Compound with molecular formula C3H6O can show
(a) positional isomerism
(b) both positional isomeris

Q: Why can't $C_3H_6O$ show positional isomerism?

Positional isomerism requires the same functional group to change its position on the carbon chain. For a 3-carbon chain, an aldehyde can only be at position 1, and a ketone can only be at position 2. Since their positions cannot vary without changing the functional group itself, positional isomerism is impossible.

Q: What exactly is metamerism and why is it wrong here?

Metamerism happens when the alkyl groups attached to a central functional group (like $-O-$ or $-CO-$) are different. For a ketone to show metamerism, you need at least 5 carbons (e.g., diethyl ketone vs methyl propyl ketone). Propanone ($CH_3-CO-CH_3$) is too small to have metamers.

Q: Can $C_3H_6O$ form an alcohol?

Yes! It can form an unsaturated alcohol like allyl alcohol ($CH_2=CH-CH_2OH$). However, allyl alcohol and propanal are still functional isomers because they have different functional groups (alkene+alcohol vs aldehyde).

Q: How do I quickly know if a formula has a double bond?

Use the Double Bond Equivalent (DBE) formula: $C + 1 - H/2 + N/2$. For $C_3H_6O$, it is $3 + 1 - 6/2 = 1$. A DBE of 1 means the molecule has exactly one double bond or one ring.

Q: Are aldehydes and ketones always functional isomers?

Yes, as long as they have the same molecular formula (same number of carbon atoms), aldehydes and ketones are classic examples of functional group isomers because they share the general formula $C_nH_{2n}O$.

Question

Compound with molecular formula $C_3H_6O$ can show

Options:

(A) positional isomerism

(B) both positional isomerism and metamerism

(C) metamerism

(D) functional group isomerism

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are given a box of molecular building blocks: 3 Carbon atoms, 6 Hydrogen atoms, and 1 Oxygen atom. Your task is to build stable molecules. The molecular formula is $C_3H_6O$.

The very first thing a chemist does is look for patterns. Notice how the number of hydrogen atoms is exactly double the number of carbon atoms? This perfectly matches the general formula $C_nH_{2n}O$.

Whenever you see this specific ratio, an alarm bell should ring in your head: **One Degree of Unsaturation**. This means our molecule must contain either one double bond or one ring. If we stick to acyclic (open-chain) structures, that double bond must be between a carbon and an oxygen atom, forming a **carbonyl group** ($-C=O$).

### The Master Equation: Aldehydes vs Ketones
Now that we know we have a carbonyl group, where can we place it? We have a 3-carbon chain to play with.

**Possibility 1: The Terminal Position**
If we place the carbonyl group at the very end of the chain, it becomes an **aldehyde**. Let's draw it out: $CH_3-CH_2-CHO$. This molecule is called **propanal**.

**Possibility 2: The Internal Position**
What if we place the carbonyl group in the middle of the chain? It is now sandwiched between two carbon atoms, making it a **ketone**. Let's draw this one: $CH_3-CO-CH_3$. This molecule is called **propanone** (or acetone, the common nail polish remover!).

### Final Calculation and Conclusion
We have successfully built two entirely different molecules from the exact same set of atoms. Propanal is an aldehyde, and propanone is a ketone. Because they belong to completely different chemical families (functional groups), they are classified as **functional group isomers**.

But wait, why not positional isomers? Positional isomerism requires the *same* functional group to be at different positions. Here, the functional groups themselves are different!

What about metamerism? Metamerism occurs when the alkyl chains on either side of a polyvalent functional group (like an ether or ketone) differ. For ketones, you need at least 5 carbon atoms (like pentan-2-one and pentan-3-one) to show metamerism. With only 3 carbons, propanone is the *only* possible ketone.

Therefore, the only correct relationship between the possible structures of $C_3H_6O$ is **functional group isomerism**.

Animated Solution for Chemistry - Organic Chemistry: Compound with molecular formula $C_{3} H_{6} O$ can show

Select Answer:

Visualized Solution

Analyzing the Molecular Formula

Identifying Functional Groups

Drawing the Aldehyde

Drawing the Ketone

Conclusion

Why not other isomerisms?

The Sigma Insight: Nomenclature and Characterisation

Analyzing the Setup Imagine you are given a box of molecular building blocks: 3 Carbon atoms, 6 Hydrogen atoms, and 1 Oxygen atom

The Master Equation

Final Calculation and Conclusion We have successfully built two entirely different molecules from the exact same set of atoms

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Animated Solution for Chemistry - Organic Chemistry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Compound with molecular formula C3​H6​O can show