Question

How to write the reaction equation for the reaction of I2/NaOH with carbonyl compound CH3CHO , CH3CH2CHO and CH3CH2COCH3?

Answer 1

The reaction is the base-catalyzed reaction of an enolate with iodine.

Explanation:

`"R"color(red)("COCH")"R"_2 + "I"_2 +"OH"^(-) → "R"color(red)("COCI")"R"_2 +"H"_2"O" +"I"^-`

If there is more than one α-hydrogen, the electron-withdrawing effect of the first iodine makes the other α-hydrogens even more acidic, and we get multiple substitutions. For example,

`"R"color(red)("COCH"_3) + "3I"_2 + "3OH"^(-) → "R"color(red)("COCI"_3) +"3H"_2"O" + 3"I"^-`

The electron-withdrawing iodine atoms make the triiodomethyl group an effective leaving group, so the ketone can undergo nucleophilic acyl substitution.

`"RCO"color(red)("CI"_3) + "OH"^(-) → "RCOOH" +color(red)("CI"_3^(-)) → "RCOO"^(-) + color(red)("CHI"_3)`

The `"CHI"_3` (iodoform) is a bright yellow precipitate with a "hospital" odour, so the iodoform reaction serves as a test for compounds containing the `"COCH"_3` group.

For your compounds, the equations are

A. `"HCOCH"_3 + "3I"_2 + "4OH"^(-) → "HCOO"^(-) + "CHI"_3 + "3I"^(-) +"3H"_2"O"`

B. `"CH"_3"CH"_2"CHO" + "2I"_2 + "2OH"^(-) → "CH"_3"CI"_2"CHO" + "2OH"^(-) + "2H"_2"O"`
(no iodoform!)

C. `"CH"_3"CH"_2"COCH"_3 + "3I"_2 + "4OH"^(-) → "CH"_3"CH"_2"COO"^(-) + "CHI"_3 + "3H"_2"O"`