Step 1: Understanding the Concept:
The haloform reaction is given by compounds containing a methyl keto group ($-CO-CH_3$) or alcohols that can be oxidized to a methyl keto group (e.g., $CH_3-CH(OH)-R$).
Step 2: Formula Application:
Check the structure for the $CH_3-CO-$ group:
- (a) Ethanal: $CH_3-CHO$ (Contains methyl keto equivalent).
- (c) Propanone: $CH_3-CO-CH_3$ (Contains methyl keto).
- (d) Butanone: $CH_3-CO-CH_2CH_3$ (Contains methyl keto).
Step 3: Explanation:
Propanal ($CH_3-CH_2-CHO$) is an aldehyde, but it lacks the specific methyl group directly attached to the carbonyl carbon ($CH_3-CO-$). Instead, it has an ethyl group attached to the formyl group. Therefore, it cannot form a haloform (like iodoform).
Step 4: Final Answer:
Propanal does not exhibit the haloform reaction.