Question

The reaction of alkoxide ion with alkyl halide to form ether is known as:
A) Wurtz reaction.
B) Kolbe’s reaction.
C) Williamson’s synthesis.
D) Perkin reaction.

Answer 1

We know that Williamson ether synthesis is the common method for the synthesis of ether. This synthesis involves nucleophilic displacement of a halide ion or other good leaving group by an alkoxide ion.

Complete answer:
We know that Perkin Reaction is a chemical reaction that gives -unsaturated aromatic acid. A reaction between aromatic aldehydes, the aliphatic acid anhydride, and the alkali salt of the acid to give cinnamic acid derivatives is called Perkin reaction. Hence option D is incorrect.
In Wurtz reaction the higher alkanes are prepared by heating the alkyl halides with sodium metal. Hence option A is incorrect.
Let us see Kolbe’s reaction,
We need to know that when sodium phenoxide is heated with carbon dioxide under pressure a carboxyl group is introduced in ortho-position to the hydroxyl group. If the potassium phenoxide is used instead of sodium phenoxide para-isomer is obtained as the primary product.
Thus, option B is incorrect.
Now, we discuss Williamson’s synthesis,
We have to remember that the reaction of alkoxide ion with alkyl halide to form ether is known as Williamson's synthesis. Williamson's synthesis is used to prepare diethyl ether. This is the industrial process for making ethers. Sodium ethoxide is heated with ethyl iodide to form diethyl ether. We can write the chemical equation for this reaction as,
${C_2}{H_5}ONa + {C_2}{H_5}I \to {C_2}{H_5} - O - {C_2}{H_5} + NaI$
Thus option C is correct.

Note:
Now we discuss the limitations of Williamson Ether Synthesis as,
-In the presence of the alkoxide, tertiary alkyl halides or sterically hindered primary or secondary alkyl halides tend to undergo ${E_2}$ elimination.
-Alkali phenoxide may undergo C-alkylation in addition to expected O-alkylation.