Phenol xrightarrow[(ii) CO_2 / 140^{circ}C]{( | vedclass

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(D) The reaction sequence is as follows:$1$. Phenol reacts with $NaOH$ followed by $CO_2$ at $140^{\circ}C$ (Kolbe-Schmitt reaction) to form sodium sa

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aspirin

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(D) The reaction sequence is as follows:$1$. Phenol reacts with $NaOH$ followed by $CO_2$ at $140^{\circ}C$ (Kolbe-Schmitt reaction) to form sodium salicylate,which upon acidification gives $B$ (salicylic acid).$2$. Salicylic acid $(C_6H_4(OH)COOH)$ reacts with acetic acid $(CH_3COOH)$ in the presence of $Al_2O_3$ (or $H_2SO_4$) to undergo acetylation of the phenolic $-OH$ group.$3$. The product $C$ is acetylsalicylic acid,commonly known as aspirin.

Phenol $\xrightarrow[(ii) CO_2 / 140^{\circ}C]{(i) NaOH} A$ $\xrightarrow{H^{+} / H_2O} B$ $\xrightarrow[CH_3COOH, \Delta]{Al_2O_3} C$
In this reaction,the end product $C$ is :

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Phenol $\xrightarrow[(ii) CO_2 / 140^{\circ}C]{(i) NaOH} A$ $\xrightarrow{H^{+} / H_2O} B$ $\xrightarrow[CH_3COOH, \Delta]{Al_2O_3} C$In this reaction,the end product $C$ is :