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(B) The starting material is $2-(2-hydroxyethyl)phenol$,which contains both a phenolic $-OH$ group and an aliphatic $-OH$ group.$K_2CO_3$ is a mild ba

Vedclass · Accepted Answer

$1-$($2$-hydroxyethyl)$-2-$methoxybenzene

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(B) The starting material is $2-(2-hydroxyethyl)phenol$,which contains both a phenolic $-OH$ group and an aliphatic $-OH$ group.$K_2CO_3$ is a mild base that selectively deprotonates the more acidic phenolic $-OH$ group $(pK_a \approx 10)$ compared to the aliphatic $-OH$ group $(pK_a \approx 16)$.The resulting phenoxide ion is resonance-stabilized by the benzene ring,making it the preferred site for nucleophilic substitution.Upon addition of $CH_3I$ $(1. \, eq.)$,the phenoxide ion acts as a nucleophile and attacks the methyl iodide via an $S_N2$ mechanism to form the methyl ether at the phenolic position.Therefore,the major product is $1-(2-hydroxyethyl)-2-methoxybenzene$.

The major product of the following reaction is
$2-(2-hydroxyethyl)phenol$ $\xrightarrow[2. \, CH_3I \, (1. \, eq.)]{1. \, K_2CO_3}$

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The major product of the following reaction is $2-(2-hydroxyethyl)phenol$ $\xrightarrow[2. \, CH_3I \, (1. \, eq.)]{1. \, K_2CO_3}$