p-Cresol reacts with chloroform in an alkalin | vedclass

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(C) $1$. $p$-Cresol reacts with $CHCl_3$ in the presence of $NaOH$ (Reimer-Tiemann reaction) to form $2$-hydroxy-$5$-methylbenzaldehyde (compound $A$

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(C) $1$. $p$-Cresol reacts with $CHCl_3$ in the presence of $NaOH$ (Reimer-Tiemann reaction) to form $2$-hydroxy-$5$-methylbenzaldehyde (compound $A$ or $X$).$2$. The aldehyde group $(-CHO)$ of $X$ reacts with $HCN$ to form a cyanohydrin $(Y)$,which is $2$-hydroxy-$5$-methylmandelonitrile.$3$. Acidic hydrolysis of the cyanohydrin $(Y)$ converts the $-CN$ group into a $-COOH$ group,resulting in $2$-hydroxy-$5$-methylmandelic acid.$4$. The structure of the final product is $2$-hydroxy-$5$-methylmandelic acid,which contains a chiral carbon atom at the alpha position relative to the carboxylic acid group.

$p$-Cresol reacts with chloroform in an alkaline medium to give compound $A$,which on addition of hydrogen cyanide forms compound $Y$. Acidic hydrolysis of $Y$ yields a chiral carboxylic acid. What is the structure of the carboxylic acid?

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