# Haloform reaction with limited bromine

We were asked the reaction between 1 mole acetone and 1 mole $\ce{Br2}$ in basic medium. Since only a limited amount of bromine was given I got the final answer as 1-bromopropan-2-one (mono bromo acetone). However, in the given solution the entire haloform reaction was shown with $\ce{CHBr3}$ and acetate ion as the final product.

I am not sure how a single mole of $\ce{Br2}$ is sufficient for this(at least 3 moles are needed, I believe).

1. $$3$$ $$\ce{H}$$ atoms out of total $$6$$ in a molecule of acetone is brominated.
2. So $$\frac{3}{2}$$ moles $$\ce{Br}$$ molecules is needed to react with 1 mol acetone.
3. Hence $$1$$ mol $$\ce{Br}$$ molecule will react with $$2/3$$ mol acetone. Thus acetone is the limiting reagent and the reaction is completed.