Question

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

• A.

  

• B.

  

• C.

  

• D.

  

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Oxidation of an aldose with hot nitric acid ($HNO_3$) converts both the top aldehyde ($-CHO$) group and the bottom primary alcohol ($-CH_2OH$) group into carboxylic acid ($-COOH$) groups, yielding an aldaric acid. If the resulting aldaric acid is optically inactive, it must possess an internal plane of symmetry (a meso compound).
Step 2: Key Formula or Approach:
For a 4-carbon aldaric acid to have a plane of symmetry, the chiral centers at C2 and C3 must be exact mirror reflections of each other across the horizontal center plane of the Fischer projection. This means both $-OH$ groups must lie on the exact same side.
Step 3: Detailed Explanation:
Let's analyze the properties required:
1. It is a D-sugar. By definition, in a Fischer projection, the $-OH$ group on the highest numbered chiral carbon (C3 for an aldotetrose) must be on the right side.
2. It yields a meso aldaric acid. After $HNO_3$ oxidation, the top and bottom groups are both $-COOH$. To have a plane of symmetry, the $-OH$ on C2 must be on the exact same side as the $-OH$ on C3.
Since the C3 $-OH$ is on the right (D-sugar), the C2 $-OH$ must also be on the right.
Therefore, the structure of the D-aldotetrose must have:
- C1: $-CHO$ (top)
- C2: $-OH$ on the right
- C3: $-OH$ on the right
- C4: $-CH_2OH$ (bottom)
This specific sugar is known as D-erythrose.
Let's evaluate the options provided in the images:
- Image a : Both $-OH$ groups are on the left. This is L-erythrose.
- Image b : C2 $-OH$ on right, C3 $-OH$ on left. This is D-threose (yields optically active aldaric acid).
- Image c : Both $-OH$ groups are on the right. This is D-erythrose.
- Image d : C2 $-OH$ on left, C3 $-OH$ on right. This is L-threose.
Step 4: Final Answer:
Option (C) is correct as it represents D-erythrose.