Question

The replacement of diazonium group by fluorine is known as:
A.) Gattermann reaction
B.) Sandmeyer reaction
C.) Balz-Schiemann reaction
D.) Etard reaction

Answer 1

Hint: When diazonium salt is reacted with the fluoroboric acid ($HB{F}_4$) then the benzene diazonium fluoroborate ($C_6{H}_5-N_{2}B{F}_4$) is obtained which again on heating produce the required fluorobenzene ($C_6{H}_5-F$).

Complete answer:
In this question, the Gattermann reaction is a chemical reaction in which when diazonium salt is treated with a halogen acid ($HBrorHCl$) in the presence of copper powder then the corresponding benzene halide is formed. This reaction is used to produce benzene chloride and benzene bromide. It can be given as:
$C_6{H}_5-{N_2}^{+}{X}^{-}\underset{(X=Cl,Br)}{\overset{Cu+HX}{\to }}{C}_6{H}_5-X+N_2$
The Sandmeyer reaction is used to obtain benzene chloride, and benzene bromide in the product when benzene diazonium salt is reacted with $C{u}_2{X}_2/HX(X=Br,Cl)$. This reaction can be represented as:
$C_6{H}_5-{N_2}^{+}{X}^{-}\underset{(X=Cl,Br)}{\overset{C{u}_2{X}_2+HX}{\to }}{C}_6{H}_5-X+N_2$
The Balz-Schiemann reaction is used for the preparation of benzene fluoride. When benzene diazonium salt is reacted with fluoroboric acid ($HB{F}_4$) and then heated to form benzene chloride. In it the diazonium group is replaced by the fluorine. This reaction can be represented as:
$C_6{H}_5-{N_2}^{+}C{l}^{-}\stackrel{HB{F}_4}{\to }{C}_6{H}_5-{N_2}^{+}B{F_4}^{-}\stackrel{\mathrm{\Delta }}{\to }{C}_6{H}_5-F+B{F}_3$
The Etard reaction is used for the preparation of benzaldehyde. In this reaction when toluene reacts with chromyl chloride ($Cr{O}_{2}C{l}_2$) then it gives corresponding benzaldehyde.
This reaction can be given as:
$C_6{H}_5-C{H}_3+Cr{O}_{2}C{l}_2\to C_6{H}_5-CHO$

Hence, option C.) is the correct answer.

Note:
Always remember that if we want to prepare benzene chloride and benzene bromide from the diazonium salt then it can be obtained by the Gattermann reaction or by Sandmeyer reaction.