p-Cresol reacts with chloroform in an alkalin | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

Vedclass · Answer

(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?

Similar Questions

$Ph-CO-CHO \xrightarrow[\Delta]{conc. NaOH} \text{Product}$

Which substance will give a positive iodoform test?

Acetaldehyde cannot give:

The major product formed in the following reaction is:

Which of the reagent is used to convert $2-Butanone$ into propanoic acid?

Vedclass Test Series

Exam Paper Generator

Online Exam Module