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(B) The given reaction sequence is the Reimer-Tiemann reaction followed by oxidation:$1$. $Phenol$ reacts with $CHCl_3$ and $NaOH$ to form $o-hydroxyb

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(B) The given reaction sequence is the Reimer-Tiemann reaction followed by oxidation:$1$. $Phenol$ reacts with $CHCl_3$ and $NaOH$ to form $o-hydroxybenzaldehyde$ (Salicylaldehyde) as the major product $(A)$.$2$. The intermediate $A$ is already $Salicylaldehyde$. The step $\xrightarrow[H_3O^+]{} B$ typically refers to the acidification of the phenoxide intermediate to obtain the final aldehyde product $B$.$3$. Oxidation of $Salicylaldehyde$ $(B)$ using an oxidizing agent converts the $-CHO$ group into a $-COOH$ group, resulting in $Salicylic \ acid$ $(C)$.Therefore, the final product $C$ is $Salicylic \ acid$.

In the following sequence of the reaction, the final product "$C$" is:
$Phenol \xrightarrow[NaOH]{CHCl_3} A \xrightarrow[H_3O^+]{} B \xrightarrow[Oxidation]{} C$

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In the following sequence of the reaction, the final product "$C$" is:$Phenol \xrightarrow[NaOH]{CHCl_3} A \xrightarrow[H_3O^+]{} B \xrightarrow[Oxidation]{} C$