Understanding Metamerism in Organic Molecules

Metamerism is a type of structural isomerism where compounds have the same molecular formula and the same functional group, but differ in the distribution of alkyl groups on either side of the functional group or polyvalent atom. This type of isomerism is commonly observed in functional groups like ethers (-O-), ketones (>C=O), esters (-COO-), and secondary amines (>NH). We need to examine the given molecular formulas to see which one can exhibit this kind of structural variation.

Analyzing Molecular Formula C₄H₁₀O for Metamerism

The molecular formula C₄H₁₀O can represent alcohols or ethers. Let's look at the possible ether structures, as metamerism is common in ethers:

• Diethyl ether: $\text{CH}_3\text{CH}_2\text{-O-}\text{CH}_2\text{CH}_3$. Here, the oxygen atom is bonded to two ethyl ($\text{C}_2\text{H}_5$) groups.
• Methyl propyl ether: $\text{CH}_3\text{-O-}\text{CH}_2\text{CH}_2\text{CH}_3$. Here, the oxygen atom is bonded to a methyl ($\text{CH}_3$) group and a propyl ($\text{C}_3\text{H}_7$) group.
• Methyl isopropyl ether: $\text{CH}_3\text{-O-}\text{CH(CH}_3)_2$. Here, the oxygen atom is bonded to a methyl ($\text{CH}_3$) group and an isopropyl group ($\text{C}_3\text{H}_7$).

Comparing diethyl ether ($\text{CH}_3\text{CH}_2\text{-O-}\text{CH}_2\text{CH}_3$) and methyl propyl ether ($\text{CH}_3\text{-O-}\text{CH}_2\text{CH}_2\text{CH}_3$), we see they have the same molecular formula ($\text{C}_4\text{H}_{10}\text{O}$) and the same functional group (ether), but the alkyl groups attached to the oxygen atom are different (ethyl/ethyl vs methyl/propyl). Therefore, $\text{C}_4\text{H}_{10}\text{O}$ can show metamerism as ethers.

Analyzing Molecular Formula C₅H₁₂ for Metamerism

The molecular formula $\text{C}_5\text{H}_{12}$ corresponds to alkanes. Alkanes do not have a functional group or a polyvalent atom (like oxygen in ethers, nitrogen in amines, or carbonyl carbon in ketones/esters) where different alkyl groups can be attached to show metamerism. Alkanes typically exhibit chain isomerism (e.g., pentane, isopentane, neopentane). Thus, $\text{C}_5\text{H}_{12}$ does not show metamerism.

Analyzing Molecular Formula C₃H₈O for Metamerism

The molecular formula $\text{C}_3\text{H}_8\text{O}$ can represent alcohols or ethers. Let's consider the possible ethers with this formula:

• Methyl ethyl ether: $\text{CH}_3\text{-O-}\text{CH}_2\text{CH}_3$.

For metamerism to exist, there must be at least two different structural isomers with the same functional group but different alkyl group distributions around the polyvalent atom. For the ether functional group in $\text{C}_3\text{H}_8\text{O}$, the only way to split the three carbon atoms is into a methyl group ($\text{C}_1$) and an ethyl group ($\text{C}_2$). There is only one possible ether structure, methyl ethyl ether. Since there is only one ether, it cannot have a metamer. Thus, $\text{C}_3\text{H}_8\text{O}$ does not show metamerism.

Conclusion on Metamerism Exhibition

Based on the analysis of the given molecular formulas:

• $\text{C}_4\text{H}_{10}\text{O}$ shows metamerism (e.g., diethyl ether and methyl propyl ether).
• $\text{C}_5\text{H}_{12}$ (alkanes) does not show metamerism.
• $\text{C}_3\text{H}_8\text{O}$ (ether) does not show metamerism.

Therefore, among the given options, only $\text{C}_4\text{H}_{10}\text{O}$ exhibits metamerism.