Iodoform reaction is given by: (a) $CH_3CH_2OH$ (b) $CH_3COCH_3$ (c) [Compound Missing] (d) $CH_3OH$

The Iodoform test is positive for compounds containing a methyl ketone group ($-COCH_3$) or a methyl carbinol group ($-CH(OH)CH_3$) attached to at least one hydrogen atom or alkyl/aryl group.

1. Ethanol ($CH_3CH_2OH$): It contains the $CH_3-CH(OH)-$ group. Upon reaction with iodine and alkali ($I_2/NaOH$), it is oxidized to acetaldehyde ($CH_3CHO$), which contains the $CH_3-C=O$ group, and subsequently yields iodoform ($CHI_3$, yellow precipitate).
2. Acetone ($CH_3COCH_3$): It contains the $CH_3-C=O$ group directly. It reacts to form iodoform.
3. Methanol ($CH_3OH$): It does not contain the requisite methyl group attached to the carbinol carbon. It gives a negative test.

Since both Ethanol (a) and Acetone (b) give the test, the correct selection is '(a) and (b)'.