Question

An organic compound \(P\) of molecular formula \( \mathrm{C_6H_{12}O_3} \) gives positive iodoform test but negative Tollens’ test. When \(P\) is treated with dilute acid, it produces \(Q\). \(Q\) gives positive Tollens’ test and also iodoform test. Identify compound \(P\).

• A. \( \mathrm{CH_3COCH(OCH_3)CH_2OCH_3} \)
• B. \( \mathrm{CH_3COCH_2CH(OCH_3)_2} \)
• C. \( \mathrm{HCOCH_2CH_2CH(OCH_3)_2} \)
• D. \( \mathrm{CH_3COC(OCH_3)_2CH_3} \)

Hint:

Remember:

• Iodoform test detects: \[ \mathrm{CH_3CO-} \] group.
• Tollens’ reagent detects aldehydes.
• Ketals/acetals hydrolyse back to carbonyl compounds in acidic medium.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem requires structural identification of organic compounds using diagnostic chemical tests. - A positive iodoform test indicates the structural presence of a methyl ketone ($\text{CH}_3\text{C=O}$) group or a methyl carbinol ($\text{CH}_3\text{CH(OH)}$) group. - A positive Tollens' test indicates the presence of an aldehyde ($\text{-CHO}$) functional group, while a negative test indicates its absence (e.g., ketones or protected aldehydes). - Acid hydrolysis of an acetal or ketal functional group ($\text{-CH(OR)}_2$ or $\text{-C(OR)}_2\text{-}$) converts it back into its corresponding aldehyde or ketone along with alcohols.
Step 2: Key Formula or Approach:
Analyze the structural clues for Compound P and Compound Q systematically: 1. Compound P ($\text{C}_6\text{H}_{12}\text{O}_3$): Has a $\text{CH}_3\text{CO-}$ group (positive iodoform) but lacks a $\text{-CHO}$ group (negative Tollens'). 2. Acid Hydrolysis of P: Dilute acid breaks down any acetal linkages present in P. 3. Compound Q: Possesses both an aldehyde group (positive Tollens') and a methyl ketone group (positive iodoform).
Step 3: Detailed Explanation:
Let's test Option (2), $\text{CH}_3\text{COCH}_2\text{CH(OCH}_3)_2$: - It contains the methyl ketone group ($\text{CH}_3\text{CO-}$), which accounts for Compound P's positive iodoform test. - The aldehyde group is protected as a dimethyl acetal group ($\text{-CH(OCH}_3)_2$). Because it contains a ketone and a protected acetal rather than a free aldehyde, it will give a negative Tollens' test. This completely matches all given behaviors for Compound P. Now, let's look at what happens when Option (2) is treated with dilute acid (hydrolysis): The dimethyl acetal group ($\text{-CH(OCH}_3)_2$) undergoes clean cleavage to restore the parent aldehyde group ($\text{-CHO}$), eliminating two molecules of methanol ($\text{CH}_3\text{OH}$): \[ \text{CH}_3\text{COCH}_2\text{CH(OCH}_3)_2 + \text{H}_2\text{O} \xrightarrow{\text{dil. H}^+} \text{CH}_3\text{COCH}_2\text{CHO} + 2\text{CH}_3\text{OH} \] The resulting hydrolyzed product is Compound Q: $\text{CH}_3\text{COCH}_2\text{CHO}$. Let's check the structural features of Compound Q ($\text{CH}_3\text{COCH}_2\text{CHO}$): 1. It contains a free terminal aldehyde group ($\text{-CHO}$), meaning it will yield a positive Tollens' test. 2. It simultaneously retains its methyl ketone segment ($\text{CH}_3\text{CO-}$), which yields a positive iodoform test. This perfectly satisfies all condition criteria outlined in the problem. Let's briefly check why other options fail: - Option (1) and (4) do not possess an acetal group that can hydrolyze into a free aldehyde, so their products will fail Tollens' test. - Option (3) contains a formyl hydrogen ($\text{HCO-}$), which behaves as an active aldehyde group, meaning P would mistakenly give a positive Tollens' test initially.
Step 4: Final Answer:
The structure of compound P is $\text{CH}_3\text{COCH}_2\text{CH(OCH}_3)_2$.