Q.Write Haloform reaction.

The haloform reaction converts a $CH_3CO-$ (or $CH_3CH(OH)-$) compound, on treatment with $X_2$/$NaOH$, into a haloform ($CHX_3$) and the sodium salt of a carboxylic acid with one fewer carbon.

Compounds containing a methyl ketone group ($CH_3-CO-R$), acetaldehyde ($CH_3CHO$), or a secondary alcohol of the type $CH_3-CH(OH)-R$ (which is first oxidised in situ by the halogen to the corresponding methyl ketone/aldehyde) react with a halogen ($Cl_2$, $Br_2$ or $I_2$) in the presence of $NaOH$ to give a haloform ($CHCl_3$, $CHBr_3$ or $CHI_3$) and the sodium salt of a carboxylic acid containing one carbon less than the original methyl-ketone chain. This is called the haloform reaction; with iodine it is specifically called the iodoform reaction/test, giving a characteristic yellow precipitate of iodoform, $CHI_3$.

Mechanism outline (using acetone + $I_2$/NaOH as example):

1. $NaOH$ generates the enolate of acetone; the three hydrogens of the $CH_3$ group alpha to the carbonyl are successively replaced by iodine (base-catalysed halogenation of the alpha carbon, repeated three times), giving $CI_3-CO-CH_3$.
2. Hydroxide ion then attacks the carbonyl carbon of this tri-iodo ketone; the strongly electron-withdrawing $CI_3$ group makes $CI_3^-$ a good enough leaving group, so the $C-C$ bond cleaves, expelling the trihalomethyl carbanion $CI_3^-$ and leaving $CH_3COOH$ (immediately deprotonated by excess $NaOH$ to $CH_3COO^-Na^+$).
3. The carbanion $CI_3^-$ picks up a proton from water/solvent to give the haloform, $CHI_3$. …