In the given reaction,the product is:C_6H_5OH | vedclass

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Question

(B) The reaction involves the nucleophilic substitution of ethyl iodide $(C_2H_5I)$ by the phenoxide ion $(C_6H_5O^-)$.$1$. Phenol $(C_6H_5OH)$ reacts

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$C_6H_5OC_2H_5$

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(B) The reaction involves the nucleophilic substitution of ethyl iodide $(C_2H_5I)$ by the phenoxide ion $(C_6H_5O^-)$.$1$. Phenol $(C_6H_5OH)$ reacts with the ethoxide ion $(C_2H_5O^-)$ in ethanol to form the phenoxide ion $(C_6H_5O^-)$.$2$. The phenoxide ion acts as a nucleophile and attacks the ethyl iodide $(C_2H_5I)$ via an $S_N2$ mechanism.$3$. The reaction is: $C_6H_5O^- + C_2H_5I \rightarrow C_6H_5OC_2H_5 + I^-$.$4$. The product formed is ethyl phenyl ether $(C_6H_5OC_2H_5)$,also known as phenetole.

In the given reaction,the product is:
$C_6H_5OH + C_2H_5I \xrightarrow{C_2H_5O^- / C_2H_5OH} \text{Product}$

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In the given reaction,the product is:$C_6H_5OH + C_2H_5I \xrightarrow{C_2H_5O^- / C_2H_5OH} \text{Product}$