Q.40 Which of the following pairs of compounds can be distinguished by iodoform test
performed in ammonium hydroxide?
(A) CH3 COCH3 and C 2 H5 OH
(B) C 2 H5 OH and CH3 OH
(C) CH3 COCH3 and C6 H5 COCH3
(D) C 6 H5 COCH3 and C2H5 OH
Acetophenone (C₆H₅COCH₃) and ethanol (C₂H₅OH) can be distinguished by the iodoform test in ammonium hydroxide, as both give a positive test in NaOH but only ethanol reacts in NH₄OH.
Iodoform Test Basics
The iodoform test detects methyl ketones (R-CO-CH₃) or alcohols oxidizable to them (R-CH(OH)-CH₃ or CH₃CH₂OH), forming yellow CHI₃ precipitate. In NaOH, aliphatic (acetone) and aromatic methyl ketones (acetophenone) both test positive due to enolization. NH₄OH provides weaker basicity, allowing ethanol oxidation to acetaldehyde (which gives iodoform) but hindering acetophenone’s reaction due to poor enolization under mild conditions.
Option Analysis
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(A) CH₃COCH₃ and C₂H₅OH: Both positive in NH₄OH—acetone via direct reaction, ethanol via oxidation. Cannot distinguish.
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(B) C₂H₅OH and CH₃OH: Ethanol positive (oxidizes to CH₃CHO); methanol negative (no suitable group). Distinguishable, but not the focus here.
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(C) CH₃COCH₃ and C₆H₅COCH₃: Acetone positive; acetophenone often negative or slow in NH₄OH (aromatic ketones less reactive). Possible distinction, but both positive in standard NaOH.
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(D) C₆H₅COCH₃ and C₂H₅OH: Correct—acetophenone negative in NH₄OH; ethanol positive. Clear distinction.
The iodoform test in ammonium hydroxide serves as a key qualitative tool in organic chemistry, especially for CSIR NET aspirants tackling compound differentiation like CH3COCH3 (acetone), C2H5OH (ethanol), and C6H5COCH3 (acetophenone). This test leverages weak basic conditions to selectively identify methyl ketones and ethanol, producing a yellow iodoform (CHI₃) precipitate with antiseptic odor.
Test Mechanism
Under NH₄OH, iodine forms hypoiodite (I₂ + NH₄OH → NH₄I + HOI). Ethanol oxidizes to acetaldehyde (CH₃CHO), which has CH₃CO- and yields CHI₃. Aliphatic methyl ketones like acetone react directly via α-iodination, but aromatic ones like acetophenone require stronger base (NaOH) for enolization—NH₄OH fails here.
Why It Distinguishes Pairs
| Pair | Acetone (CH₃COCH₃) | Ethanol (C₂H₅OH) | Acetophenone (C₆H₅COCH₃) | Distinguishable? |
|---|---|---|---|---|
| (A) CH₃COCH₃ & C₂H₅OH | Positive (yellow ppt) | Positive (yellow ppt) | – | No |
| (B) C₂H₅OH & CH₃OH | – | Positive | – | Yes, but not queried |
| (C) CH₃COCH₃ & C₆H₅COCH₃ | Positive | – | Negative/slow | Yes |
| (D) C₆H₅COCH₃ & C₂H₅OH | – | Positive | Negative | Yes |
Answer: (D)—Ideal for exams, as NH₄OH exploits reactivity differences.
