Q.An organic compound with the molecular formula $C_9H_{10}O$ forms 2,4-DNP derivative, reduces Tollens' reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound.

The key idea is that a compound reducing Tollens' reagent must be an aldehyde, and undergoing Cannizzaro reaction means it has no $\alpha$-hydrogen — which for an aldehyde almost always means the $-CHO$ is attached directly to an aromatic ring or another carbon with no H. Vigorous oxidation yielding 1,2-benzenedicarboxylic acid (phthalic acid) shows the aldehyde group is directly attached to a benzene ring that also carries a second, oxidisable side-chain at the ortho position.

Reasoning steps:

1. Functional group clues: Reduces Tollens' reagent → aldehyde ($-CHO$). Undergoes Cannizzaro reaction → no $\alpha$-hydrogen, so the $-CHO$ must be attached directly to the aromatic ring (an aryl aldehyde).

2. Oxidation product: Vigorous oxidation gives 1,2-benzenedicarboxylic acid (phthalic acid), i.e. a benzene ring with two $-COOH$ groups at ortho positions. One $-COOH$ comes from the $-CHO$ itself (oxidised in place); the other must come from a second ring substituent that is also fully oxidisable to $-COOH$ — any alkyl side-chain on a benzene ring is oxidised this way under vigorous conditions, regardless of chain length. …

The problem gives you four clues about a compound $\ce{C9H10O}$. Let’s unpack each one and see how they fit together.

Why the Cannizzaro reaction is the key

The Cannizzaro reaction is a disproportionation of an aldehyde that has no $\alpha$-hydrogen atoms — that is, no hydrogen on the carbon next to the $-\ce{CHO}$ group. In strong base, one molecule gets oxidised to a carboxylic acid and another gets reduced to an alcohol. So if your compound undergoes Cannizzaro, it must be an aldehyde with the $-\ce{CHO}$ group attached either to an aromatic ring or to a tertiary carbon (like $(\ce{CH3})3\ce{CCHO}$). That immediately narrows the field.

Now let’s walk through the clues step by step.

1. Forms a 2,4-DNP derivative

   This tells you the compound has a carbonyl group ($\ce{C=O}$) — either an aldehyde or a ketone. 2,4-dinitrophenylhydrazine reacts with any carbonyl to give an orange/red precipitate.

2. Reduces Tollens’ reagent

   Tollens’ reagent ($\ce{[Ag(NH3)2]+}$) is reduced only by aldehydes (and some $\alpha$-hydroxy ketones, but that’s rare). Ketones do not react. So the compound is an aldehyde.

3. Undergoes Cannizzaro reaction

   As we said, this means the aldehyde has no $\alpha$-hydrogen. For an aromatic aldehyde, that’s automatic — the $\alpha$-carbon is part of the benzene ring. So the compound is an aromatic aldehyde.

4. On vigorous oxidation, gives 1,2-benzenedicarboxylic acid

   That’s phthalic acid — a benzene ring with two carboxylic acid groups in the ortho positions. Vigorous oxidation (like hot $\ce{KMnO4}$ or $\ce{HNO3}$) will oxidise any alkyl side chain on a benzene ring to $-\ce{COOH}$, and it will also oxidise $-\ce{CHO}$ to $-\ce{COOH}$. So the original compound must have had two substituents on the benzene ring that could be oxidised to $-\ce{COOH}$ groups, and they must be ortho to each other.

   One of those substituents is the aldehyde group ($-\ce{CHO}$). The other must be a carbon chain that, after oxidation, becomes the second $-\ce{COOH}$. Since the molecular formula is $\ce{C9H10O}$, and we already have $\ce{C6H4}$ (benzene ring) + $\ce{CHO}$ (1 carbon), that accounts for $\ce{C7H5O}$. The remaining $\ce{C2H5}$ must be the other side chain — an ethyl group ($-\ce{CH2CH3}$).

   So the compound is an ethylbenzaldehyde with the two groups ortho to each other. …

Method: Retrospective Analysis (Reverse Engineering) from Chemical Tests

This method works backwards from the given reactions and final oxidation product to deduce the structure.

