The iodoform test is used for the detection of

$COCH_3$

The correct answer is Option (1) → $COCH_3$

Iodoform test is positive for compounds that have:

• CH₃–CO– (methyl ketone group) → e.g., acetaldehyde, acetone, acetophenone
• CH₃–CH(OH)– (methyl carbinol group) → e.g., ethanol, secondary alcohols with CH₃CH(OH)– structure

These give yellow precipitate of CHI₃ (iodoform) with I₂ + NaOH (or I₂ + base).

Why others are wrong:

• COO⁻ (carboxylate): No iodoform test
• CHO (aldehyde): Only acetaldehyde (CH₃CHO) gives it because it has CH₃CO– structure; other aldehydes don’t
• OH (alcohol): Only specific ones like ethanol, not all alcohols