Alcohols containing $\mathrm{CH}_3 \mathrm{CH}(\mathrm{OH})-$ group and carbonyl compounds containing $\mathrm{CH}_3 \mathrm{CO}-$ group give yellow precipitate with iodine and $\mathrm{NaOH}$ solution. This reaction is called iodoform test.
Thus, $\mathrm{CH}_3 \mathrm{CHO}$ due to presence of $\mathrm{CH}_3 \mathrm{CO}$ group gives yellow ppt with $\mathrm{I}_2$ and $\mathrm{NaOH}$.
$\mathrm{CH}_3 \mathrm{CHO}+3 \mathrm{I}_2+4 \mathrm{NaOH} \longrightarrow \mathrm{CHI}_3 +\mathrm{HCOONa}+3 \mathrm{NaI}+2 \mathrm{H}_2 \mathrm{O}$