Q.Acetone is treated with $\text{I}_2$ and $\text{NaOH}$ (the iodoform test). Give the products and explain, step by step, why the reaction stops after three successive halogenations at the same carbon.

Base removes one of acetone's methyl $\alpha$-hydrogens to give an enolate, which reacts with $\text{I}_2$ to install one iodine, giving $\text{ICH}_2\text{COCH}_3$. The electron-withdrawing iodine now already installed makes the remaining two hydrogens on that same carbon even more acidic than the original three were, so the second and then third iodinations at that carbon happen progressively faster than the first, giving the fully substituted $\text{CI}_3\text{COCH}_3$. Hydroxide then attacks this trihalomethyl ketone's carbonyl carbon and expels the highly stabilised $\text{CI}_3^-$ carbanion (stabilised by all three electron-withdrawing iodines), which is protonated to iodoform, $\text{CHI}_3$, while the rest of the molecule is left as sodium acetate, $\t …

Track why each successive halogenation on the SAME carbon becomes progressively faster (increasing acidity from the electron-withdrawing halogens already installed), …

Thinking the halogenation spreads to other carbons in the molecule -- it does not; all three halogens go onto the SAME methyl carbon, because that carbon's remaining hydrogens be …