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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

Background

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

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

My thoughts

enter image description here

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

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

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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

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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  • $\begingroup$ Oxidative decarboxylation is one consequence of nitric acid oxidation. If tautomerization occurs, D-idose (3c) and D-gulose (3d) would be formed. HNO3 oxidation of D-gulose would afford L-glutaric acid (aka, D-gularic acid) as 1, not D-glutaric acid as shown. Your structure as drawn id D-gularic acid. Rotated by 180o, It is L-glutaric acid. $\endgroup$ – user55119 Sep 26 '19 at 22:53
  • $\begingroup$ This is essentially correct. The product 3d and 1 are enantiomers. This answer could expand on the fact more. $\endgroup$ – Jan Oct 17 '19 at 6:36

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