Question

Given below are two statements:

Statement I: 2,6-diethylcyclohexanone and 6-methyl-2-n-propylcyclohexanone are metamers.
Statement II: 2,2,6,6 - tetramethylcyclohexanone exhibits keto-enol tautomerism.

In the light of the above statements, choose the correct answer from the options given below

• A. Both Statement I and Statement II are true
• B. Both Statement I and Statement II are false
• C. Statement I is true but Statement II is false
• D. Statement I is false but Statement II is true

Hint:

Metamers differ in alkyl groups around the functional group. Keto-enol tautomerism requires at least one hydrogen atom on an alpha-carbon.

Answer 1

The Correct Option is C

The concept of isomerism allows us to classify compounds based on structural differences. Metamerism is a type of structural isomerism where compounds with the same molecular formula differ in the distribution of alkyl groups around a polyvalent functional group, such as a ketone ($R-CO-R'$).

Let's evaluate Statement I:
The first compound is 2,6-diethylcyclohexanone. It consists of a cyclohexanone ring with two ethyl groups attached to the $\alpha$-carbons (at positions 2 and 6). Its molecular formula is $C_{10}H_{18}O$.
The second compound is 6-methyl-2-n-propylcyclohexanone. It has a methyl group at position 6 and an n-propyl group at position 2 on the cyclohexanone ring. Its molecular formula is also $C_{10}H_{18}O$.
In these cyclic ketones, the distribution of carbon atoms on either side of the carbonyl group (via the alpha positions) changes from (ethyl, ethyl) to (methyl, propyl). Since the alkyl distribution around the functional group is different while the molecular formula remains the same, they are indeed metamers. Thus, Statement I is true.

Now, let's evaluate Statement II:
Keto-enol tautomerism is a chemical equilibrium between a keto form (a ketone or an aldehyde) and an enol (an alcohol with a double bond). This process requires the presence of at least one acidic hydrogen atom on the carbon atom adjacent to the carbonyl group, known as the $\alpha$-carbon.
In 2,2,6,6-tetramethylcyclohexanone, the $\alpha$-carbons are at positions 2 and 6. At each of these positions, there are two methyl groups ($CH_3$) attached. Because the carbon at position 2 is bonded to the $C_1$ (carbonyl), $C_3$ (ring), and two methyl groups, it has no remaining bonds for hydrogen. The same applies to the carbon at position 6. Since there are zero $\alpha$-hydrogen atoms available to migrate to the oxygen atom, the molecule cannot form an enol. Therefore, Statement II is false.

Conclusion: Statement I is true and Statement II is false.