C_2H_5OH xrightarrow{Na} P xrightarrow{CH_3-C | vedclass

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Question

(B) The reaction proceeds as follows:$1$. Ethanol $(C_2H_5OH)$ reacts with sodium $(Na)$ to form sodium ethoxide $(P)$: $C_2H_5OH + Na \rightarrow C_2

Vedclass · Accepted Answer

$CH_3-CH_2-O-CH_3$

Vedclass · Answer

(B) The reaction proceeds as follows:$1$. Ethanol $(C_2H_5OH)$ reacts with sodium $(Na)$ to form sodium ethoxide $(P)$: $C_2H_5OH + Na \rightarrow C_2H_5ONa + \frac{1}{2}H_2$.$2$. Sodium ethoxide $(C_2H_5ONa)$ then reacts with methyl chloride $(CH_3-Cl)$ via the Williamson ether synthesis to form ethyl methyl ether $(Q)$: $C_2H_5ONa + CH_3-Cl \rightarrow C_2H_5-O-CH_3 + NaCl$.Thus,$Q$ is $CH_3-CH_2-O-CH_3$.

$C_2H_5OH$ $\xrightarrow{Na} P$ $\xrightarrow{CH_3-Cl} Q$; $Q$ is

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