In the reactionPhenol xrightarrow[text{Reimer | vedclass

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(C) The Reimer-Tiemann reaction of phenol with $CHCl_3$ and $NaOH$ yields salicylaldehyde ($X = 2$-hydroxybenzaldehyde).When salicylaldehyde is treate

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(C) The Reimer-Tiemann reaction of phenol with $CHCl_3$ and $NaOH$ yields salicylaldehyde ($X = 2$-hydroxybenzaldehyde).When salicylaldehyde is treated with concentrated $NaOH$,it undergoes the Cannizzaro reaction. Since salicylaldehyde is an aromatic aldehyde without $\alpha$-hydrogens,it undergoes self-oxidation and reduction.The aldehyde group $(-CHO)$ is reduced to a primary alcohol group $(-CH_2OH)$ and oxidized to a carboxylate group $(-COO^-)$.Upon subsequent acidification with $H^+$,the carboxylate group is converted to a carboxylic acid group $(-COOH)$.Thus,the final product $Z$ is a mixture of $2$-hydroxybenzyl alcohol and $2$-hydroxybenzoic acid (salicylic acid).

In the reaction
$Phenol$ $\xrightarrow[\text{Reimer-Tiemann}]{} X$ $\xrightarrow[(ii) H^+]{(i) NaOH \text{ (conc.)}} Z$
Product $Z$ is

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In the reaction$Phenol$ $\xrightarrow[\text{Reimer-Tiemann}]{} X$ $\xrightarrow[(ii) H^+]{(i) NaOH \text{ (conc.)}} Z$Product $Z$ is