Iodoform cannot be prepared from:

C H_{3} O H

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C_{2} H_{5} O H

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C H_{3} C H O

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C H_{3} C O C H_{3}

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Solution

The correct option is C $C H_{3} O H$
The formation of iodoform $C H I_{3}$ depends on previous formation of $C H_{3} - C O -$ group which next undergoes iodination to $C I_{3} - C O -$ before finally producing $C I_{3}$ (iodoform).
This means that ethanal (acetaldehyde)- option (C) and propanone (acetone)-option (D) will give positive results since they both contain the necessary $C H_{3} - C O -$ group.
Ethanol gives positive results because it is oxidised (by $I_{2}$ ) to ethanal. Any Alkan-2-ol $H_{3} C C H (O H) - R$ will give positive results due to oxidation.
Methanol $(C H_{3} - O H)$ -option (A) does not meet any of the criteria, it neither contains $C H_{3} - C O -$ group, nor it is oxidised to $(H C H O)$ in presence of $I_{2}$
Thus methanol will give negative iodoform test.
Option A is correct.

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