Question

Iodoethane reacts with X to form diethyl ether. What is X?
A. NaOH
B. ${H_2}S{O_4}$
C. ${C_2}{H_5}ONa$
D. $N{a_2}{S_2}{O_3}$

Answer 1

The Williamson Ether synthesis is the easiest, and perhaps the fastest, way to create ethers. The Williamson ether reaction follows an SN2 bimolecular nucleophilic substitution mechanism. In this reaction in the question, iodoethane reacts with sodium ethoxide and diethyl ether is formed.

Complete step by step answer:
Williamson Ether Reactions involve an alkoxide which reacts with a primary haloalkane or a sulfonate ester. Alkoxides consist of the conjugate base of an alcohol and are composed of an R group bonded to an oxygen atom. They are often written as $R - {O^-}$, where R is the organic substituent.
$R - {O^ - } + {\text{ }}R’ {\text{ }} - {\text{ }}X \to R{\text{ }}-{\text{ }}O{\text{ }}-{\text{ }}R’{\text{ }} + {\text{ }}{X^ - }$
The reaction of sodium Ethoxide $\left( {{C_2}{H_5}ONa} \right)$ with chloroethane $({C_2}{H_5}I)$ to form diethyl ether $\left( {{C_2}{H_5}} \right)$ and sodium chloride $(NaI)$
So, this is Williamson's synthesis.

The reaction is:
${C_2}{H_5}ONa{\text{ }} + {\text{ }}{C_2}{H_5}I \to {C_2}{H_5}-{\text{ }}O{\text{ }} - {\text{ }}{C_2}{H_5} + {\text{ }}NaI$

Therefore, the correct answer is option (C).

Note: The Williamson ether reaction follows an ${S_N}2$ bimolecular nucleophilic substitution mechanism in which there is a backside attack of an electrophile by a nucleophile and it happens all at once. In order for the ${S_N}2$ reaction to take place there must be a good leaving group which is strongly electronegative, a halide. In this reaction, there is an alkoxide ion $\left( {R{O^ - }} \right)$ which acts as the nucleophile and attack the electrophilic carbon with the leaving group, which in most cases is an alkyl halide. Also, this reaction does not favour the formation of bulky ethers because of steric hindrance and predominant formation of alkenes instead.