Question

Compound with molecular formula C₃H₆O can show :

• A. Positional isomerism
• B. Functional group isomerism
• C. Metamerism
• D. Both positional isomerism and metamerism

Hint:

Aldehydes and Ketones with the same carbon count are almost always functional isomers.

Answer 1

The Correct Option is B

Step 1: Calculate Degree of Unsaturation: $DU = \frac{2(3)+2-6}{2} = 1$.
Step 2: Possible structures include **Propanal** ($CH_3CH_2CHO$) and **Propanone** ($CH_3COCH_3$).
Step 3: Since propanal (aldehyde) and propanone (ketone) have different functional groups, they exhibit **Functional group isomerism**.
Step 4: Metamerism is not possible as it requires at least 4 carbons for ethers/ketones to have different alkyl groups around the functional group.