Question

Write the structure of the major organic product in each of the following reactions:
${C_6}{H_5}ONa + {C_2}{H_5}Cl \to$

Answer 1

The reaction of an organohalide with a deprotonated alcohol yields an ether and this process is known as Williamson’s ether synthesis. It is a nucleophilic substitution reaction.
The reaction is as follows: $R - X + R - ONa \to R - O - R + NaX$ where R-X is alkyl halide, R-O-R is an ether and R-O-Na is an alkoxide .

Complete step-by-step answer: We know that Williamson ether synthesis is a SN2 reaction in which an alkoxide ion displaces a halide ion from an alkyl halide to give an ether. It mainly occurs for primary alkyl halide due to less steric hindrance and occurs with inversion of configuration at chiral centres. So the reaction between sodium phenoxide ion which is ${C_6}{H_5}ONa$ and ethyl chloride which is ${C_2}{H_5}Cl$ will be as follows
${C_6}{H_5}ONa + {C_2}{H_5}Cl \to {C_6}{H_5} - O - {C_2}{H_5} + NaCl$. It is a nucleophilic substitution reaction where phenoxide ion attacks the alkyl halide . The C-Br bond is broken and a new C-O bond is formed . Then the ether is formed and we get ethyl phenyl ether as the major product and sodium chloride as the side product.

Hence the major organic product of this reaction is ethyl phenyl ether that is ${C_6}{H_5} - O - {C_2}{H_5}$

Note: In Williamson’s ether synthesis the nucleophile used is $R{O^ - }$ instead of ROH as the former is a better nucleophile as it is good at holding negative charge and the reaction will be much faster with it due to the higher electron density on the nucleophile. As this reaction is an SN2 reaction primary alkyl halides work well due to the steric hindrance factor.