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(A) $1$. The reaction between sodium methoxide $(CH_3ONa)$ and bromoethane $(CH_3CH_2Br)$ is a Williamson ether synthesis.$2$. $CH_3ONa$ acts as a str

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Formation of $CH_3OCH_2CH_3 + NaBr$

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(A) $1$. The reaction between sodium methoxide $(CH_3ONa)$ and bromoethane $(CH_3CH_2Br)$ is a Williamson ether synthesis.$2$. $CH_3ONa$ acts as a strong nucleophile $(CH_3O^-)$.$3$. The nucleophile attacks the electrophilic carbon of the ethyl bromide $(CH_3CH_2Br)$ via an $S_N2$ mechanism.$4$. The bromide ion $(Br^-)$ acts as a leaving group, resulting in the formation of ethyl methyl ether $(CH_3OCH_2CH_3)$ and sodium bromide $(NaBr)$.$5$. The balanced chemical equation is: $CH_3ONa + CH_3CH_2Br \rightarrow CH_3OCH_2CH_3 + NaBr$.

What is the product of the reaction between $CH_3ONa$ and $CH_3CH_2Br$?
$CH_3ONa + CH_3CH_2Br \rightarrow$ ?

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What is the product of the reaction between $CH_3ONa$ and $CH_3CH_2Br$?$CH_3ONa + CH_3CH_2Br \rightarrow$ ?