrel{+}{N}}_2{X}^{-}\underset{HBr}{\overset{CuBr}{\to }}ArBr$.

This is Sandmeyer reaction in which the diazonium group is replaced by bromine.

$Ar{\stackrel{+}{N}}_2{X}^{-}\underset{\Delta }{\overset{HB{F}_4}{\to }}ArF$.

This is Balz-Schiemann's reaction in which the diazonium group is replaced by fluorine.

$Ar{\stackrel{+}{N}}_2{X}^{-}\stackrel{C_2{H}_{5}OH}{\to }ArH$.

In this reaction, the diazonium group is replaced by hydrogen.

$Ar{\stackrel{+}{N}}_2{X}^{-}\stackrel{H_{2}O}{\to }ArOH$.

In this reaction, the diazonium group is replaced by -OH group.

Hence, option A is correct.