[tangent1.57]

In my organic chemistry book, they have the oxidation of sugars.  And they show that the oxidation of 
     CHO
     |
HO-C-H
HO-C-H           
 H -C-OH
     |
     CH2OH

     |
     |  OXIDATION TO become an ALDARIC ACID
     |
    \/

    COOH
     |
HO-C-H
HO-C-H
 H -C-OH
     |
     COOH

I realize that there are no planes of symmetry, and there are 3 chiral centers at C2, C3, and C4.  However, this molecule can be rotated 180 degrees to the left to be:

     COOH
     |
HO-C-H
H  -C-OH
H  -C-OH
     |
     COOH

Therefore, the D-oxidation of arabinose and D-lyxose with nitric acid yields the exact same aldaric molecule which is optically inert. 

[thfmag]

Hi !
The oxidation of both molecules that you have drawn give the same product. That is correct.
But to decide if a molecule is chiral or not, you have to look at the molecule only and its symmetry element.

If you draw the mirror image of the molecule, you will see that they are nonsuperposable. They are enantiomers (like your hands). The molecule is asymmetric so optically active.

The molecule has only 2 chiral centres (and not 3 as you said), carbons 2 and 4. Carbon 3 bears two identical substituents with the same configuration (carbon 2 and 4 have a S configuration) so it is not a chiral centre. Maybe if you make a model of the molecule that will be easier to see.

[sjb]

...
The molecule has only 2 chiral centres (and not 3 as you said), carbons 2 and 4. Carbon 3 bears two identical substituents with the same configuration (carbon 2 and 4 have a S configuration) so it is not a chiral centre. Maybe if you make a model of the molecule that will be easier to see.

It may be true that C₂ and C₄ are both S, but this does not stop C₃ existing in R and S forms. If one were R and the other S, then you'd have a mirror plane running through the H-C-OH of C₃, and then I don't think you'd have isomers.

[thfmag]

Hi back!
See attachment.