Question Details

The compound which shows metamerism is :

Options

C₅H₁₂

C₃H₈O

C₃H₆O

C₄H₁₀O

Show Answer

Correct Answer :

Option D

C₄H₁₀O

Solution :

The correct answer is C₄H₁₀O.

Understanding Metamerism:
Metamerism is a type of structural isomerism that arises due to the unequal distribution of alkyl groups on either side of a functional group in the same homologous series. It is typically exhibited by polyvalent functional groups such as ethers (-O-), thioethers (-S-), ketones (-CO-), secondary amines (-NH-), and esters (-COO-).

Let's analyze the options to see which molecular formula can exhibit metamerism:

1. C₅H₁₂:
This is an alkane (general formula C_nH_2n+2). Alkanes do not have a functional group or heteroatom to show metamerism. They only exhibit chain isomerism (e.g., n-pentane, isopentane, and neopentane).

2. C₃H₈O:
This represents a saturated monohydric alcohol or an ether. The general formula is C_nH_2n+2O.
Let's look at the possible ether structures for this formula:
$C H_{3} - O - C H_{2} C H_{3}$ (Methoxyethane)
There is only one possible ether isomer. Since we cannot have another ether with a different distribution of alkyl groups around the oxygen atom, C₃H₈O cannot exhibit metamerism.

3. C₃H₆O:
This formula has one double bond equivalent (degree of unsaturation = 1). It can represent aldehydes, ketones, or unsaturated alcohols/ethers. Let's look at the ketone with 3 carbons:
$C H_{3} - C O - C H_{3}$ (Propanone)
Since there is only one ketone possible with 3 carbon atoms, we cannot distribute the alkyl groups differently around the carbonyl carbon. Hence, it does not show metamerism.

4. C₄H₁₀O:
This is a saturated compound containing one oxygen atom, which can form ethers. Let's write the different ether isomers for C₄H₁₀O:
Isomer I: $C H_{3} - O - C H_{2} - C H_{2} - C H_{3}$ (1-Methoxypropane / Methyl propyl ether)
Isomer II: $C H_{3} C H_{2} - O - C H_{2} C H_{3}$ (Ethoxyethane / Diethyl ether)
Isomer III: $C H_{3} - O - C H (C H_{3})_{2}$ (2-Methoxypropane / Methyl isopropyl ether)

Comparing Isomer I and Isomer II, we see that the alkyl groups attached to the divalent oxygen atom are different (methyl and propyl groups in Isomer I, versus ethyl and ethyl groups in Isomer II). Therefore, these compounds are metamers of each other, confirming that C₄H₁₀O shows metamerism.