All structures given below are of vitamin C. Most stable of them is :

Solution

Chemistry ma’am · 3 steps

Look for the tautomer where the C=C, the OH groups and the lactone C=O all form one conjugated system.

1. Same atoms, different tautomers. All four drawings are keto–enol tautomers of the same five-membered lactone ring. They differ only in where the C=O and C=C sit. The most stable one has the most conjugation and keeps the ester (lactone) group intact.

   Option	What is drawn	Conjugation and resonance	Stability
   (a)	C-3 ketone, C-2 is CH(OH)	No C=C, so the two C=O groups are isolated	less stable
   (b)	2,3-enediol with the lactone C=O	O–C=C–C=O conjugated, and both OH lone pairs push into it	most stable
   (c)	C-2 ketone next to the lactone C=O	No C=C, and two C=O dipoles sit side by side and repel	less stable
   (d)	C-1 enolised, C-3 ketone	Lactone C=O is gone, so the ester resonance is lost	less stable

2. Why the enediol wins. In (b) the C=C is conjugated with the lactone C=O, and the lone pairs on the two OH oxygens delocalise across this system. The ring stays flat and the ester group is kept, so this form gains extra stabilisation that the plain keto forms cannot get.
3. So. Structure (b), the 2,3-enediol lactone, is the most stable. It is the form in which vitamin C (ascorbic acid) actually exists, and its enediol OH groups make it acidic.

The idea to keep. Keto forms usually beat enols, but an enol wins when its C=C is conjugated with a C=O. Vitamin C is exactly this case: a conjugated enediol lactone.

Answer from the NTA final answer key. Solution by PrepWiser.