Question

Heating together sodium ethoxide and ethyl chloride will give?

Answer 1

We need to know that ethoxide is an organic anion that is the conjugate base of ethanol. Sodium ethoxide may either deprotonate the α-position of an ester molecule, forming an enolate, or the ester molecule may undergo a nucleophilic substitution. Ethyl chloride is a halohalide.

Complete answer:
Let’s start the question. It is asked to heat two compounds which are sodium ethoxide and ethyl chloride. First let’s see what is the chemical formula for both the structures.
${C_2}{H_5}Cl$: Ethyl chloride
${C_2}{H_5}ONa$: Sodium ethoxide
Now when we heat these two compounds together the reaction can be represented as below:
${C_2}{H_5}Cl + {C_2}{H_5}ONa\xrightarrow{\Delta }{C_2}{H_5}O{C_2}{H_5} + NaCl$
The reaction represents Williamson’s synthesis.
An alkyl halide is to be primary, so a nucleophilic substitution reaction occurs because with tertiary alkyl halides, alkenes are formed. Alkoxides are nucleophiles and strong bases. The Williamson ether synthesis is an $S{N_2}$ reaction in which an alkoxide ion is a nucleophile that displaces a halide ion from an alkyl halide to give an ether. $S{N_2}$ reactions are the nucleophilic substitution reactions having order two since two reactants are involved. Nucleophile attacks from one side leading to the production of another compound or racemic mixture is formed. The stability order for these types of reactions is primary, secondary and then tertiary.

Note:
We need to know that on reacting ethoxide and alkyl halide together it will lead to the formation of new functional groups i.e. ether. In the case of sodium ethoxide and ethyl chloride, diethyl ether is formed. In $S{N_2}$ reaction since there is no intermediate formed. There is a formation of a transition state in these types of reaction.