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(A) The reaction of phenol with $CHCl_3$ and $KOH$ is the Reimer-Tiemann reaction,which yields salicylaldehyde ($2$-hydroxybenzaldehyde) as the interm

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(A) The reaction of phenol with $CHCl_3$ and $KOH$ is the Reimer-Tiemann reaction,which yields salicylaldehyde ($2$-hydroxybenzaldehyde) as the intermediate $X$. Salicylaldehyde contains an aldehyde group but no $\alpha$-hydrogen atoms. When treated with $50 \%$ $KOH$ solution,it undergoes the Cannizzaro reaction,a disproportionation reaction. One molecule of salicylaldehyde is reduced to the corresponding alcohol ($2$-hydroxybenzyl alcohol),and another molecule is oxidized to the corresponding carboxylic acid salt (potassium $2-$hydroxybenzoate).

The final product of the following reaction is/are:

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

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