In the given reaction,the compound C is:C_6H_ | vedclass

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(D) The reaction sequence is known as Kolbe's reaction.$1$. Phenol $(C_6H_5OH)$ reacts with $NaOH$ to form sodium phenoxide $(A)$: $C_6H_5OH + NaOH \r

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Salicylic acid

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(D) The reaction sequence is known as Kolbe's reaction.$1$. Phenol $(C_6H_5OH)$ reacts with $NaOH$ to form sodium phenoxide $(A)$: $C_6H_5OH + NaOH \rightarrow C_6H_5ONa + H_2O$.$2$. Sodium phenoxide reacts with $CO_2$ at $140 \ ^oC$ and $4-7 \ atm$ pressure to form sodium salicylate $(B)$: $C_6H_5ONa + CO_2 \rightarrow C_6H_4(OH)COONa$.$3$. Acidification of sodium salicylate with $HCl$ yields salicylic acid $(C)$: $C_6H_4(OH)COONa + HCl \rightarrow C_6H_4(OH)COOH + NaCl$.Thus,the final product $C$ is salicylic acid.

In the given reaction,the compound $C$ is:
$C_6H_5OH$ $\xrightarrow{NaOH} (A)$ $\xrightarrow{CO_2, 140 \ ^oC, (4-7 \ atm)} (B)$ $\xrightarrow{HCl} (C)$

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In the given reaction,the compound $C$ is:$C_6H_5OH$ $\xrightarrow{NaOH} (A)$ $\xrightarrow{CO_2, 140 \ ^oC, (4-7 \ atm)} (B)$ $\xrightarrow{HCl} (C)$