Which L-sugar on oxidation with HNO_3 gives a | vedclass

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

Question

(A) Oxidation of an aldose with $HNO_3$ converts both the terminal $-CHO$ and $-CH_2OH$ groups into $-COOH$ groups,resulting in a dicarboxylic acid (a

Vedclass · Accepted Answer

Vedclass · Answer

(A) Oxidation of an aldose with $HNO_3$ converts both the terminal $-CHO$ and $-CH_2OH$ groups into $-COOH$ groups,resulting in a dicarboxylic acid (aldaric acid).For the resulting acid to be optically active,it must lack a plane of symmetry (i.e.,it should not be a meso compound).Structure $(a)$ is $L$-gulose. Upon oxidation,it forms $L$-gularic acid,which is optically active.Structure $(d)$ is $L$-galactose. Upon oxidation,it forms mucic acid (galactaric acid),which is a meso compound (optically inactive due to an internal plane of symmetry).Therefore,the $L$-sugar that yields an optically active dibasic acid is $(a)$.

Which $L$-sugar on oxidation with $HNO_3$ gives an optically active dibasic acid ($2$ $COOH$ groups)?

Similar Questions

Two statements are given below.
Statement $I$: Milk sugar is a disaccharide of $\alpha-D$-galactose and $\beta-D$-glucose.
Statement $II$: Sucrose is a disaccharide of $\alpha-D$-glucose and $\beta-D$-fructose.
Correct answer is

Which among the following carbohydrate molecules yields the highest number of galactose molecules when one mole of each is hydrolyzed?

Glucose on reaction with Fehling solution gives

The correct corresponding names of the four aldoses with the configurations given below,respectively,are:

Anomers of glucose are defined as:

Vedclass Test Series

Exam Paper Generator

Online Exam Module