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SolutionsHalogenation

1. The haloform reaction

Predict the result of the following reaction:

Acetophenone with hydroxide and iodine
Mechanism: triiodination of the methyl group through the enolates, attack of hydroxide on the carbonyl, loss of the CI₃⁻ anion and formation of benzoic acid and iodoform

In base the methyl group is iodinated three times: each iodine that enters makes the remaining hydrogens more acidic. Hydroxide then attacks the carbonyl and the CI3− anion, stabilized by the three iodines, leaves. Benzoate (benzoic acid on acidification) and iodoform, CHI3, a yellow precipitate with a very characteristic smell, are obtained.

This is the iodoform test, positive for methyl ketones (and for acetaldehyde and the alcohols that are oxidized to them).

2. Acid or base

Why does the bromination of acetone in acid stop at bromoacetone, whereas in base it continues?

In acid the slow step is the formation of the enol, and the electron-withdrawing bromine makes the protonation of the carbonyl and the enolization more difficult: bromoacetone reacts more slowly than acetone. In base the bromine increases the acidity of the α hydrogens and the brominated enolate forms faster than the original one.

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