Statements on Isomerism and Tautomerism of Cyclohexanone Derivatives

To determine the correctness of the given statements, let us analyze them individually:

Analysis of Statement I:

Statement I: 2,6-diethylcyclohexanone and 6-methyl-2-n-propylcyclohexanone are metamers.

• Metamerism is a form of structural isomerism where compounds possessing the same molecular formula and functional group differ in the distribution of alkyl groups attached to the polyvalent functional group (in this case, the carbonyl group, $-\text{C}(=\text{O})-$).
• 2,6-diethylcyclohexanone has the molecular formula $\text{C}_{10}\text{H}_{18}\text{O}$. The alkyl substituents attached at the positions adjacent to the carbonyl carbon are two ethyl groups ($-\text{C}_2\text{H}_5$).
• 6-methyl-2-n-propylcyclohexanone also has the same molecular formula $\text{C}_{10}\text{H}_{18}\text{O}$. The alkyl substituents attached at the positions adjacent to the carbonyl carbon are one methyl group ($-\text{CH}_3$) and one $n$-propyl group ($-\text{C}_3\text{H}_7$).

Since both molecules have the same functional group (ketone) and molecular formula, but differ in the arrangement of carbon chains attached to either side of the divalent carbonyl group, they are metamers.

Therefore, Statement I is TRUE.

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Analysis of Statement II:

Statement II: 2,2,6,6-tetramethylcyclohexanone exhibits keto-enol tautomerism.

• For a ketone to undergo keto-enol tautomerism, it must possess at least one $\alpha$-hydrogen atom (a hydrogen atom attached to an $\alpha$-carbon adjacent to the carbonyl carbon, $\text{C}=\text{O}$).
• In 2,2,6,6-tetramethylcyclohexanone:
  • Position 1 ($\text{C}1$) is the carbonyl carbon atom ($\text{C}=\text{O}$).
  • Positions 2 ($\text{C}2$) and 6 ($\text{C}6$) are the two $\alpha$-carbons. Both $\text{C}2$ and $\text{C}6$ are bonded to two methyl groups each, making them fully substituted quaternary carbon atoms: $\text{C}2: -\text{C}(\text{CH}_3)_2-$ $\text{C}6: -\text{C}(\text{CH}_3)_2-$
• Since both $\alpha$-carbon atoms ($\text{C}2$ and $\text{C}6$) are bonded to four carbon atoms, there are zero $\alpha$-hydrogen atoms in the molecule.

Without any $\alpha$-hydrogens, proton transfer cannot occur to form the enol tautomer. Thus, 2,2,6,6-tetramethylcyclohexanone does not exhibit keto-enol tautomerism.

Therefore, Statement II is FALSE.

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Conclusion:

• Statement I is true.
• Statement II is false.

Hence, the correct option is C (Statement I is true but Statement II is false).