The major product formed in the following rea | vedclass

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(C) The reaction is: $CH_{3}Br + CH_{3}CH_{2}ONa \rightarrow CH_{3}CH_{2}OCH_{3} + NaBr$. This reaction is known as the Williamson ether synthesis. In

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$CH_{3}CH_{2}OCH_{3}$

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(C) The reaction is: $CH_{3}Br + CH_{3}CH_{2}ONa \rightarrow CH_{3}CH_{2}OCH_{3} + NaBr$. This reaction is known as the Williamson ether synthesis. In this reaction,a primary alkyl halide $(CH_{3}Br)$ reacts with a sodium alkoxide $(CH_{3}CH_{2}ONa)$ via an $S_{N}2$ mechanism to form an ether $(CH_{3}CH_{2}OCH_{3})$,which is ethyl methyl ether. Since $CH_{3}Br$ is a primary halide,the $S_{N}2$ pathway is highly favorable.

The major product formed in the following reaction is: $CH_{3}Br + CH_{3}CH_{2}ONa \longrightarrow$

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