Which of the following will not form a yellow precipitate on heating with an alkaline solution of iodine :-
A
$CH_3CH_2CH(OH)CH_3$
B
$CH_3OH$
C
$CH_3CH_2OH$
D
$CH_3CH(OH)CH_3$
The answer is B.
Solution
Heating with iodine and alkali gives a yellow precipitate of iodoform, $CHI_3$, only if the compound contains the $CH_3CO-$ group or the $CH_3CH(OH)-$ group (which is first oxidised to $CH_3CO-$). $CH_3CH_2CH(OH)CH_3$ (butan-2-ol) has the $CH_3CH(OH)-$ group; it gives iodoform. $CH_3CH_2OH$ (ethanol) has $CH_3CH(OH)-$ with H as the other group; it is oxidised to $CH_3CHO$ and gives iodoform. $CH_3CH(OH)CH_3$ (propan-2-ol) has the $CH_3CH(OH)-$ group; it gives iodoform. $CH_3OH$ (methanol) has only one carbon. It cannot give a $CH_3CO-$ group (it is oxidised to HCHO), so it does not give the iodoform test. So methanol, $CH_3OH$, will not form a yellow precipitate.
ConceptStructural requirement for iodoform test
Aldehydes, Ketones and Carboxylic Acids in past papers
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