A D-aldotetrose on oxidation with concentrate | vedclass

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

Question

(C) -aldotetrose has the general structure $CHO-(CHOH)_2-CH_2OH$.Upon oxidation with concentrated $HNO_3$,the terminal aldehyde $(-CHO)$ and primary a

Vedclass · Accepted Answer

Vedclass · Answer

(C) -aldotetrose has the general structure $CHO-(CHOH)_2-CH_2OH$.Upon oxidation with concentrated $HNO_3$,the terminal aldehyde $(-CHO)$ and primary alcohol $(-CH_2OH)$ groups are oxidized to carboxylic acid $(-COOH)$ groups,resulting in a tartaric acid derivative $(HOOC-(CHOH)_2-COOH)$.For this dicarboxylic acid to be optically inactive,it must contain a plane of symmetry,making it a meso compound.In the Fischer projection of the $D$-aldotetrose,the $D$-configuration implies that the hydroxyl group at the chiral carbon furthest from the aldehyde group (i.e.,$C-3$) is on the right side.For the resulting dicarboxylic acid to be meso,the hydroxyl groups must be on the same side (erythro configuration) so that a plane of symmetry exists.Therefore,the $D$-aldotetrose must have both hydroxyl groups on the right side,which corresponds to $D$-erythrose.

$A$ $D$-aldotetrose on oxidation with concentrated $HNO_3$ resulted in an optically inactive dicarboxylic acid. The structure of the $D$-aldotetrose is:

Similar Questions

Which of the following is a non-reducing sugar?

Which of the following molecules represents the $D$-form?

Name the macromolecules that are chosen as drug targets.

Assertion $(A)$: Both Glucose and Fructose have the same $D$-configuration.
Reason $(R)$: Both Glucose and Fructose are dextrorotatory.
The correct option among the following is:

The incorrect statement from the following with respect to carbohydrates is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module