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The reaction given below is known as
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{ONa}+I \mathrm{C}_2 \mathrm{H}_5 \longrightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{OC}_2 \mathrm{H}_5+\mathrm{Nal}$
Options:
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{ONa}+I \mathrm{C}_2 \mathrm{H}_5 \longrightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{OC}_2 \mathrm{H}_5+\mathrm{Nal}$
Solution:
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Verified Answer
The correct answer is:
Williamson's synthesis
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{ONa}+\mathrm{IC}_2 \mathrm{H}_5 \rightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{OC}_2 \mathrm{H}_5+\mathrm{NaI}$
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