Given molecular formula

$C_3H_6O$

First, find the degree of unsaturation:

$\text{DU} = \frac{2C+2-H}{2} = \frac{2(3)+2-6}{2} = \frac{2}{2}=1$

So the compound has one unsaturation, which may be due to:

• one double bond, or
• one ring.

Check possible functional groups for $C_3H_6O$

Some possible structures are:

• Propanal: $CH_3CH_2CHO$
• Propanone: $CH_3COCH_3$
• Allyl alcohol: $CH_2=CHCH_2OH$
• Cyclopropanol, etc.

In standard isomerism questions, the important pair is:

• Aldehyde: propanal
• Ketone: propanone

These have the same molecular formula $C_3H_6O$ but different functional groups.

Hence, $C_3H_6O$ shows functional group isomerism.

Check positional isomerism

Positional isomerism means same carbon skeleton and same functional group, but different position of the functional group or multiple bond.

For a 3-carbon carbonyl compound:

• Aldehyde group cannot shift position freely; it must be terminal.
• Ketone on a 3-carbon chain can only be at carbon-2.

So among the usual carbonyl compounds of formula $C_3H_6O$, positional isomerism is not possible.

Check metamerism

Metamerism is shown by compounds having a divalent functional group like:

• ethers $R-O-R'$
• secondary amines $R-NH-R'$
• ketones $R-CO-R'$

But for $C_3H_6O$, only one ketone is possible:

• $CH_3COCH_3$

There is no other distribution of carbon atoms across the carbonyl group possible with only 3 carbons.

Therefore, metamerism is not shown.

Evaluate options

• A: Positional isomerism — Incorrect
• B: Both positional isomerism and metamerism — Incorrect
• C: Metamerism — Incorrect
• D: Functional group isomerism — Correct

Final answer

The molecular formula $C_3H_6O$ can show functional group isomerism due to the pair:

$CH_3CH_2CHO \quad \text{and} \quad CH_3COCH_3$