Question:
Which one of the following is not expected to undergo iodoform reaction?
Which one of the following is not expected to undergo iodoform reaction?
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The "Methyl rule" is absolute: if the carbon with the -OH doesn't have a methyl ($-CH_3$) group directly attached to it, the iodoform test will be negative.
Updated On: May 6, 2026
- Propan-2-ol
- 1-Phenylethanol
- 2-Butanol
- Ethanol
- Diphenyl methanol
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The Correct Option is
Solution and Explanation
Concept: The iodoform test identifies the presence of specific structural units that can be oxidized/halogenated to form $CHI_3$ (yellow precipitate).
• Required Units: A molecule must contain either a methyl ketone group ($CH_3-CO-$) or a methyl carbinol group ($CH_3-CH(OH)-$).
• Mechanism: Sodium hypoiodite ($NaOI$) oxidizes the alcohol to a carbonyl and then halogenates the methyl group.
Step 1: Evaluate each option for the $CH_3-CH(OH)-$ unit.
• (A) Propan-2-ol: $CH_3-CH(OH)-CH_3$ (Has the unit)
• (B) 1-Phenylethanol: $C_6H_5-CH(OH)-CH_3$ (Has the unit)
• (C) 2-Butanol: $CH_3-CH(OH)-CH_2CH_3$ (Has the unit)
• (D) Ethanol: $H-CH(OH)-CH_3$ (Has the unit)
Step 2: Identify the failing compound. (E) Diphenyl methanol has the structure $(C_6H_5)_2CH-OH$. In this molecule, the central carbon is attached to two phenyl rings, not a methyl group. Because it lacks the essential $-CH_3$ group attached to the carbinol carbon, it cannot form iodoform.
• Required Units: A molecule must contain either a methyl ketone group ($CH_3-CO-$) or a methyl carbinol group ($CH_3-CH(OH)-$).
• Mechanism: Sodium hypoiodite ($NaOI$) oxidizes the alcohol to a carbonyl and then halogenates the methyl group.
Step 1: Evaluate each option for the $CH_3-CH(OH)-$ unit.
• (A) Propan-2-ol: $CH_3-CH(OH)-CH_3$ (Has the unit)
• (B) 1-Phenylethanol: $C_6H_5-CH(OH)-CH_3$ (Has the unit)
• (C) 2-Butanol: $CH_3-CH(OH)-CH_2CH_3$ (Has the unit)
• (D) Ethanol: $H-CH(OH)-CH_3$ (Has the unit)
Step 2: Identify the failing compound. (E) Diphenyl methanol has the structure $(C_6H_5)_2CH-OH$. In this molecule, the central carbon is attached to two phenyl rings, not a methyl group. Because it lacks the essential $-CH_3$ group attached to the carbinol carbon, it cannot form iodoform.
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