Question

$+C_{2}H_{5}I \xrightarrow[{{\text{Anhydrous}} (C_{2}H_{5}OH)}] {{^{-}OC_{2}H_{5}}}$?

• A. $C_{6}H_{5}OC_{2}H_{5}$
• B. $C_{2}H_{5}OC_{2}H_{5}$
• C. $C_{6}H_{5}OC_{6}H_{5}$
• D. $C_{6}H_{5}I$

Answer 1

Answer: (B)

This reaction is an example of Williamson’s synthesis
$C_{2}H_{5}O^{-}$ will extract proton from phenol converting the latter into phenoxide ion But $C_{2}H_{5}O^{-}$ is a better nucleophile than $C_{6}H_{5}O^{-}$ (phenoxide) ion because in the former the negative charge is localised over oxygen, while in the latter it is delocalized over the whole molecular framework. So, it is $C_{2}H_{5}O^{-}$ on that would make nucleophilic attack at ethyl iodide to give diethyl ether (Williamson’s synthesis)