The iodoform test is used to detect the presence of compounds containing a methyl ketone group (\( \text{CH}_3\text{CO}- \)) or compounds that can be oxidized to form such a group, such as alcohols with the structure \( \text{CH}_3\text{CH(OH)}- \). The test involves the reaction with iodine and a base, producing a yellow precipitate of iodoform (\( \text{CHI}_3 \)).
Let’s analyze each option:
- Methanol (\( \text{CH}_3\text{OH} \)): A primary alcohol with no \( \text{CH}_3\text{CH(OH)}- \) or methyl ketone group. It does not give a positive iodoform test.
- Ethanol (\( \text{CH}_3\text{CH}_2\text{OH} \)): A primary alcohol with the structure \( \text{CH}_3\text{CH}_2\text{OH} \). Upon oxidation, it forms acetaldehyde (\( \text{CH}_3\text{CHO} \)), which can further oxidize to a compound that gives a positive iodoform test due to the \( \text{CH}_3\text{CO}- \)-like structure after reaction.
- Propan-1-ol (\( \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \)): A primary alcohol. Upon oxidation, it forms propanal (\( \text{CH}_3\text{CH}_2\text{CHO} \)), which does not contain the required \( \text{CH}_3\text{CO}- \) or \( \text{CH}_3\text{CH(OH)}- \) structure for the iodoform test.
- Propan-2-ol (\( \text{CH}_3\text{CH(OH)CH}_3 \)): A secondary alcohol with the structure \( \text{CH}_3\text{CH(OH)CH}_3 \). Upon oxidation, it forms acetone (\( \text{CH}_3\text{COCH}_3 \)), a methyl ketone, which gives a positive iodoform test.
Both ethanol and propan-2-ol can give a positive iodoform test, but propan-2-ol directly forms a methyl ketone (acetone), making it the more definitive choice in the context of MHTCET, where secondary alcohols like propan-2-ol are often emphasized for this test.
Thus, the compound that gives a positive iodoform test is propan-2-ol.