Q.An aromatic compound 'A' (molecular formula $C_8H_8O$) gives a positive 2,4-DNP test. It gives a yellow precipitate of compound 'B' on treatment with iodine and sodium hydroxide solution. Compound 'A' does not give the Tollens or Fehling's test. On drastic oxidation with potassium permanganate it forms a carboxylic acid 'C' (molecular formula $C_7H_6O_2$), which is also formed along with the yellow compound in the above reaction. Identify A, B and C and write all the reactions involved.

Concept: IUPAC Nomenclature & Iodoform Reaction

The positive 2,4-DNP test confirms a carbonyl group ($\ce{C=O}$). The iodoform test (yellow precipitate of $\ce{CHI3}$) indicates a methyl ketone ($\ce{COCH3}$) or a methyl carbinol ($\ce{CH(OH)CH3}$) that oxidises to a methyl ketone under the reaction conditions. Since Tollens/Fehling tests are negative, the compound is a ketone, not an aldehyde.

Step 1 – Identify the methyl ketone

Molecular formula $\ce{C8H8O}$ with a benzene ring (aromatic) and a $\ce{COCH3}$ group gives $\ce{C6H5COCH3}$ (acetophenone). This fits: $\ce{C6H5}$ ($\ce{C6H5}$) + $\ce{COCH3}$ ($\ce{C2H3O}$) = $\ce{C8H8O}$.

Step 2 – Iodoform reaction

Acetophenone with $\ce{I2/NaOH}$ gives yellow iodoform ($\ce{CHI3}$, compound B) and sodium benzoate ($\ce{C6H5COONa}$). Acidification yields benzoic acid ($\ce{C7H6O2}$, compound C).

Step 3 – Drastic oxidation …

The problem is a classic organic identification puzzle from the IUPAC nomenclature and reactions chapter. The clues are all about functional group tests and oxidation behaviour. Let’s decode them one by one.

1. Molecular formula $\ce{C8H8O}$ and aromatic nature

The formula has 8 carbons, 8 hydrogens, and 1 oxygen. For an aromatic compound, the benzene ring itself accounts for $\ce{C6H5-}$ (6C, 5H). That leaves $\ce{C2H3O}$ for the side chain. The side chain must contain the oxygen — so it’s either a carbonyl group ($\ce{C=O}$) or an alcohol/ether. But the tests will tell us which.

2. Positive 2,4-DNP test

This test (with 2,4-dinitrophenylhydrazine) gives an orange-red precipitate for any carbonyl compound — aldehyde or ketone. So A has a $\ce{C=O}$ group. The side chain is therefore an acyl group, not an alcohol or ether.

3. Positive iodoform test (iodine + NaOH gives yellow precipitate)

The iodoform test is specific for methyl ketones ($\ce{R-CO-CH3}$) or compounds that can be oxidised to a methyl ketone (like ethanol or secondary alcohols with a $\ce{CH3CH(OH)-}$ group). The yellow precipitate is iodoform, $\ce{CHI3}$. So A must contain the $\ce{COCH3}$ (acetyl) group. That fits the leftover $\ce{C2H3O}$ perfectly: $\ce{-COCH3}$.

4. Negative Tollens and Fehling’s tests

These tests are positive for aldehydes (and some α-hydroxy ketones). A fails both, so it is not an aldehyde. This confirms A is a ketone — specifically an aromatic methyl ketone.

5. Drastic oxidation with $\ce{KMnO4}$ gives carboxylic acid C ($\ce{C7H6O2}$)

Drastic oxidation (hot, alkaline $\ce{KMnO4}$) cleaves alkyl side chains on benzene rings down to the ring, turning any carbon chain attached to the ring into a carboxyl group ($\ce{-COOH}$). The product $\ce{C7H6O2}$ is benzoic acid ($\ce{C6H5COOH}$). This tells us the benzene ring has a single carbon side chain that gets fully oxidised to $\ce{-COOH}$. Since A is $\ce{C6H5COCH3}$, oxidation removes the methyl carbon and converts the carbonyl carbon to carboxyl — exactly giving benzoic acid.

