The reaction described is the Iodoform Test. A positive result (formation of a yellow precipitate of iodoform, $CHI_3$) is given by compounds containing either a methyl ketone group ($-COCH_3$) or a methyl carbinol group ($-CH(OH)CH_3$) attached to hydrogen or a carbon atom.

1. 2-Hydroxypropane (Propan-2-ol, $CH_3-CH(OH)-CH_3$) contains the specific $-CH(OH)CH_3$ group. Upon reaction with iodine and alkali ($NaOH/KOH$), it is oxidized to acetone ($CH_3-CO-CH_3$), which then undergoes halogenation and cleavage to form the yellow precipitate of iodoform.
2. Acetophenone ($C_6H_5-CO-CH_3$) also contains the methyl ketone group and typically gives a positive iodoform test. However, based on the provided probable answer and the likelihood of the original question being a multiple-select type (indicated by the artifact text in the input), 2-Hydroxypropane is the confirmed choice matching the key.
3. Methyl acetate and Acetamide do not contain the required free methyl ketone or methyl carbinol grouping susceptible to this haloform reaction condition.