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Question #45103

Difficulty: 8/10 | 🟢 Free

The Question

In standard biomolecular systems, the keto tautomer is generally favored over the enol form due to the high bond enthalpy of the $\ce{C=O}$ bond. For the synthetic nucleobase analog 1,3-dimethylpyrimidine-2,4,6(1H,3H,5H)-trione, the tautomeric equilibrium constant $K_{eq} = \frac{[\text{Enol}]}{[\text{Keto}]}$ in a non-polar solvent at $298\text{ K}$ is $10^{-5}$ .

To study conformational locking, a bulky tert-butyl group is introduced at the C5 position, yielding 5-(tert-butyl)-1,3-dimethylpyrimidine-2,4,6(1H,3H,5H)-trione. The introduction of this group induces severe B-strain (steric hindrance) in the $sp^3$ -hybridized state of the C5 carbon, destabilizing the keto form's enthalpy by $39.9\text{ kJ/mol}$ .

Assuming the enthalpy of the enol form and the standard entropies of both forms remain unaffected by the substitution, what is the new equilibrium constant $K'_{eq}$ for the 5-tert-butyl derivative at $298\text{ K}$ ?

(Given: $2.303 RT = 5.7\text{ kJ/mol}$ at $298\text{ K}$ )

Options

A.
$1.0 \times 10^2$
B.
$1.0 \times 10^{-12}$
C.
$1.0 \times 10^7$
D.
$1.0 \times 10^{-2}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Steric Hindrance (B-Strain)|Nitrogenous Base Keto-Enol Tautomerism|Nucleoside vs Nucleotide

Why This Question Is Tricky

Common Mistakes

This problem tests a critical 'Chemical Exception' where standard bond enthalpy rules (Keto > Enol) are overridden by steric factors (B-strain).

Students often memorize that the keto form of nucleobases and beta-diketones is overwhelmingly stable due to the C=O bond strength.

They fail to account for the fact that introducing a bulky group (like tert-butyl) on an sp3 alpha-carbon creates massive internal strain.

Tautomerization to the enol form rehybridizes the alpha-carbon to sp2, widening the bond angles (~109.5° to ~120°) and placing the bulky group in a less sterically congested environment.

The trap options catch students who misapply the thermodynamic formula (e.g., adding the destabilization energy instead of subtracting it, leading to 10^-12) or who ignore the initial equilibrium state.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Organic Chemistry – Some Basic Principles and Techniques - Class 11 Part 2
Concept	Steric Hindrance (B-Strain)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free