The iodoform test is a test for the presence of carbonyl compounds with the structure ##”RCOCH”_3## and alcohols with the structure ##”RCH(OH)CH”_3##.
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.
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