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: $CH_{3}Br + CH_{3}CH_{2}ONa \longrightarrow$

Similar Questions

Predict the major products of the following reaction:
$m$-methoxyphenethyl$-1-$phenylethyl ether + $HI$ (excess) $\xrightarrow{\text{heat}}$ ?

The following compound can readily be prepared by Williamson ether synthesis by reaction between:

$(CH_3)_3C-CH(O)CH_2 + CH_3SNa \xrightarrow{C_2H_5OH} ?$
The product formed is:

What is $X$ in the following reaction?
$C_2H_5OC_2H_5 + 4[H] \xrightarrow{Red \ P + HI} 2X + H_2O$

The acidic hydrolysis of ether $(X)$ shown below is fastest when :

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