Concept:The prerequisite for a positive haloform (iodoform) test is the presence of a methyl group bonded directly to a carbonyl carbon (\( CH_3-CO- \)), or an alcohol that oxidizes to this structure (\( CH_3-CH(OH)- \)).
Solution:- A "Methyl ketone" explicitly contains the required \( CH_3-CO- \) structural fragment.
- In the presence of iodine and a base, the three hydrogen atoms on the methyl group are replaced by iodine, followed by cleavage to yield a yellow precipitate of Iodoform (\( CHI_3 \)).
Why other options are incorrect:Methanol (\( CH_3OH \)) lacks the required \( CH_3-C-O \) linkage. 3-Hexanol oxidizes to 3-hexanone (ethyl propyl ketone), which lacks a terminal methyl ketone. Propionaldehyde (\( CH_3CH_2CHO \)) lacks a terminal methyl group directly attached to the carbonyl carbon.
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