Question

The compound which shows metamerism is :

• A. $C _{5} H _{12}$
• B. $C _{3} H _{8} O$
• C. $C _{3} H _{6} O$
• D. $C _{4} H _{10} O$

Answer 1

The Correct Option is D

To understand which compound among the options shows metamerism, we first need to clarify what metamerism in organic chemistry is. Metamerism is a type of isomerism that occurs in compounds with the same molecular formula but different alkyl groups on either side of a functional group. Typically, it involves ethers, secondary amines, thioethers, ketones, etc.

Let's evaluate the given compounds:

1. \(C _{5} H _{12}\) represents an alkane (specifically pentane or isomers of pentane). Alkanes do not have a functional group that allows for the existence of metamers.
2. \(C _{3} H _{8} O\) could represent an alcohol or an ether, such as propanol or methoxyethane. However, due to the low carbon count, it doesn't provide an opportunity for the formation of metamerism.
3. \(C _{3} H _{6} O\) can represent aldehydes or ketones, such as propanal or acetone, but similarly, there are no opportunities for metamerism because these structures are too small to have different alkyl groups on either side of a functional group.
4. \(C _{4} H _{10} O\) is an ether, represented generally as R-O-R'. This can be either methoxypropane (or propoxymethane) or ethoxyethane. Here, because of the different possible alkyl groups on either side of the oxygen, this compound displays metamerism.

Hence, the compound \(C _{4} H _{10} O\) is the one that shows metamerism. It can have different structural formulas where the alkyl groups differ in composition on either side of the ether functional group. This correctly illustrates the concept of metamerism.

Thus, the correct answer is \(C _{4} H _{10} O\).