A solution of "I"_2 is added to a small amount of your unknown, followed by just enough "NaOH" to remove the colour.
The formation of a pale yellow precipitate of iodoform (with a characteristic "antiseptic" smell) is a positive result.
color(red)"MECHANISM:"
1. OH⁻ removes an acidic α-hydrogen.
"RCOCH"_3 + color(aqua)("OH"^"⁻") ⇌ color(green)("RCOCH"_2)^"-" + "H"_2"O"
2. The enolate ion displaces an "I"^"-" from "I"_2.
color(green)"RCOCH"_2^"-" + "I-I" ⇌ color(orange)"RCOCH"_2"I" + "I"^-"
The process repeats twice more.
3. color(orange)("RCOCH"_2"I") + color(aqua)("OH"^"-") ⇌ color(blue)"RCOCHI"^"-" +"H"_2"O"
4. color(blue)"RCOCHI"^"⁻" + "I-I" → color(maroon)"RCOCHI"_2 + "I"^"⁻"
5. color(maroon)"RCOCHI"_2 + color(aqua)("OH"^"⁻") ⇌ color(purple)"RCOCI₂"^"⁻" + "H"_2"O"
6. color(purple)"RCOCI"_2^"⁻" + "I-I" → color(fuchsia)"RCOCI"_3 + "I⁻"
7. A hydroxide ion adds to the base of the carbonyl.
color(fuchsia)("RCOCI"_3) + color(aqua)("OH"^"⁻") ⇌ color(lime)("RC"stackrel("-")"O""(OH)CI"_3)
8. The carbonyl group re-forms and eliminates a "CI"_3⁻ ion.
color(lime)("RC"stackrel("-")("O")"(OH)CI"_3) ⇌ "RCOOH" + color(olive)("CI"_3^"-")
9. The acidic "RCOOH" and the strongly basic "CI"_3^"-" ion are neutralized.
"RCOOH(aq)" + color(olive)("CI"_3^"-")("aq") → "RCOO"^"-""(aq)" + color(red)("HCI"_3)"(s)"
The overall equation is
color(red)("RCOCH"_3 + "3I"_2 + "4OH"^"-" → "RCOO"^"-" + "3I"^"-" + "3H"_2"O" + "HCI"_3)
The reaction also gives a positive test with secondary alcohols of the type "RCH(OH)CH"_3, because they are oxidized under the reaction conditions to methyl ketones.
"RCH(OH)CH"_3 + "I"_2 + "2OH"^"-" → "RCOCH"_3 + "2I"^"-" + "2H"_2"O"
If "R = H", the reactants are "HCOCH"_3 (acetaldehyde) and "HOCH"_2"CH"_3 (ethanol). They also give the iodoform test.
