The synthesis of alkyl fluorides is best accomplished by:
A ) Free radical fluorination
B ) sandmeyer’s reaction
C ) Finkelstein reaction
D ) Swarts reaction
Answer
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Hint: To prepare alkyl fluorides, consider the reaction between alkyl halide (chloride or bromide) and silver fluoride (or antimony trifluoride or antimony pentafluoride or cobalt difluoride or mercurous fluoride) . In this reaction, the halogen atom is replaced with fluorine atoms.
Complete step by step answer:
The general formula of alkyl fluorides is \[{\text{R}} - {\text{F}}\]. The general formula of alkyl halides is \[{\text{R}} - {\text{X}}\].
Here, X is either fluorine, or chlorine, or bromine or iodine. R is the alkyl group.
A ) Free radical fluorination reaction is not used for fluorine. Only free radical chlorination or bromination occurs.
\[{\text{C}}{{\text{H}}_4}{\text{ }}\xrightarrow[{h\upsilon }]{{{\text{C}}{{\text{l}}_2}}}{\text{ CH}_3} - {\text{Cl}} \\
{\text{C}}{{\text{H}}_4}{\text{}}\xrightarrow[{h\upsilon}]{{{\text{B}}{{\text{r}}_{\text{2}}}}}{\text{ CH}_3} - {\text{Br}} \\ \]
Hence, free radical fluorination is not used for preparation of alkyl fluorides.
B ) Sandmeyer’s reaction involves heating aryl azide with cuprous halide to form halobenzene.
Thus, sandmeyer’s reaction is used for the preparation of halobenzene. It cannot be used for the preparation of alkyl fluorides.
C ) Finkelstein reaction involves reaction of alkyl halide with sodium iodide to prepare alkyl iodide.
\[{\text{R}} - {\text{X + NaI }} \to {\text{ R}} - {\text{I + NaX}}\]
Instead of sodium iodide, you can also use sodium bromide or sodium chloride. But you cannot use sodium fluoride as fluoride ion will not displace chloride, bromide or iodide ion as fluoride ion is better leaving the group than chloride, bromide or iodide ion. Better leaving group than fluoride ion is needed for this reaction to occur. Hence, the Finkelstein reaction cannot be used to prepare alkyl fluorides.
D ) Swarts reaction involves the reaction between alkyl halide and silver fluoride (or antimony trifluoride or antimony pentafluoride or cobalt difluoride or mercurous fluoride) to form alkyl fluoride.
\[{\text{R}} - {\text{X }}\xrightarrow[{{\text{or Sb}}{{\text{F}}_3}}]{{{\text{AgF}}}}{\text{ R}} - {\text{F}}\]
Hence, swarts reaction can be used for the synthesis of alkyl fluorides.
Hence, the correct answer is the option D ) Swarts reaction.
Note: The yield of swarts reaction is low. The bond between metal and fluorine breaks and a new bond between carbon and fluorine is formed. The leaving group (chlorine or bromine) then forms a bond with metal. Swart reagent is the mixture of antimony trifluoride and chlorine.
Complete step by step answer:
The general formula of alkyl fluorides is \[{\text{R}} - {\text{F}}\]. The general formula of alkyl halides is \[{\text{R}} - {\text{X}}\].
Here, X is either fluorine, or chlorine, or bromine or iodine. R is the alkyl group.
A ) Free radical fluorination reaction is not used for fluorine. Only free radical chlorination or bromination occurs.
\[{\text{C}}{{\text{H}}_4}{\text{ }}\xrightarrow[{h\upsilon }]{{{\text{C}}{{\text{l}}_2}}}{\text{ CH}_3} - {\text{Cl}} \\
{\text{C}}{{\text{H}}_4}{\text{}}\xrightarrow[{h\upsilon}]{{{\text{B}}{{\text{r}}_{\text{2}}}}}{\text{ CH}_3} - {\text{Br}} \\ \]
Hence, free radical fluorination is not used for preparation of alkyl fluorides.
B ) Sandmeyer’s reaction involves heating aryl azide with cuprous halide to form halobenzene.
Thus, sandmeyer’s reaction is used for the preparation of halobenzene. It cannot be used for the preparation of alkyl fluorides.
C ) Finkelstein reaction involves reaction of alkyl halide with sodium iodide to prepare alkyl iodide.
\[{\text{R}} - {\text{X + NaI }} \to {\text{ R}} - {\text{I + NaX}}\]
Instead of sodium iodide, you can also use sodium bromide or sodium chloride. But you cannot use sodium fluoride as fluoride ion will not displace chloride, bromide or iodide ion as fluoride ion is better leaving the group than chloride, bromide or iodide ion. Better leaving group than fluoride ion is needed for this reaction to occur. Hence, the Finkelstein reaction cannot be used to prepare alkyl fluorides.
D ) Swarts reaction involves the reaction between alkyl halide and silver fluoride (or antimony trifluoride or antimony pentafluoride or cobalt difluoride or mercurous fluoride) to form alkyl fluoride.
\[{\text{R}} - {\text{X }}\xrightarrow[{{\text{or Sb}}{{\text{F}}_3}}]{{{\text{AgF}}}}{\text{ R}} - {\text{F}}\]
Hence, swarts reaction can be used for the synthesis of alkyl fluorides.
Hence, the correct answer is the option D ) Swarts reaction.
Note: The yield of swarts reaction is low. The bond between metal and fluorine breaks and a new bond between carbon and fluorine is formed. The leaving group (chlorine or bromine) then forms a bond with metal. Swart reagent is the mixture of antimony trifluoride and chlorine.
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