The iodoform test identifies compounds with a methyl ketone group (\( \text{CH}_3\text{CO}- \)) or those oxidizable to this group, such as \( \text{CH}_3\text{CH(OH)}- \) alcohols. The reaction with iodine and base yields a yellow iodoform precipitate (\( \text{CHI}_3 \)).Analysis of options:- Methanol (\( \text{CH}_3\text{OH} \)): Primary alcohol lacking the required \( \text{CH}_3\text{CH(OH)}- \) or methyl ketone moiety. Negative iodoform test result.- Ethanol (\( \text{CH}_3\text{CH}_2\text{OH} \)): Primary alcohol whose oxidation product, acetaldehyde (\( \text{CH}_3\text{CHO} \)), can be further oxidized to a structure yielding a positive iodoform test due to a \( \text{CH}_3\text{CO}- \)-like intermediate.- Propan-1-ol (\( \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \)): Primary alcohol. Oxidation yields propanal (\( \text{CH}_3\text{CH}_2\text{CHO} \)), which lacks the necessary \( \text{CH}_3\text{CO}- \) or \( \text{CH}_3\text{CH(OH)}- \) groups for a positive iodoform test.- Propan-2-ol (\( \text{CH}_3\text{CH(OH)CH}_3 \)): Secondary alcohol with the structure \( \text{CH}_3\text{CH(OH)CH}_3 \). Oxidation produces acetone (\( \text{CH}_3\text{COCH}_3 \)), a methyl ketone, resulting in a positive iodoform test.Both ethanol and propan-2-ol can give a positive iodoform test. However, propan-2-ol directly forms acetone, a methyl ketone, making it the more commonly emphasized compound for this test in contexts like MHTCET, particularly for secondary alcohols.Therefore, propan-2-ol is the compound that yields a positive iodoform test.