C₆H₁₂O₃ gives positive iodoform test on hydrolysis with dil. acid. The hydrolysis product formed gives Tollens’ and iodoform test both. Find the structure of C₆H₁₂O₃.

The correct answer is: CH₃COCH₂CH₂CH(OH)OCH₃

Step-by-Step Analysis:

1. Understanding the Functional Groups & Tests:
- Iodoform Test: A positive iodoform test is given by compounds containing either a methyl ketone group (CH₃CO-) or a methyl carbinol group (CH₃CH(OH)-).
- Tollens' Test: A positive Tollens' test (formation of a silver mirror) is characteristic of aldehydes (-CHO) and other easily oxidizable functional groups like hemiacetals.

2. Analyzing the Compound C₆H₁₂O₃:
The given compound CH₃COCH₂CH₂CH(OH)OCH₃ contains:
- A methyl ketone group (CH₃CO-) at one end, which is responsible for giving a positive iodoform test directly.
- A hemiacetal group (-CH(OH)OCH₃) at the other end.

3. Hydrolysis with Dilute Acid:
Hemiacetals are unstable in the presence of dilute acids and readily undergo hydrolysis to form the corresponding aldehyde and alcohol.
The hydrolysis of the hemiacetal group in CH₃COCH₂CH₂CH(OH)OCH₃ proceeds as follows:

CH₃COCH₂CH₂CH(OH)OCH₃ + H₂O (dil. acid) → CH₃COCH₂CH₂CHO + CH₃OH

The hydrolysis products formed are 4-oxopentanal (CH₃COCH₂CH₂CHO) and methanol (CH₃OH).

4. Characterizing the Hydrolysis Product:
Let us analyze the major organic product, 4-oxopentanal (CH₃COCH₂CH₂CHO):
- It contains an aldehyde group (-CHO), which reduces Tollens' reagent to give a positive Tollens' test.
- It also retains the methyl ketone group (CH₃CO-), which reacts with iodine and sodium hydroxide to give a yellow precipitate of iodoform (positive iodoform test).
Therefore, the hydrolysis product satisfies both conditions, confirming the initial structure.