Step 1: Analyse the molecular formula

$C_9H_{10}O$

Degree of unsaturation (DoU) = $\frac{2C + 2 - H}{2} = \frac{2(9) + 2 - 10}{2} = \frac{10}{2} = 5$

Four unsaturations likely come from a benzene ring, leaving one more — possibly a carbonyl ($C=O$) or an alkene.

Step 2: Interpret the chemical tests

• Forms 2,4-DNP derivative → Contains a carbonyl group ($C=O$), either aldehyde or ketone.
• Reduces Tollens' reagent → Must be an aldehyde (ketones do not reduce Tollens').
• Undergoes Cannizzaro reaction → Aldehyde with no $\alpha$-hydrogen (i.e., $\alpha$-carbon has no H attached). This is the key restriction.

So the compound is an aromatic aldehyde with no $\alpha$-hydrogen.

Step 3: Use the oxidation product

Vigorous oxidation gives 1,2-benzenedicarboxylic acid (phthalic acid):

This means the original compound has a benzene ring with two substituents in ortho (1,2) positions, both of which can be oxidised to $-COOH$ groups.

One substituent is already $-CHO$ (the aldehyde). The other must be a side chain that oxidises to $-COOH$ — likely an alkyl group.

Step 4: Assemble the structure

• Benzene ring with two ortho substituents.
• One is $-CHO$ (aldehyde, no $\alpha$-H).
• The other must be a group that gives total $C_9H_{10}O$.

Count carbons: …

Common Mistakes & How to Avoid Them (Cannizzaro Reaction Problem)

Mistake 1: Forgetting that Cannizzaro requires no α-hydrogen

• The error: Students often pick an aldehyde that has α-hydrogens (e.g., propanal, butanal) and still apply Cannizzaro.
• Why it's wrong: Cannizzaro reaction is specific to aldehydes without α-hydrogen (like HCHO, Ar–CHO, or $R_3C$–CHO). If α-hydrogens exist, the compound undergoes aldol condensation instead.
• How to avoid: Before writing Cannizzaro, check the α-carbon. If it has even one H, Cannizzaro is impossible.

Mistake 2: Ignoring the 2,4-DNP test implication

• The error: Students conclude the compound is an aldehyde from Tollens' test, but forget that 2,4-DNP also confirms a carbonyl group (aldehyde or ketone).
• Why it's wrong: 2,4-DNP alone doesn't distinguish aldehyde vs ketone — but combined with Tollens' (positive) and Cannizzaro (positive), it narrows to an aldehyde without α-H.
• How to avoid: Always list what each test tells you:
  • 2,4-DNP positive → C=O present
  • Tollens' positive → aldehyde (not ketone)
  • Cannizzaro positive → aldehyde with no α-H

Mistake 3: Misinterpreting "vigorous oxidation gives 1,2-benzenedicarboxylic acid"

• The error: Students think the compound itself contains two –COOH groups, or that it's already phthalic acid.
• Why it's wrong: Vigorous oxidation (e.g., hot KMnO₄) converts alkyl side chains on benzene to –COOH. Getting phthalic acid means the original compound had two substituents in ortho positions on the benzene ring.
• How to avoid: Draw the product (phthalic acid) and work backwards:
  • Two –COOH groups at ortho positions → original had two alkyl/aldehyde groups at ortho positions.
  • One of them must be –CHO (from Tollens' and Cannizzaro).

Mistake 4: Forgetting the molecular formula constraint ($C_9H_{10}O$)

• The error: Students propose a structure that doesn't match the formula (e.g., adding too many carbons or oxygens).
• Why it's wrong: The formula is a hard constraint. $C_9H_{10}O$ has degree of unsaturation = $\frac{2(9)+2-10}{2} = 5$. A benzene ring accounts for 4, so there's one more double bond or ring — which is the C=O of the aldehyde.
• How to avoid: Always calculate unsaturation first. Then check: benzene (4) + one C=O (1) = 5. Perfect fit. …