${\text{CH}}_3\text{CHO}$ and ${\text{C}}_6{\text{H}}_5{\text{CH}}_2\text{CHO}$ can be distinguished chemically by:

The Iodoform test is given by aldehydes and ketones that contain the methyl keto group (${\text{CH}}_3-\text{C}=\text{O}$) or by alcohols that can be oxidized to this group.${\text{CH}}_3\text{CHO}$ (Acetaldehyde) contains the methyl group attached to the carbonyl carbon, so it gives a positive iodoform test, resulting in a yellow precipitate of ${\text{CHI}}_3$.${\text{C}}_6{\text{H}}_5{\text{CH}}_2\text{CHO}$ (Phenylacetaldehyde) does not have a ${\text{CH}}_3$ group directly attached to the carbonyl carbon; it has a $-{\text{CH}}_2{\text{C}}_6{\text{H}}_5$ group. Therefore, it gives a negative iodoform test.The reaction for ${\text{CH}}_3\text{CHO}$:
$${\text{CH}}_3\text{CHO}+3{\text{I}}_2+4\text{NaOH}\to {\text{CHI}}_3\downarrow (\text{yellow})+\text{HCOONa}+3\text{NaI}+3{\text{H}}_2\text{O}$$