
The replacement of diazonium group by fluorine is known as:
A.) Gattermann reaction
B.) Sandmeyer reaction
C.) Balz-Schiemann reaction
D.) Etard reaction
Answer
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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 ($HBr{\text{ }}or{\text{ }}HCl$) 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:
${C_6}{H_5} - {N_2}^ + {X^ - }\xrightarrow[{(X = Cl,Br)}]{{Cu + HX}}{C_6}{H_5} - X + {N_2}$
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:
${C_6}{H_5} - {N_2}^ + {X^ - }\xrightarrow[{(X = Cl,Br)}]{{C{u_2}{X_2} + HX}}{C_6}{H_5} - X + {N_2}$
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:
${C_6}{H_5} - {N_2}^ + C{l^ - }\xrightarrow{{HB{F_4}}}{C_6}{H_5} - {N_2}^ + B{F_4}^ - \xrightarrow{\Delta }{C_6}{H_5} - F + B{F_3}$
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.
Complete answer:
In this question, the Gattermann reaction is a chemical reaction in which when diazonium salt is treated with a halogen acid ($HBr{\text{ }}or{\text{ }}HCl$) 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:
${C_6}{H_5} - {N_2}^ + {X^ - }\xrightarrow[{(X = Cl,Br)}]{{Cu + HX}}{C_6}{H_5} - X + {N_2}$
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:
${C_6}{H_5} - {N_2}^ + {X^ - }\xrightarrow[{(X = Cl,Br)}]{{C{u_2}{X_2} + HX}}{C_6}{H_5} - X + {N_2}$
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:
${C_6}{H_5} - {N_2}^ + C{l^ - }\xrightarrow{{HB{F_4}}}{C_6}{H_5} - {N_2}^ + B{F_4}^ - \xrightarrow{\Delta }{C_6}{H_5} - F + B{F_3}$
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.
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