Which compound react with alkaline aqueous iodine to give a pale-yellow precipitate of tri-iodomethane?
1.butanone
2.ethanal
3.propan-2-ol
A. 1, 2 and 3 are correct
B. 1 and 2 only are correct
C. 2 and 3 only are correct
D. 1 only is correct

Hint: Any compounds containing the $C{H_3}C = 0$ group or the $C{H_3}CH(OH)$ group will give a positive result with the iodoform test. When ${I_2}$ and NaOH is added to a compound containing one of these groups, then a pale-yellow precipitate of iodoform (triiodomethane) is formed.

Complete step by step answer:
When Iodine and sodium hydroxide are added to a compound that contains either a methyl ketone or a secondary alcohol with a methyl group in the alpha position, it gives a positive result - a pale yellow precipitate of iodoform or triiodomethane is formed.
Tri-iodomethane (iodoform) reaction shows a positive result – a pale yellow precipitate of triiodomethane (iodoform) is given by an aldehyde or ketone containing the grouping:

“R” can be a hydrogen atom or a hydrocarbon group (for example, an alkyl group).
If R is hydrogen, then we have the aldehyde ethanal, $C{H_3}CHO$ . Ethanal is the only aldehyde to give a triiodomethane reaction.
If R is a hydrocarbon group, then we have a ketone. Many ketones give this reaction, but all have a methyl group on one side of the carbon-oxygen bond. These are known as methyl ketones. And butanone is a methyl ethyl ketone $(C{H_3}(C = O)C{H_2}C{H_3})$ .
This test can also be used to identify alcohols $(C{H_3}C{H_2}OH)$ . If the alcohol is a tertiary alcohol then it does not give yellow precipitate as it cannot be oxidised. If the alcohol is a primary alcohol then it must be ethanol (as this is oxidised to ethanal, which is the only aldehyde that gives a positive result with the iodoform test). All secondary alcohols give a positive result, as they are oxidised to ketones. And propan-2-ol is a secondary alcohol.
 So, all the compounds above i.e. butanone, Ethanal and propan-2-ol react with alkaline aqueous iodine to give a pale-yellow precipitate of tri-iodomethane.

Therefore, the correct answer is option (A).

Note: The iodoform test can be used to identify only ethanal if it is an aldehyde (this is the only aldehyde with the $C{H_3}CHO$ group). This occurs as three I atoms replace the H atoms of $C{H_3}C = OR$ , and the \[C - C\] bond breaks due to the electron withdrawing effect of the three I atoms (as I is more electronegative than C) forming $C{H_3}I$ and the salt anion of a carboxylic acid (depending on the R group of the original compound, which influences the length of the carbon chain of the anion \[RCOO - \] that is formed).