Question

Identify the incorrect match with respect to compounds to be distinguished and reagent used

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

To distinguish aldehydes from ketones, remember these key tests: - Tollens' Test (Silver Mirror): Aldehyde (+), Ketone (-). - Fehling's/Benedict's Test (Red Precipitate): Aliphatic Aldehyde (+), Ketone (-). Brady's test is a general test for the carbonyl group (C=O) and is positive for both aldehydes and ketones.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This question asks to identify which pair of compounds cannot be distinguished by the given chemical test. This requires knowledge of common qualitative tests for different organic functional groups.
Step 2: Analyzing Each Match:
- (A) CH₃OH (Methanol), CH₃CH₂OH (Ethanol) with I₂ + NaOH (Iodoform Test): The iodoform test gives a positive result (yellow precipitate of iodoform, CHI₃) for compounds containing a CH₃-CH(OH)- group or a CH₃-C(=O)- group.
- Ethanol (CH₃-CH₂OH) has the required CH₃-CH(OH)- group and will give a positive iodoform test.
- Methanol (CH₃OH) does not have this group and will give a negative test.
- Therefore, this reagent can be used to distinguish between them. This match is correct.
- (B) CH₃CH₂OH (Ethanol, a 1° alcohol), CH₃-C(CH₃)₂-OH (tert-Butyl alcohol, a 3° alcohol) with Anhydrous ZnCl₂ + Conc. HCl (Lucas Test): The Lucas test is used to distinguish between primary, secondary, and tertiary alcohols.
- Tertiary alcohols react immediately to give a cloudy solution (due to the formation of the insoluble alkyl chloride).
- Secondary alcohols react within a few minutes.
- Primary alcohols react only upon heating.
- Since we have a primary alcohol (ethanol) and a tertiary alcohol (tert-butyl alcohol), they will react at very different rates with the Lucas reagent, allowing them to be easily distinguished. This match is correct.
- (C) CH₃-C≡CH (Propyne, a terminal alkyne), CH₃-C≡C-CH₃ (But-2-yne, an internal alkyne) with Na (Sodium metal): Terminal alkynes have an acidic hydrogen atom attached to the sp-hybridized carbon. This acidic hydrogen can react with active metals like sodium to liberate hydrogen gas.
- Propyne will react: \(2\text{CH}_3\text{-C}\equiv\text{CH} + 2\text{Na} \rightarrow 2\text{CH}_3\text{-C}\equiv\text{C}^-\text{Na}^+ + \text{H}_2\uparrow\).
- Internal alkynes like but-2-yne do not have an acidic hydrogen and will not react with sodium metal.
- Therefore, sodium metal can distinguish between them. This match is correct.
- (D) CH₃-CHO (Ethanal, an aldehyde), (CH₃)₂CO (Propanone, a ketone) with H₂N-NH-Ph-NO₂ (2,4-Dinitrophenylhydrazine, or Brady's reagent): Brady's reagent reacts with both aldehydes and ketones to form brightly colored (orange, red, or yellow) precipitates called 2,4-dinitrophenylhydrazones.
- Since both ethanal and propanone are carbonyl compounds, they will both give a positive test with Brady's reagent.
- Therefore, this reagent cannot be used to distinguish between them. This match is incorrect. To distinguish an aldehyde from a ketone, one would use a mild oxidizing agent like Tollens' reagent or Fehling's solution, which react with aldehydes but not ketones.
Step 3: Final Answer:
The incorrect match is the use of Brady's reagent to distinguish between an aldehyde and a ketone, as both give a positive result. Therefore, option (D) is the correct answer.