Compound A(C9H10O) shows positive iodoform test. Oxidation of A with KMnO4/KOH gives acid B(C8H6O4).

Q: Why does the iodoform test specifically indicate a methyl ketone?

The iodoform reaction requires a methyl group adjacent to a carbonyl carbon. The base removes the acidic alpha-hydrogens of the methyl group, replacing them with iodine, eventually cleaving the $C-C$ bond to form $CHI_3$ (iodoform).

Q: Does $KMnO_4$ oxidize the benzene ring itself?

No, under normal conditions, the benzene ring is highly stable due to aromaticity. $KMnO_4$ selectively oxidizes the alkyl or acyl side chains at the benzylic position.

Q: What if compound A had an ethyl group instead of a methyl and a methyl ketone?

An ethyl group would also be oxidized to a $-COOH$ group by $KMnO_4$. However, the molecular formula $C_9H_{10}O$ and the positive iodoform test strictly constrain the side chains to be a methyl and a methyl ketone.

Q: Why is phthalic acid the only dicarboxylic acid that readily forms an anhydride?

Phthalic acid has its two $-COOH$ groups at the ortho position. This close proximity allows them to easily eliminate a water molecule upon heating to form a stable 5-membered cyclic anhydride. Meta and para isomers cannot form cyclic anhydrides easily due to steric strain and distance.

Q: Could compound A be an alcohol?

While some alcohols (like secondary alcohols with a methyl group) give a positive iodoform test, the formula $C_9H_{10}O$ has a degree of unsaturation of 5 (4 for the benzene ring + 1 for the carbonyl double bond). An alcohol would require a double bond elsewhere, which doesn't fit the oxidation products perfectly without additional clues. The methyl ketone is the most direct and correct interpretation.

Question

Compound $A(C_9H_{10}O)$ shows positive iodoform test. Oxidation of A with $KMnO_4/KOH$ gives acid $B(C_8H_6O_4)$. Anhydride of B is used for the preparation of phenolphthalein. Compound A is

Sigma Percentile · Accepted Answer

### The Iodoform Clue

Let's decode the structure of compound A step by step. We are given its molecular formula, $C_9H_{10}O$, and a series of chemical reactions it undergoes. The first major clue is that compound A gives a positive iodoform test.

What does this tell us? A positive iodoform test is a classic indicator for the presence of a methyl ketone group. So, our compound must have a carbonyl group attached to a methyl group, represented as $-C(=O)CH_3$.

### The Power of Strong Oxidation

Next, we subject compound A to strong oxidation using potassium permanganate ($KMnO_4$) and potassium hydroxide ($KOH$). The result is an acid B with the formula $C_8H_6O_4$. Notice the four oxygen atoms? This indicates a dicarboxylic acid.

The strong oxidizing agent converts both the methyl ketone group and any other alkyl side chains directly into carboxylic acid ($-COOH$) groups. Since compound A has 9 carbon atoms, and the benzene ring takes up 6, while the methyl ketone group takes up 2, we are left with exactly 1 carbon atom. This must be a methyl group ($-CH_3$) attached to the ring. Both the methyl group and the methyl ketone group get oxidized to form the two $-COOH$ groups in compound B.

### The Phenolphthalein Connection

Now for the final, most revealing clue. The anhydride of acid B is used to prepare phenolphthalein. Do you remember which anhydride is used for this? Yes, it's **phthalic anhydride**!

Phthalic anhydride is formed from phthalic acid, which is benzene-1,2-dicarboxylic acid. This means the two carboxylic acid groups must be **ortho** to each other on the benzene ring.

### Bringing It All Together

If the two $-COOH$ groups in compound B are at the ortho position, then the original groups in compound A—the methyl group and the methyl ketone group—must also be ortho to each other.

Therefore, compound A is **ortho-methylacetophenone**. Looking at our options, this matches perfectly with option (c). In such road-map problems, every reagent is a clue. Iodoform revealed the functional group, oxidation confirmed the carbon count, and the anhydride gave us the exact position. Keep connecting the dots!

Animated Solution for Chemistry - Aldehydes and Ketones: Compound $A (C_{9} H_{10} O)$ shows positive iodoform test. Oxidation of A with $K M n O_{4} / K O H$ gives acid $B (C_{8} H_{6} O_{4})$ . Anhydride of B is used for the preparation of phenolphthalein. Compound A is

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Visualized Solution

$C_{9} H_{10} O$

$Iodoform Test$

$Oxidation$

$Carbon Count$

$Anhydride Formation$

$Final Conclusion$

$Summary$

The Sigma Insight: Chemical Reactions of Aldehydes and Ketones

The Iodoform Clue

The Power of Strong Oxidation

The Phenolphthalein Connection

Bringing It All Together

Similar Questions

The compound, A in the following :

Identify (A) in the following reaction sequence :