Will (3R,4S,5R)-1,3,4,5,6-pentahydroxyhexan-2-one tautomerise in nitric acid?

Question

The following question was sent to me .The question is " Which of the following is A ?"

The given answer was a,b,c

Background

If both ends of an aldose chain are oxidized to carboxylic acids the product is called an aldaric acid.

source : https://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/carbhyd.htm#carb2

My thoughts

• Options a and c are direct reactions.
• Option b gives 2 and it can only rotated in the plane it is drawn (2D) by 180 degrees (http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch07/ch7-7.html) to give 1.
• How ever option d in acidic medium could tautomerise to give 3d , that on oxidation again gives 1.

Therefore answer should be a,b,c,d.

My answer hinges on tautomerisation of ketone is acidic medium. Would this compound (3R,4S,5R)-1,3,4,5,6-pentahydroxyhexan-2-one tautomerise in nitric acid ?

Any help in this regard is very much appreciated. Thankyou.

reference :

1 http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch07/ch7-7.html

Answer 1

@Sir Arthur7 did a fine job of discrediting D-ketose (d) as a viable candidate for structure A. Belatedly, I will expand upon my comment above at the request of @Jan. [In my comment above, "glutaric" should read "glucaric".]
Assuming that $\ce{HNO3}$ is capable of tautomerizing the D-ketose via the enediol and accomplishing terminal oxidation, two epimeric aldaric acids could arise: D-gularic acid and D-idaric acid. Note that they only differ in the hydroxyl configuration at C₂. Rotation of D-gularic by 180^o about an axis perpendicular to the page results in a structure that is also recognized as L-glucaric acid. Thus, D-gularic and L-glucaric acid are identical. When the same rotation is performed on D-idaric acid, D-idaric acid is returned. Choices (a), (b) and (c) all produce D-glucaric acid (red box) while option (d) produces the enantiomer, L-glutaric acid, and the irrelevant D-idaric acid. Option (d) is not a viable structure for A.

Answer 2

Well I am just a high school student, so I may be wrong, and I beg pardon in advance, but I would like to share my notions about this topic.

@ChakravarthyKalyan I would like to ask you to review your last step in the conversion of 3d to 1. 3d on oxidation should yield this product (in the image)--which is by my knowledge not same as 1.

So even your theory does not yield (d) as the correct option, I assume. Also I doubt the double tautomerism of the compound in option (d) in acidic medium, though your mechanisms were convincing. This closely resembles Lobry-Debryn-van-Eckenstein rearrangement that occurs only in presence of alkaline medium.

As per my knowledge, ketoses in presence of strong oxidizing agents like conc. HNO3, undergo oxidative cleavage forming 2 carboxylic acids(acc. to Popoff's rule), or as @user55119 had pointed out to me, that alpha ketoacids can undergo oxidative decarboxylation and form CO2 and an acid(though its mechanism is quite unclear to me as well).