Question

What is action of \(\text{CH}_3\text{ONa}\) on \(\text{CH}_3\text{CH}_2\text{Br}\)
\(\text{CH}_3\text{ONa}+\text{CH}_3\text{CH}_2\text{Br}\rightarrow\) ?

• A. Formation of \(\text{CH}_3\text{OCH}_2\text{CH}_3+\text{NaBr}\)
• B. Formation of \(\text{CH}_3\text{OH}+\text{NaBr}\)
• C. Formation of \(\text{CH}_3\text{CH}_2\text{OH}\)
• D. Formation of \((\text{CH}_3)_2\text{O}\)

Hint:

This is the Williamson ether synthesis.

Answer 1

The Correct Option is A

Step 1: Recognise the reaction
Sodium methoxide is a strong nucleophile and ethyl bromide is a primary alkyl halide, which suits an SN2 reaction. This is the Williamson ether synthesis.

Step 2: Product
\(CH_3ONa + CH_3CH_2Br \rightarrow CH_3OCH_2CH_3 + NaBr\).

Step 3: Other options
The alcohols and dimethyl ether would need the wrong halide or water, and do not form here.

Final Answer:
The product is ethyl methyl ether and NaBr. \[ \boxed{\text{(A)}\ CH_3OCH_2CH_3 + NaBr} \]