6. The iodoform reaction also produces C

In the iodoform reaction, a methyl ketone $\ce{R-COCH3}$ reacts with $\ce{I2/NaOH}$ to give $\ce{RCOONa}$ (the sodium salt of the carboxylic acid) and $\ce{CHI3}$ (yellow precipitate). For A, $\ce{R = C6H5-}$, so the salt is sodium benzoate, which on acidification gives benzoic acid (C). This matches perfectly. …

Method: Retro-synthesis with Functional Group Analysis

This method works backwards from the chemical tests and degradation products to deduce the structure.

Step 1 — Analyse the molecular formula and unsaturation

• Formula of A: $\mathrm{C_8H_8O}$
• Saturation formula for 8 carbons: $\mathrm{C_8H_{18}}$
• Degree of unsaturation = $\frac{2(8) + 2 - 8}{2} = 5$

Five degrees of unsaturation strongly suggests a benzene ring (4 unsaturations) plus one more double bond or ring.

Step 2 — Interpret the chemical tests

So A is an aromatic methyl ketone.

Step 3 — Identify compound B (iodoform reaction)

The iodoform reaction:

• B is the yellow precipitate: iodoform ($\mathrm{CHI_3}$)

Step 4 — Identify compound C (oxidation product)

• Drastic oxidation with $\mathrm{KMnO_4}$ gives C: $\mathrm{C_7H_6O_2}$
• This formula matches benzoic acid ($\mathrm{C_6H_5COOH}$)
• Also formed alongside iodoform in the iodoform test (as sodium benzoate, then acidified)

This means the aromatic ring has a methyl ketone side chain that gets oxidised to a carboxyl group.

Step 5 — Deduce compound A

• A = aromatic methyl ketone = $\mathrm{C_6H_5COCH_3}$
• Check formula: $\mathrm{C_8H_8O}$ ✓
• This is acetophenone

Final Answer

| Compound | Name | Structure | …

Step 1: Understanding the clues

• Molecular formula of A: $C_8H_8O$

  Degree of unsaturation = $\frac{2C + 2 - H}{2} = \frac{16+2-8}{2} = 5$ → aromatic ring (4) + one more double bond (likely C=O).

• Positive 2,4-DNP test → A has a carbonyl group (aldehyde or ketone).

• Yellow precipitate with $I_2/NaOH$ → Iodoform test positive → A must have a $CH_3CO-$ group (methyl ketone) or $CH_3CH(OH)-$ group (but here it’s a ketone, since no alcohol).

• No Tollens or Fehling’s test → A is not an aldehyde → it’s a ketone.

• Drastic oxidation with $KMnO_4$ gives $C_7H_6O_2$ → benzoic acid ($C_6H_5COOH$).

  So A has a benzene ring with a side chain that gets oxidised to $COOH$.

• Same carboxylic acid C is also formed along with the yellow compound → the iodoform reaction on A gives benzoic acid + iodoform (yellow).

Step 2: Identifying A, B, C

• A must be acetophenone ($C_6H_5COCH_3$).

  • IUPAC name: 1-phenylethanone (or phenyl methyl ketone).
  • Iodoform test: $C_6H_5COCH_3 + 3I_2 + 4NaOH \rightarrow C_6H_5COONa + CHI_3 \downarrow + 3NaI + 3H_2O$ Yellow precipitate B = iodoform ($CHI_3$). Carboxylic acid C = benzoic acid ($C_6H_5COOH$).

• Oxidation with $KMnO_4$:

  $C_6H_5COCH_3 \xrightarrow[KMnO_4, \Delta]{} C_6H_5COOH + CO_2 + H_2O$

Step 3: Common mistakes & how to avoid them