Predict the products of the following reactions.

1. cyclopentyl methyl ketone + excess CI₂ + excess NaOH
2. 1-cyclopentylethanol + excess I₂ + excess NaOH
3. propiophenone + excess Br₂ + excess NaOH

Haloform reaction ishalogenation reaction of a methyl ketone in presence of a base.

In a haloform reactionthe group RC=O of methyl ketone will change into respectiveacid salt$\left(RC=O^{-}N{a}^{+}\right)$, and methyl group change intohaloform $\left({\text{CHX}}_{\text{3}}\right)$

In the haloform reaction three moles of halogen are required to react with one mole of methyl ketone in the presence of excess of base.

Cyclopentyl methyl ketone, when reacted with excess chlorine under strongly basic conditions (excess NaOH), gives cyclopentane carboxylate and chloroform as a products.

formation of chloroform

In the first step of the reaction 1-cyclopentylethanol is oxidised to cyclopentyl methyl ketone. Then it reacts with excess of iodine in the presence of base to give sodium cyclopentane carboxylate and iodoform as products.

Formation of iodoform

When reacted with excess halogen bromine under strongly basic conditions (excess NaOH), propiophenone replace its both alpha hydrogen with bromine, thus forming

as a 2,2-dibromo-1phenylpropan-1-one as a product .

Formation of 2,2-dibromo-1phenylpropan-1-one