Explain the iodoform test.
(N/A) Aim: The iodoform test is used to identify compounds containing the $CH_3CO-$ group (methyl ketones) or compounds that can be oxidized to this group,such as $CH_3CH(OH)-$ (secondary alcohols) and ethanol.
Procedure: The compound is treated with iodine $(I_2)$ in the presence of sodium hydroxide $(NaOH)$. This generates sodium hypoiodite $(NaOI)$ in situ.
Reaction: $R-CO-CH_3 + 3I_2 + 4NaOH \rightarrow R-COONa + CHI_3 + 3NaI + 3H_2O$
Observation: The formation of a yellow precipitate of iodoform $(CHI_3)$ indicates a positive test result.
Note: Ethanol $(CH_3CH_2OH)$ also gives this test because it is oxidized to acetaldehyde $(CH_3CHO)$ by sodium hypoiodite,which then undergoes the iodoform reaction.
Procedure: The compound is treated with iodine $(I_2)$ in the presence of sodium hydroxide $(NaOH)$. This generates sodium hypoiodite $(NaOI)$ in situ.
Reaction: $R-CO-CH_3 + 3I_2 + 4NaOH \rightarrow R-COONa + CHI_3 + 3NaI + 3H_2O$
Observation: The formation of a yellow precipitate of iodoform $(CHI_3)$ indicates a positive test result.
Note: Ethanol $(CH_3CH_2OH)$ also gives this test because it is oxidized to acetaldehyde $(CH_3CHO)$ by sodium hypoiodite,which then undergoes the iodoform reaction.
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