Which of the following molecules is a tautomer of the structure shown?
The correct answer is Option D. The structure in option D is the proper tautomer of the given pyrimidine ring because it is related by proton transfer and concomitant shift of a double bond (keto–enol / lactam–lactim type tautomerism), without any change in connectivity.
Introduction
In many competitive exams, you will meet questions like “which of the following molecules is a tautomer of the structure shown”, especially for heterocyclic systems such as pyrimidines and nucleobases. Tautomerism in these rings usually involves proton transfer between nitrogen or oxygen and the corresponding carbonyl or double‑bond position, giving closely related structures called lactam–lactim or keto–enol tautomers.
Concept of tautomerism in this question
The parent structure in the slide is a six‑membered pyrimidine ring bearing two carbonyl groups and two ring nitrogens, analogous to uracil‑type or cytosine‑type systems. In such compounds, tautomerism occurs when a proton shifts (for example from N–H to O or from carbon adjacent to C=O to O/N) with a simultaneous relocation of a C=C or C=N bond, but the ring and substituent skeleton remain unchanged.
Key checks for a true tautomer of this structure:
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Same number and type of atoms (same molecular formula).
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Same connectivity of the ring and substituents (no bonds broken between heavy atoms).
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Difference only in position of H atoms and location of double bonds (keto vs enol, lactam vs lactim).
Option A explanation
In option A, both carbonyl groups of the parent ring have been converted into C–OH (enol or lactim) functions, and the double bonds in the ring are rearranged to keep conjugation with these OH groups. This corresponds to a bis‑enol / double lactim form of the same pyrimidine skeleton.
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The ring remains six‑membered with two nitrogens in the same positions, so connectivity is preserved.
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Only proton positions (N–H vs O–H) and C=O vs C–OH are interconverted, which is exactly what happens in lactam–lactim tautomerism of uracil‑like bases.
Therefore, Option A is a valid tautomer of the original structure.
Option B explanation
Option B shows a pyrimidine‑type ring in which one of the carbonyl groups has been replaced by an exocyclic C≡N (cyano) or similar nitrile‑like unit instead of C=O. This introduces a new atom and breaks the original C–O double‑bond framework.
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The overall connectivity and functional groups are no longer the same as in the original molecule.
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Because tautomerism cannot change the elemental composition or convert a carbonyl into a nitrile, option B is not a tautomer but a different compound altogether.
Option C explanation
Option C depicts a heterocycle that lacks the two carbonyl groups and instead has only one C–OH group on the ring, more like a partially saturated pyridinol‑type structure.
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The pattern of heteroatoms and the oxidation state of the ring are different from the parent diketone pyrimidine.
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Removal of a carbonyl and change from a di‑carbonyl to mono‑hydroxy system is not achievable by a simple proton shift and double‑bond migration, so option C is not a tautomer of the given molecule.
Option D explanation (correct tautomer)
Option D shows the same pyrimidine ring where one carbonyl remains as C=O while the other has become C–OH, with corresponding redistribution of the ring double bonds and N–H positions. This matches the classic keto–enol (or lactam–lactim) pair known for pyrimidine bases such as uracil and cytosine, where one carbonyl is in keto form and the other is in its enolic/lactim form.
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The ring and all atoms are identical to the original; only one hydrogen has migrated between N and O with a shift in one C=C/C=O bond.
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Such mono‑lactim forms coexist with the fully keto (di‑lactam) form in many pyrimidine systems and are classic examples of tautomerism.
Therefore, among the options, Option D best represents a true tautomer of the structure shown and is the correct answer.
