Alcohol identification via sodium and iodoform tests

Steps:

• Compound Z reacts with Na to evolve H₂ (combustible gas), confirming it's an alcohol (R-OH + Na → R-ONa + ½H₂).
• Iodoform test uses alcoholic aqueous I₂; yellow CHI₃ precipitate forms with alcohols having CH₃CH(OH)- or oxidizable to CH₃CO- groups.
• Z gives no yellow precipitate, so it lacks CH₃CH(OH)R (secondary) or CH₃CH₂OH (primary ethanol) structure.
• Possible Z: primary alcohol like propan-1-ol (CH₃CH₂CH₂OH), without required methyl carbinol group.

Why A is correct:

• A is propan-1-ol, a primary alcohol without CH₃CH(OH)- group, so negative iodoform test; reacts with Na to give H₂.

Why the others are wrong:

• B is ethanol, gives yellow iodoform precipitate (CH₃CH₂OH oxidizes to CH₃CHO).
• C is a methyl ketone (e.g., acetone), gives iodoform but not Na/H₂ reaction (no -OH).
• D is secondary alcohol like propan-2-ol (CH₃CH(OH)CH₃), gives yellow iodoform precipitate.

Final answer: A