Answer
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Hint:Here metallic fluoride means ‘MF’ where M is the metal. Preparation of alkyl fluoride from alkyl chloride reaction is a nucleophilic substitution reaction.
Complete step by step solution:
> Williamson synthesis: It is an organic reaction, forming ether from an organohalide and a de-protonated alcohol (alkoxide). An example:
\[~{{\left[ Na \right]}^{+}}{{[{{C}_{2}}{{H}_{5}}O]}^{-}}~+\text{ }{{C}_{2}}{{H}_{5}}Cl\text{ }\to \text{ }{{C}_{2}}{{H}_{5}}O{{C}_{2}}{{H}_{5}}~+\text{ }{{\left[ Na \right]}^{+}}\left[ Cl \right]\]
> Finkelstein reaction: A\[{{S}_{N}}2\] reaction in which one halogen atom (the leaving group) is replaced by another halogen atom (the nucleophile). In this example of the Finkelstein reaction, 1-chloro-2-phenylethane is treated with sodium iodide to produce 1-iodo-2-phenylethane.
> Swarts Reaction: It is generally used to get alkyl fluorides from alkyl chlorides or alkyl bromides. This is done by heating of the alkyl chloride/bromide in the presence of the fluoride of some heavy metals (\[AgF\], \[H{{g}_{2}}{{F}_{2}}\]).
\[R-X+AgF\to R-F+AgX\]
> Wurtz reaction: It is an organic chemical coupling reaction wherein sodium metal (Na) is reacted with two alkyl halides in the environment provided by a solution of dry ether in order to form a higher alkane along with a compound containing sodium and the halogen.
The main difference between Finkelstein reaction and Swarts reaction is that the end product of Finkelstein reaction is alkyl iodide whereas the end product of Swarts reaction is alkyl fluoride.
So, from the above explanation we can say that the correct option is “C”.
Note: And here the Wurtz reaction is limited to the synthesis of symmetric alkanes. If two dissimilar alkyl halides are taken as reactants, then the product is a mixture of alkanes. Don’t get confused between Finkelstein reaction and Swarts Reactions because they have different products.
Complete step by step solution:
> Williamson synthesis: It is an organic reaction, forming ether from an organohalide and a de-protonated alcohol (alkoxide). An example:
\[~{{\left[ Na \right]}^{+}}{{[{{C}_{2}}{{H}_{5}}O]}^{-}}~+\text{ }{{C}_{2}}{{H}_{5}}Cl\text{ }\to \text{ }{{C}_{2}}{{H}_{5}}O{{C}_{2}}{{H}_{5}}~+\text{ }{{\left[ Na \right]}^{+}}\left[ Cl \right]\]
> Finkelstein reaction: A\[{{S}_{N}}2\] reaction in which one halogen atom (the leaving group) is replaced by another halogen atom (the nucleophile). In this example of the Finkelstein reaction, 1-chloro-2-phenylethane is treated with sodium iodide to produce 1-iodo-2-phenylethane.
> Swarts Reaction: It is generally used to get alkyl fluorides from alkyl chlorides or alkyl bromides. This is done by heating of the alkyl chloride/bromide in the presence of the fluoride of some heavy metals (\[AgF\], \[H{{g}_{2}}{{F}_{2}}\]).
\[R-X+AgF\to R-F+AgX\]
> Wurtz reaction: It is an organic chemical coupling reaction wherein sodium metal (Na) is reacted with two alkyl halides in the environment provided by a solution of dry ether in order to form a higher alkane along with a compound containing sodium and the halogen.
The main difference between Finkelstein reaction and Swarts reaction is that the end product of Finkelstein reaction is alkyl iodide whereas the end product of Swarts reaction is alkyl fluoride.
So, from the above explanation we can say that the correct option is “C”.
Note: And here the Wurtz reaction is limited to the synthesis of symmetric alkanes. If two dissimilar alkyl halides are taken as reactants, then the product is a mixture of alkanes. Don’t get confused between Finkelstein reaction and Swarts Reactions because they have different products.
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