Concept:The iodoform test gives a positive result for specific structural motifs: compounds containing a methyl ketone group (\(\text{CH}_3\text{CO-}\)), or alcohols that can be oxidized to a methyl ketone by the reagent (specifically containing the \(\text{CH}_3\text{CH}(\text{OH})-\) structural unit).
Formula:$$ \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{NaOI}} \text{CH}_3\text{CHO} \xrightarrow{\text{NaOI}} \text{CHI}_3\downarrow $$
Solution:- Looking at primary alcohols, only ethanol (\(\text{CH}_3\text{CH}_2\text{OH}\)) possesses the required terminal methyl group adjacent to the carbon bearing the hydroxyl group.
- It oxidizes to acetaldehyde, which has the necessary \(\text{CH}_3\text{CO-}\) group, subsequently reacting to form yellow iodoform crystals.
Why other options are incorrect:Methanol oxidizes to formaldehyde (no methyl group). 1-Propanol oxidizes to propanal (ethyl group attached, no methyl group adjacent to carbonyl). 1-Butanol behaves similarly. None of them possess the \(\text{CH}_3\text{CH}(\text{OH})-\) substructure.
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