The product formed in the following reaction | vedclass

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

Question

(B) The reaction involves the oxidative cleavage of a vicinal diol (or $\alpha$-hydroxy carbonyl compound) by periodic acid $(HIO_4)$.$1$. The startin

Vedclass · Accepted Answer

$HCHO, 2HCO_2H$

Vedclass · Answer

(B) The reaction involves the oxidative cleavage of a vicinal diol (or $\alpha$-hydroxy carbonyl compound) by periodic acid $(HIO_4)$.$1$. The starting material is glyceraldehyde $(CHO-CH(OH)-CH_2OH)$.$2$. The first $HIO_4$ molecule cleaves the bond between the $C_1$ (aldehyde) and $C_2$ (hydroxyl) carbons. This produces $HCOOH$ (formic acid) and $HOCH_2CHO$ (glycolaldehyde).$3$. The second $HIO_4$ molecule cleaves the glycolaldehyde $(HOCH_2CHO)$ into $HCHO$ (formaldehyde) and $HCOOH$ (formic acid).$4$. Summing the products: $HCOOH + HCOOH + HCHO = 2HCOOH + HCHO$.Therefore,the final products are $HCHO$ and $2HCO_2H$.

The product formed in the following reaction is:
$CHO-CH(OH)-CH_2OH \xrightarrow{2HIO_4} ?$

Similar Questions

In the reaction,the acid obtained will be
$CH_3CHO + HCN \to CH_3CH(OH)CN \xrightarrow{H_2O} CH_3CH(OH)COOH$

The product of the following reaction is :

Least reactive towards nucleophilic addition is

$C_6H_5-CO-CHO$ $\xrightarrow{conc. OH^-} (A)$ $\xrightarrow{H_3O^+} (B)$ $\xrightarrow[2. NaOH/CaO/\Delta]{1. KMnO_4} (C)$
Product $(C)$ is:

What is the structure of $C$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The product formed in the following reaction is: $CHO-CH(OH)-CH_2OH \xrightarrow{2HIO_4} ?$