Correct Option: (d) Iodoform test

Step 1 (Governing Principle): [0.5 Mark]

The Iodoform test is specific for compounds containing the methyl ketone group ($CH_3CO-$) or a secondary alcohol group that can be oxidized to a methyl ketone ($CH_3CH(OH)-$). The reaction involves the treatment of the compound with $I_2$ and $NaOH$ (or $NaOI$), resulting in the formation of a yellow precipitate of iodoform ($CHI_3$).

Step 2 (Step-by-Step Chemical Analysis): [0.25 Mark]

• Pentan-2-one ($CH_3COCH_2CH_2CH_3$): Contains a $CH_3CO-$ group. It reacts with $I_2/NaOH$ to give a yellow precipitate of iodoform. $$CH_3COCH_2CH_2CH_3 + 3I_2 + 4NaOH \rightarrow CHI_3 \downarrow + CH_3CH_2CH_2COONa + 3NaI + 3H_2O$$
• Pentan-3-one ($CH_3CH_2COCH_2CH_3$): Does not contain a $CH_3CO-$ group. Therefore, it does not give the iodoform test.

Step 3 (Final Concluded Answer): [0.25 Mark]

Since Pentan-2-one gives a positive iodoform test (yellow precipitate) and Pentan-3-one does not, the Iodoform test is the correct method to distinguish between them.

💡 EL10 Concept Insight:

Think of the Iodoform test as a "Methyl-Ketone Detector"—it only reacts if there is a specific $CH_3$ group attached directly to the $C=O$ (carbonyl) carbon. Pentan-2-one has this "handle," but Pentan-3-one is symmetrical and lacks this specific structure, so it stays clear.