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

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


**The given answer was a,b,c**


[![enter image description here](https://i.stack.imgur.com/lWNUJ.png)](https://i.stack.imgur.com/lWNUJ.png)


**Background**


If both ends of an aldose chain are oxidized to carboxylic acids the product is called an aldaric acid.
[![enter image description here](https://i.stack.imgur.com/FM6y7.jpg)](https://i.stack.imgur.com/FM6y7.jpg)


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


**My thoughts**


[![enter image description here](https://i.stack.imgur.com/spyQP.png)](https://i.stack.imgur.com/spyQP.png)


* 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](https://i.stack.imgur.com/lWNUJ.png) <http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch07/ch7-7.html>




@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 180o 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-glucaric acid, and the irrelevant D-idaric acid. Option (d) is not a viable structure for **A**.  



![](https://i.stack.imgur.com/qyTTE.jpg)
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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.


[![Oxidation of 3d](https://i.stack.imgur.com/Rp8lE.png)](https://i.stack.imgur.com/Rp8lE.png) 


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).


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